$\prod_{1\leq l\leq n-1}a_{l}=\prod_{1\leq l\leq
n-1}(a_{l}-a_{n})\left(1+\sum_{1\leq j\leq n-1}\frac{a_{n}\prod_{k\not=j,1\leq
k\leq n-1}a_{k}}{\prod_{k\not=j,1\leq k\leq
n-1}(a_{k}-a_{j})}\frac{1}{(a_{j}-a_{n})}\right),$

which is

$\prod_{1\leq l\leq n-1}a_{l}=\prod_{1\leq l\leq n-1}(a_{l}-a_{n})+\sum_{1\leq
j\leq n-1}\frac{a_{n}\prod_{k\not=j,1\leq k\leq
n-1}a_{k}}{\prod_{k\not=j,1\leq k\leq n-1}(a_{k}-a_{j})}\frac{\prod_{1\leq
l\leq n-1}(a_{l}-a_{n})}{(a_{j}-a_{n})},$

i.e.

(3.3.14) $\displaystyle\prod_{1\leq l\leq n-1}a_{l}=\prod_{1\leq l\leq
n-1}(a_{l}-a_{n})+\sum_{1\leq j\leq n-1}\frac{a_{n}\prod_{k\not=j,1\leq k\leq
n-1}a_{k}(a_{k}-a_{n})}{\prod_{k\not=j,1\leq k\leq n-1}(a_{k}-a_{j})}.$

Let us reformulate (3.3.14) as an equality between polynomials (to be proven)
with

(3.3.15) $\displaystyle\prod_{1\leq l\leq n-1}(a_{l}-X)+\sum_{1\leq j\leq
n-1}\frac{X\prod_{k\not=j,1\leq k\leq n-1}a_{k}(a_{k}-X)}{\prod_{k\not=j,1\leq
k\leq n-1}(a_{k}-a_{j})}-\prod_{1\leq l\leq n-1}a_{l}=0,$

and let us assume that the $(a_{j})_{1\leq j\leq n-1}$ are distinct and
different from $0$. The polynomial $\mathcal{Q}$ on the left-hand-side has
degree less than $n-1$ and we have

$\mathcal{Q}(0)=0,\qquad\text{and \quad$\forall j\in\llbracket 1\mkern
1.5mu..\mkern 1.5mun-1\rrbracket,$}$
$\mathcal{Q}(a_{j})=\frac{a_{j}\prod_{k\not=j,1\leq k\leq
n-1}a_{k}(a_{k}-a_{j})}{\prod_{k\not=j,1\leq k\leq
n-1}(a_{k}-a_{j})}-\prod_{1\leq l\leq n-1}a_{l}=0,$

so that $\mathcal{Q}$ has degree less than $n-1$ with $n$ distinct roots and
this proves the identity (3.3.15) when the $(a_{j})_{1\leq j\leq n-1}$ are
distinct and all different from $0$, proving (3.3.13) in that case; of course
we may assume that all $a_{j}$ are positive and noting from (3.3.5) that
$K_{\alpha}$ is continuous on $({\mathbb{R}_{+}^{*}})^{n}$, we get Formula
(3.3.13) in all cases where all the $a_{j}$ are positive, concluding the proof
of the lemma. ∎

######  Lemma 3.22.

With the notations of Theorem 3.13, we have, assuming

$0<a_{1}\leq\dots\leq a_{n},$

the inequality

(3.3.16) $K_{0\in\mathbb{N}^{n}}(a_{1},\dots,a_{n})\geq\sum_{1\leq j\leq
n}e^{-a_{j}}\frac{\prod_{1\leq l<j}a_{l}}{(j-1)!}\geq e^{-\min_{1\leq j\leq
n}a_{j}}=\max_{1\leq j\leq n}e^{-a_{j}}.$

######  Remark 3.23.

The above estimate is sharp in the sense that when all the $a_{j}$ are equal
to the same $a>0$, we have proven in (3.2.4) that

$K_{0}(a)=\frac{e^{-a}}{(n-1)!}\int_{0}^{+{\infty}}e^{-s}(s+a)^{n-1}ds=e^{-a}\sum_{0\leq
l\leq n-1}\frac{a^{l}}{(n-1-l)!l!}\Gamma(n-l)\\\ =e^{-a}\sum_{0\leq l\leq
n-1}\frac{a^{l}}{l!}=e^{-a}\sum_{1\leq j\leq
n}\frac{a^{j-1}}{(j-1)!}=\sum_{1\leq j\leq n}e^{-a_{j}}\frac{\prod_{1\leq
l<j}a_{l}}{(j-1)!}_{|a_{1}=\dots=a_{n}=a}.$

###### Proof.

The property is true for $n=1$ since $K_{0}(a_{1})=e^{-a_{1}}$. We check the
case $n=2$ with $a_{1}<a_{2}$, and we find

$K_{(0,0)}(a_{1},a_{2})=e^{-a_{1}}+\int_{0}^{a_{1}}e^{-t_{1}}e^{-a_{2}(1-t_{1}/a_{1})}dt_{1}\\\
=e^{-a_{1}}+e^{-a_{2}}\frac{e^{a_{2}-a_{1}}-1}{\frac{a_{2}}{a_{1}}-1}=e^{-a_{1}}+e^{-a_{2}}a_{1}\frac{e^{a_{2}-a_{1}}-1}{{a_{2}}-{a_{1}}}\geq
e^{-a_{1}}+e^{-a_{2}}a_{1}.$

Let us consider for some $n\geq 3$, $0<a_{1}<\dots<a_{n}$ and inductively,

$\displaystyle K_{0\in\mathbb{N}^{n}}$ $\displaystyle(a_{1},\dots,a_{n})$
$\displaystyle=e^{-a_{1}}P_{0}(a_{1})+\int_{0}^{a_{1}}e^{-t_{1}}K_{0\in\mathbb{N}^{n-1}}\bigl{(}a_{2}(1-t_{1}/a_{1}),\dots,a_{n}(1-t_{1}/a_{1})\bigr{)}dt_{1}$
$\displaystyle=e^{-a_{1}}P_{0}(a_{1})+a_{1}\int_{0}^{1}e^{-a_{1}\theta}K_{0\in\mathbb{N}^{n-1}}\bigl{(}a_{2}(1-\theta),\dots,a_{n}(1-\theta)\bigr{)}d\theta$
$\displaystyle\geq e^{-a_{1}}+a_{1}\int_{0}^{1}e^{-a_{1}\theta}\sum_{2\leq
j\leq n}e^{-a_{j}(1-\theta)}\frac{\prod_{2\leq
l<j}a_{l}}{(j-2)!}(1-\theta)^{j-2}d\theta$
$\displaystyle=e^{-a_{1}}+\sum_{2\leq j\leq
n}e^{-a_{j}}\Bigl{(}\underbrace{a_{1}\prod_{2\leq l<j}a_{l}}_{\prod_{1\leq
k<j}a_{k}}\Bigr{)}\int_{0}^{1}e^{(a_{j}-a_{1})\theta}\frac{1}{(j-2)!}(1-\theta)^{j-2}d\theta$
$\displaystyle\geq e^{-a_{1}}+\sum_{2\leq j\leq
n}e^{-a_{j}}\Bigl{(}\prod_{1\leq
k<j}a_{k}\Bigr{)}\int_{0}^{1}\frac{1}{(j-2)!}(1-\theta)^{j-2}d\theta$
$\displaystyle=e^{-a_{1}}+\sum_{2\leq j\leq n}e^{-a_{j}}\Bigl{(}\prod_{1\leq
k<j}a_{k}\Bigr{)}\frac{1}{(j-1)!},$

concluding the proof of the lemma. ∎

######  Remark 3.24.

The reader may have noticed that it is not obvious on Formula (3.3.13)

$K_{0,\dots,0}(a_{1},\dots,a_{n})=\sum_{1\leq j\leq
n}e^{-a_{j}}\frac{\prod_{k\not=j}a_{k}}{\prod_{k\not=j}(a_{k}-a_{j})},$

that $K_{0}$ is an entire function. Let us start with taking a look at

$\displaystyle K_{0,0}(a_{1},a_{2})$
$\displaystyle=\frac{e^{-a_{1}}a_{2}}{a_{2}-a_{1}}+\frac{e^{-a_{2}}a_{1}}{a_{1}-a_{2}}=\frac{a_{2}e^{-a_{1}}-a_{1}e^{-a_{2}}}{a_{2}-a_{1}}$
$\displaystyle=e^{-\frac{(a_{1}+a_{2})}{2}}\frac{a_{2}e^{-\frac{a_{1}}{2}+\frac{a_{2}}{2}}-a_{1}e^{-\frac{a_{2}}{2}+\frac{a_{1}}{2}}}{a_{2}-a_{1}}$
$\displaystyle=e^{-\frac{(a_{1}+a_{2})}{2}}\frac{a_{2}(\cosh\frac{a_{2}-a_{1}}{2}+\sinh\frac{a_{2}-a_{1}}{2})-a_{1}(\cosh\frac{a_{1}-a_{2}}{2}+\sinh\frac{a_{1}-a_{2}}{2})}{a_{2}-a_{1}}$
$\displaystyle=e^{-\frac{(a_{1}+a_{2})}{2}}\Bigl{[}\cosh(\frac{a_{2}-a_{1}}{2})+\frac{(a_{2}+a_{1})\sinh(\frac{a_{2}-a_{1}}{2})}{a_{2}-a_{1}}\Bigr{]}$
$\displaystyle=e^{-\frac{(a_{1}+a_{2})}{2}}\Bigl{[}\cosh(\frac{a_{2}-a_{1}}{2})+\frac{\frac{1}{2}(a_{2}+a_{1})\sinh(\frac{a_{2}-a_{1}}{2})}{\frac{a_{2}-a_{1}}{2}}\Bigr{]}$
(3.3.17)
$\displaystyle=e^{-\frac{(a_{1}+a_{2})}{2}}\Bigl{[}\cosh(\frac{a_{2}-a_{1}}{2})+\frac{1}{2}(a_{2}+a_{1})\operatorname{shc}(\frac{a_{2}-a_{1}}{2})\Bigr{]},$

where $\operatorname{shc}$ stands for the even entire function defined by

(3.3.18) $\operatorname{shc}t=\frac{\sinh t}{t}.$

We have also from Lemma 3.16

(3.3.19)
$F_{\alpha}(a)=\int_{\mathbb{R}}\frac{\sin\tau}{\pi\tau}\Bigl{(}\prod_{1\leq
j\leq
n}\frac{(1+i\tau\mu_{j})^{\alpha_{j}}}{(1-i\tau\mu_{j})^{\alpha_{j}+1}}\Bigr{)}d\tau,$

and defining the function $F_{\alpha}(a,\lambda)$ as the absolutely converging
integral,

(3.3.20)
$F_{\alpha}(a,\lambda)=\int_{\mathbb{R}}\frac{\sin(\lambda\tau)}{\pi\tau}\Bigl{(}\prod_{1\leq
j\leq
n}\frac{(1+i\tau\mu_{j})^{\alpha_{j}}}{(1-i\tau\mu_{j})^{\alpha_{j}+1}}\Bigr{)}d\tau,\quad
F_{\alpha}(a)=F_{\alpha}(a,1),$

we get

$\displaystyle\frac{\partial
F_{\alpha}}{\partial\lambda}(a,\lambda)=\frac{1}{\pi}\int_{\mathbb{R}}{\cos(\lambda\tau)}\Bigl{(}\prod_{1\leq
j\leq
n}\frac{(1+i\tau\mu_{j})^{\alpha_{j}}}{(1-i\tau\mu_{j})^{\alpha_{j}+1}}\Bigr{)}d\tau$
$\displaystyle=\frac{1}{2\pi}\int_{\mathbb{R}}e^{i\lambda\tau}\Bigl{(}\prod_{1\leq
j\leq
n}\frac{(1+i\tau\mu_{j})^{\alpha_{j}}}{(1-i\tau\mu_{j})^{\alpha_{j}+1}}\Bigr{)}d\tau$
$\displaystyle\hskip
44.0pt+\frac{1}{2\pi}\int_{\mathbb{R}}e^{i\lambda\tau}\Bigl{(}\prod_{1\leq
j\leq
n}\frac{(1-i\tau\mu_{j})^{\alpha_{j}}}{(1+i\tau\mu_{j})^{\alpha_{j}+1}}\Bigr{)}d\tau$
$\displaystyle=\frac{1}{2\pi}\int_{\mathbb{R}}e^{i\lambda\tau}\Bigl{(}\prod_{1\leq
j\leq
n}\frac{(1+i\tau\mu_{j})^{\alpha_{j}}}{(1-i\tau\mu_{j})^{\alpha_{j}+1}}+\prod_{1\leq
j\leq
n}\frac{(1-i\tau\mu_{j})^{\alpha_{j}}}{(1+i\tau\mu_{j})^{\alpha_{j}+1}}\Bigr{)}d\tau$
$\displaystyle=i\sum_{1\leq j\leq
n}\text{Res}\left(e^{i\lambda\tau}\prod_{1\leq j\leq
n}\frac{(1-i\tau\mu_{j})^{\alpha_{j}}}{(1+i\tau\mu_{j})^{\alpha_{j}+1}};\tau=i/\mu_{j}=ia_{j}\right)$
$\displaystyle=i\sum_{1\leq j\leq
n}\text{Res}\left(e^{i\lambda\tau}\prod_{1\leq j\leq
n}\frac{(-i\mu_{j})^{\alpha_{j}}(ia_{j}+\tau)^{\alpha_{j}}}{(i\mu_{j})^{\alpha_{j}+1}(-ia_{j}+\tau)^{\alpha_{j}+1}};\tau=ia_{j}\right)$
$\displaystyle=\frac{1}{i^{n-1}}\sum_{1\leq j\leq
n}\text{Res}\left(e^{i\lambda\tau}\prod_{1\leq j\leq
n}(-1)^{\alpha_{j}}\frac{a_{j}(ia_{j}+\tau)^{\alpha_{j}}}{(\tau-
ia_{j})^{\alpha_{j}+1}};\tau=ia_{j}\right),$

so that assuming that the $a_{j}$ are positive and distinct, we get

$\displaystyle\frac{\partial
F_{\alpha}}{\partial\lambda}(a,\lambda)=\frac{1}{i^{n-1}}(\prod a_{k})$
$\displaystyle\times\sum_{1\leq j\leq
n}\frac{1}{\alpha_{j}!}\left(\frac{d}{d\tau}\right)^{\alpha_{j}}\left(e^{i\lambda\tau}(-1)^{\alpha_{j}}(ia_{j}+\tau)^{\alpha_{j}}\prod_{1\leq
k\leq n,k\not=j}(-1)^{\alpha_{k}}\frac{(ia_{k}+\tau)^{\alpha_{k}}}{(\tau-
ia_{k})^{\alpha_{k}+1}}\right)_{|\tau=ia_{j}}$
$\displaystyle=\frac{1}{i^{n-1}}(\prod_{1\leq k\leq n}a_{k})\sum_{1\leq j\leq
n}\frac{1}{\alpha_{j}!}$ $\displaystyle\hskip
44.0pt\times\left(\frac{d}{id\sigma}\right)^{\alpha_{j}}\left(e^{-\lambda\sigma}(-1)^{\alpha_{j}}(ia_{j}+i\sigma)^{\alpha_{j}}\hskip-12.0pt\prod_{1\leq
k\leq
n,k\not=j}(-1)^{\alpha_{k}}\frac{(ia_{k}+i\sigma)^{\alpha_{k}}}{(i\sigma-
ia_{k})^{\alpha_{k}+1}}\right)_{|\sigma=a_{j}}$
$\displaystyle=(-1)^{n-1+|\alpha|}(\prod_{1\leq k\leq n}a_{k})\sum_{1\leq
j\leq n}\frac{1}{\alpha_{j}!}$ $\displaystyle\hskip
44.0pt\times\left(\frac{d}{d\sigma}\right)^{\alpha_{j}}\left(e^{-\lambda\sigma}(a_{j}+\sigma)^{\alpha_{j}}\hskip-12.0pt\prod_{1\leq
k\leq n,k\not=j}\frac{(a_{k}+\sigma)^{\alpha_{k}}}{(\sigma-
a_{k})^{\alpha_{k}+1}}\right)_{|\sigma=a_{j}}$ $\displaystyle=(\prod_{1\leq
k\leq n}a_{k})\sum_{1\leq j\leq
n}\frac{(-1)^{\alpha_{j}}}{\alpha_{j}!}\left(\frac{d}{d\sigma}\right)^{\alpha_{j}}\left(e^{-\lambda\sigma}(a_{j}+\sigma)^{\alpha_{j}}\hskip-12.0pt\prod_{1\leq
k\leq
n,k\not=j}\frac{(a_{k}+\sigma)^{\alpha_{k}}}{(a_{k}-\sigma)^{\alpha_{k}+1}}\right)_{|\sigma=a_{j}}.$

Since $F_{\alpha}(a,+{\infty})=1$, thanks to Lemma 9.7, we find eventually
that

$\displaystyle
F_{\alpha}(a)=F_{\alpha}(a,1)=\int_{+{\infty}}^{1}\frac{\partial
F_{\alpha}}{\partial\lambda}(a,\lambda)d\lambda+1=1-K_{\alpha}(a),$
$\displaystyle K_{\alpha}(a)=(\prod_{1\leq k\leq n}a_{k})\sum_{1\leq j\leq
n}\frac{(-1)^{\alpha_{j}}}{\alpha_{j}!}$ $\displaystyle\hskip
75.0pt\int^{+{\infty}}_{1}\left(\frac{d}{d\sigma}\right)^{\alpha_{j}}\left(e^{-\lambda\sigma}(a_{j}+\sigma)^{\alpha_{j}}\hskip-12.0pt\prod_{1\leq
k\leq
n,k\not=j}\frac{(a_{k}+\sigma)^{\alpha_{k}}}{(a_{k}-\sigma)^{\alpha_{k}+1}}\right)_{|\sigma=a_{j}}d\lambda$
$\displaystyle=\sum_{1\leq j\leq n}\frac{(-1)^{\alpha_{j}}}{\alpha_{j}!}$
$\displaystyle\int^{+{\infty}}_{1}e^{-\lambda
a_{j}}\\!\\!\\!\left(\frac{d}{d\sigma}-\lambda\right)^{\alpha_{j}}\left((a_{j}+\sigma)^{\alpha_{j}}a_{j}\hskip-12.0pt\prod_{1\leq
k\leq
n,k\not=j}\frac{(a_{k}+\sigma)^{\alpha_{k}}a_{k}}{(a_{k}-\sigma)^{\alpha_{k}+1}}\right)_{|\sigma=a_{j}}d\lambda.$
$\displaystyle=\sum_{1\leq j\leq
n}\frac{(-1)^{\alpha_{j}}}{\alpha_{j}!}\int^{+{\infty}}_{1}e^{-\lambda
a_{j}}\left(\frac{d}{d\sigma}-\lambda\right)^{\alpha_{j}}\left((a_{j}+\sigma)^{\alpha_{j}}\hskip-12.0pt\prod_{1\leq
k\leq
n,k\not=j}\frac{(a_{k}+\sigma)^{\alpha_{k}}}{(a_{k}-\sigma)^{\alpha_{k}+1}}\right)_{|\sigma=a_{j}}d\lambda$
$\displaystyle=\sum_{1\leq j\leq
n}\frac{(-1)^{\alpha_{j}}}{\alpha_{j}!}\int^{+{\infty}}_{a_{j}}\\!\\!\\!e^{-t_{j}}$
$\displaystyle\hskip
32.0pt\left(\frac{d}{da_{j}s}-\frac{t_{j}}{a_{j}}\right)^{\alpha_{j}}\left((a_{j}+a_{j}s)^{\alpha_{j}}\hskip-12.0pt\prod_{1\leq
k\leq
n,k\not=j}\frac{a_{k}(a_{k}+a_{j}s)^{\alpha_{k}}}{(a_{k}-a_{j}s)^{\alpha_{k}+1}}\right)_{|s=1}{dt_{j}}$
$\displaystyle=\sum_{1\leq j\leq
n}\frac{(-1)^{\alpha_{j}}}{\alpha_{j}!}\int^{+{\infty}}_{a_{j}}e^{-t}\left(\frac{d}{ds}-{t}\right)^{\alpha_{j}}\left((1+s)^{\alpha_{j}}\hskip-12.0pt\prod_{1\leq
k\leq
n,k\not=j}\frac{a_{k}(a_{k}+a_{j}s)^{\alpha_{k}}}{(a_{k}-a_{j}s)^{\alpha_{k}+1}}\right)_{|s=1}{dt}$
$\displaystyle=\sum_{1\leq j\leq
n}\frac{(-1)^{\alpha_{j}}}{\alpha_{j}!}\int^{+{\infty}}_{a_{j}}e^{-t}\left(\frac{d}{ds}-{1}\right)^{\alpha_{j}}\left((t+s)^{\alpha_{j}}\hskip-12.0pt\prod_{1\leq
k\leq
n,k\not=j}\frac{a_{k}(a_{k}+a_{j}s/t)^{\alpha_{k}}}{(a_{k}-a_{j}s/t)^{\alpha_{k}+1}}\right)_{|s=t}{dt}$
$\displaystyle=\sum_{1\leq j\leq
n}\frac{(-1)^{\alpha_{j}}}{\alpha_{j}!}\int^{+{\infty}}_{a_{j}}e^{-t}$
$\displaystyle\hskip
44.0pt\left(\frac{d}{d(s+t)}-{1}\right)^{\alpha_{j}}\left((t+s)^{\alpha_{j}}\hskip-12.0pt\prod_{1\leq
k\leq
n,k\not=j}\frac{ta_{k}(t(a_{k}-a_{j})+a_{j}(s+t))^{\alpha_{k}}}{(t(a_{k}+a_{j})-a_{j}(s+t))^{\alpha_{k}+1}}\right)_{|s+t=2t}{dt}$
$\displaystyle=\sum_{1\leq j\leq n}{(-1)^{\alpha_{j}}}$
$\displaystyle\times\int^{+{\infty}}_{a_{j}}e^{-t}\left(\frac{d}{ds}-{1}\right)^{\alpha_{j}}\left(\frac{s^{\alpha_{j}}}{\alpha_{j}!}\prod_{1\leq
k\leq
n,k\not=j}\frac{ta_{k}(t(a_{k}-a_{j})+a_{j}s)^{\alpha_{k}}}{(t(a_{k}+a_{j})-a_{j}s)^{\alpha_{k}+1}}\right)_{|s=2t}{dt}$
$\displaystyle=\sum_{1\leq j\leq
n}{(-1)^{\alpha_{j}}}e^{-a_{j}}\int^{+{\infty}}_{0}e^{-t}$
$\displaystyle\times\left(\frac{d}{ds}-{1}\right)^{\alpha_{j}}\left(\frac{s^{\alpha_{j}}}{\alpha_{j}!}\prod_{1\leq
k\leq
n,k\not=j}\frac{(t+a_{j})a_{k}\bigl{(}(t+a_{j})(a_{k}-a_{j})+a_{j}s\bigr{)}^{\alpha_{k}}}{\bigl{(}(t+a_{j})(a_{k}+a_{j})-a_{j}s\bigr{)}^{\alpha_{k}+1}}\right)_{|s=2t+2a_{j}}{dt}.$

We have also to deal with

$\prod_{1\leq k\leq
n,k\not=j}\frac{(t+a_{j})a_{k}\bigl{(}(t+a_{j})(a_{k}-a_{j})+a_{j}s\bigr{)}^{\alpha_{k}}}{\bigl{(}(t+a_{j})(a_{k}+a_{j})-a_{j}s\bigr{)}^{\alpha_{k}+1}}$

and

$\bigl{(}(t+a_{j})(a_{k}+a_{j})-a_{j}(2t+2a_{j})\bigr{)}=a_{j}(a_{k}+a_{j})-2a_{j}^{2}+t(a_{k}-a_{j})=(t+a_{j})(a_{k}-a_{j})$
$(t+a_{j})(a_{k}+a_{j})-a_{j}s=(t+a_{j})(a_{k}-a_{j})+a_{j}(2t+2a_{j}-s)$

so that

(3.3.21) $K_{\alpha}(a)=\sum_{1\leq j\leq
n}{(-1)^{\alpha_{j}}}e^{-a_{j}}\int^{+{\infty}}_{0}e^{-t}\\\
\times\left(\frac{d}{ds}-{1}\right)^{\alpha_{j}}\\!\\!\\!\biggl{(}\frac{s^{\alpha_{j}}}{\alpha_{j}!}\\!\\!\prod_{1\leq
k\leq
n,k\not=j}\\!\\!\\!\\!\\!\\!\\!\\!\\!\frac{(t+a_{j})a_{k}\bigl{(}(t+a_{j})(a_{k}+a_{j})+a_{j}(s-2t-2a_{j})\bigr{)}^{\alpha_{k}}}{\bigl{(}(t+a_{j})(a_{k}-a_{j})-a_{j}(s-2t-2a_{j})\bigr{)}^{\alpha_{k}+1}}\biggr{)}_{|s=2t+2a_{j}}\hskip-20.0pt{dt}.$

###  3.4. A conjecture on integrals of products of Laguerre polynomials

We formulate in this section a conjecture on the behaviour of the functions
$K_{\alpha}(a)$; as displayed in the previous sections, we know several useful
elements for the analysis of these functions, including some quite explicit
expression. However, in the non-isotropic case, we were not able to prove the
estimate $F_{\alpha}(a)\leq 1$, equivalent to $K_{\alpha}(a)\geq 0$, except
for the case $\alpha=0$. We are thus reduced to conjectural statements.

######  Conjecture 3.25.

Let $n\geq 1$ be an integer and let
$\alpha=(\alpha_{1},\dots,\alpha_{n})\in\mathbb{N}^{n}$. For
$a=(a_{1},\dots,a_{n})\in(0,+{\infty})^{n}$ we define

(3.4.1)
$K_{\alpha}(a)=\int_{\begin{subarray}{c}t=(t_{1},\dots,t_{n})\in\mathbb{R}_{+}^{n}\\\
\sum_{1\leq j\leq n}t_{j}/a_{j}\geq
1\end{subarray}}e^{-(t_{1}+\dots+t_{n})}\prod_{1\leq j\leq
n}(-1)^{\alpha_{j}}L_{\alpha_{j}}(2t_{j})dt,$

where $L_{k}$ stands for the classical Laguerre polynomial

(3.4.2) $L_{k}(X)=\Bigl{(}\frac{d}{dX}-1\Bigr{)}^{k}\frac{X^{k}}{k!}.$

Then we conjecture that, assuming $0<a_{1}\leq\dots\leq a_{n}$, we have

(3.4.3) $K_{\alpha}(a)\geq\sum_{1\leq j\leq n}e^{-a_{j}}\frac{\prod_{1\leq
l<j}a_{l}}{(j-1)!}.$

######  Remark 3.26.

A slightly stronger and more symmetrical version of the above conjecture is
that for $n,\alpha,a,K_{\alpha}$ as above, we have

(3.4.4) $K_{\alpha}(a)\geq K_{0}(a).$

It is indeed stronger since we have proven in Lemma 3.22 that $K_{0}(a)$ is
greater than the right-hand-side of (3.4.3).

######  Theorem 3.27.

The previous conjecture is a proven theorem in the following cases.

  * (1)

When $n=1$.

  * (2)

For all $n\geq 1$, when all the $a_{j}$ are equal.

  * (3)

For all $n\geq 1$, when $\alpha=0_{\mathbb{N}^{n}}$.

  * (4)

When $n=2$ and $\min(\alpha_{1},\alpha_{2})=0$.

###### Proof.

$\mathbf{(1)}$ When $n=1$, we have proven above (in Proposition 3.3) that for
$\alpha\in\mathbb{N}$, $a>0$,

(3.4.5) $K_{\alpha}(a)=e^{-a}P_{\alpha}(a)\geq e^{-a},$

which is indeed (3.4.4) in that case. With the notations of Theorem 3.5 (and
in particular where $D_{a}$ is defined in (3.1.22)) this implies

(3.4.6) ${\text{\rm Op}_{\text{\rm w}}(\mathbf{1}_{D_{a}})}\leq 1-e^{-a},$

an inequality due to P. Flandrin in the 1988 paper [11].

$\mathbf{(2)}$ Assuming that all the $a_{j}$ are equal to $a>0$, we have
proven in Theorem 3.7 that for $\alpha\in\mathbb{N}^{n}$,
$|\alpha|=\sum_{1\leq j\leq n}\alpha_{j}$,

(3.4.7)
$K_{\alpha}(a,\dots,a)\geq\frac{\Gamma(n,a)}{\Gamma(n)}=e^{-a}\sum_{1\leq
j\leq n}\frac{a^{j-1}}{(j-1)!}=K_{0}(a,\dots,a)$

since from (3.3.5), we have

$\displaystyle K_{0}(a,\dots,a)$ $\displaystyle=\int_{\begin{subarray}{c}\sum
t_{j}\geq a\\\ t_{j}\geq 0\end{subarray}}e^{-(t_{1}+\dots+t_{n})}dt$
$\displaystyle=\int_{\begin{subarray}{c}t_{n}\geq a\\\ t_{j}\geq
0\end{subarray}}e^{-(t_{1}+\dots+t_{n})}dt+\int_{0}^{a}e^{-t_{n}}\int_{\sum
t_{j}\geq a-t_{n}}e^{-(t_{1}+\dots+t_{n-1})}dt$ (inductively)
$\displaystyle=e^{-a}+\int_{0}^{a}e^{-t_{n}}e^{-(a-t_{n})}\sum_{1\leq j\leq
n-1}\frac{(a-t_{n})^{j-1}}{(j-1)!}dt_{n}$
$\displaystyle=e^{-a}\Bigl{(}1+\sum_{1\leq j\leq
n-1}\frac{a^{j}}{j!}\Bigr{)}=e^{-a}\sum_{1\leq j\leq
n}\frac{a^{j-1}}{(j-1)!},$

proving (3.4.4) in that case. With

(3.4.8)
$D_{(a)}=\\{(x,\xi)\in\mathbb{R}^{2n},2\pi\frac{|x|^{2}+|\xi|^{2}}{a}\leq
1\\},$

this implies that

(3.4.9) ${\text{\rm Op}_{\text{\rm w}}(\mathbf{1}_{D_{(a)}})}\leq
1-e^{-a}\sum_{1\leq j\leq n}\frac{a^{j-1}}{(j-1)!},$

an inequality proven in the 2010 article [33] by E. Lieb and Y. Ostrover.

$\mathbf{(3)}$ When $\alpha=0_{\mathbb{N}^{n}}$, we have proven (3.4.3) in
Lemma 3.22.

$\mathbf{(4)}$ When $n=2$, from the case $n=1$ we have
$K_{\alpha_{2}}(a_{2})=e^{-a_{2}}P_{\alpha_{2}}(a_{2}),$ so that from Lemma
3.20, we obtain

$K_{\alpha_{1},\alpha_{2}}(a_{1},a_{2})=e^{-a_{1}}P_{\alpha_{1}}(a_{1})+a_{1}\int_{0}^{1}e^{-\theta
a_{1}-(1-\theta)a_{2}}(-1)^{\alpha_{1}}L_{\alpha_{1}}(2\theta
a_{1})P_{\alpha_{2}}(a_{2}(1-\theta))d\theta,$

and if $\alpha_{1}$=0, it means that

$K_{0,\alpha_{2}}(a_{1},a_{2})=e^{-a_{1}}+a_{1}\int_{0}^{1}e^{-\theta
a_{1}-(1-\theta)a_{2}}P_{\alpha_{2}}(a_{2}(1-\theta))d\theta\\\ \geq
e^{-a_{1}}+a_{1}\int_{0}^{1}e^{-\theta
a_{1}-(1-\theta)a_{2}}d\theta=K_{0,0}(a_{1},a_{2}),$

and the reasoning is identical for $\alpha_{2}=0$, concluding the proof of the
theorem. ∎

We are interested in the Weyl quantization of the indicatrix of

(3.4.10) $D_{a_{1},\dots,a_{n}}=\\{(x,\xi)\in\mathbb{R}^{2n},2\pi\sum_{1\leq
j\leq n}\frac{x_{j}^{2}+\xi_{j}^{2}}{a_{j}}\leq 1\\},\quad a_{j}>0,$

and we have a weaker conjecture.

######  Conjecture 3.28 (A weak form of Conjecture 3.25).

With $n,\alpha,a,K_{\alpha}$ as in Conjecture 3.25, we conjecture that

(3.4.11) $K_{\alpha}(a)\geq 0.$

Note that Inequality (3.4.11) is equivalent to

(3.4.12) ${\text{\rm Op}_{\text{\rm
w}}(\mathbf{1}_{D_{a_{1},\dots,a_{n}}})}\leq 1.$

######  Remark 3.29.

In the first place, although the second conjecture is much weaker than the
first, there is no reason to believe that the weak conjecture should be easier
to prove than the first: in particular, in the known cases, it is indeed the
proof of the precise statement (3.4.3) which leads to (3.4.11) and we are not
aware of a direct proof of (3.4.11), even in one dimension.

A summary of our knowledge on the functions $K_{\alpha}$. As proven in Remarks
3.14 and 3.15, the functions $K_{\alpha}$ are entire functions given on the
open subset (3.3.6) by Formula (3.3.5) (see also Formula (3.3.17)). Moreover
the function $F_{\alpha}(a)=1-K_{\alpha}(a)$ can be expressed as a simple
integral for $a_{j}>0$,

(3.4.13)
$F_{\alpha}(a_{1},\dots,a_{n})=\int_{\mathbb{R}}\frac{\sin\tau}{\pi\tau}\Bigl{(}\prod_{1\leq
j\leq
n}\frac{(1+i\tau\mu_{j})^{\alpha_{j}}}{(1-i\tau\mu_{j})^{\alpha_{j}+1}}\Bigr{)}d\tau,\quad\mu_{j}=\frac{1}{a_{j}},$

and we have an explicit expression of the function $K_{\alpha}$ as a sum of
simple integrals in (3.3.21). However, having an explicit expression does not
mean much and for instance, we do have several explicit expressions for the
Laguerre polynomials but Inequality (9.3.3) remains very hard work, requiring
a deep understanding of these polynomials. We have also an induction formula
in Lemma 3.20. As a further remark, we have the following

######  Lemma 3.30.

Let $n,\alpha,a,K_{\alpha}$ as in Conjecture 3.25. Then we have

(3.4.14)
$\displaystyle\lim_{a_{n}\rightarrow+{\infty}}K_{\alpha_{1},\dots,\alpha_{n-1},\alpha_{n}}(a_{1},\dots,a_{n-1},a_{n})=K_{\alpha_{1},\dots,\alpha_{n-1}}(a_{1},\dots,a_{n-1}),$
(3.4.15)
$\displaystyle\lim_{a_{1}\rightarrow{0}_{+}}K_{\alpha_{1},\alpha_{2},\dots,\alpha_{n}}(a_{1},a_{2},\dots,a_{n})=1.$

###### Proof.

Formula (3.3.5) and Lebesgue Dominated Convergence Theorem imply the first
equality (3.4.14). Lemma 3.20, in which we may swap the variables $a_{1}$ and
$a_{n}$ gives for $a_{1}>0$

$K_{\alpha_{1},\alpha_{2},\dots,\alpha_{n}}(a_{1},a_{2},\dots,a_{n})=e^{-a_{1}}P_{\alpha_{1}}(a_{1})\\\
+a_{1}\int_{0}^{1}e^{-\theta
a_{1}}(-1)^{\alpha_{1}}L_{\alpha_{1}}(2a_{1}\theta)K_{\alpha_{2},\dots,\alpha_{n}}\bigl{(}a_{2}(1-\theta),\dots,a_{n}(1-\theta)\bigr{)}d\theta,$

and since $P_{\alpha_{1}}$ is a polynomial such that $P_{\alpha_{1}}(0)=1$, we
get (3.4.15). ∎

Reasons to believe in the conjecture. This is true in one dimension, also in
$n$ dimensions for spheres and it is a quadratic problem in the sense that
ellipsoids are convex subsets of $\mathbb{R}^{2n}$ characterized by an
inequality

$\\{X\in\mathbb{R}^{2n},p(X)\leq 0\\},$

where $p$ is a polynomial of degree 2 with a positive-definite quadratic part.
We shall see below in this paper that convexity of a set $A$ does not
guarantee that the quantization ${\text{\rm Op}_{\text{\rm
w}}(\mathbf{1}_{A})}$ is smaller than 1 as an operator and that Flandrin’s
conjecture is not true, but it is hard to believe that such a phenomenon could
occur for ellipsoids. We must point out a specific feature of anisotropy
related to Mehler’s formula (2.2.1): if all the $\mu_{j}$ are equal to the
same $\mu>0$ (this is the isotropic case), then, with
$q_{\mu}(x,\xi)=\mu(|x|^{2}+|\xi|^{2})$, we have

${\text{\rm Op}_{\text{\rm w}}(e^{2i\pi\tau
q_{\mu}(x,\xi)})}=\phi(\tau\mu)e^{2i\arctan(\tau\mu)\sum_{1\leq j\leq
n}\pi(x_{j}^{2}+D_{j}^{2})},$

where $\phi(\tau\mu)$ is a scalar quantity. As a consequence, if we quantize
$F(q_{\mu}(x,\xi))$, we get

${\text{\rm Op}_{\text{\rm
w}}\bigl{(}F\bigl{(}q_{\mu}(x,\xi)\bigr{)}\bigr{)}}=\int_{\mathbb{R}}\hat{F}(\tau)\phi(\tau\mu)e^{2i\frac{\arctan(\tau\mu)}{\mu}\pi{\text{\rm
Op}_{\text{\rm w}}(q_{\mu})}}d\tau,$

and thus

${\text{\rm Op}_{\text{\rm
w}}\bigl{(}F\bigl{(}q_{\mu}(x,\xi)\bigr{)}\bigr{)}}=\widetilde{F}({\text{\rm
Op}_{\text{\rm
w}}(q_{\mu})}),\qquad\widetilde{F}(\lambda)=\int_{\mathbb{R}}\hat{F}(\tau)\phi(\tau\mu)e^{2i\pi\frac{\arctan(\tau\mu)}{\mu}\lambda}d\tau,$

and ${\text{\rm Op}_{\text{\rm
w}}\bigl{(}F\bigl{(}q_{\mu}(x,\xi)\bigr{)}\bigr{)}}$ appears as a function of
the self-adjoint operator ${\text{\rm Op}_{\text{\rm w}}(q_{\mu})}$. Following
the same route in the anisotropic case, we get, with

(3.4.16) $\displaystyle q_{\mu}(x,\xi)$ $\displaystyle=\sum_{1\leq j\leq
n}\mu_{j}(x_{j}^{2}+\xi_{j}^{2}),$ (3.4.17) $\displaystyle{\text{\rm
Op}_{\text{\rm w}}\bigl{(}F\bigl{(}q_{\mu}(x,\xi)\bigr{)}\bigr{)}}$
$\displaystyle=\int_{\mathbb{R}}\hat{F}(\tau)\phi(\tau\mu)e^{2i\pi\sum_{1\leq
j\leq
n}(\frac{\arctan(\tau\mu_{j}}{\mu_{j}})\mu_{j}(x_{j}^{2}+D_{j}^{2})}d\tau,$

and since $\frac{1}{\mu_{j}}\arctan(\tau\mu_{j})$ does depend on $\mu_{j}$
(and not only on $\tau$), the operator ${\text{\rm Op}_{\text{\rm
w}}\bigl{(}F\bigl{(}q_{\mu}(x,\xi)\bigr{)}\bigr{)}}$ is not a function of the
self-adjoint operator ${\text{\rm Op}_{\text{\rm w}}(q_{\mu})}$. As a final
comment on the strongest form of the Conjecture (3.4.4), we would say that it
could be seen as a property of the Laguerre polynomials, known in the case
$n=1$, where it stands as follows: we define for $k\in\mathbb{N}$, the
polynomial $P_{k}$ by

(3.4.18) $P_{k}(x)=\int_{0}^{+{\infty}}e^{-t}(-1)^{k}L_{k}(2x+2t)dt,$

and we have $P_{k}(0)=1$ from (9.3.5). Moreover, we have the inequality
(equivalent to (3.4.4) for $n=1$)

(3.4.19) $\forall x\geq 0,\quad P_{k}(x)\geq P_{k}(0).$

We note that $e^{-x}P_{k}(x)=\int_{x}^{+{\infty}}e^{-s}(-1)^{k}L_{k}(2s)ds,$
so that the unique solution $P_{k}$ of the Initial Value Problem for the ODE

(3.4.20) $P_{k}(x)-P^{\prime}_{k}(x)=(-1)^{k}L_{k}(2x),\quad P_{k}(0)=1,$

does satisfy (3.4.19). We note that from Lemma 3.2, we have

$P^{\prime}_{k}(X)=2\sum_{0\leq l<k}(-1)^{l}L_{l}(2X),$

so that (3.4.19) is a consequence of Feldheim Inequality (9.3.3). Let us
reformulate (3.4.4), using the polynomials $P_{k}$: for $a_{j}\geq 0$,

(3.4.21)
$K_{\alpha}(a)=\int_{\begin{subarray}{c}t=(t_{1},\dots,t_{n})\in\mathbb{R}_{+}^{n}\\\
\sum_{1\leq j\leq n}t_{j}/a_{j}\geq 1\end{subarray}}\prod_{1\leq j\leq
n}\frac{\partial}{\partial
t_{j}}\Bigl{\\{}-e^{-t_{j}}P_{\alpha_{j}}(t_{j})\Bigr{\\}}dt\\\ \geq
K_{0}(a)=\int_{\begin{subarray}{c}t=(t_{1},\dots,t_{n})\in\mathbb{R}_{+}^{n}\\\
\sum_{1\leq j\leq n}t_{j}/a_{j}\geq 1\end{subarray}}\prod_{1\leq j\leq
n}\frac{\partial}{\partial t_{j}}\Bigl{\\{}-e^{-t_{j}}\Bigr{\\}}dt,$

which is equivalent to

(3.4.22) $\int H\bigl{(}1-\sum_{1\leq j\leq n}s_{j}\bigr{)}\prod_{1\leq j\leq
n}H(s_{j})\frac{\partial}{\partial
s_{j}}\Bigl{\\{}-e^{-a_{j}s_{j}}P_{\alpha_{j}}(a_{j}s_{j})\Bigr{\\}}ds\\\
\leq\int H\bigl{(}1-\sum_{1\leq j\leq n}s_{j}\bigr{)}\prod_{1\leq j\leq
n}a_{j}H(s_{j})e^{-a_{j}s_{j}}ds,$

where $H=\mathbf{1}_{\mathbb{R}_{+}}$ (Heaviside function). This is equivalent
to

(3.4.23) $\int H\bigl{(}1-\sum_{1\leq j\leq n}s_{j}\bigr{)}\prod_{1\leq j\leq
n}H(s_{j})e^{-a_{j}s_{j}}\left(a_{j}-\frac{\partial}{\partial
s_{j}}\right)\Bigl{\\{}P_{\alpha_{j}}(a_{j}s_{j})\Bigr{\\}}ds\\\ \leq\int
H\bigl{(}1-\sum_{1\leq j\leq n}s_{j}\bigr{)}\prod_{1\leq j\leq
n}a_{j}H(s_{j})e^{-a_{j}s_{j}}ds,$

i.e. to

(3.4.24) $\int H\bigl{(}1-\sum_{1\leq j\leq n}s_{j}\bigr{)}\prod_{1\leq j\leq
n}H(s_{j})e^{-a_{j}s_{j}}\\\ \times\left(\prod_{1\leq j\leq
n}a_{j}-\prod_{1\leq j\leq n}\left(a_{j}-\frac{\partial}{\partial
s_{j}}\right)\Bigl{\\{}P_{\alpha_{j}}(a_{j}s_{j})\Bigr{\\}}\right)ds\geq 0.$

Note that for $n=1$, it means for $a\geq 0$,

$\int_{0}^{1}e^{-as}\bigl{(}a-aP_{k}(as)+aP^{\prime}_{k}(as)\bigr{)}ds={1-e^{-a}}+\int_{0}^{1}\frac{d}{ds}\bigl{\\{}e^{-as}P_{k}(as)\bigr{\\}}\\\
={1-e^{-a}}+e^{-a}P_{k}(a)-P_{k}(0)=e^{-a}\bigl{(}P_{k}(a)-1\bigr{)}\geq 0,$

which holds true from (3.4.19).

######  Remark 3.31.

There are several classical results on products of Laguerre polynomials, in
particular the article [5], On some expansions in Laguerre polynomials by A.
Erdélyi and also the paper [34], Linearization of the products of the
generalized Lauricella polynomials and the multivariate Laguerre polynomials
via their integral representations by Shuoh-Jung Liu, Shy-Der Lin, Han-Chun Lu
and H. M. Srivastava. However it seems that the non-negativity of the
polynomials $P_{\alpha;1},P^{\prime}_{\alpha;1}$ do not suffice to tackle the
conjecture in two dimensions and more.

##  4\. Parabolas

###  4.1. Preliminary remarks

We start with a picture, demonstrating that the epigraph of a parabola is an
increasing union of ellipses.

Figure 4. The epigraph of a parabola is an increasing union of ellipses.

It is easy to see that the epigraph of a parabola, i.e. the set
$\\{(x,\xi)\in\mathbb{R}^{2},\xi>x^{2}\\}$ is a countable increasing union of
ellipses in the sense that

(4.1.1) $\mathcal{P}=\\{(x,\xi)\in\mathbb{R}^{2},\xi>x^{2}\\}=\cup_{k\geq
1}\underbrace{\\{(x,\xi)\in\mathbb{R}^{2},\xi>x^{2}+k^{-2}\xi^{2}\\}}_{\mathcal{E}_{k}}.$

Note that for $k\geq 1$ we have
$\mathcal{E}_{k}\subset\mathcal{E}_{k+1}\subset\mathcal{P}$ since
$x^{2}+k^{-2}\xi^{2}\geq x^{2}+(k+1)^{-2}\xi^{2}>x^{2},$ from the fact that
$\xi>0$ on $\mathcal{E}_{k}$. Moreover, if $\xi>x^{2}$ and
$k>{\xi}/{\sqrt{\xi-x^{2}}}$, we get $(x,\xi)\in\mathcal{E}_{k}$.

######  Remark 4.1.

The ellipse $\mathcal{E}_{k}$ is symplectically equivalent to a circle with
area $\frac{\pi k^{3}}{4}$ since

$x^{2}+k^{-2}\xi^{2}-\xi=x^{2}+k^{-2}\bigl{(}\xi-\frac{k^{2}}{2}\bigr{)}^{2}-\frac{k^{2}}{4}=(\lambda^{-1}y)^{2}+k^{-2}\bigl{(}\lambda\eta-\frac{k^{2}}{2}\bigr{)}^{2}-\frac{k^{2}}{4}\\\
=\lambda^{-2}y^{2}+\lambda^{2}k^{-2}\bigl{(}\eta-\frac{k^{2}}{2\lambda}\bigr{)}^{2}-\frac{k^{2}}{4},$

so that choosing $\lambda$ such that $\lambda^{-2}=\lambda^{2}k^{-2}$, e.g.
$\lambda=\sqrt{k}$, we get

$x^{2}+k^{-2}\xi^{2}-\xi=k^{-1}\bigl{(}y^{2}+(\eta-\frac{k^{2}}{2\lambda})^{2}\bigr{)}-\frac{k^{2}}{4},$

and
$\mathcal{E}_{k}=\\{(y,\zeta)\in\mathbb{R}^{2},y^{2}+\zeta^{2}<\frac{k^{3}}{4}\\},$
where $(y,\zeta)$ are the affine symplectic coordinates

$y=xk^{1/2},\quad\zeta=\xi k^{-1/2}-\frac{k^{3/2}}{2}.$

######  Lemma 4.2.

Let $u\in\mathscr{S}(\mathbb{R})$. Then $\mathcal{W}(u,u)$ belongs to
$\mathscr{S}(\mathbb{R}^{2})$ and with $\mathcal{E},\mathcal{E}_{k}$ defined
by (4.1.1), we have

$\iint_{\xi>x^{2}}\mathcal{W}(u,u)(x,\xi)dxd\xi=\lim_{k\rightarrow+{\infty}}\iint_{\mathcal{E}_{k}}\mathcal{W}(u,u)(x,\xi)dxd\xi\leq\|u\|_{L^{2}(\mathbb{R})}^{2}.$

###### Proof.

Since $\mathcal{W}(u,u)$ belongs to $\mathscr{S}(\mathbb{R}^{2n})\subset
L^{1}(\mathbb{R}^{2n})$, we may apply the Lebesgue Dominated Convergence
Theorem and (4.1.1) to obtain the equality in the lemma. On the other hand
Theorem 3.5 and Remark 4.1 imply

$\iint_{\mathcal{E}_{k}}\mathcal{W}(u,u)(x,\xi)dxd\xi=\langle{\text{\rm
Op}_{\text{\rm
w}}(\mathbf{1}_{\mathcal{E}_{k}})}u,u\rangle\leq(1-e^{-\frac{\pi
k^{3}}{2}})\|u\|_{L^{2}(\mathbb{R})}^{2}\leq\|u\|_{L^{2}(\mathbb{R})}^{2},$

and the sought result. ∎

######  Remark 4.3.

Moreover, Theorem 3.5 and the expression of $F_{0}(a)=1-e^{-a}$ imply that
with $\psi_{0}$ defined in (9.1.31), we have

$\iint_{\mathcal{E}_{k}}\mathcal{W}(\psi_{0},\psi_{0})(x,\xi)dxd\xi=\langle{\text{\rm
Op}_{\text{\rm
w}}(\mathbf{1}_{\mathcal{E}_{k}})}\psi_{0},\psi_{0}\rangle=\|\psi_{0}\|_{L^{2}(\mathbb{R})}^{2}(1-e^{-\pi
k^{3}/3}),$

so that from Lemma 4.2, we have
$\iint_{\mathcal{P}}\mathcal{W}(\psi_{0},\psi_{0})(x,\xi)dxd\xi=\|\psi_{0}\|_{L^{2}(\mathbb{R})}^{2},$
entailing

(4.1.2)
$\sup_{\phi\in\mathscr{S}(\mathbb{R}),\|\phi\|_{L^{2}(\mathbb{R})}=1}\iint_{\mathcal{P}}\mathcal{W}(\phi,\phi)(x,\xi)dxd\xi=1.$

######  Remark 4.4.

We want to study the operator with Weyl symbol $H(\xi-x^{2})$
($H=\mathbf{1}_{\mathbb{R}_{+}}$ is the Heaviside function) and since
$\xi-x^{2}$ is a polynomial with degree less than 2, see from (1.2.5) that
${\text{\rm Op}_{\text{\rm w}}\bigl{(}H(\xi-x^{2})\bigr{)}}$ commutes with
$D_{x}-x^{2}=e^{2\pi ix^{3}/3}D_{x}e^{-2\pi ix^{3}/3}$, and the latter has
(continuous) spectrum $\mathbb{R}$: we expect thus that ${\text{\rm
Op}_{\text{\rm w}}\bigl{(}H(\xi-x^{2})\bigr{)}}$ should have continuous
spectrum and be conjugated to a Fourier multiplier.

###  4.2. Calculation of the kernel

The Weyl symbol of the operator ${\text{\rm Op}_{\text{\rm
w}}(\mathbf{1}_{\mathcal{P}})}$ is

$H(\xi-x^{2}),$

($\mathcal{P}$ is defined in (4.1.1), $H$ is the Heaviside function
$H=\mathbf{1}_{\mathbb{R}_{+}}$), corresponding to the distribution kernel
$k_{\mathcal{P}}(x,y)$ obtained from Proposition 1.9 by (we use freely
integrals meaning only Fourier transform in the distribution sense),

$k_{\mathcal{P}}(x,y)=\int
e^{2i\pi(x-y)\xi}H(\xi-(\frac{x+y}{2})^{2})d\xi=\int
e^{2i\pi(x-y)(\xi+(\frac{x+y}{2})^{2})}H(\xi)d\xi\\\
=e^{2i\pi(x-y)(\frac{x+y}{2})^{2}}\frac{1}{2}\bigl{(}\delta_{0}(y-x)+\frac{1}{i\pi(y-x)}\bigr{)}\\\
=\frac{\delta_{0}(y-x)}{2}+\frac{e^{2i\pi(x-y)(\frac{x+y}{2})^{2}}}{2i\pi(y-x)}.$

We have

$4(x-y)(\frac{x+y}{2})^{2}=(x^{2}-y^{2})(x+y)=x^{3}-y^{3}+x^{2}y-y^{2}x\\\
=\frac{4}{3}(x^{3}-y^{3})+\frac{1}{3}(y-x)^{3},$

so that

(4.2.1)
$k_{\mathcal{P}}(x,y)=e^{i\frac{2\pi}{3}x^{3}}\left(\frac{\delta_{0}(y-x)}{2}+\frac{e^{i\frac{\pi}{2}\frac{1}{3}(y-x)^{3}}}{2i\pi(y-x)}\right)e^{-i\frac{2\pi}{3}y^{3}},$

and the operator ${\text{\rm Op}_{\text{\rm w}}(\mathbf{1}_{\mathcal{P}})}$ is
unitarily equivalent to the operator with kernel

(4.2.2)
$\tilde{k}(x,y)=\frac{\delta_{0}(y-x)}{2}+\frac{e^{i\frac{\pi}{6}(y-x)^{3}}}{2i\pi(y-x)}.$

We have proven the following result.

######  Lemma 4.5.

The operator with Weyl symbol
$\mathbb{R}^{2}\ni(x,\xi)\mapsto\mathbf{1}_{\mathbb{R}_{+}}(\xi-x^{2})$ has
the distribution kernel

$k_{\mathcal{P}}(x,y)=e^{i\frac{2\pi}{3}x^{3}}\left(\frac{\delta_{0}(y-x)}{2}+\frac{e^{i\frac{\pi}{6}(y-x)^{3}}}{2i\pi(y-x)}\right)e^{-i\frac{2\pi}{3}y^{3}},$

and is thus unitarily equivalent to

(4.2.3) $\frac{\operatorname{Id}}{2}+\text{convolution
with\quad}\frac{ie^{-i\pi t^{3}/6}}{2\pi}\textrm{\rm pv}\frac{1}{t}.$

######  Lemma 4.6.

The distribution $\frac{ie^{-i\pi t^{3}/6}}{2\pi}\textrm{\rm pv}\frac{1}{t}$
has the Fourier transform

(4.2.4) $\frac{1}{2\pi}\int\frac{\sin(2\pi as\tau+\frac{s^{3}}{3})}{s}ds,\quad
a=(2/\pi)^{1/3}.$

The operator (4.2.3) is the Fourier multiplier $\omega(D_{t})$ with

(4.2.5)
$\omega(\tau)=\frac{1}{2}\left(1+\frac{1}{\pi}\int_{-{\infty}}^{+{\infty}}\frac{\sin(s\eta+\frac{s^{3}}{3})}{s}ds\right),\quad\eta=2^{4/3}\pi^{2/3}\tau.$

###### Proof.

We calculate in the distribution sense ($t=as,a=(2/\pi)^{1/3}$),

$\int e^{-2i\pi t\tau}i\frac{e^{-i\pi t^{3}/6}}{2\pi t}dt=\frac{i}{2\pi}\int
e^{-2i\pi as\tau}\frac{e^{-i\pi
a^{3}s^{3}/6}}{s}ds=\frac{i}{2\pi}\int\frac{(-i)\sin(\frac{s^{3}}{3}+2\pi
as\tau)}{s}ds\\\ =\frac{1}{2\pi}\int\frac{\sin(2\pi
as\tau+\frac{s^{3}}{3})}{s}ds,$

so that with $\eta=2\pi a\tau$, we get

$\omega(\tau)=\frac{1}{2}\left(1+\frac{1}{\pi}\int_{-{\infty}}^{+{\infty}}\frac{\sin(s\eta+\frac{s^{3}}{3})}{s}ds\right)=\frac{1}{2}\bigl{(}1-F(\eta)\bigr{)}=G(\eta),$

proving the lemma. ∎

######  Lemma 4.7.

We have, with $\eta=2^{4/3}\pi^{2/3}\tau$,

(4.2.6)
$\displaystyle\omega(\tau)=\frac{1}{2}\left(1+\frac{1}{\pi}\int_{-{\infty}}^{+{\infty}}\frac{\sin(s\eta+\frac{s^{3}}{3})}{s}ds\right)=G(\eta),\quad\omega(0)=\frac{2}{3}=G(0),$
(4.2.7) $\displaystyle
G^{\prime}(\eta)=\frac{1}{2\pi}\int_{\mathbb{R}}\cos(s\eta+\frac{s^{3}}{3})ds=\operatorname{Re}\frac{1}{2\pi}\int_{\mathbb{R}}\exp{i(s\eta+\frac{s^{3}}{3})}ds=\operatorname{\tt
A\\!i}(\eta),$ (4.2.8) $\displaystyle
G(\eta)=\frac{2}{3}+\int_{0}^{\eta}\operatorname{\tt A\\!i}(\xi)d\xi,$

where $\operatorname{\tt A\\!i}$ is the Airy function defined as the inverse
Fourier transform of $t\mapsto e^{i(2\pi t)^{3}/3}$.

###### Proof.

We have

(4.2.9)
$\frac{1}{\pi}\int_{-{\infty}}^{+{\infty}}\frac{\sin(\frac{s^{3}}{3})}{s}ds=\frac{1}{\pi}\int_{-{\infty}}^{+{\infty}}\frac{\sin(\sigma)}{3^{1/3}\sigma^{1/3}}3^{1/3}\frac{1}{3}\sigma^{-2/3}d\sigma\\\
=\frac{1}{3\pi}\int_{-{\infty}}^{+{\infty}}\frac{\sin\sigma}{\sigma}d\sigma=\frac{1}{3},$

proving (4.2.6). We have also

$G(\eta)=\frac{1}{2}+\operatorname{Im}\Bigl{\\{}\text{Inverse Fourier
Transform}\big{\\{}y\mapsto e^{i(2\pi y)^{3}/3}\text{pv}(\frac{1}{2\pi
y})\bigr{\\}}\Bigr{\\}},$

and thus

$G^{\prime}(\eta)=\operatorname{Im}\Bigl{\\{}\text{Inverse Fourier
Transform}\big{\\{}y\mapsto e^{i(2\pi y)^{3}/3}i\bigr{\\}}\Bigr{\\}}\\\
=\operatorname{Im}\left(\int e^{2i\pi y\eta}e^{i(2\pi
y)^{3}/3}idy\right)=\operatorname{Im}\left(\frac{1}{2\pi}\int
e^{it\eta}e^{it^{3}/3}idt\right)=\operatorname{\tt A\\!i}(\eta),$

which is (4.2.7), implying (4.2.8). ∎

######  Lemma 4.8.

With $G$ defined in Lemma 4.7, we get that $G$ is an entire function, real-
valued on the real line such that

(4.2.10)
$\lim_{\eta\rightarrow+{\infty}}G(\eta)=1,\quad\lim_{\eta\rightarrow-{\infty}}G(\eta)=0,$

and moreover with $\eta_{0}$ the largest zero of the Airy function
($\eta_{0}\approx-2.33811$), the function $G$ has an absolute minimum at
$\eta_{0}$ with $G(\eta_{0})\approx-0.274352$,

(4.2.11) $\forall\eta\in\mathbb{R},\quad G(\eta_{0})\leq G(\eta)<1.$

###### Proof.

The first statements follow from Lemma 4.7 and (4.2.10) is implied by (4.2.8)
and (9.6.34), (9.6.38). The strict inequality in (4.2.11) follows for
$\eta\geq 0$ from (4.2.7) since $\operatorname{\tt A\\!i}$ is positive on
$[0,+{\infty})$ so that $G$ is strictly increasing there from $G(0)=2/3$ to
$G(+{\infty})=1$. The other statements are proven in Section 9.6 of the
Appendix. ∎

###  4.3. The main result

Collecting the results of Lemmas 4.5, 4.6, 4.7, 4.8 and of Section 9.6 in the
Appendix, we have proven the following theorem.

######  Theorem 4.9.

Let $H(\xi-x^{2})=\mathbf{1}\\{(x,\xi)\in\mathbb{R}^{2},\xi\geq x^{2}\\}$ be
the indicatrix of the epigraph of the parabola with equation $\xi=x^{2}$. Then
the operator with Weyl symbol $H(\xi-x^{2})$ is unitary equivalent to the
Fourier multiplier $G(2^{4/3}\pi^{2/3}\tau)$ where

(4.3.1) $G(\eta)=\frac{2}{3}+\int_{0}^{\eta}\operatorname{\tt
A\\!i}(\xi)d\xi=\int_{-{\infty}}^{\eta}\operatorname{\tt
A\\!i}(\xi)d\xi,\qquad\text{($\operatorname{\tt A\\!i}$ is the Airy
function)}.$

The function $G$ is entire on $\mathbb{C}$, real valued on the real line and
such that

$G(\mathbb{R})=[G(\eta_{0}),1),$

where $\eta_{0}$ is the largest zero of the Airy function

(4.3.2) we have $\eta_{0}\approx-2.338107410,\qquad
G(\eta_{0})\approx-0.2743520591$.

The operator with Weyl symbol $H(\xi-x^{2})$ is self-adjoint bounded on
$L^{2}(\mathbb{R})$ with norm $1$, with spectrum equal to $[G(\eta_{0}),1]$
(continuous spectrum) and

(4.3.3) $\forall u\in L^{2}(\mathbb{R}),\qquad
G(\eta_{0})\|u\|_{L^{2}(\mathbb{R})}^{2}\leq\iint_{\xi\geq
x^{2}}\mathcal{W}(u,u)(x,\xi)dxd\xi\leq\|u\|_{L^{2}(\mathbb{R})}^{2}.$

Figure 5. The function $G$. More details on $G$ are given in our Appendix 9.6.

###  4.4. Paraboloids, a conjecture

We are interested now in multi-dimensional versions of the previous results,
namely, we would like to find a bound for integrals of the Wigner distribution
on paraboloids of $\mathbb{R}^{2n}$ for $n\geq 2$. Let us start with recalling
Theorem 21.5.3 in [22], a version of which was given in our Theorem 3.12 in
the positive-definite case.

####  4.4.1. On non-negative quadratic forms

######  Theorem 4.10 (Symplectic reduction of quadratic forms, Theorem 21.5.3
in [22]).

Let $q$ be a non-negative quadratic form on
$\mathbb{R}^{n}\times\mathbb{R}^{n}$ equipped with the canonical symplectic
form (1.2.21). Then there exists $S$ in the symplectic group
$Sp(n,\mathbb{R})$ of $\mathbb{R}^{2n}$,

$r\in\\{0,\dots,n\\}$, $\mu_{1},\dots,\mu_{r}$ positive, and $s\in\mathbb{N}$
such that $r+s\leq n$,

so that for all $X=(x,\xi)\in\mathbb{R}^{n}\times\mathbb{R}^{n}$,

(4.4.1) $q(SX)=\sum_{1\leq j\leq
r}\mu_{j}(x_{j}^{2}+\xi_{j}^{2})+\sum_{r+1\leq j\leq r+s}x_{j}^{2}.$

######  Definition 4.11.

Let $n\in\mathbb{N}^{*}$ and let $\mathbb{R}^{2n}$ be equipped with the
canonical symplectic form (1.2.21). Let $q$ be a non-negative quadratic form
on $\mathbb{R}^{2n}$ with rank $2n-1$ and $T$ be a non-zero vector in
$\mathbb{R}^{2n}$ such that $q(\sigma T)=0$. A paraboloid $\mathcal{P}$ of
$\mathbb{R}^{2n}$ with vertex $0$ and shape $(q,T)$ is defined by

(4.4.2) $\mathcal{P}=\\{X\in\mathbb{R}^{2n},q(X)\leq[X,T]\\}.$

A paraboloid $\mathcal{Q}$ with vertex $m\in\mathbb{R}^{2n}$ and shape $(q,T)$
is defined as

(4.4.3) $\mathcal{Q}=\mathcal{P}+m,$

where $\mathcal{P}$ is a paraboloid with vertex $0$ and shape $(q,T)$.

######  Remark 4.12.

We can find some symplectic coordinates such that

$q(X)-[X,T]=\sum_{1\leq j\leq r}\mu_{j}(x_{j}^{2}+\xi_{j}^{2})+\sum_{r+1\leq
j\leq r+s}x_{j}^{2}+\sum_{1\leq j\leq n}(x_{j}\tau_{j}-\xi_{j}t_{j}),$

with $2r+s=2n-1$. We can get rid of the linear terms
$x_{j}\tau_{j}-\xi_{j}t_{j}$ when $1\leq j\leq r$ by writing

$\mu_{j}(x_{j}^{2}+\xi_{j}^{2})+x_{j}\tau_{j}-\xi_{j}t_{j}=\mu_{j}\bigl{(}x_{j}+\frac{\tau_{j}}{2\mu_{j}}\bigr{)}^{2}+\mu_{j}\bigl{(}\xi_{j}-\frac{t_{j}}{2\mu_{j}}\bigr{)}^{2}-\frac{1}{4\mu_{j}}(t_{j}^{2}+\tau_{j}^{2}),$

and also of $x_{j}\tau_{j}$ for $r+1\leq j\leq r+s$, since

$x_{j}^{2}+x_{j}\tau_{j}=(x_{j}+\frac{\tau_{j}}{2})^{2}-\frac{\tau_{j}^{2}}{4}.$

We are left with using affine symplectic coordinates $(y,\eta)$ so that

$q(X)-[X,T]=\sum_{1\leq j\leq r}\mu_{j}(y_{j}^{2}+\eta_{j}^{2})+\sum_{r+1\leq
j\leq r+s}y_{j}^{2}\\\ -\sum_{r+1\leq j\leq r+s}\eta_{j}t_{j}+\sum_{r+s+1\leq
j\leq n}(y_{j}\tau_{j}-\eta_{j}t_{j})-a.$

Since we have $2r+s=2n-1$, we get $r+s+1=2n-r$: we cannot have $r+s+1\leq n$
since it would imply that $2n-r\leq n$ and thus $r\geq n$, which is
incompatible with $2r+s=2n-1,r,s\geq 0.$ We get then that $s=2l+1,r=n-1-l$ and
since $r+s\leq n,1\leq s$, we have $l=0$, $s=1$, $r=n-1$, and

$q(X)-[X,T]=\sum_{1\leq j\leq
n-1}\mu_{j}(y_{j}^{2}+\eta_{j}^{2})+y_{n}^{2}-\eta_{n}t_{n}-a,$

and $t_{n}\in\mathbb{R}^{*}$. With
$y_{n}=t^{1/3}\tilde{y}_{n},\eta_{n}=t^{-1/3}\tilde{\eta}_{n}$, we get

$q(X)-[X,T]=\sum_{1\leq j\leq
n-1}\mu_{j}(y_{j}^{2}+\eta_{j}^{2})+t^{2/3}(\tilde{y}_{n}^{2}-\tilde{\eta}_{n}-at^{-2/3}),$

and the inequality $q(X)-[X,T]\leq 0$ is equivalent to

$\sum_{1\leq j\leq
n-1}t^{-2/3}\mu_{j}(y_{j}^{2}+\eta_{j}^{2})+\tilde{y}_{n}^{2}\leq\tilde{\eta}_{n}+at^{-2/3}.$

We can thus assume ab initio that our paraboloid is given by the inequality

(4.4.4) $\sum_{1\leq j\leq
n-1}\nu_{j}(x_{j}^{2}+\xi_{j}^{2})+x_{n}^{2}\leq\xi_{n}.$

####  4.4.2. On the kernel for the paraboloid

We shall consider the paraboloid

(4.4.5) $\mathcal{P}_{n}=\\{(x,\xi)\in\mathbb{R}^{2n},\ x_{n}^{2}+\sum_{1\leq
j\leq n-1}(x_{j}^{2}+\xi_{j}^{2})\leq\xi_{n}\\}.$

We have, with
$X^{\prime}=(x^{\prime};\xi^{\prime})=(x_{1},\dots,x_{n-1};\xi_{1},\dots,\xi_{n-1})$,

$\displaystyle P={\text{\rm Op}_{\text{\rm
w}}\bigl{(}H(\xi_{n}-x_{n}^{2}-|X^{\prime}|^{2})\bigr{)}}=\int_{\mathbb{R}}\hat{H}(\tau){\text{\rm
Op}_{\text{\rm w}}(e^{2i\pi\tau(\xi_{n}-x_{n}^{2})})}{\text{\rm Op}_{\text{\rm
w}}(e^{-2i\pi\tau|X^{\prime}|^{2}})}d\tau$ $\displaystyle=\sum_{k\geq
0}\int_{\mathbb{R}}\hat{H}(\tau)\mathbb{P}_{k;n-1}\otimes{\text{\rm
Op}_{\text{\rm
w}}(e^{2i\pi\tau(\xi_{n}-x_{n}^{2})})}e^{-i(\arctan\tau)(2k+n-1)}(1+\tau^{2})^{-\frac{(n-1)}{2}}d\tau$
$\displaystyle=\frac{1}{2}\operatorname{Id}$ $\displaystyle{\hskip
15.0pt}{\hskip 15.0pt}+\frac{1}{2i\pi}\sum_{k\geq
0}\mathbb{P}_{k;n-1}\otimes\int_{\mathbb{R}}{\text{\rm Op}_{\text{\rm
w}}(e^{2i\pi\tau(\xi_{n}-x_{n}^{2})})}\frac{1}{\tau}\Bigl{(}\frac{1-i\tau}{(1+\tau^{2})^{1/2}}\Bigr{)}^{2k+{n-1}}(1+\tau^{2})^{-\frac{(n-1)}{2}}d\tau$
$\displaystyle=\frac{1}{2}\operatorname{Id}+\frac{1}{2}\sum_{k\geq
0}\mathbb{P}_{k;n-1}\otimes\int_{\mathbb{R}}{\text{\rm Op}_{\text{\rm
w}}(e^{2i\pi\tau(\xi_{n}-x_{n}^{2})})}\frac{(1-i\tau)^{k}}{i\pi\tau(1+i\tau)^{k+n-1}}d\tau.$

Let $k(x_{n},y_{n})$ be the kernel of the operator in the integral: we have

$k(x_{n},y_{n})=e^{\frac{2i\pi}{3}(x_{n}^{3}-y_{n}^{3})}e^{-\frac{i\pi}{6}(x_{n}-y_{n})^{3}}\frac{i}{\pi(x_{n}-y_{n})}\frac{(1+i(x_{n}-y_{n}))^{k}}{(1-i(x_{n}-y_{n}))^{k+n-1}}.$

As a result, we find that $P$ is unitarily equivalent to $\tilde{P}$, with

(4.4.6) $2\tilde{P}=\sum_{k\geq
0}\mathbb{P}_{k;n-1}\otimes\bigl{(}I_{n}+\text{convolution with
}\frac{ie^{-\frac{i\pi}{6}x_{n}^{3}}}{\pi
x_{n}}\frac{(1+ix_{n})^{k}}{(1-ix_{n})^{k+n-1}}\bigr{)}.$

We define

(4.4.7)
$\omega_{k,n-1}(\tau)=\frac{1}{2}+\int\frac{ie^{-\frac{i\pi}{6}t^{3}}}{2\pi
t}\frac{(1+it)^{k}}{(1-it)^{k+n-1}}e^{-2i\pi t\tau}dt\\\
=\frac{1}{2}+\int\frac{e^{\frac{i\pi}{6}t^{3}}}{2i\pi
t}\frac{(1-it)^{k}}{(1+it)^{k+n-1}}e^{2i\pi t\tau}dt,$

and we get that

(4.4.8) $\tilde{P}=\sum_{k\geq
0}\mathbb{P}_{k;n-1}\otimes\omega_{k,n-1}(D_{x_{n}}).$

We note that for $n=1$, the sum is reduced to $k=0$ with $\mathbb{P}_{0;0}=I$,
so that we recover Formula (4.2.6) with $\omega_{0,0}=\omega$. We find also
that

(4.4.9) $\omega_{k,n-1}^{\prime}(\tau)=\int
e^{\frac{i\pi}{6}t^{3}}\frac{(1-it)^{k}}{(1+it)^{k+n-1}}e^{2i\pi t\tau}dt,$

in the sense that the inverse Fourier transform of $t\mapsto
e^{\frac{i\pi}{6}t^{3}}\frac{(1-it)^{k}}{(1+it)^{k+n-1}}$ is the distribution
derivative of $\omega_{k,n-1}$. Going back to the normalization of Lemma 4.7,
we have, with $\eta=2^{4/3}\pi^{2/3}\tau$,

(4.4.10) $\displaystyle G_{k,n-1}(\eta)$ $\displaystyle=\omega_{k,n-1}(\tau),$
$\displaystyle G_{k,n-1}^{\prime}(\eta)$ $\displaystyle=2^{-4/3}\pi^{-2/3}\int
e^{\frac{i\pi}{6}t^{3}}\frac{(1-it)^{k}}{(1+it)^{k+n-1}}e^{2^{-\frac{1}{3}}i\pi^{\frac{1}{3}}t\eta}dt,$
(4.4.11)
$\displaystyle\underbrace{=}_{t=\pi^{-\frac{1}{3}}2^{\frac{1}{3}}s}\frac{1}{2\pi}\int
e^{\frac{is^{3}}{3}}\frac{(1-i\pi^{-1/3}2^{1/3}s)^{k}}{(1+i\pi^{-1/3}2^{1/3}s)^{k+n-1}}e^{is\eta}ds:=A_{k,n-1}(\eta).$

We have $A_{0,0}=\operatorname{\tt A\\!i}$ and $A_{k,n-1}$ is an entire
function, real-valued on the real line; we have

$G_{k,n-1}(\eta)=\int_{-\infty}^{\eta}A_{k,n-1}(\xi)d\xi,\quad
G_{k,n-1}(+{\infty})=1.$

######  Remark 4.13.

We claim that the asymptotic properties of the functions $A_{k,n-1}$ are
analogous to the properties of the standard Airy function and we have indeed
from (4.4.9),

(4.4.12)
$\omega_{k,n-1}^{\prime}(\tau)=(1-iD)^{k}(1+iD)^{-k-n+1}\mathcal{F}^{-1}(e^{\frac{i\pi}{6}t^{3}}).$

We claim as well that

$-\frac{1}{2}<\inf_{k\geq 0,\eta\in\mathbb{R}}G(\eta)<0,\quad\sup_{k\geq
0,\eta\in\mathbb{R}}G(\eta)=1,$

so that $\tilde{P}$ is bounded on $L^{2}(\mathbb{R}^{n})$ and

(4.4.13) $\int_{\xi_{n}\geq x_{n}^{2}+\sum_{1\leq j\leq
n-1}(x_{j}^{2}+\xi_{j}^{2})}\mathcal{W}(u,u)(x,\xi)dxd\xi\leq\|u\|_{L^{2}(\mathbb{R}^{n})}^{2}.$

##  5\. Conics with eccentricity greater than 1

We want to consider now integrals of the Wigner distribution on “hyperbolic”
convex subsets of the plane such as

(5.0.1) $\mathcal{C}_{\sigma}=\\{(x,\xi)\in\mathbb{R}^{2},x\xi\geq\sigma,x\geq
0\\},$

where $\sigma$ is a non-negative parameter. It is convenient to start with the
limit-case where $\sigma=0$ and
$\mathcal{C}_{0}=\\{(x,\xi)\in\mathbb{R}^{2},x\geq 0,\xi\geq 0\\}$ (we will
label $\mathcal{C}_{0}$ as the _quarter-plane_). The indicatrix function of
$\mathcal{C}_{0}$ is $H(x)H(\xi)$ where $H=\mathbf{1}_{\mathbb{R}_{+}}$ is the
Heaviside function.

Acknowledgements. The author is grateful to Thomas Duyckaerts for sharp
comments on a first version of this section.

######  N.B.

The reader will see a great similarity between our calculations below in this
section and the J.G. Wood & A.J. Bracken paper [48]. This article is very
important for the problem at stake – Integrating the Wigner distribution on
subsets of the phase space – and was a wealthy source of information for us,
although as a mathematician, the author has a quite rigid relationship with
calculations, and feels the need to justify formal manipulations; for
instance, we may point out that the test functions used in [48] are
homogeneous distributions of type

$x_{\pm}^{-\frac{1}{2}+i\omega},\quad\omega\in\mathbb{R},$

which are not in $L^{2}(\mathbb{R})$ (not even in $L^{2}_{\textrm{loc}}$), a
situation which raises some difficulties, first when you try to normalize in
$L^{2}$ these test functions and also when trying to give a non-formal meaning
to their images under the operator with Weyl symbol $H(x)H(\xi)$, images which
are not clearly defined. In our joint paper [4] with B. Delourme and T.
Duyckaerts, proving that Flandrin’s conjecture is not true, we followed
numerical arguments which were quite apart from the arguments of [48].
However, in this article, we do follow many of the arguments of [48], along
with avoiding formal calculations.

###  5.1. The quarter-plane, a counterexample to Flandrin’s conjecture

####  5.1.1. Preliminaries

We study in this section the operator

(5.1.1) $A_{0}={\text{\rm Op}_{\text{\rm w}}(H(x)H(\xi))}$

where $H=\mathbf{1}_{\mathbb{R}_{+}}$, that is the Weyl quantization of the
characteristic function of the first quarter of the plane.

######  Lemma 5.1.

The operator $A_{0}$ given by (5.1.1) is bounded self-adjoint on
$L^{2}(\mathbb{R})$.

###### Proof.

Since the Weyl symbol of $A_{0}$ is real-valued, $A_{0}$ is formally self-
adjoint and it is enough to prove that $A_{0}$ is bounded on
$L^{2}(\mathbb{R})$. Let us start with recalling the classical formulas

(5.1.2) $\hat{H}(t)=\frac{\delta_{0}(t)}{2}+\frac{1}{2i\pi}\text{\rm
pv}\left(\frac{1}{t}\right),\quad\widehat{\operatorname{sign}}=\frac{1}{i\pi}\text{\rm
pv}\left(\frac{1}{t}\right),$

useful below. The kernel888There is no difficulty at defining the product
$S\bigl{(}(x+y)/2\bigr{)}T(x-y)$ for $S,T$ tempered distributions on the real
line since we may use the tensor product with  $\langle
S(\frac{x+y}{2})T(x-y),\Phi(x,y)\rangle_{\mathscr{S}^{\prime}(\mathbb{R}^{2}),\mathscr{S}(\mathbb{R}^{2})}=\langle
S(x_{1})\otimes
T(x_{2}),\Phi(x_{1}+\frac{x_{2}}{2},x_{1}-\frac{x_{2}}{2})\rangle_{\mathscr{S}^{\prime}(\mathbb{R}^{2}),\mathscr{S}(\mathbb{R}^{2})}.$
However, we shall not use directly Formula (5.1.3), since want to avoid formal
manipulation involving for instance meaningless products such as
$H(x)H(y)k_{0}(x,y)$. We refer the reader to footnote 9 on page 9 and to
Remark 5.2 for more details on this matter.  of $A_{0}$ is

(5.1.3)
$k_{0}(x,y)=H(x+y)\hat{H}(y-x)=H(x+y)\frac{1}{2}\Bigl{(}\delta_{0}(y-x)+\frac{1}{i\pi}\text{pv}\frac{1}{y-x}\Bigr{)}.$

For $\lambda>0$, we define
$A_{0,\lambda}=\left(H(x)\mathbf{1}_{[0,\lambda]}(\xi)\right)^{w},$ whose
distribution-kernel is the $L^{{\infty}}(\mathbb{R}^{2n})$ function

$k_{0,\lambda}(x,y)=H(x+y)e^{i\pi(x-y)\lambda}\frac{\sin(\pi(x-y)\lambda)}{\pi(x-y)}.$

We can thus notice that

(5.1.4)
$k_{0,\lambda}(x,y)=\overbracket{H(x)H(y)e^{i\pi(x-y)\lambda}\frac{\sin(\pi(x-y)\lambda)}{\pi(x-y)}}^{k_{0,\lambda}^{\flat}(x,y)}\\\
+\underbracket{H(x+y)\bigl{(}H(-x)H(y)+H(x)H(-y)\bigr{)}\frac{\sin(\pi(x-y)\lambda)}{\pi(x-y)}e^{i\pi(x-y)\lambda}}_{k_{0,\lambda}^{\sharp}(x,y)},$

and the operator with distribution-kernel $k_{0,\lambda}^{\flat}$ is

$H{\text{\rm Op}_{\text{\rm w}}(\mathbf{1}_{[0,\lambda]}(\xi))}H,\text{\ that
is\quad$H\mathbf{1}_{[0,\lambda]}(D)H$,}$

where $H$ stands for the operator of multiplication by the Heaviside function
$H$. On the other hand, the operator with distribution kernel
$k_{0,\lambda}^{\sharp}$ is such that

$|k_{0,\lambda}^{\sharp}(x,y)|\leq
H(x+y)\frac{H(-x)H(y)+H(x)H(-y)}{\pi|x-y|}\\\
=H(x+y)\frac{H(-x)H(y)}{\pi(y-x)}+H(x+y)\frac{H(x)H(-y)}{\pi(x-y)}.$

According to Proposition 9.13 in our Appendix, the Hardy operator and the
modified Hardy operators are bounded on $L^{2}(\mathbb{R})$ and we obtain
that, for $\phi,\psi\in\mathscr{S}(\mathbb{R}^{n})$, with
$H=H(x),\check{H}=H(-x)$,

(5.1.5) $\left|\iint
H(x)\mathbf{1}_{[0,\lambda]}(\xi)W(\phi,\psi)(x,\xi)dxd\xi\right|\\\
\leq\|H\phi\|_{L^{2}(\mathbb{R})}\|H\psi\|_{L^{2}(\mathbb{R})}+\frac{1}{2}\|H\phi\|_{L^{2}(\mathbb{R})}\|\check{H}\psi\|_{L^{2}(\mathbb{R})}+\frac{1}{2}\|\check{H}\phi\|_{L^{2}(\mathbb{R})}\|H\psi\|_{L^{2}(\mathbb{R})},$

so that

(5.1.6) $|\langle
A_{0}\phi,\psi\rangle_{\mathscr{S}^{*}(\mathbb{R}),\mathscr{S}(\mathbb{R})}|\\\
=\Bigl{|}\iint
H(x)H(\xi)\overbracket{W(\phi,\psi)(x,\xi)}^{\in\mathscr{S}(\mathbb{R}^{2})}dxd\xi\Bigr{|}=\lim_{\lambda\rightarrow+{\infty}}\left|\iint
H(x)\mathbf{1}_{[0,\lambda]}(\xi)W(\phi,\psi)(x,\xi)dxd\xi\right|\\\
\leq\|H\phi\|_{L^{2}(\mathbb{R})}\|H\psi\|_{L^{2}(\mathbb{R})}+\frac{1}{2}\|H\phi\|_{L^{2}(\mathbb{R})}\|\check{H}\psi\|_{L^{2}(\mathbb{R})}+\frac{1}{2}\|\check{H}\phi\|_{L^{2}(\mathbb{R})}\|H\psi\|_{L^{2}(\mathbb{R})},$

yielding the $L^{2}$-boundedness of the operator $A_{0}$, and this concludes
the proof of the lemma. ∎

######  Remark 5.2.

That cumbersome detour with the operator $A_{0,\lambda}$ is useful to ensure
that the operator $A$ is indeed bounded on $L^{2}(\mathbb{R})$. The kernel
$k_{0}$ of $A_{0}$ is a distribution of order 1 and the product
$H(x)H(y)k_{0}(x,y)$ is not a priori meaningful, even when $k$ is a Radon
measure999Even a wave-front-set approach, which would allow the product
$H(x)\text{pv}(1/(y-x))$, does not offer a meaning for the product
$H(x)H(y)\text{pv}(1/(y-x))$ since the wave-front-set of $\text{pv}(1/(y-x))$
is located on the conormal of the first diagonal (i.e.
$\\{(x,x;\xi,-\xi)\\}_{x\in\mathbb{R},\xi\in\mathbb{R}^{*}}$), whereas the
wave-front set at $(0,0)$ of $H(x)H(y)$ contains all directions and in
particular is antipodal to the conormal of the diagonal at $(0,0)$. . However
with the proven $L^{2}$-boundedness of $A_{0}$, the products of operators
$HA_{0}H$, $\check{H}A_{0}H$, $HA_{0}\check{H}$, $\check{H}A_{0}\check{H}$
make sense and for instance we may approximate in the strong-operator-topology
the operator $HA_{0}H$ by the operator
$\chi(\cdot/\varepsilon)A\chi(\cdot/\varepsilon),$ where $\chi$ is a smooth
function supported in $[1,+{\infty})$ and equal to $1$ on $[2,+{\infty})$. We
have indeed

$HAH=\bigl{(}H-\chi(\cdot/\varepsilon)\bigr{)}AH+\chi(\cdot/\varepsilon)A\bigl{(}H-\chi(\cdot/\varepsilon)\bigr{)}+\chi(\cdot/\varepsilon)A\chi(\cdot/\varepsilon),$

so that for $u\in L^{2}(\mathbb{R})$, $HAHu=\lim_{\varepsilon\rightarrow
0_{+}}\chi(\cdot/\varepsilon)A\chi(\cdot/\varepsilon)u.$ The operator with
kernel

$H(x+y)\chi(x/\varepsilon)\chi(y/\varepsilon)\textrm{\rm
pv}\frac{1}{i\pi(y-x)}=\chi(x/\varepsilon)\chi(y/\varepsilon)\text{\rm
pv}\frac{1}{i\pi(y-x)},$

converges strongly towards the operator $H(\operatorname{sign}D)H$.

######  Proposition 5.3.

Let $A_{0}={\text{\rm Op}_{\text{\rm w}}(H(x)H(\xi))}$ be the operator with
Weyl symbol $H(x)H(\xi)$, a priori sending $\mathscr{S}(\mathbb{R})$ into
$\mathscr{S}^{\prime}(\mathbb{R})$. Then $A_{0}$ can be uniquely extended to a
self-adjoint bounded operator on $L^{2}(\mathbb{R})$ with

(5.1.7)
$\|A_{0}\|_{\mathcal{B}(L^{2}(\mathbb{R}))}\leq\frac{1+\sqrt{2}}{2}\approx
1.207$

######  N.B.

The bound above can be significantly improved (see Proposition 5.30 for
optimal bounds) and moreover we will show below that the spectrum of $A_{0}$
actually intersects $(1,+{\infty})$. In fact it is easier to start with the
information that $A_{0}$ is indeed bounded on $L^{2}(\mathbb{R})$.

###### Proof.

The $L^{2}(\mathbb{R})$-boundedness of $A_{0}$ is given by Lemma 5.1. We are
left with proving the bound (5.1.7): we note that (5.1.6) implies

$|\langle
A_{0}u,u\rangle_{L^{2}(\mathbb{R})}|\leq\|Hu\|_{L^{2}(\mathbb{R})}^{2}+\|Hu\|_{L^{2}(\mathbb{R})}\|\check{H}u\|_{L^{2}(\mathbb{R})},$

proving the proposition, since the eigenvalues of the quadratic form
$\mathbb{R}^{2}\ni(x_{1},x_{2})\mapsto x_{1}^{2}+x_{1}x_{2}$ are
$(1\pm\sqrt{2})/2$. ∎

We can do much better and actually diagonalize the operator $A_{0}$, using as
in Proposition 9.13 logarithmic coordinates on each half-line. We state a
lemma on “diagonal” terms whose proof is already given above.

######  Lemma 5.4 (Diagonal terms).

Let $A_{0}$ be the operator with Weyl symbol $H(x)H(\xi)$. With $H$ standing
as well for the operator of multiplication by $H(x)$, we have

(5.1.8) $\displaystyle
HA_{0}H=HH(D)H=H\frac{(\operatorname{Id}+\operatorname{sign}D)}{2}H.$

######  Lemma 5.5 (Off-diagonal terms).

Let $B_{0}=2\operatorname{Re}\check{H}A_{0}H=\check{H}A_{0}H+HA_{0}\check{H}$.
Then we have for all $u\in L^{2}(\mathbb{R})$,

(5.1.9) $|\langle
B_{0}u,u\rangle_{L^{2}(\mathbb{R})}|\leq\frac{1}{2}\|Hu\|_{L^{2}(\mathbb{R})}\|\check{H}u\|_{L^{2}(\mathbb{R})}.$

###### Proof of the Lemma.

For $u\in\mathscr{S}(\mathbb{R})$ such that $0\notin\operatorname{supp}u$, we
define for $t\in\mathbb{R}$,

(5.1.10) $\phi_{1}(t)=u(e^{t})e^{t/2},\quad\phi_{2}(t)=u(-e^{t})e^{t/2},$

so that

(5.1.11)
$\|Hu\|_{L^{2}(\mathbb{R})}^{2}=\|\phi_{1}\|_{L^{2}(\mathbb{R})}^{2},\quad\|\check{H}u\|_{L^{2}(\mathbb{R})}^{2}=\|\phi_{2}\|_{L^{2}(\mathbb{R})}^{2}.$

We have

$\displaystyle\langle
B_{0}u,u\rangle_{L^{2}(\mathbb{R})}=\iint\frac{H(x+y)\bigl{(}\check{H}(x)H(y)+H(x)\check{H}(y)\bigr{)}}{2i\pi(y-x)}u(y)\bar{u}(x)dydx$
$\displaystyle\hskip
22.0pt=\iint\frac{H(-e^{s}+e^{t})e^{\frac{s+t}{2}}}{2i\pi(e^{t}+e^{s})}\phi_{1}(t)\bar{\phi}_{2}(s)dsdt-\iint\frac{H(e^{s}-e^{t})e^{\frac{s+t}{2}}}{2i\pi(e^{t}+e^{s})}\phi_{2}(t)\bar{\phi}_{1}(s)dsdt$
$\displaystyle\hskip
22.0pt=\iint\frac{H(t-s)}{4i\pi\cosh(\frac{s-t}{2})}\phi_{1}(t)\bar{\phi}_{2}(s)dsdt-\iint\frac{H(s-t)}{4i\pi\cosh(\frac{s-t}{2})}\phi_{2}(t)\bar{\phi}_{1}(s)dsdt$

so that

(5.1.12) $\displaystyle\langle
B_{0}u,u\rangle_{L^{2}(\mathbb{R})}=\langle\tilde{S}_{0}\ast\phi_{1},\phi_{2}\rangle_{L^{2}(\mathbb{R})}+\langle
S_{0}\ast\phi_{2},\phi_{1}\rangle_{L^{2}(\mathbb{R})},$ (5.1.13)
$\displaystyle\tilde{S}_{0}(t)=\frac{\check{H}(t)}{4i\pi\cosh(t/2)},\quad
S_{0}(t)=\frac{iH(t)}{4\pi\cosh(t/2)}.$

We calculate

$\int_{0}^{+{\infty}}\frac{dt}{4\pi\cosh(t/2)}=\frac{1}{2\pi}[\arctan(\sinh(t/2))]_{0}^{+{\infty}}=\frac{1}{4}=\int_{-{\infty}}^{0}\frac{dt}{4\pi\cosh(t/2)},$

so that

(5.1.14) $|\langle
B_{0}u,u\rangle_{L^{2}(\mathbb{R})}|\leq\frac{1}{2}\|\phi_{1}\|_{L^{2}(\mathbb{R})}\|\phi_{2}\|_{L^{2}(\mathbb{R})}=\frac{1}{2}\|Hu\|_{L^{2}(\mathbb{R})}\|\check{H}u\|_{L^{2}(\mathbb{R})},$

proving the estimate of the lemma for $u\in\mathscr{S}(\mathbb{R})$ such that
$0\notin\operatorname{supp}u$. We use now that we already know that $B_{0}$ is
a bounded self-adjoint operator on $L^{2}(\mathbb{R})$: let $u$ be a function
in $L^{2}(\mathbb{R})$ and let $(\phi_{k})_{k\geq 1}$ be a sequence101010Such
a sequence is easy to find: a first step is to find a sequence
$(\tilde{\phi}_{k})_{k\geq 1}$ in the Schwartz space converging in
$L^{2}(\mathbb{R})$ towards $u$, then consider with a given $\omega\in
C^{{\infty}}(\mathbb{R};[0,1])$ such that $\omega(t)=0$ for $|t|\leq 1$ and
$\omega(t)=1$ for $|t|\geq 2$, $\phi_{k}(x)=\omega(kx)\tilde{\phi}_{k}(x).$ in
$\mathscr{S}(\mathbb{R})$ such that each $\phi_{k}$ vanishes in a neighborhood
of 0 so that $\lim_{k}\phi_{k}=u$ in $L^{2}(\mathbb{R})$. We find that

$|\langle B_{0}u,u\rangle_{L^{2}(\mathbb{R})}|\leq|\langle
B_{0}(u-\phi_{k}),u\rangle_{L^{2}(\mathbb{R})}|+|\langle
B_{0}\phi_{k},u-\phi_{k}\rangle_{L^{2}(\mathbb{R})}|+|\langle
B_{0}\phi_{k},\phi_{k}\rangle_{L^{2}(\mathbb{R})}|\\\
\leq\|B_{0}\|_{\mathcal{B}(L^{2}(\mathbb{R}))}\bigl{(}\|u-\phi_{k}\|_{L^{2}(\mathbb{R})}\|u\|_{L^{2}(\mathbb{R})}+\|u-\phi_{k}\|_{L^{2}(\mathbb{R})}\|\phi_{k}\|_{L^{2}(\mathbb{R})}\bigr{)}\\\
+\frac{1}{2}\|H\phi_{k}\|_{L^{2}(\mathbb{R})}\|\check{H}\phi_{k}\|_{L^{2}(\mathbb{R})},$

providing readily the result of the lemma since the multiplication by $H$ and
$\check{H}$ are bounded operators on $L^{2}(\mathbb{R})$. ∎

######  Remark 5.6.

The estimate (5.1.9) and Lemma 5.4 are already improving (5.1.7), since the
eigenvalues of the quadratic form $\mathbb{R}^{2}\ni(x_{1},x_{2})\mapsto
x_{1}^{2}+\frac{1}{2}x_{1}x_{2}$ are $(2\pm\sqrt{5})/4$, so that the right-
hand-side of (5.1.7) can be replaced by $(2+\sqrt{5})/4\approx 1.059$. Anyhow,
we shall provide below a diagonalization of $A_{0}$ and optimal bounds.

######  N.B.

We shall be a little faster in the sequel on the “cumbersome” detours to avoid
formal multiplication of kernels by Heaviside functions but the reader should
keep in mind that it is an important point to secure
$L^{2}(\mathbb{R})$-boundedness _before_ any further manipulation of the
kernels.

####  5.1.2. An isometric isomorphism

######  Remark 5.7.

The mapping $\Psi$ defined by

(5.1.15) $\begin{matrix}\Psi:\
L^{2}(\mathbb{R})&\longrightarrow&L^{2}(\mathbb{R};\mathbb{C}^{2})\\\
u&\mapsto&\Bigl{(}(Hu)(e^{t})e^{t/2},(\check{H}u)(-e^{t})e^{t/2}\Bigr{)}\end{matrix}$

is an isometric isomorphism of Hilbert spaces: indeed we have

$\|u\|^{2}_{L^{2}(\mathbb{R})}=\int_{\mathbb{R}}|u(e^{t})|^{2}e^{t}dt+\int_{\mathbb{R}}|u(-e^{t})|^{2}e^{t}dt.$

Moreover if $(\phi_{1},\phi_{2})\in L^{2}(\mathbb{R};\mathbb{C}^{2})$, we may
define for $x\in\mathbb{R}^{*}$

$u(x)=H(x)\phi_{1}(\ln x)x^{-1/2}+\check{H}(x)\phi_{2}(\ln|x|)|x|^{-1/2},$

and we have $\Psi(u)(t)=\bigl{(}\phi_{1}(t),\phi_{2}(t)\bigr{)}.$

######  Remark 5.8.

Using Lemma 5.4 and Notations (5.1.10) we see that

$\displaystyle\langle HA_{0}Hu,u\rangle_{L^{2}(\mathbb{R})}$
$\displaystyle=\frac{1}{2}\|\phi_{1}\|^{2}_{L^{2}(\mathbb{R})}+\iint\frac{1}{2i\pi}\text{pv}\frac{e^{(s+t)/2}}{e^{t}-e^{s}}\phi_{1}(t)\bar{\phi}_{1}(s)dsdt$
$\displaystyle=\frac{1}{2}\|\phi_{1}\|^{2}_{L^{2}(\mathbb{R})}+\iint\frac{1}{4i\pi}\text{pv}\frac{1}{\sinh(\frac{t-s}{2})}\phi_{1}(t)\bar{\phi}_{1}(s)dsdt$
(5.1.16)
$\displaystyle=\int_{\mathbb{R}}|\hat{\phi}_{1}(\tau)|^{2}\bigl{(}\frac{1}{2}+\hat{T}_{0}(\tau)\bigr{)}d\tau,$

with

(5.1.17) $T_{0}(t)=\frac{t}{4\sinh(t/2)}\text{pv}\frac{i}{\pi t}.$

We have

(5.1.18)
$\displaystyle\hat{T}_{0}=\operatorname{sign}\ast\rho_{0},\quad\text{with\quad}\rho_{0}(\tau)=\int\frac{t}{4\sinh(t/2)}e^{-2i\pi
t\tau}dt,$

and we note that the function $\rho_{0}$ belongs to $\mathscr{S}(\mathbb{R})$,
as the Fourier transform of a function in $\mathscr{S}(\mathbb{R})$. Also we
have

$\int\rho_{0}(\tau)d\tau=\hat{\rho}_{0}(0)=\frac{1}{2},$

and this yields with
$\frac{d}{d\tau}\left\\{\frac{1}{2}+\hat{T}_{0}\right\\}=2\rho_{0}$ (which
follows from (5.1.18)) and

(5.1.19)
$\frac{1}{2}+\hat{T}_{0}(\tau)=1-\int_{\tau}^{+{\infty}}2\rho_{0}(\tau^{\prime})d\tau^{\prime},$

since

$\frac{d}{d\tau}\left\\{\frac{1}{2}+\hat{T}_{0}+\int_{\tau}^{+{\infty}}2\rho_{0}(\tau^{\prime})d\tau^{\prime}\right\\}=0\quad\text{and\quad}\lim_{\tau\rightarrow+{\infty}}(\operatorname{sign}\ast\rho_{0})(\tau)=\frac{1}{2}.$

######  Theorem 5.9.

Let $A_{0}$ be the operator with Weyl symbol $H(x)H(\xi)$. The operator
$A_{0}$ is bounded self-adjoint on $L^{2}(\mathbb{R})$ so that we may define,
with $\Psi$ defined in (5.1.15),

(5.1.20) $\widetilde{A}_{0}=\Psi A_{0}\Psi^{-1}.$

The operator $\widetilde{A}_{0}$ is the Fourier multiplier on
$L^{2}(\mathbb{R};\mathbb{C}^{2})$ given by the matrix

(5.1.21)
$\mathcal{M}_{0}(\tau)=\begin{pmatrix}\frac{1}{2}+\hat{T}_{0}(\tau)&\hat{S}_{0}(\tau)\\\
\overline{\hat{S}_{0}(\tau)}&0\end{pmatrix},$

where $T_{0},S_{0}$ are defined respectively in (5.1.17), (5.1.13). In
particular we have with $\Phi=(\phi_{1},\phi_{2})\in
L^{2}(\mathbb{R};\mathbb{C}^{2})$,

(5.1.22)
$\langle\widetilde{A}_{0}\Phi,\Phi\rangle_{L^{2}(\mathbb{R};\mathbb{C}^{2})}=\int_{\mathbb{R}}e^{2i\pi
t\tau}\langle\mathcal{M}_{0}(\tau)\hat{\Phi}(\tau),\hat{\Phi}(\tau)\rangle_{\mathbb{C}^{2}}d\tau.$

######  Remark 5.10.

As a consequence of Theorem 5.9, we find that the spectrum of the self-adjoint
bounded operator $A_{0}$ is the closure of the set of eigenvalues of the
matrices $\mathcal{M}_{0}(\tau)$ when $\tau$ runs on the real line.

###### Proof.

The proof follows readily from Remarks 5.7, 5.8 and Lemmas 5.4, 5.5. ∎

######  Lemma 5.11.

Let $\mathcal{N}$ be a $2\times 2$ Hermitian matrix

$\mathcal{N}=\begin{pmatrix}a_{11}&a_{12}\\\
\overline{a_{12}}&0\end{pmatrix}.$

Then the eigenvalues $\lambda_{-}\leq\lambda_{+}$ of $\mathcal{N}$ are such
that

(5.1.23) $\lambda_{-}<0<1<\lambda_{+},$

if and only if

(5.1.24) $a_{12}\not=0\text{ \quad and \quad}|a_{12}|^{2}>1-a_{11}.$

###### Proof.

The characteristic polynomial of $\mathcal{N}$ is
$p(\lambda)=\lambda^{2}-a_{11}\lambda-|a_{12}|^{2}$ and since $a_{11}$ is
real-valued, has two real roots $\lambda_{-}\leq\lambda_{+}$. If (5.1.24)
holds true, the roots are distinct and

$p(0)=-|a_{12}|^{2}<0,\quad p(1)=1-a_{11}-|a_{12}|^{2}<0,$

implying (5.1.23). Conversely, if (5.1.23) is satisfied, then $p(0),p(1)$ are
both negative, implying (5.1.24), completing the proof of the lemma. ∎

######  Lemma 5.12.

Let us define for $\omega\in\mathbb{R}$,

(5.1.25)
$I(\omega)=\frac{1}{4\pi}\int_{0}^{+{\infty}}\frac{\sin(t\omega)}{\cosh(t/2)}dt.$

Then we have

(5.1.26)
$I(\omega)=\frac{1}{4\pi\omega}+O(\omega^{-3}),\quad|\omega|\rightarrow+{\infty}.$

###### Proof.

Indeed we have for $\omega\in\mathbb{R}^{*}$,

$I(\omega)=-\frac{1}{4\pi\omega}\int_{0}^{+{\infty}}\frac{\frac{d}{dt}\cos(t\omega)}{\cosh(t/2)}dt=\frac{1}{4\pi\omega}\Bigl{(}1-\int_{0}^{+{\infty}}\frac{\cos(t\omega)}{(\cosh(t/2))^{2}}\frac{1}{2}\sinh(t/2)dt\Bigr{)}\\\
=\frac{1}{4\pi\omega}\bigl{(}1+g(\omega)\bigr{)},$

with

$\displaystyle g(\omega)$ $\displaystyle=-\int_{0}^{+{\infty}}{\frac{d}{\omega
dt}\\{\sin(t\omega)\\}}{\operatorname{sech}(t/2)}\frac{1}{2}\tanh(t/2)dt$
$\displaystyle=\frac{1}{2\omega}\int_{0}^{+{\infty}}\sin(t\omega)\frac{d}{dt}\bigl{\\{}{\operatorname{sech}(t/2)}\tanh(t/2)\bigr{\\}}dt$
$\displaystyle=-\frac{1}{2\omega^{2}}\int_{0}^{+{\infty}}\frac{d}{dt}\bigl{\\{}\cos(t\omega)\bigr{\\}}\frac{d}{dt}\bigl{\\{}{\operatorname{sech}(t/2)}\tanh(t/2)\bigr{\\}}dt$
$\displaystyle=\frac{1}{2\omega^{2}}\left\\{\int_{0}^{+{\infty}}\cos(t\omega)\frac{d^{2}}{dt^{2}}\bigl{\\{}{\operatorname{sech}(t/2)}\tanh(t/2)\bigr{\\}}dt+\frac{1}{2}\right\\}=O(\omega^{-2}),$

proving the lemma. ∎

######  Proposition 5.13.

The matrix $\mathcal{M}_{0}(\tau)$ defined in (5.1.21) is equal to

(5.1.27) $\mathcal{M}_{0}(\tau)=\begin{pmatrix}a_{11}(\tau)&a_{12}(\tau)\\\
\overline{a_{12}(\tau)}&0\end{pmatrix}$

with

(5.1.28)
$1-a_{11}(\tau)=\int_{\tau}^{+{\infty}}2\rho_{0}(\tau^{\prime})d\tau^{\prime},\quad
a_{12}(\tau)=\frac{i}{4\pi}\int_{0}^{+{\infty}}\frac{1}{\cosh(t/2)}e^{-2i\pi\tau
t}dt.$

We have

(5.1.29) $\displaystyle 1-a_{11}(\tau)$
$\displaystyle=O(\tau^{-N})\quad\text{for any $N$ when
$\tau\rightarrow+{\infty}$},$ (5.1.30)
$\displaystyle\operatorname{Re}(a_{12}(\tau))$
$\displaystyle=\frac{1}{8\pi^{2}\tau}+O(\tau^{-3})\quad\text{when
$\tau\rightarrow+{\infty}$.}$

###### Proof.

Formulas (5.1.27), (5.1.28) follow from Theorem 5.9, (5.1.19) and (5.1.13).
The estimates (5.1.29) follow from the fact that $\rho_{0}$ belongs to the
Schwartz class and (5.1.30) is a reformulation of Lemma 5.12. ∎

######  Theorem 5.14.

Let $A_{0}$ be the operator with Weyl symbol $H(x)H(\xi)$, where $H$ is the
Heaviside function. Then $A_{0}$ is a bounded self-adjoint operator on
$L^{2}(\mathbb{R})$ such that

(5.1.31) $\inf\bigl{(}\text{\rm
spectrum}(A_{0})\bigr{)}<0<1<\sup\bigl{(}\text{\rm spectrum}(A_{0})\bigr{)}.$

###### Proof.

Using Remark 5.10 and Proposition 5.13 we find that for $\tau$ large enough,
Conditions (5.1.24) are satisfied, proving readily (5.1.31). ∎

######  Corollary 5.15 (A counterexample to Flandrin’s conjecture).

There exists a function $\phi_{0}\in\mathscr{S}(\mathbb{R})$, with
$L^{2}(\mathbb{R})$ norm equal to 1 such that

(5.1.32) $\iint_{x\geq 0,\xi\geq
0}\mathcal{W}(\phi_{0},\phi_{0})(x,\xi)dxd\xi>1.$

There exists $a>0$ such that $\iint_{0\leq x\leq a,0\leq\xi\leq
a}\mathcal{W}(\phi_{0},\phi_{0})(x,\xi)dxd\xi>1.$

######  Remark 5.16.

On page 2178 of [11], we find the sentence “it is conjectured that

(5.1.33) $\forall u\in
L^{2}(\mathbb{R}),\quad\iint_{\mathcal{C}}\mathcal{W}(u,u)(x,\xi)dxd\xi\leq\|u\|_{L^{2}(\mathbb{R})}^{2},$

is true for any convex domain $\mathcal{C}$”, a quite mild commitment for the
validity of (5.1.33), although that statement was referred to later on as
Flandrin’s conjecture in the literature. The second part of the above
corollary is providing a disproof of that conjecture based upon an “abstract”
argument used in the proof of Theorem 5.14; the result of that corollary was
already known via a numerical analysis argument after our joint work [4] with
B. Delourme and T. Duyckaerts.

###### Proof.

From Theorem 5.14, we find $u_{0}\in L^{2}(\mathbb{R})$ such that

$\|u_{0}\|_{L^{2}(\mathbb{R})}^{2}<\langle A_{0}u_{0},u_{0}\rangle.$

Let $\psi\in\mathscr{S}(\mathbb{R})$: we have

$|\langle A_{0}u_{0},u_{0}\rangle-\langle A_{0}\psi,\psi\rangle|=|\langle
A_{0}(u_{0}-\psi),u_{0}\rangle+\langle A_{0}\psi,u_{0}-\psi\rangle|\\\
\leq\|A_{0}\|_{\mathcal{B}(L^{2}(\mathbb{R}))}\|u_{0}-\psi\|_{L^{2}(\mathbb{R})}\bigl{(}\|u_{0}\|_{L^{2}(\mathbb{R})}+\|\psi\|_{L^{2}(\mathbb{R})}\bigr{)},$

and thus if $(\psi_{k})_{k\geq 1}$ is a sequence of $\mathscr{S}(\mathbb{R})$
converging towards $u_{0}$ in $L^{2}(\mathbb{R})$, we get

$\|u_{0}\|_{L^{2}(\mathbb{R})}^{2}<\langle A_{0}u_{0},u_{0}\rangle\\\
\leq\langle
A_{0}\psi_{k},\psi_{k}\rangle+\underbrace{\|A_{0}\|_{\mathcal{B}(L^{2}(\mathbb{R}))}\|u_{0}-\psi_{k}\|_{L^{2}(\mathbb{R})}\bigl{(}\|u_{0}\|_{L^{2}(\mathbb{R})}+\|\psi_{k}\|_{L^{2}(\mathbb{R})}\bigr{)}}_{=\sigma_{k},\text{
goes to 0 when $k\rightarrow+{\infty}$.}}.$

There exists $k_{0}\geq 1$ such that for $k\geq k_{0}$, we have

$0\leq\sigma_{k}\leq\frac{1}{2}\bigl{(}\langle
A_{0}u_{0},u_{0}\rangle-\|u_{0}\|_{L^{2}(\mathbb{R})}^{2}\bigr{)}=\frac{\varepsilon_{0}}{2},\quad\varepsilon_{0}>0.$

We obtain that for $k\geq k_{0}$,

$\|u_{0}\|_{L^{2}(\mathbb{R})}^{2}<\langle A_{0}u_{0},u_{0}\rangle\leq\langle
A_{0}\psi_{k},\psi_{k}\rangle+\frac{\varepsilon_{0}}{2},$

and thus

$\|\psi_{k}\|_{L^{2}(\mathbb{R})}^{2}=\underbrace{\|\psi_{k}\|_{L^{2}(\mathbb{R})}^{2}-\|u_{0}\|_{L^{2}(\mathbb{R})}^{2}}_{=\theta_{k},\text{
goes to 0 when $k\rightarrow+{\infty}$}}+\|u_{0}\|_{L^{2}(\mathbb{R})}^{2}\\\
=\theta_{k}+\langle
A_{0}u_{0},u_{0}\rangle-\varepsilon_{0}\leq\theta_{k}+\langle
A_{0}\psi_{k},\psi_{k}\rangle+\frac{\varepsilon_{0}}{2}-\varepsilon_{0}\\\
=\langle A_{0}\psi_{k},\psi_{k}\rangle+\theta_{k}-\frac{\varepsilon_{0}}{2}.$

Choosing now $k\geq k_{0}$ and $k$ large enough to have
$\theta_{k}<\varepsilon_{0}/4$, we get

$\|\psi_{k}\|_{L^{2}(\mathbb{R})}^{2}\leq\langle
A_{0}\psi_{k},\psi_{k}\rangle-\frac{\varepsilon_{0}}{4}<\langle
A_{0}\psi_{k},\psi_{k}\rangle,$

and since for $\tilde{\phi}=\psi_{k}$, the Wigner distribution
$\mathcal{W}(\tilde{\phi},\tilde{\phi})$ belongs to
$\mathscr{S}(\mathbb{R}^{2})$, we have

$\|\tilde{\phi}\|_{L^{2}(\mathbb{R})}^{2}<\langle
A_{0}\tilde{\phi},\tilde{\phi}\rangle=\iint
H(x)H(\xi)\mathcal{W}(\tilde{\phi},\tilde{\phi})(x,\xi)dxd\xi,$

and noting that this strict inequality above implies that
$\tilde{\phi}\not=0$, we may set $\phi_{0}=\tilde{\phi}/\|\tilde{\phi}\|$ and
get the first statement in the corollary.

######  N.B.

The proof above is complicated by the fact that the identity

$\langle
a^{w}u,u\rangle_{L^{2}(\mathbb{R}^{n})}=\iint_{\mathbb{R}^{2n}}a(x,\xi)\mathcal{W}(u,u)(x,\xi)dxd\xi,$

is valid a priori for $u\in\mathscr{S}(\mathbb{R}^{n})$ (and in that case
$\mathcal{W}(u,u)$ belongs to $\mathscr{S}(\mathbb{R}^{2n})$), but could be
meaningless as a Lebesgue integral even for ${\text{\rm Op}_{\text{\rm
w}}(a)}$ bounded on $L^{2}(\mathbb{R}^{n})$ and $u\in L^{2}(\mathbb{R}^{n})$,
since we shall have $\mathcal{W}(u,u)\in L^{2}(\mathbb{R}^{2n})$ but not in
$L^{1}(\mathbb{R}^{2n})$ (we shall see in Section 6 that generically the
Wigner distribution of a pulse $u$ in $L^{2}(\mathbb{R}^{n})$ does not belong
to $L^{1}(\mathbb{R}^{2n})$).

Since $\mathcal{W}(\phi,\phi)$ belongs to the Schwartz space of
$\mathbb{R}^{2}$, the Lebesgue Dominated Convergence Theorem provides the last
statement in the Corollary. ∎

######  N.B.

The reader will notice that the results of the incoming Section 5.2 in the
special case $\sigma=0$ imply the results of Section 5.1, which could be then
erased, say at the second reading. However, as far as the first – and maybe
only – reading is concerned, we checked that most of the computational
arguments in the next section are much more involved and it seemed worth while
to the author to avoid unnecessary complications for the disproof of
Flandrin’s conjecture via the quarter-plane example and set apart the more
involved examples of the hyperbolic regions tackled in Section 5.2.

###  5.2. Hyperbolic regions

We consider in this section the (5.0.1) set $\mathcal{C}_{\sigma}$ with a non-
negative $\sigma$.

####  5.2.1. A preliminary observation

We want to consider the operator $A_{\sigma}$ with Weyl symbol
$H(x)H(x\xi-\sigma)$ and as in Section 5.1.1, we would like to secure the fact
that $A_{\sigma}$ is bounded on $L^{2}(\mathbb{R})$.

######  Claim 5.17.

For all $\sigma\geq 0$ the operator $A_{\sigma}$ is bounded self-adjoint on
$L^{2}(\mathbb{R})$.

###### Proof of the Claim.

Let us choose

(5.2.1) $\text{$\chi_{0}\in C^{{\infty}}(\mathbb{R};[0,1])$ with
}\begin{cases}\text{$\chi_{0}(t)=0$, for $t\leq 1$,}\\\ \text{$\chi_{0}(t)=1$,
for $t\geq 2$.}\end{cases}$

For $\phi,\psi\in\mathscr{S}(\mathbb{R})$, we have

(5.2.2)
$\langle(A_{0}-A_{\sigma})\phi,\psi\rangle_{\mathscr{S}^{*}(\mathbb{R}),\mathscr{S}(\mathbb{R})}=\iint
H(x)H(\xi)H(\sigma-x\xi)\underbrace{\mathcal{W}(\phi,\psi)(x,\xi)}_{\in\mathscr{S}(\mathbb{R}^{2})}dxd\xi\\\
=\lim_{\epsilon\rightarrow
0_{+}}\iint\chi_{0}(x/\epsilon)H(\xi)H(\sigma-x\xi)\mathcal{W}(\phi,\psi)(x,\xi)dxd\xi.$

The kernel $k_{\sigma,\epsilon}$ of the operator with Weyl symbol
$\chi_{0}(x/\epsilon)H(\xi)H(\sigma-x\xi)$ is

(5.2.3)
$\ell_{\sigma,\epsilon}(x,y)=\chi_{0}\bigl{(}\frac{x+y}{2\epsilon}\bigr{)}e^{2i\pi\sigma\frac{x-y}{x+y}}\frac{\sin(\frac{2\pi\sigma(x-y)}{x+y})}{\pi(x-y)},$

and we have

$\displaystyle\iint\ell_{\sigma,\epsilon}(x,y)$
$\displaystyle\phi(y)\bar{\psi}(x)dydx=\iint\chi_{0}\bigl{(}\frac{x+y}{2\epsilon}\bigr{)}e^{2i\pi\sigma\frac{x-y}{x+y}}\frac{\sin(\frac{2\pi\sigma(x-y)}{x+y})}{\pi(x-y)}\phi(y)\bar{\psi}(x)dxdy$
$\displaystyle=\iint\chi_{0}\bigl{(}\frac{x+y}{2}\bigr{)}e^{2i\pi\sigma\frac{x-y}{x+y}}\frac{\sin(\frac{2\pi\sigma(x-y)}{x+y})}{\pi\epsilon(x-y)}\phi(\epsilon
y)\bar{\psi}(\epsilon x)\epsilon^{2}dxdy$ (5.2.4)
$\displaystyle=\iint\underbrace{\chi_{0}\bigl{(}\frac{x+y}{2}\bigr{)}e^{2i\pi\sigma\frac{x-y}{x+y}}\frac{\sin(\frac{2\pi\sigma(x-y)}{x+y})}{\pi(x-y)}}_{m_{\sigma}(x,y)}\underbrace{\phi(\epsilon
y)\epsilon^{1/2}}_{\phi_{\epsilon}(y)}\underbrace{\bar{\psi}(\epsilon
x)\epsilon^{1/2}}_{\bar{\psi}_{\epsilon}(x)}dydx.$

We note that, assuming as we may that $\sigma>0$,

(5.2.5)
$|m_{\sigma}(x,y)H(x)H(y)|=\chi_{0}\bigl{(}\frac{x+y}{2}\bigr{)}\left|\frac{\sin(\frac{2\pi\sigma(x-y)}{x+y})}{\frac{2\pi\sigma(x-y)}{x+y}}\right|\frac{2\sigma
H(x)H(y)}{x+y}\\\ \leq\frac{2\sigma H(x)H(y)}{x+y},$

and

(5.2.6)
$|m_{\sigma}(x,y)\check{H}(x)H(y)|=\chi_{0}\bigl{(}\frac{x+y}{2}\bigr{)}\left|\frac{\sin(\frac{2\pi\sigma(x-y)}{x+y})}{{\pi(x-y)}}\right|{\check{H}(x)H(y)}\\\
\leq\frac{\check{H}(x)H(y)}{\pi(y-x)},$

as well as

(5.2.7)
$|m_{\sigma}(x,y)\check{H}(y)H(x)|=\chi_{0}\bigl{(}\frac{x+y}{2}\bigr{)}\left|\frac{\sin(\frac{2\pi\sigma(x-y)}{x+y})}{{\pi(x-y)}}\right|{\check{H}(y)H(x)}\\\
\leq\frac{\check{H}(y)H(x)}{\pi(x-y)}.$

As a consequence, since we have also
$m_{\sigma}(x,y)\check{H}(x)\check{H}(y)\equiv 0,$ the inequalities (5.2.5),
(5.2.6), (5.2.7), the identities (5.2.4), (5.2.2) and Proposition 9.13 imply
that

$|\langle(A_{0}-A_{\sigma})\phi,\psi\rangle_{\mathscr{S}^{*}(\mathbb{R}),\mathscr{S}(\mathbb{R})}|\leq
2\pi\sigma\underbrace{\|H\phi_{\epsilon}\|_{L^{2}(\mathbb{R})}}_{\|H\phi\|_{L^{2}(\mathbb{R})}}\|H\psi_{\epsilon}\|_{L^{2}(\mathbb{R})}+\underbrace{\|\check{H}\phi_{\epsilon}\|_{L^{2}(\mathbb{R})}}_{\|\check{H}\phi\|_{L^{2}(\mathbb{R})}}\|H\psi_{\epsilon}\|_{L^{2}(\mathbb{R})}\\\
+\|H\phi_{\epsilon}\|_{L^{2}(\mathbb{R})}\|\check{H}\psi_{\epsilon}\|_{L^{2}(\mathbb{R})},$

proving that $A_{0}-A_{\sigma}$ is bounded on $L^{2}(\mathbb{R})$; with
Proposition 5.3, this implies that $A_{\sigma}$ is also bounded on
$L^{2}(\mathbb{R})$, proving the claim. ∎

######  N.B.

With that important piece of information in Claim 5.17, we shall be less
strict in our manipulations of the kernels and accept below some abuse of
language in these matters.

The Weyl quantization of $\mathbf{1}_{\mathcal{C}_{\sigma}}$ has the kernel

(5.2.8)
$k_{\sigma}(x,y)=H(x+y)e^{4i\pi\sigma(\frac{x-y}{x+y})}\frac{1}{2}\Bigl{(}\delta_{0}(y-x)+\frac{1}{i\pi}\text{pv}\frac{1}{y-x}\Bigr{)},$

a formula to be compared to (5.1.3). Using the Schwartz function $\phi_{0}$ of
Corollary 5.15, we get from the Lebesgue Dominated Convergence Theorem that
for $\sigma$ small enough,

(5.2.9) $\langle{\text{\rm Op}_{\text{\rm
w}}(\mathbf{1}_{\mathcal{C}_{\sigma}})}\phi_{0},\phi_{0}\rangle_{L^{2}(\mathbb{R})}=\iint_{x\xi\geq\sigma,x>0}\mathcal{W}(\phi_{0},\phi_{0})(x,\xi)dxd\xi>1.$

However, this argument does not work for large positive $\sigma$ and we must
go back to a direct calculation.

####  5.2.2. Diagonal terms

Denoting by $A_{\sigma}$ the operator with kernel (5.2.8) (and Weyl symbol
$H(x\xi-\sigma)H(x)$), we find that for $u\in\mathscr{S}(\mathbb{R})$,
$u_{+}=Hu$, we have

$\displaystyle\langle A_{\sigma}Hu,Hu\rangle_{L^{2}(\mathbb{R})}$
$\displaystyle=\iint
e^{4i\pi\sigma(\frac{x-y}{x+y})}\frac{1}{2}\Bigl{(}\delta_{0}(y-x)+\frac{1}{i\pi}\text{pv}\frac{1}{y-x}\Bigr{)}u_{+}(y)\bar{u}_{+}(x)dydx$
$\displaystyle=\frac{1}{2}\|u_{+}\|^{2}_{L^{2}(\mathbb{R}_{+})}+\iint_{\mathbb{R}^{2}}e^{4i\pi\sigma(\frac{e^{s}-e^{t}}{e^{s}+e^{t}})}\frac{1}{2i\pi}\text{pv}\frac{1}{e^{t}-e^{s}}u_{+}(e^{t})\bar{u}_{+}(e^{s})e^{s+t}dsdt$
$\displaystyle=\frac{1}{2}\|u_{+}\|^{2}_{L^{2}(\mathbb{R}_{+})}+\iint_{\mathbb{R}^{2}}e^{4i\pi\sigma\tanh(\frac{s-t}{2})}\frac{1}{2i\pi}\text{pv}\frac{e^{(s+t)/2}}{e^{t}-e^{s}}\phi_{1}(t)\bar{\phi}_{1}(s)dsdt,$

with

(5.2.10) $\phi_{1}(t)=u_{+}(e^{t})e^{t/2},\qquad\text{so that
$\|\phi_{1}\|_{L^{2}(\mathbb{R})}=\|u_{+}\|_{L^{2}(\mathbb{R}_{+})}.$}$

We get

$\langle
A_{\sigma}Hu,Hu\rangle_{L^{2}(\mathbb{R})}=\frac{1}{2}\|u\|^{2}_{L^{2}(\mathbb{R}_{+})}+\frac{1}{4i\pi}\iint_{\mathbb{R}^{2}}\frac{e^{4i\pi\sigma\tanh(\frac{s-t}{2})}}{\sinh(\frac{t-s}{2})}\phi(t)\bar{\phi}(s)dsdt,$

and noting that $\sinh x=xC(x),$ with $C$ even such that
$1/C\in\mathscr{S}(\mathbb{R})$, we find

$\displaystyle\langle A_{\sigma}Hu,Hu\rangle_{L^{2}(\mathbb{R})}$
$\displaystyle=\frac{1}{2}\|\phi_{1}\|^{2}_{L^{2}(\mathbb{R})}-\frac{1}{2i\pi}\iint_{\mathbb{R}^{2}}\frac{e^{4i\pi\sigma\tanh(\frac{s-t}{2})}}{(s-t)C(\frac{s-t}{2})}\phi(t)\bar{\phi}(s)dsdt$
$\displaystyle=\frac{1}{2}\|\phi_{1}\|^{2}_{L^{2}(\mathbb{R})}+\langle
T_{\sigma}\ast\phi_{1},\phi_{1}\rangle_{L^{2}(\mathbb{R})}$ (5.2.11)
$\displaystyle=\int_{\mathbb{R}}|\hat{\phi}_{1}(\tau)|^{2}\bigl{(}\frac{1}{2}+\hat{T}_{\sigma}(\tau)\bigr{)}d\tau,$

with

(5.2.12)
$T_{\sigma}(t)=\frac{1}{4}\frac{te^{4i\pi\sigma\tanh(\frac{t}{2})}}{\sinh(t/2)}\text{pv}{\frac{i}{\pi
t}}.$

We note that

(5.2.13)
$\displaystyle\hat{T}_{\sigma}(\tau)=\operatorname{sign}\ast\rho_{\sigma},\quad\text{with}$
(5.2.14)
$\displaystyle\rho_{\sigma}(\tau)=\frac{1}{4}\int\frac{te^{4i\pi\sigma\tanh(\frac{t}{2})}}{\sinh(t/2)}e^{-2i\pi
t\tau}dt,\quad\rho_{\sigma}\in\mathscr{S}(\mathbb{R}),$

since the function $\mathbb{R}\ni
t\mapsto\frac{te^{4i\pi\sigma\tanh(\frac{t}{2})}}{\sinh(t/2)}$ belongs to the
Schwartz space111111Indeed, the iterated derivatives of $\tanh$ are
polynomials of $\tanh$ (check this by induction on the order of derivatives)
and thus bounded on the real line; since the function $t\mapsto t/\sinh(t/2)$
belongs to the Schwartz space, this proves that the above product is in
$\mathscr{S}(\mathbb{R})$.. Note also that the function $\rho_{\sigma}$ is
real-valued on the real line. This entails that

(5.2.15)
$\frac{d}{d\tau}\Bigl{\\{}\frac{1}{2}+\hat{T}_{\sigma}\Bigr{\\}}=2\rho_{\sigma},$

and since

$\rho_{\sigma}(\tau)=\frac{1}{4}\mathcal{F}\Bigl{\\{}t\mapsto\frac{te^{4i\pi\sigma\tanh(t/2)}}{\sinh({t}/2)}\Bigr{\\}},\quad\text{implying
}\quad\int_{\mathbb{R}}\rho_{\sigma}(\tau)d\tau=\frac{1}{2},$

we get that

(5.2.16)
$\lim_{\tau\rightarrow\pm{\infty}}\hat{T}_{\sigma}(\tau)=\pm\frac{1}{2}.$

This yields that

(5.2.17)
$\frac{1}{2}+\hat{T}_{\sigma}(\tau)-1=\int_{+{\infty}}^{\tau}2\rho_{\sigma}(\tau^{\prime})d\tau^{\prime}=-1+\int_{-{\infty}}^{\tau}2\rho_{\sigma}(\tau^{\prime})d\tau^{\prime},$

where the last equality follows from (5.2.16): indeed we have for $\tau>0$,
from (5.2.15),

(5.2.18)
$\frac{1}{2}+\hat{T}_{\sigma}(\tau)-1=\int_{+{\infty}}^{\tau}2\rho_{\sigma}(\tau^{\prime})d\tau^{\prime}=-1+\int_{-{\infty}}^{\tau}2\rho_{\sigma}(\tau^{\prime})d\tau^{\prime},$

and for $\tau<0$,

$\frac{1}{2}+\hat{T}_{\sigma}(\tau)=\int_{-{\infty}}^{\tau}2\rho_{\sigma}(\tau^{\prime})d\tau^{\prime}=1+\int_{+{\infty}}^{\tau}2\rho_{\sigma}(\tau^{\prime})d\tau^{\prime}.$

We note that

(5.2.19) $\forall
N\in\mathbb{N},\quad\sup_{\tau\in\mathbb{R}}|\tau|^{N}\left|\frac{1}{2}+\hat{T}_{\sigma}(\tau)-H(\tau)\right|<+{\infty}.$

Indeed for $\tau>0$, we have, using $\rho_{\sigma}\in\mathscr{S}(\mathbb{R})$,

$\left|\tau^{N}\int_{+{\infty}}^{\tau}\rho_{\sigma}(\tau^{\prime})d\tau^{\prime}\right|\leq\int_{\tau}^{+{\infty}}|\rho_{\sigma}(\tau^{\prime})|{\tau^{\prime}}^{N}d\tau^{\prime}\leq\int_{0}^{+{\infty}}|\rho_{\sigma}(\tau^{\prime})|{\tau^{\prime}}^{N}d\tau<+{\infty}.$

Also, for $\tau<0$, we have

$\left|\tau^{N}\int_{-{\infty}}^{\tau}\rho_{\sigma}(\tau^{\prime})d\tau^{\prime}\right|\leq\int_{-{\infty}}^{\tau}|\rho_{\sigma}(\tau^{\prime})||\tau^{\prime}|^{N}d\tau^{\prime}\leq\int_{-{\infty}}^{0}|\rho_{\sigma}(\tau^{\prime})||\tau^{\prime}|^{N}d\tau<+{\infty}.$

This means that the Fourier multiplier $\frac{1}{2}+\hat{T}_{\sigma}(\tau)$ is
somehow “exponentially close” to $H(\tau)$ for large values of $|\tau|$ and in
particular close to 1 for large positive values of $\tau$. We have also

(5.2.20) $\hat{T}_{\sigma}(\tau)=\frac{i}{4\pi}\int_{\mathbb{R}}e^{-2i\pi\tau
t}\frac{e^{4i\pi\sigma\tanh(\frac{t}{2})}}{\sinh(t/2)}dt\\\
=\frac{1}{2\pi}\int_{0}^{+{\infty}}\frac{\sin(2\pi
t\tau-4\pi\sigma\tanh(t/2))}{\sinh(t/2)}dt.$

The next lemma provides more precise estimates than (5.2.19).

######  Lemma 5.18.

Let $\tau>0,\sigma\geq 0$. Defining
$a_{11}(\tau,\sigma)=\frac{1}{2}+\hat{T}_{\sigma}(\tau)$ as given by (5.2.12),
we have

(5.2.21) $|1-a_{11}(\tau,\sigma)|\leq 2e^{-{\pi^{2}\tau}}e^{4\pi\sigma}.$

###### Proof.

Using (5.2.18) and Lemma 9.16, we find that for $\tau>0$,

$|1-a_{11}(\tau,\sigma)|\leq
2\int_{\tau}^{+{\infty}}|\rho_{\sigma}(\tau^{\prime})|d\tau^{\prime}\leq
2\int_{\tau}^{+{\infty}}|\rho_{\sigma}(\tau^{\prime})|d\tau^{\prime}\\\ \leq
12e^{4\pi\sigma}\int_{\tau}^{+{\infty}}e^{-{\pi^{2}\tau^{\prime}}}d\tau^{\prime}=e^{4\pi\sigma}\frac{12}{\pi^{2}}e^{-{\pi^{2}\tau}},$

entailing the sought result. ∎

####  5.2.3. Off-diagonal terms

We want now to check the off-diagonal terms: we have with
$u\in\mathscr{S}(\mathbb{R})$,

(5.2.22) $\displaystyle u_{+}=Hu,\ u_{-}=\check{H}u,$ (5.2.23)
$\displaystyle\phi_{1}(t)=u_{+}(e^{t})e^{t/2},\quad\phi_{2}(t)=u_{-}(-e^{t})e^{t/2}$
$\displaystyle\langle A_{\sigma}\check{H}u,Hu\rangle_{L^{2}(\mathbb{R})}$
$\displaystyle=\iint
e^{4i\pi\sigma(\frac{x-y}{x+y})}\frac{H(x+y)\check{H}(y)H(x)}{2i\pi}\text{pv}\frac{1}{y-x}u_{-}(y)\bar{u}_{+}(x)dydx$
$\displaystyle=\iint
e^{4i\pi\sigma(\frac{e^{s}+e^{t}}{e^{s}-e^{t}})}\frac{H(e^{s}-e^{t})}{2i\pi}\text{pv}\frac{1}{-e^{t}-e^{s}}\phi_{2}(t)\bar{\phi}_{1}(s)e^{\frac{t+s}{2}}dtds$
$\displaystyle=\iint e^{4i\pi\sigma\text{coth}(\frac{s-t}{2})}\
\frac{iH(s-t)}{4\pi}\frac{1}{\cosh(\frac{t-s}{2})}\phi_{2}(t)\bar{\phi}_{1}(s)dtds$
$\displaystyle=\frac{i}{4\pi}\iint e^{4i\pi\sigma\text{coth}(\frac{s-t}{2})}\
{H(s-t)}\frac{1}{\cosh(\frac{s-t}{2})}\phi_{2}(t)\bar{\phi}_{1}(s)dtds$
(5.2.24) $\displaystyle=\langle
S_{\sigma}\ast\phi_{2},\phi_{1}\rangle_{L^{2}(\mathbb{R})},$

with

(5.2.25)
$S_{\sigma}(t)=\frac{i}{4\pi}H(t)\frac{e^{4i\pi\sigma\text{coth}(\frac{t}{2})}}{\cosh(\frac{t}{2})},$

and

(5.2.26) $\hat{S}_{\sigma}(\tau)=\frac{i}{4\pi}\int
H(t)\frac{e^{4i\pi\sigma\text{coth}(\frac{t}{2})}}{\cosh(\frac{t}{2})}e^{-2i\pi
t\tau}dt\\\
=\frac{i}{4\pi}\int_{0}^{+{\infty}}\frac{\cos(4\pi\sigma\operatorname{coth}(t/2)-2\pi
t\tau)}{\cosh(\frac{t}{2})}dt-\frac{1}{4\pi}\int_{0}^{+{\infty}}\frac{\sin(4\pi\sigma\operatorname{coth}(t/2)-2\pi
t\tau)}{\cosh(\frac{t}{2})}dt\\\
=\frac{i}{4\pi}\int_{0}^{+{\infty}}\frac{\cos(2\pi
t\tau-4\pi\sigma\operatorname{coth}(t/2))}{\cosh(\frac{t}{2})}dt+\frac{1}{4\pi}\int_{0}^{+{\infty}}\frac{\sin(2\pi
t\tau-4\pi\sigma\operatorname{coth}(t/2))}{\cosh({t}/2)}dt.$

Note that from (5.2.12), (5.2.14), we have

$\displaystyle\hat{T}_{\sigma}(\tau)=\frac{i}{4\pi}\int\frac{e^{4i\pi\sigma\tanh(\frac{t}{2})}}{\sinh(t/2)}e^{-2i\pi
t\tau}dt=\frac{1}{2\pi}\int_{0}^{+{\infty}}\frac{\sin(2\pi
t\tau-4\pi\sigma\tanh(t/2))}{\sinh(t/2)}dt.$

####  5.2.4. An isometric isomorphism

######  Theorem 5.19.

Let $\sigma\geq 0$ be given, let $\mathcal{C}_{\sigma}$ be the set defined by
(5.0.1) and let $A_{\sigma}$ be the operator with Weyl symbol
$\mathbf{1}_{\mathcal{C}_{\sigma}}$, (whose kernel is given by (5.2.8)). The
operator $A_{\sigma}$ is bounded self-adjoint on $L^{2}(\mathbb{R})$ so that
we may define, with $\Psi$ defined in (5.1.15),

(5.2.27) $\widetilde{A}_{\sigma}=\Psi A_{\sigma}\Psi^{-1}.$

The operator $\widetilde{A}_{\sigma}$ is the Fourier multiplier on
$L^{2}(\mathbb{R};\mathbb{C}^{2})$ given by the matrix

(5.2.28)
$\mathcal{M}_{\sigma}(\tau)=\begin{pmatrix}\frac{1}{2}+\hat{T}_{\sigma}(\tau)&\hat{S}_{\sigma}(\tau)\\\
\overline{\hat{S}_{\sigma}(\tau)}&0\end{pmatrix},$

where $T_{\sigma},S_{\sigma}$ are defined respectively in (5.2.12), (5.2.20),
(5.2.25). In particular we have with $\Phi=(\phi_{1},\phi_{2})\in
L^{2}(\mathbb{R};\mathbb{C}^{2})$,

(5.2.29)
$\langle\widetilde{A}_{\sigma}\Phi,\Phi\rangle_{L^{2}(\mathbb{R};\mathbb{C}^{2})}=\int_{\mathbb{R}}e^{2i\pi
t\tau}\langle\mathcal{M}_{\sigma}(\tau)\hat{\Phi}(\tau),\hat{\Phi}(\tau)\rangle_{\mathbb{C}^{2}}d\tau.$

###### Proof.

We have

$\displaystyle\text{kernel}(HA_{\sigma}H)$
$\displaystyle=e^{4i\pi\sigma\frac{x-y}{x+y}}H(x)H(y)\hat{H}(y-x),$
$\displaystyle\text{kernel}(\check{H}A_{\sigma}H+HA_{\sigma}\check{H})$
$\displaystyle=e^{4i\pi\sigma\frac{x-y}{x+y}}H(x+y)\bigl{(}\check{H}(x)H(y)+H(x)\check{H}(y)\bigr{)}\frac{1}{2i\pi(y-x)},$
$\displaystyle\check{H}A_{\sigma}\check{H}$ $\displaystyle=0.$

Proposition 9.13 in our Appendix is readily giving the $L^{2}$-boundedness
(and self-adjointness) of $\check{H}A_{\sigma}H+HA_{\sigma}\check{H}$. We find
also that $HA_{\sigma}H-\frac{H}{2}$ has kernel

$e^{4i\pi\sigma\frac{x-y}{x+y}}H(x)H(y)\frac{1}{2i\pi(y-x)},$

and thus it is enough to study the operator with kernel

$e^{4i\pi\sigma\frac{e^{s}-e^{t}}{e^{s}+e^{t}}}\frac{e^{\frac{s+t}{2}}}{2i\pi(e^{t}-e^{s})}=e^{4i\pi\sigma\tanh(\frac{s-t}{2})}\frac{1}{4i\pi\sinh(\frac{t-s}{2})},$

which is a convolution operator by

$T_{\sigma}(t)=e^{4i\pi\sigma\tanh(\frac{t}{2})}\frac{t}{4\sinh(\frac{t}{2})}\text{pv}\frac{i}{\pi
t},$

given by (5.2.12). Formula (5.2.14) implies in particular that
$\hat{T}_{\sigma}$ is bounded (and real-valued) on the real line, entailing
eventually the boundedness and self-adjointness of $A_{\sigma}$. Formulas
(5.2.11), (5.2.24) and (5.2.25) are providing (5.2.29), completing the proof
of the theorem. ∎

####  5.2.5. The main result on hyperbolic regions

######  Theorem 5.20.

Let $\sigma\geq 0$ be given and let $A_{\sigma}$ be the operator defined in
Theorem 5.19. Then $A_{\sigma}$ is a bounded self-adjoint operator on
$L^{2}(\mathbb{R})$ such that

(5.2.30) $\inf\bigl{(}\text{\rm
spectrum}(A_{\sigma})\bigr{)}<0<1<\sup\bigl{(}\text{\rm
spectrum}(A_{\sigma})\bigr{)}.$

The spectrum of $A_{\sigma}$ is the closure of the set of eigenvalues of
$\mathcal{M}_{\sigma}(\tau)$ for $\tau$ running on the real line.

######  Remark 5.21.

It is enough to prove that, with a given $\sigma\geq 0$, there exists
$\tau\in\mathbb{R}$ such that $\mathcal{M}_{\sigma}(\tau)$ satisfies (5.1.24).

###### Proof.

We have from (5.2.28), (5.2.20), (5.2.26),

(5.2.31)
$\displaystyle\mathcal{M}_{\sigma}(\tau)=\begin{pmatrix}\frac{1}{2}+\frac{1}{2\pi}\int_{0}^{+{\infty}}\frac{\sin(2\pi
t\tau-4\pi\sigma\tanh(t/2))}{\sinh(t/2)}dt&\cdot&\frac{i}{4\pi}\int_{0}^{+{\infty}}\frac{e^{{-2i\pi}(t\tau-\frac{2\sigma}{\tanh(t/2)})}}{\cosh({t}/2)}dt\\\
\frac{1}{4i\pi}\int_{0}^{+{\infty}}\frac{e^{{2i\pi}(t\tau-\frac{2\sigma}{\tanh(t/2)})}}{\cosh({t}/2)}dt&\cdot&0\end{pmatrix}$
$\displaystyle=\begin{pmatrix}a_{11}(\tau,\sigma)&a_{12}(\tau,\sigma)\\\
a_{21}(\tau,\sigma)&a_{22}(\tau,\sigma)\end{pmatrix}.$

On the other hand we have

(5.2.32)
$\overline{a_{12}}=a_{21}=\frac{1}{4i\pi}\int_{0}^{+{\infty}}\frac{e^{{2i\pi}(t\tau-\frac{2\sigma}{\tanh(t/2)})}}{\cosh({t}/2)}dt,$

so that

(5.2.33)
$\operatorname{Re}a_{12}(\tau,\sigma)=\frac{1}{4\pi}\int_{0}^{+{\infty}}\frac{\sin[{{2\pi}(t\tau-\frac{2\sigma}{\tanh({\frac{t}{2}})})]}}{\cosh(\frac{t}{2})}dt.$

We note that the function
$t\mapsto\frac{e^{{2i\pi}(t\tau-\frac{2\sigma}{\tanh(t/2)})}}{\cosh({t}/2)}$
is holomorphic on $\mathbb{C}\backslash i\pi\mathbb{Z}$, with simple poles at
$(2\mathbb{Z}+1)i\pi$ (zeroes of $\cosh(t/2)$) and essential singularities at
$2\mathbb{Z}i\pi$ (zeroes of $\sinh(t/2)$). We shall need a more explicit
quantitative expression for $a_{21}$ to obtain a precise asymptotic result
which could be compared to the estimate (5.2.21). The next lemma is proven in
[48]; we provide a proof here for the convenience of the reader.

######  Lemma 5.22.

Let $\tau>0,\sigma\geq 0$ be given and let $a_{21}(\tau,\sigma)$ be given by
(5.2.32). We have

(5.2.34) $\displaystyle\operatorname{Re}a_{21}(\tau,\sigma)$
$\displaystyle=\frac{e^{-2\pi^{2}\tau}}{4\pi}\biggl{\\{}\int_{0}^{\pi}\Bigl{(}\frac{e^{2\pi(t\tau-2\sigma\tan(t/2))}-1}{\sin(t/2)}+\frac{\sinh(t/2)-\sin(t/2)}{\sinh(t/2)\sin(t/2)}\Bigr{)}dt$
$\displaystyle\hskip 75.0pt+\int_{0}^{\pi}\frac{1-\cos
2\pi(t\tau-2\sigma\tanh(t/2))}{\sinh(t/2)}dt$ $\displaystyle\hskip
110.0pt-\int_{\pi}^{+{\infty}}\frac{\cos
2\pi(t\tau-2\sigma\tanh(t/2))}{\sinh(t/2)}dt\biggr{\\}}.$

###### Proof of Lemma 5.22.

Let $0<\epsilon<\pi/2<\pi<R$ be given. We consider the closed path
$\gamma_{\epsilon,R}$ of $\mathbb{C}\backslash i\pi\mathbb{Z}$ with
$\text{index}_{\gamma_{\epsilon,R}}(i\pi\mathbb{Z})\equiv 0$,

(5.2.35)
$\gamma_{\epsilon,R}=[\epsilon,R]\cup[R,R+i\pi]\cup[R+i\pi,\epsilon+i\pi]\\\
\cup\\{i\pi+\epsilon e^{i\theta}\\}_{0\geq\theta\geq-\pi/2}\cup
i[\pi-\epsilon,\epsilon]\cup\\{\epsilon e^{i\theta}\\}_{\pi/2\geq\theta\geq
0},$

and we have

(5.2.36)
$\oint_{\gamma_{\epsilon,R}}\frac{e^{{2i\pi}(z\tau-\frac{2\sigma}{\tanh(z/2)})}}{\cosh({z}/2)}dz=0.$

We note as well that

(5.2.37)
$I_{2}=\oint_{[R,R+i\pi]}\frac{e^{{2i\pi}(z\tau-\frac{2\sigma}{\tanh(z/2)})}}{\cosh({z}/2)}dz=i\int_{0}^{\pi}\frac{e^{{2i\pi}((R+it)\tau-\frac{2\sigma}{\tanh(\frac{R+it}{2})})}}{\cosh(\frac{R+it}{2})}dt\\\
=ie^{2i\pi R\tau}\int_{0}^{\pi}e^{-2\pi
t\tau}e^{-4i\pi\sigma\frac{1+e^{-R-it}}{1-e^{-R-it}}}\frac{2dt}{e^{\frac{R+it}{2}}(1+e^{-R-it})},$

so that

$|I_{2}|\leq
2e^{-R/2}\int_{0}^{\pi}e^{4\pi\sigma\operatorname{Im}\left(\frac{1+e^{-R-it}}{1-e^{-R-it}}\right)}\frac{dt}{|1-e^{-R}|},$

and since

$\operatorname{Im}\left(\frac{1+e^{-R-it}}{1-e^{-R-it}}\right)=\operatorname{Im}\frac{(1+e^{-R-it})(1-e^{-R+it})}{|1-e^{-R-it}|^{2}}=\frac{-2e^{-R}\sin
t}{|1-e^{-R-it}|^{2}}\leq 0,$

we get

(5.2.38) $|I_{2}|\leq e^{-R/2}\frac{2\pi}{1-e^{-R}},\quad\text{where $I_{2}$
is defined in \eqref{I22222}}.$

Let us now check121212Let us note for future reference the standard formulas
(5.2.39) $\cosh\bigl{(}\frac{i\pi}{2}+z\bigr{)}=i\sinh
z,\quad\sinh\bigl{(}\frac{i\pi}{2}+z\bigr{)}=i\cosh
z,\quad\tanh\bigl{(}\frac{i\pi}{2}+z\bigr{)}=\operatorname{coth}z.$

(5.2.40) $I_{4}=-\int_{-\pi/2}^{0}\frac{e^{{2i\pi}((i\pi+\epsilon
e^{i\theta})\tau-{2\sigma}{\operatorname{coth}(\frac{i\pi+\epsilon
e^{i\theta}}{2})})}}{\cosh\frac{i\pi+\epsilon e^{i\theta}}{2}}i\epsilon
e^{i\theta}d\theta\\\
=-e^{-2\pi^{2}\tau}\int_{-\pi/2}^{0}\frac{e^{{2i\pi}\left(\epsilon
e^{i\theta}\tau-{2\sigma}{\tanh(\frac{\epsilon
e^{i\theta}}{2})}\right)}}{i\sinh\frac{\epsilon e^{i\theta}}{2}}i\epsilon
e^{i\theta}d\theta,$

and since

$\left|\frac{e^{{2i\pi}\left(\epsilon
e^{i\theta}\tau-{2\sigma}{\tanh(\frac{\epsilon
e^{i\theta}}{2})}\right)}}{i\sinh\frac{\epsilon e^{i\theta}}{2}}i\epsilon
e^{i\theta}\right|\leq 2\max_{|z|\leq\pi/2}|\frac{z}{\sinh
z}|e^{\pi^{2}\tau}e^{4\pi\sigma\sup_{|z|\leq\pi/4}\left|\frac{\sinh z}{\cosh
z}\right|},$

the Lebesgue Dominated Convergence Theorem gives

(5.2.41) $\lim_{\epsilon\rightarrow 0_{+}}I_{4}=-\pi e^{-2\pi^{2}\tau}.$

Defining now

(5.2.42) $I_{6}=-\int_{0}^{\pi/2}\frac{e^{{2i\pi}(\epsilon
e^{i\theta}\tau-{2\sigma}{\operatorname{coth}(\frac{\epsilon
e^{i\theta}}{2})})}}{\cosh\frac{\epsilon e^{i\theta}}{2}}i\epsilon
e^{i\theta}d\theta,$

and noting that

$\displaystyle 4\pi\sigma\operatorname{Im}$
$\displaystyle\operatorname{coth}(\frac{\epsilon
e^{i\theta}}{2})=4\pi\sigma\operatorname{Im}\frac{1+e^{-\epsilon
e^{i\theta}}}{1-e^{-\epsilon
e^{i\theta}}}=4\pi\sigma\operatorname{Im}\frac{(1+e^{-\epsilon
e^{i\theta}})(1-e^{-\epsilon e^{-i\theta}})}{|1-e^{-\epsilon
e^{i\theta}}|^{2}}$
$\displaystyle=4\pi\sigma\operatorname{Im}\frac{e^{-\epsilon
e^{i\theta}}-e^{-\epsilon e^{-i\theta}}}{|1-e^{-\epsilon
e^{i\theta}}|^{2}}=4\pi\sigma\operatorname{Im}\frac{e^{-\epsilon\cos\theta}(e^{-i\epsilon\sin\theta}-e^{i\epsilon\sin\theta})}{|1-e^{-\epsilon
e^{i\theta}}|^{2}}$ $\displaystyle=4\pi\sigma
e^{-\epsilon\cos\theta}\operatorname{Im}\frac{(-2i)\sin(\epsilon\sin\theta)}{|1-e^{-\epsilon
e^{i\theta}}|^{2}}=-4\pi\sigma
e^{-\epsilon\cos\theta}\frac{2\sin(\epsilon\sin\theta)}{|1-e^{-\epsilon
e^{i\theta}}|^{2}}\leq 0,$

we get that

$|I_{6}|\leq\int_{0}^{\pi/2}\frac{e^{-2\pi\epsilon\tau\sin\theta}}{\min_{|z|\leq\pi/4}|\cosh
z|}d\theta\epsilon\leq\epsilon\frac{\pi/2}{\min_{|z|\leq\pi/4}|\cosh z|},$

entailing

(5.2.43) $\lim_{\epsilon\rightarrow 0_{+}}I_{6}=0.$

With

(5.2.44)
$I_{1}=\oint_{[\epsilon,R]}\frac{e^{{2i\pi}(z\tau-\frac{2\sigma}{\tanh(z/2)})}}{\cosh({z}/2)}dz,$

we have from (5.2.32)

(5.2.45) $\lim_{\begin{subarray}{c}\epsilon\rightarrow 0_{+}\\\
R\rightarrow+{\infty}\end{subarray}}I_{1}=4i\pi a_{21}.$

We define now

$\displaystyle
I_{5}=-\oint_{[i\epsilon,i(\pi-\epsilon)]}\frac{e^{{2i\pi}(z\tau-\frac{2\sigma}{\tanh(z/2)})}}{\cosh({z}/2)}dz=-\int_{\epsilon}^{\pi-\epsilon}\frac{e^{{2i\pi}(it\tau-\frac{2\sigma}{\tanh(it/2)})}}{\cosh({it}/2)}idt$
$\displaystyle=-\int_{\epsilon}^{\pi-\epsilon}e^{-2\pi
t\tau}\frac{e^{\frac{-4i\pi\sigma}{i\tan(t/2)}}}{\cos({t}/2)}idt=-i\int_{\epsilon}^{\pi-\epsilon}e^{-2\pi
t\tau}\frac{e^{\frac{-4\pi\sigma}{\tan(t/2)}}}{\cos({t}/2)}dt$
$\displaystyle=-i\int_{\epsilon}^{\pi-\epsilon}e^{-2\pi(\pi-s)\tau}\frac{e^{-\frac{4\pi\sigma}{\tan((\pi-s)/2)}}}{\cos({(\pi-s)}/2)}ds=-ie^{-2\pi^{2}\tau}\int_{\epsilon}^{\pi-\epsilon}e^{2\pi
s\tau}\frac{e^{-\frac{4\pi\sigma\sin(s/2)}{\cos(s/2)}}}{\sin({s}/2)}ds,$

so that

(5.2.46) $\displaystyle I_{5}=$ $\displaystyle-
ie^{-2\pi^{2}\tau}\int_{\epsilon}^{\pi-\epsilon}e^{2\pi
s\tau}\frac{e^{-{4\pi\sigma\tan(s/2)}{}}}{\sin({s}/2)}ds.$

We have also

(5.2.47)
$I_{3}=\oint_{[R+i\pi,\epsilon+i\pi]}\frac{e^{{2i\pi}(z\tau-\frac{2\sigma}{\tanh(z/2)})}}{\cosh({z}/2)}dz=-\int_{\epsilon}^{R}\frac{e^{{2i\pi}((t+i\pi)\tau-\frac{2\sigma}{\tanh((t+i\pi)/2)})}}{\cosh({(t+i\pi)}/2)}dt,$

so that using Formulas (5.2.39), we get

$I_{3}=-e^{-2\pi^{2}\tau}\int_{\epsilon}^{R}\frac{e^{{2i\pi}(t\tau-{2\sigma}{\tanh(t/2)})}}{i\sinh(t/2)}dt,$

and

(5.2.48)
$I_{3}+I_{5}=ie^{-2\pi^{2}\tau}\left(\int_{\epsilon}^{R}\frac{e^{{2i\pi}(t\tau-{2\sigma}{\tanh(t/2)})}}{\sinh(t/2)}dt-\int_{\epsilon}^{\pi-\epsilon}e^{2\pi
t\tau}\frac{e^{-{4\pi\sigma\tan(t/2)}{}}}{\sin({t}/2)}dt\right)\\\
=ie^{-2\pi^{2}\tau}\bigg{\\{}\int_{\epsilon}^{\pi-\epsilon}\Bigl{(}\frac{e^{{2i\pi}(t\tau-{2\sigma}{\tanh(t/2)})}}{\sinh(t/2)}-\frac{e^{2\pi(t\tau-2\sigma\tan(t/2))}}{\sin({t}/2)}\Bigr{)}dt\\\
+\int_{\pi-\epsilon}^{R}\frac{e^{{2i\pi}(t\tau-{2\sigma}{\tanh(t/2)})}}{\sinh(t/2)}dt\biggr{\\}}.$

From (5.2.36), (5.2.35), (5.2.37), (5.2.40), (5.2.42), (5.2.44), (LABEL:4040),
(5.2.47), we find that

$I_{1}=-I_{2}-(I_{3}+I_{5})-I_{4}-I_{6},$

so that taking the limit of both
sides131313$I_{1},I_{2},I_{4},I_{6},I_{3}+I_{5}$ do have limits when
$\epsilon\rightarrow 0_{+},R\rightarrow+{\infty}$. when $\epsilon\rightarrow
0_{+},R\rightarrow+{\infty}$ we get, thanks to (5.2.45), (5.2.38), (5.2.48),
(5.2.41), (5.2.43),

(5.2.49) $4i\pi a_{21}=\\\
-ie^{-2\pi^{2}\tau}\bigg{\\{}\int_{0}^{\pi}\Bigl{(}\frac{e^{{2i\pi}(t\tau-{2\sigma}{\tanh(t/2)})}}{\sinh(t/2)}-\frac{e^{2\pi(t\tau-2\sigma\tan(t/2))}}{\sin({t}/2)}\Bigr{)}dt+\int_{\pi}^{+{\infty}}\frac{e^{{2i\pi}(t\tau-{2\sigma}{\tanh(t/2)})}}{\sinh(t/2)}dt\biggr{\\}}\\\
+\pi e^{-2\pi^{2}\tau},$

implying that

$a_{21}=\\\
\frac{e^{-2\pi^{2}\tau}}{4\pi}\bigg{\\{}\int_{0}^{\pi}\Bigl{(}-\frac{e^{{2i\pi}(t\tau-{2\sigma}{\tanh(t/2)})}}{\sinh(t/2)}+\frac{e^{2\pi(t\tau-2\sigma\tan(t/2))}}{\sin({t}/2)}\Bigr{)}dt-\int_{\pi}^{+{\infty}}\frac{e^{{2i\pi}(t\tau-{2\sigma}{\tanh(t/2)})}}{\sinh(t/2)}dt\biggr{\\}}\\\
-\frac{i}{4}e^{-2\pi^{2}\tau}$

that is

(5.2.50)
$a_{21}=\frac{e^{-2\pi^{2}\tau}}{4\pi}\int_{0}^{\pi}\Bigl{(}\frac{e^{2\pi(t\tau-2\sigma\tan(t/2))}}{\sin({t}/2)}-\frac{\cos{{2\pi}(t\tau-{2\sigma}{\tanh(t/2)})}}{\sinh(t/2)}\Bigr{)}dt\\\
-\frac{e^{-2\pi^{2}\tau}}{4\pi}\int_{\pi}^{+{\infty}}\frac{\cos{{2\pi}(t\tau-{2\sigma}{\tanh(t/2)})}}{\sinh(t/2)}dt\\\
-i\frac{e^{-2\pi^{2}\tau}}{4\pi}\int_{0}^{\pi}\frac{\sin{{2\pi}(t\tau-{2\sigma}{\tanh(t/2)})}}{\sinh(t/2)}dt-\frac{i}{4}e^{-2\pi^{2}\tau}\\\
-i\frac{e^{-2\pi^{2}\tau}}{4\pi}\int_{\pi}^{+{\infty}}\frac{\sin{{2\pi}(t\tau-{2\sigma}{\tanh(t/2)})}}{\sinh(t/2)}dt,$

yielding

(5.2.51)
$\operatorname{Re}a_{21}=\frac{e^{-2\pi^{2}\tau}}{4\pi}\int_{0}^{\pi}\Bigl{(}\frac{e^{2\pi(t\tau-2\sigma\tan(t/2))}}{\sin({t}/2)}-\frac{\cos{{2\pi}(t\tau-{2\sigma}{\tanh(t/2)})}}{\sinh(t/2)}\Bigr{)}dt\\\
-\frac{e^{-2\pi^{2}\tau}}{4\pi}\int_{\pi}^{+{\infty}}\frac{\cos{{2\pi}(t\tau-{2\sigma}{\tanh(t/2)})}}{\sinh(t/2)}dt,$

completing the proof of Lemma 5.22. ∎

######  Remark 5.23.

Formula (5.2.50) also yields

$\operatorname{Im}a_{12}=-\operatorname{Im}a_{21}=\frac{e^{-2\pi^{2}\tau}}{4\pi}\Bigl{\\{}\int_{0}^{\pi}\frac{\sin{{2\pi}(t\tau-{2\sigma}{\tanh(t/2)})}}{\sinh(t/2)}dt+\pi\\\
+\int_{\pi}^{+{\infty}}\frac{\sin{{2\pi}(t\tau-{2\sigma}{\tanh(t/2)})}}{\sinh(t/2)}dt\Bigr{\\}},$

and since from (5.2.31), we have

(5.2.52) $a_{11}=\frac{1}{2}+\frac{1}{2\pi}\int_{0}^{+{\infty}}\frac{\sin(2\pi
t\tau-4\pi\sigma\tanh(t/2))}{\sinh(t/2)}dt,$

this gives

(5.2.53)
$\operatorname{Im}a_{12}=\frac{e^{-2\pi^{2}\tau}}{4\pi}\bigl{(}2\pi(a_{11}-\frac{1}{2})+\pi\bigr{)}=\frac{e^{-2\pi^{2}\tau}}{2}a_{11}.$

To complete the proof of Theorem 5.20, it will be enough, according to Lemma
5.11, to prove that, for $\tau\rightarrow+{\infty}$, $|a_{12}|^{2}\gg
1-a_{11}.$ To achieve that, we note from (5.2.53) that the imaginary part of
$a_{12}$ is useless and we shall prove simply that

${(\operatorname{Re}a_{12})}^{2}\gg 1-a_{11}.$

To get this we are going to use (5.2.21) and a precise asymptotic behavior for
${(\operatorname{Re}a_{12})}^{2}$ displayed in the next lemma and issued from
the explicit formula (5.2.34).

######  Lemma 5.24.

Let $\tau\geq 1,\sigma\geq 0$ be given and let $a_{21}(\tau,\sigma)$ be given
by (5.2.32). We have then

(5.2.54)
$\displaystyle\operatorname{Re}a_{21}(\tau,\sigma)\geq\frac{e^{-8\pi\sqrt{\tau}\sqrt{\sigma}}}{8\pi^{3}\tau}-\frac{1}{2\pi}e^{-2\pi^{2}\tau}.$

###### Proof of the lemma.

Since for $t\geq 0$ we have $\sinh(t/2)-\sin(t/2)\geq 0$, we get from
(5.2.34),

(5.2.55)
$\operatorname{Re}a_{21}(\tau,\sigma)\geq\frac{e^{-2\pi^{2}\tau}}{4\pi}\biggl{\\{}\int_{0}^{\pi}\frac{e^{2\pi(t\tau-2\sigma\tan(t/2))}-1}{\sin(t/2)}dt-\int_{\pi}^{+{\infty}}\frac{1}{\sinh(t/2)}dt\biggr{\\}}\\\
=\frac{e^{-2\pi^{2}\tau}}{4\pi}\int_{0}^{\pi}\frac{e^{2\pi(t\tau-2\sigma\tan(t/2))}-1}{\sin(t/2)}dt-\frac{e^{-2\pi^{2}\tau}}{2\pi}\ln\bigl{(}\operatorname{coth}\frac{\pi}{4}\bigr{)}.$

Let us define

(5.2.56)
$\omega=2\pi\tau,\quad\kappa=2\pi\sigma,\quad\nu=\kappa^{1/2}\omega^{-1/2},\quad\phi_{\nu}(s)=s-\nu^{2}\tan
s.$

We have

$2\pi\bigl{(}t\tau-2\sigma\tan(t/2))=2\pi\tau\bigl{(}t-2\nu^{2}\tan(t/2)\bigr{)}=4\pi\tau\bigl{(}\frac{t}{2}-\nu^{2}\tan\frac{t}{2}\bigr{)}=2\omega\phi_{\nu}(t/2).$

We have thus

(5.2.57)
$\operatorname{Re}a_{21}(\tau,\sigma)\geq\frac{e^{-\pi\omega}}{2\pi}\int_{0}^{\pi/2}\frac{e^{2\omega\phi_{\nu}(s)}-1}{\sin
s}ds-\frac{e^{-\pi\omega}}{2\pi}\underbrace{\ln\bigl{(}\operatorname{coth}\frac{\pi}{4}\bigr{)}}_{\approx
0.421908}.$

Defining

(5.2.58)
$\psi_{\nu}(\omega)=\frac{e^{-\pi\omega}}{2\pi}\int_{0}^{\pi/2}\frac{e^{2\omega\phi_{\nu}(s)}-1}{\sin
s}ds,$

we can use (5.2.56), (5.2.57) and (9.5.22) to get whenever $\tau>0$,

$\displaystyle
2\pi\operatorname{Re}a_{21}(\tau,\sigma)\geq\frac{e^{-8\pi\sqrt{\tau}\sqrt{\sigma}}}{\pi^{2}\tau}\Bigl{(}\frac{1}{2}-\frac{1}{4\tau}\Bigr{)}-e^{-2\pi^{2}\tau},$

so that for $\tau\geq 1$ we find

(5.2.59) $\displaystyle
2\pi\operatorname{Re}a_{21}(\tau,\sigma)\geq\frac{e^{-8\pi\sqrt{\tau}\sqrt{\sigma}}}{4\pi^{2}\tau}-e^{-2\pi^{2}\tau},$

yielding the lemma. ∎

We eventually go back to the proof of Theorem 5.20: let $\sigma>0$ be given.
From Lemma 5.24 and (5.2.21), we have for $\tau\geq 1$,

$\displaystyle|1-a_{11}(\tau,\sigma)|\leq 2e^{-{\pi^{2}\tau}}e^{4\pi\sigma},$
$\displaystyle\operatorname{Re}a_{21}(\tau,\sigma)\geq\frac{e^{-8\pi\sqrt{\tau}\sqrt{\sigma}}}{8\pi^{3}\tau}-\frac{1}{2\pi}e^{-2\pi^{2}\tau}=\frac{e^{-8\pi\sqrt{\tau}\sqrt{\sigma}}}{8\pi^{3}\tau}\Bigl{(}1-\frac{4\pi^{2}\tau
e^{8\pi\sqrt{\tau}\sqrt{\sigma}}}{e^{2\pi^{2}\tau}}\Bigr{)}.$

This entails that for $\tau\geq\tau_{0}(\sigma)$, we have

(5.2.60)
$\operatorname{Re}a_{21}(\tau,\sigma)\geq\frac{e^{-8\pi\sqrt{\tau}\sqrt{\sigma}}}{16\pi^{3}\tau},$

and thus $a_{21}\not=0$ and

(5.2.61)
$|a_{21}(\sigma,\tau)|^{2}\geq\frac{e^{-16\pi\sqrt{\tau}\sqrt{\sigma}}}{2^{8}\pi^{6}\tau^{2}}>|1-a_{11}(\tau,\sigma)|,$

where the last inequality above holds true (thanks to (5.2.21)) whenever

$2e^{-{\pi^{2}\tau}}e^{4\pi\sigma}<\frac{e^{-16\pi\sqrt{\tau}\sqrt{\sigma}}}{2^{8}\pi^{6}\tau^{2}},$

which is indeed true for $\tau\geq\tau_{1}(\sigma)$. As a result for
$\tau\geq\max(4\sigma,4,\tau_{0}(\sigma),\tau_{1}(\sigma))$, we obtain that
(5.2.61) is satisfied so that Remark 5.21 implies the result of Theorem 5.20,
completing our proof. ∎

######  Remark 5.25.

The functions $\tau_{0}(\sigma),\tau_{1}(\sigma)$ can be determined rather
easily, the first one by the condition

$\tau\geq\tau_{0}(\sigma)\Longrightarrow\frac{4\pi^{2}\tau
e^{8\pi\sqrt{\tau}\sqrt{\sigma}}}{e^{2\pi^{2}\tau}}\leq\frac{1}{2},$

whereas the second one should satisfy

$\tau\geq\tau_{1}(\sigma)\Longrightarrow
e^{4\pi\sigma}2^{9}\pi^{6}\tau^{2}e^{16\pi\sqrt{\tau}\sqrt{\sigma}}<e^{{\pi^{2}\tau}}.$

###  5.3. Comments and further results

####  5.3.1. Qualitative explanations on the various computations

We would like to go back to our proofs that

(5.3.1)
$|a_{12}(\tau,\sigma)|^{2}\gg|1-a_{11}(\tau,\sigma)|,\quad\tau\rightarrow+{\infty},$

which is our key argument via Lemma 5.11 and give a couple of qualitative
explanations which may enlighten the calculations. It is of course much
simpler to begin with the case $\sigma=0$: in that case, according to
Proposition 5.13 and (5.1.18), we have

(5.3.2)
$1-a_{11}(\tau,0)=\int_{\tau}^{+{\infty}}2\rho_{0}(\tau^{\prime})d\tau,\quad
2\rho_{0}(\tau)=\int\underbrace{\left(\frac{t/2}{\sinh(t/2)}\right)}_{\begin{subarray}{c}=f_{0}(t),\
\text{$f_{0}\in\mathscr{S}(\mathbb{R})$}\\\ \text{holomorphic }\\\ \text{on
$|\operatorname{Im}t|<2\pi.$}\end{subarray}}e^{-2i\pi t\tau}ds,$

so that $2\rho_{0}(\tau)=\hat{f}_{0}(\tau).$ We get thus readily that
$\rho_{0}$ belongs to the Schwartz space, as the Fourier transform of a
function in the Schwartz space and this implies in particular that
$1-a_{11}(\tau,0)$ has fast decay towards 0 when $\tau\rightarrow+{\infty}$,
as proven in Proposition 5.13. We note also that (5.2.53) gives
$\operatorname{Im}a_{12}(\tau,0)^{2}={e^{-4\pi^{2}\tau}}a_{11}(\tau,0)^{2}/4,$
and since the limit of $a_{11}$ is 1, we do not expect any help from the
imaginary part of $a_{12}$ to proving (5.3.1). Turning our attention to
$\operatorname{Re}a_{12}$ in (5.1.30), we have,

(5.3.3)
$4\pi\operatorname{Re}a_{21}(\tau,0)=\int_{0}^{+{\infty}}\frac{\sin{(2\pi
t\tau)}}{\cosh({t}/2)}dt,$

which is the sine-Fourier transform of the function $t\mapsto
H(t)\operatorname{sech}(t/2)=g_{0}(t)$, which has a singularity at $t=0$: as a
consequence, thanks to Lemma 9.1, the Fourier transform $\hat{g}_{0}$ cannot
be rapidly decreasing, cannot even belong to $L^{1}(\mathbb{R})$ (that would
imply that $g_{0}$ is continuous). Moreover the sine-Fourier transform above
is the Fourier transform of the odd part of $g_{0}$,
$g_{\text{odd}}(t)=\operatorname{sech}(t/2)\operatorname{sign}t$, which is
also singular at 0, thus $\widehat{g_{\text{odd}}}$ cannot be rapidly
decreasing and is an odd function, which is enough to prove, without more
calculations, that (5.3.1) holds true. In Section 5.1, we used a more explicit
argument, with providing an equivalent of (5.3.3) equal to $1/(2\pi\tau)$ near
$+{\infty}$. Summing-up, (5.3.1) in the case $\sigma=0$ follows from the
existence of a singularity of the function $g_{0}$ above, which is
discontinuous at 0.

Let us now take a look at the case $\sigma>0$, which turns out to be more
computationally involved. We have from (5.2.32)

(5.3.4) $\displaystyle 4\pi i{a_{21}}(\tau,\sigma)$
$\displaystyle=\int_{\mathbb{R}}H(t)\operatorname{sech}(t/2)e^{-i4\pi\sigma\operatorname{coth}(t/2)}e^{2i\pi
t\tau}dt=\check{\widehat{g_{\sigma}}}(\tau),$ (5.3.5) $\displaystyle
g_{\sigma}(t)$
$\displaystyle=H(t)\operatorname{sech}(t/2)e^{-i4\pi\sigma\operatorname{coth}(t/2)}.$

The single discontinuity at $t=0$ of $g_{\sigma}$ when $\sigma>0$ is much
wilder than for $\sigma=0$: in the latter case, we had only a jump
discontinuity with different limits on both sides, whereas when $\sigma>0$, we
have an essential discontinuity with an oscillatory behaviour in $(-1,+1)$
when $t\rightarrow 0_{+}$ for the real and imaginary parts of $a_{12}$.
However, $g_{\sigma}$ belongs to all $L^{p}(\mathbb{R}),p\in[1,+{\infty}]$, so
that its Fourier transform belongs to $L^{p}(\mathbb{R}),p\in[2,+{\infty}]$:
we expect then that both sides of (5.3.1) have limit $0$ for
$\tau\rightarrow+{\infty}$ and we must prove that $1-a_{11}$ decays much
faster than $a_{12}$. Looking at a slightly simplified model and using the
notations (5.2.56), we define for $\omega,\nu$ positive, a function $\alpha$
presumably close to $4\pi ia_{21}$, given by

(5.3.6)
$\alpha(\omega,\nu)=\int_{0}^{+{\infty}}e^{i2\omega\mu_{\nu}(s)}\operatorname{sech}(s)ds,\quad\mu_{\nu}(s)=s-\frac{\nu^{2}}{s},\quad\mu^{\prime}_{\nu}(s)=1+\frac{\nu^{2}}{s^{2}}.$

Trying our hand with the stationary phase method, we look at

$\alpha(\omega,\nu)=\frac{1}{2i\omega}\int_{0}^{+{\infty}}\frac{d}{ds}\left\\{e^{i2\omega\mu_{\nu}(s)}\right\\}\frac{\operatorname{sech}(s)}{\mu^{\prime}_{\nu}(s)}ds\\\
=\frac{1}{2i\omega}\int_{0}^{+{\infty}}\frac{d}{ds}\left\\{e^{i2\omega\mu_{\nu}(s)}\right\\}\frac{s^{2}\operatorname{sech}(s)}{s^{2}+\nu^{2}}ds\\\
=\frac{i}{2\omega}\int_{0}^{+{\infty}}e^{i2\omega\mu_{\nu}(s)}\frac{d}{ds}\left\\{\frac{s^{2}\operatorname{sech}(s)}{s^{2}+\nu^{2}}\right\\},$

since the boundary term vanishes. Iterating that computation shows that
$\alpha(\omega,\nu)=O_{\sigma}(\omega^{-N})$ for all $N$ when
$\omega\rightarrow+{\infty}$, meaning that the information of fast decay for
$1-a_{11}$ will not suffice to get (5.3.1). Also, it is worth noticing that no
fast decay of the function $\alpha$ occurs when $\omega\rightarrow-{\infty}$,
otherwise Lemma 9.1 would give smoothness for the function $s\mapsto
e^{-2i\kappa/s}H(s)\operatorname{sech}s$: in fact we see also that for
$\sigma>0$, $\tau=-\lambda$, $\lambda>0$, we have

$2\pi
i{a_{21}}(-\lambda,\sigma)=\int_{0}^{+{\infty}}\operatorname{sech}(s)e^{-i4\pi\sigma\operatorname{coth}(s)}e^{-4i\pi
s\lambda}ds,$

and the phase function is
$\tilde{\mu}(s)=-4i\pi(s\lambda+\sigma\operatorname{coth}(s))$ and we have

$\frac{d}{ds}\bigl{\\{}s\lambda+\sigma\operatorname{coth}(s)\bigr{\\}}=\lambda-\frac{\sigma(1-\tanh^{2}s)}{\tanh^{2}s}=\frac{(\lambda+\sigma)\tanh^{2}s-\sigma}{\tanh^{2}s},$

which does vanish at $\tanh s=\sigma/(\lambda+\sigma)$. As a result we could
say that, for $\sigma>0$, the $C^{{\infty}}$ wave-front-set (see e.g. Section
8.1 in [21]) of the function $g_{\sigma}$ is reduced to
$\\{0\\}\times(-{\infty},0)$. It turns out that we can show that the Gevrey-2
wave-front-set of $g_{\sigma}$ is $\\{0\\}\times\mathbb{R}^{*}$, and it is
expressed via the lowerbound estimate (5.2.54); the route that we took for
proving this was an explicit calculation of $\operatorname{Re}a_{12}$,
following the paper [48]. Finally the upper bound (5.2.21) can be improved as

(5.3.7) $|1-a_{11}(\tau,\sigma)|\leq
C_{\sigma,\epsilon}e^{-{(\pi-\epsilon)2\pi\tau}},\quad\epsilon>0,$

and is expressing the fact the the function
$t\mapsto\frac{te^{4i\pi\sigma\tanh(\frac{t}{2})}}{\sinh(t/2)}$ is analytic on
the real line, with a radius of convergence on the real line bounded below by
$\pi$ (cf. Proposition 9.2).

####  5.3.2. More results and examples: $\ell^{p}$ balls, corners

For $a,\phi_{0}$ like in Corollary 5.15, defining

$\Omega_{p}=\\{(x,\xi)\in\mathbb{R}^{2},|x-\frac{a}{2}|^{p}+|\xi-\frac{a}{2}|^{p}<\bigl{(}\frac{a}{2}\bigr{)}^{p}\\},$

since $\mathcal{W}(\phi_{0},\phi_{0})\in\mathscr{S}(\mathbb{R}^{2})$, we get

$\lim_{p\rightarrow+{\infty}}\iint_{\Omega_{p}}\mathcal{W}(\phi_{0},\phi_{0})(x,\xi)dxd\xi=\iint_{[0,a]^{2}}\mathcal{W}(\phi_{0},\phi_{0})(x,\xi)dxd\xi>\|\phi_{0}\|_{L^{2}(\mathbb{R})}^{2},$

proving that the spectrum of ${\text{\rm Op}_{\text{\rm
w}}(\mathbf{1}_{\Omega_{p}})}$ intersects $(1,+{\infty})$ for $p$ large
enough, showing that a counterexample to Flandrin’s conjecture can be a convex
analytic open bounded set. Moreover, defining

$Q_{a}=\\{(x,\xi)\in\mathbb{R}^{2},|x|+|\xi|\leq a/\sqrt{2}\\},$

we note that $Q_{a}$ is obtained by rotation and translation of $[0,a]^{2}$ so
that we can find $\phi_{1}$ in the Schwartz space such that

$\iint_{Q_{a}}\mathcal{W}(\phi_{1},\phi_{1})(x,\xi)dxd\xi>\|\phi_{1}\|_{L^{2}(\mathbb{R})}^{2}.$

Since we have

$\lim_{p\rightarrow
1}\hskip-13.0pt\iint_{\begin{subarray}{c}|x|^{p}+|\xi|^{p}\leq(a/\sqrt{2})^{p}\end{subarray}}\hskip-12.0pt\mathcal{W}(\phi_{1},\phi_{1})(x,\xi)dxd\xi=\iint_{Q_{a}}\mathcal{W}(\phi_{1},\phi_{1})(x,\xi)dxd\xi>\|\phi_{1}\|_{L^{2}(\mathbb{R})}^{2},$

we get that for $p-1$ small enough we have

(5.3.8)
$\iint_{\begin{subarray}{c}|x|^{p}+|\xi|^{p}\leq(a/\sqrt{2})^{p}\end{subarray}}\hskip-12.0pt\mathcal{W}(\phi_{1},\phi_{1})(x,\xi)dxd\xi>\|\phi_{1}\|_{L^{2}(\mathbb{R})}^{2},$

proving that $\ell^{p}$ balls are counterexamples to Flandrin’s conjecture for
$p-1$ or $1/p$ small enough. Convex affine cones with aperture strictly less
than $\pi$ of $\mathbb{R}^{2}$ are translations and rotations of

(5.3.9)
$\Sigma_{\theta_{0}}=\\{(x,\xi)\in\mathbb{R}^{2}\backslash(\mathbb{R}_{-}\times\\{0\\}),\
\arg(x+i\xi)\in(0,\theta_{0})\\},\quad\text{for some $\theta_{0}\in(0,\pi)$.}$

The vertex of $\Sigma_{\theta_{0}}$ and its rotations is defined as 0 and the
vertex of the translation of vector $T_{0}$ of $\Sigma_{\theta_{0}}$ is
defined as $T_{0}$. We note that all convex affine cones with aperture
strictly less than $\pi$ are symplectically equivalent in $\mathbb{R}^{2}$,
since $\Sigma_{\theta_{0}}$ is symplectically equivalent to (the interior of)
the quarter plane $\Sigma_{\pi/2}$: indeed let $\theta_{0}$ be in $(0,\pi)$;
the symplectic matrix $M_{\theta_{0}}$ defined by

$M_{\theta_{0}}=\begin{pmatrix}1&-\operatorname{cotan}\theta_{0}\\\
0&1\end{pmatrix},$

is such that $M_{\theta_{0}}\begin{pmatrix}1\\\
0\end{pmatrix}=\begin{pmatrix}1\\\ 0\end{pmatrix},\quad
M_{\theta_{0}}\begin{pmatrix}\cos\theta_{0}\\\
\sin\theta_{0}\end{pmatrix}=\begin{pmatrix}0\\\ \sin\theta_{0}\end{pmatrix},$
proving that

$M_{\theta_{0}}\Sigma_{\theta_{0}}=\Sigma_{\pi/2}.$

The next result follows from Theorem 1.3 in [4] and shows that many
counterexamples to Flandrin’s conjecture can be be obtained.

######  Theorem 5.26.

Let $K$ be a subset of the closure of a convex affine cone with aperture
strictly less than $\pi$ and vertex $X_{0}$ such that $K$ contains a
neighborhood of the vertex in the cone141414We shall say that the set $K$ has
a corner.. Then there exists $\lambda>0$ such that, with

$K_{\lambda}=X_{0}+\lambda(K-X_{0}),$

there exists $\phi\in\mathscr{S}(\mathbb{R})$ such that

(5.3.10)
$\displaystyle\iint_{K_{\lambda}}\mathcal{W}(\phi,\phi)(x,\xi)dxd\xi>\|\phi\|_{L^{2}(\mathbb{R})}^{2}.$

######  N.B.

Note that (5.3.10) implies that $\phi$ is not the zero function. Also, taking
$K$ convex produces another counterexample to Flandrin’s conjecture since
$K_{\lambda}$ will be then convex, but we do not need that assumption to
proving the result.

###### Proof.

There is no loss of generality at assuming $X_{0}=0$ and

$[0,\rho_{0}]^{2}\subset K\subset\overline{\Sigma}_{\pi/2},\quad\rho_{0}>0.$

Using Corollary 5.15, we find $\phi_{0}\in\mathscr{S}(\mathbb{R})$ (so that
$\mathcal{W}(\phi_{0},\phi_{0})\in\mathscr{S}(\mathbb{R}^{2})$) such that

$\lim_{\lambda\rightarrow+{\infty}}\iint_{K_{\lambda}}\mathcal{W}(\phi_{0},\phi_{0})(x,\xi)dxd\xi=\iint_{{\Sigma}_{\pi/2}}\mathcal{W}(\phi_{0},\phi_{0})(x,\xi)dxd\xi>\|\phi_{0}\|_{L^{2}(\mathbb{R})}^{2},$

implying for $\lambda$ large enough that
$\iint_{K_{\lambda}}\mathcal{W}(\phi_{0},\phi_{0})(x,\xi)dxd\xi>\|\phi_{0}\|_{L^{2}(\mathbb{R})}^{2},$
which is the sought result. ∎

###  5.4. Numerics

######  Definition 5.27.

Let $\sigma\geq 0$ be given. With the $2\times 2$ Hermitian matrix
$\mathcal{M}_{\sigma}$ given by (5.2.31), we define for $\tau\in\mathbb{R}$,

(5.4.1) $\displaystyle\lambda_{+}(\tau,\sigma)$
$\displaystyle=\frac{1}{2}\left(a_{11}(\tau,\sigma)\\!+\\!\sqrt{a_{11}^{2}(\tau,\sigma)+4|a_{12}(\tau,\sigma)|^{2}}\right),$
(5.4.2) $\displaystyle\lambda_{-}(\tau,\sigma)$
$\displaystyle=\frac{1}{2}\left(a_{11}(\tau,\sigma)\\!-\\!\sqrt{a_{11}^{2}(\tau,\sigma)+4|a_{12}(\tau,\sigma)|^{2}}\right).$

######  Remark 5.28.

According to (5.2.53), we have

(5.4.3) $\displaystyle\lambda_{+}(\tau,\sigma)$
$\displaystyle=\frac{1}{2}\left(a_{11}(\tau,\sigma)\\!+\\!\sqrt{a_{11}^{2}(\tau,\sigma)\bigl{(}1+e^{-4\pi^{2}\tau}\bigr{)}+4{(\operatorname{Re}a_{12}(\tau,\sigma)})^{2}}\right),$
(5.4.4) $\displaystyle\lambda_{-}(\tau,\sigma)$
$\displaystyle=\frac{1}{2}\left(a_{11}(\tau,\sigma)-\sqrt{a_{11}^{2}(\tau,\sigma)\bigl{(}1+e^{-4\pi^{2}\tau}\bigr{)}+4{(\operatorname{Re}a_{12}(\tau,\sigma)})^{2}}\right),$

so that the knowledge of $a_{11}$ and $\operatorname{Re}{a_{12}}$ suffices for
expressing $\lambda_{\pm}$.

An immediate consequence of Theorem 5.20 is

######  Theorem 5.29.

Let $\sigma\geq 0$ be given and let $A_{\sigma}$ be the self-adjoint operator
bounded in $L^{2}(\mathbb{R})$ defined in Theorem 5.20. With the notations of
Definition 5.27, we have

(5.4.5) $\displaystyle M_{\sigma}:=\sup\\{\text{\rm spectrum}(A_{\sigma})\\}$
$\displaystyle=\sup_{\tau\in\mathbb{R}}\lambda_{+}(\tau,\sigma),$ (5.4.6)
$\displaystyle m_{\sigma}:=\inf\\{\text{\rm spectrum}(A_{\sigma})\\}$
$\displaystyle=\inf_{\tau\in\mathbb{R}}\lambda_{-}(\tau,\sigma).$

Moreover for all $\sigma\geq 0$ we have

(5.4.7) $m_{\sigma}<0<1<M_{\sigma}.$

####  5.4.1. The quarter-plane: $\sigma=0$

Of course, as shown by the respective calculations of Sections 5.1 and 5.2,
the case $\sigma=0$, dealing with the quarter-plane is much simpler than the
cases where $\sigma>0$. Nonetheless we know explicitly a spectral
decomposition of the operator with Weyl symbol $H(x)H(\xi)$ from Theorem 5.19,
but we can calculate without difficulty numerical expressions of $M_{0},m_{0}$
as defined in (5.4.5), (5.4.6).

######  Proposition 5.30.

We have from (9.5.37), (5.2.33),

$a_{11}(\tau,0)=\frac{1}{1+e^{-4\pi^{2}\tau}},\quad\operatorname{Re}a_{12}(\tau,0)=\frac{1}{4\pi}\int_{0}^{+{\infty}}\sin(2\pi
t\tau)\operatorname{sech}(t/2)dt,$

and we can use these formulas and (5.4.3), (5.4.4), (5.4.5), (5.4.6) to
calculate numerically

(5.4.8) $\displaystyle{\color[rgb]{1,0,0}M_{0}}$
$\displaystyle{\color[rgb]{1,0,0}\approx\mathbf{1.00767997007003}},\quad(\text{$\lambda_{+}(\tau,0)$
at $\tau\approx 0.138815397930141$}),$ (5.4.9) $\displaystyle m_{0}$
$\displaystyle\approx-0.155939843191243,\quad(\text{$\lambda_{-}(\tau,0)$ at
$\tau\approx-0.0566304954736227).$}$

Figure 6.  The function $\tau\mapsto\lambda_{+}(\tau,0)$ near its maximum,
well above 1.

Figure 7.  The functions $\tau\mapsto\lambda_{+}(\tau,0),\lambda_{-}(\tau,0)$.

####  5.4.2. On hyperbolic regions

We want now to tackle the case $\sigma>0$. In order to use the expressions
(9.5.37), (5.2.34) respectively for $a_{11}$ and $a_{12}$, we need first to
evaluate the residue term in (9.5.37). The mapping $z\mapsto\tanh z$ is a
biholomorphism of neighborhoods of $0$ in the complex plane, so that we have
for $z$ near the origin,

(5.4.10) $\displaystyle\zeta=\tanh z,\ d\zeta=(1-\zeta^{2})dz,\quad
z=\operatorname{arcth}\zeta=\frac{1}{2}\ln\left(\frac{1+\zeta}{1-\zeta}\right),$
$\displaystyle\frac{e^{2i\omega z-2i\kappa\operatorname{coth}z}}{\cosh
z}dz=\left(\frac{1+\zeta}{1-\zeta}\right)^{i\omega}e^{-2i\frac{\kappa}{\zeta}}\
\frac{2}{\left(\frac{1+\zeta}{1-\zeta}\right)^{1/2}+\left(\frac{1-\zeta}{1+\zeta}\right)^{1/2}}\frac{d\zeta}{(1-\zeta^{2})}$
(5.4.11) $\displaystyle\hskip
88.0pt=(1+\zeta)^{-\frac{1}{2}+i\omega}(1-\zeta)^{-\frac{1}{2}-i\omega}e^{-2i\frac{\kappa}{\zeta}}d\zeta,$

so that

(5.4.12) $\text{\tt Res}\left(\frac{e^{2i\omega
z-2i\kappa\operatorname{coth}z}}{\cosh z},0\right)=\text{\tt
Res}\left((1+\zeta)^{-\frac{1}{2}+i\omega}(1-\zeta)^{-\frac{1}{2}-i\omega}e^{-2i\frac{\kappa}{\zeta}},0\right).$

######  Proposition 5.31.

Let $\sigma\geq 0$ be given. Then for any $\tau\in\mathbb{R}$, with the
notations $\omega=2\pi\tau$, $\kappa=2\pi\sigma$, we have, for any
$\rho\in(0,1)$,

(5.4.13) $a_{11}(\tau,\sigma)=\frac{1}{1+e^{-2\pi\omega}}\\\
+\frac{e^{-\pi\omega}}{1+e^{-2\pi\omega}}\frac{\rho}{2\pi}\operatorname{Im}\left\\{\int_{-\pi}^{\pi}\exp\left(i\omega\operatorname{Log}\Bigl{(}\frac{1+\rho
e^{i\theta}}{1-\rho e^{i\theta}}\Bigr{)}\right)\frac{e^{-\frac{2i\kappa
e^{-i\theta}}{\rho}}e^{i\theta}}{\sqrt{1-\rho^{2}e^{2i\theta}}}\
d\theta\right\\}.$ (5.4.14)
$\operatorname{Re}a_{21}(\tau,\sigma)=\frac{e^{-\pi\omega}}{2\pi}\biggl{\\{}2\int_{0}^{\pi/2}\frac{e^{(s\omega-\kappa\tan
s)}\sinh(s\omega-\kappa\tan s)}{\sin s}ds\\\
+\ln\bigl{(}\operatorname{coth}\frac{\pi}{4}\bigr{)}+2\int_{0}^{\pi/2}\frac{\sin^{2}(s\omega-\kappa\tanh
s)}{\sinh s}ds-\int_{\pi/2}^{+{\infty}}\frac{\cos 2(s\omega-\kappa\tanh
s)}{\sinh s}ds\biggr{\\}},$ (5.4.15)
$\displaystyle\operatorname{Im}a_{12}(\tau,\sigma)$
$\displaystyle=\frac{e^{-\pi\omega}}{2}a_{11}(\tau,\sigma).\hskip 33.0pt$

###### Proof.

Formula (5.4.13) follows from (5.4.12) and (9.5.37) whereas (5.4.14) is
(5.2.34) after a change of variable $t=2s$, where the second integral term
inside the brackets is evaluated (cf. Lemma 9.14); Formula (5.4.15) is a
reminder of (5.2.53). ∎

######  N.B.

Our choice for $\rho$ in the numerical calculations of (5.4.13) is $\rho=3/4$,
which is a good compromise between using a value of $\rho$ clearly away from
$1$ (to avoid singularities coming from small denominators in the
$\operatorname{Log}$ term) and minimize the oscillations and size coming from
the term $\exp(-2i\kappa\rho^{-1}e^{-i\theta})$; note that the modulus of the
latter is

$\exp(-2\kappa\rho^{-1}\sin\theta),$

which is a smooth function of $\rho$ (flat at 0) when $\theta\in[0,\pi]$, but
is unbounded for $\rho\rightarrow 0_{+}$ when $\theta\in(-\pi,0)$. There is no
surprise here since although the residue does not depend on the choice of
$\rho\in(0,1)$, we cannot get the value of that residue by letting $\rho$ go
to 0 because of the part of the path in the lower half-plane. The argument of
$\exp(-2i\kappa\rho^{-1}e^{-i\theta})$ is $-2\kappa\rho^{-1}\cos\theta$ and
taking $\rho$ too small would be devastating for the calculations because of
the strong oscillations triggered by the term
$\exp(-2i\kappa\rho^{-1}\cos\theta)$ all over the circle. Of course for the
evaluation of $\operatorname{Log}\Bigl{(}\frac{1+\rho e^{i\theta}}{1-\rho
e^{i\theta}}\Bigr{)}$ is easier for $\rho$ small, but we have to take into
account the constraints in that direction mentioned above.

######  Remark 5.32.

It seems easier numerically for the evaluation of $a_{11}$ to use (5.4.13)
rather than any other expression (see e.g. Lemma 5.18, (5.2.31), (9.5.23)).
However the following formula could be interesting, theoretically and
numerically: recalling that $\operatorname{sinc}x=\frac{\sin x}{x}$, we have
from (5.2.31)

(5.4.16)
$a_{11}(\tau,\sigma)=\frac{1}{2}+\frac{2\omega}{\pi}\int_{0}^{+{\infty}}\operatorname{sinc}(2\omega
s)\frac{s}{\sinh s}\cos(2\kappa\tanh s)ds\\\
-\frac{2\kappa}{\pi}\int_{0}^{+{\infty}}\operatorname{sinc}(2\kappa
s)\frac{1}{\cosh s}\cos(2\omega s)ds,$

but it turns out that numerical calculations involving (5.4.16) seem to be
less reliable than the methods using (5.4.13).

We can also take a look at the following curves.

Figure 8.  Functions $\lambda_{+}(\tau,\kappa/2\pi)$ with $\kappa=1,2,3$:
their maxima are strictly greater than 1.

######  Remark 5.33.

In the above figure, in order to put the three curves on the same picture, we
have used three different logarithmic scales on the vertical axis, namely we
have drawn

$\tau\mapsto
1+\alpha_{j}\operatorname{Log}\bigl{(}\lambda_{+}(\tau,\sigma_{j})\bigr{)},\quad
1\leq j\leq 3,\sigma_{j}=j/2\pi,\alpha_{1}=20,\alpha_{2}=100,\alpha_{3}=500.$

Of course we have

$1+\alpha_{j}\operatorname{Log}\bigl{(}\lambda_{+}(\tau,\sigma_{j})\bigr{)}>1\Longleftrightarrow\operatorname{Log}\bigl{(}\lambda_{+}(\tau,\sigma_{j})\bigr{)}>0\Longleftrightarrow\lambda_{+}(\tau,\sigma_{j})>1,$

so that the piece of curves in Figure 8 which are above 1 are indeed
corresponding to curves of $\tau\mapsto\lambda_{+}(\tau,\sigma_{j})$ which go
strictly above the threshold 1. We have also

$\displaystyle\max_{\tau}\lambda_{+}(\tau,\sigma_{1})$ $\displaystyle\approx
1+55{\scriptstyle\times}10^{-5}{\hskip 15.0pt}\text{ at $\tau\approx
0.402030$, }$ $\displaystyle\max_{\tau}\lambda_{+}(\tau,\sigma_{2})$
$\displaystyle\approx 1+8{\scriptstyle\times}10^{-5}{\hskip 15.0pt}\text{ at
$\tau\approx 0.613262$,}$
$\displaystyle\max_{\tau}\lambda_{+}(\tau,\sigma_{3})$ $\displaystyle\approx
1+10^{-5}{\hskip 15.0pt}\text{ at $\tau\approx 0.854746$.}$

We are glad to have a theoretical proof of Theorem 5.20 since the numerical
analysis of cases where $\sigma$ is large, say larger than 10, seem to be very
difficult to achieve, at least through a standard use of Mathematica. The
reason for that is quite clear since using our Lemma 5.11, we did study the
function $\beta$ defined by

(5.4.17) $\beta(\tau,\sigma)=|a_{12}(\tau,\sigma)|^{2}+a_{11}(\tau,\sigma)-1,$

and proved that for each $\sigma\geq 0$ there exists $T_{0}(\sigma)$ such that
for all $\tau\geq T_{0}(\sigma)$ we have $\beta(\tau,\sigma)>0$ and
$a_{12}(\tau,\sigma)\not=0$. Thanks to Lemma 5.18 and (5.2.60) we knew that
for $\tau\geq T_{0}(\sigma)$, we had

$|1-a_{11}|\leq
2e^{-{\pi^{2}\tau}}e^{4\pi\sigma}\ll\frac{e^{-16\pi\sqrt{\tau}\sqrt{\sigma}}}{2^{8}\pi^{6}\tau^{2}}\leq(\operatorname{Re}a_{21})^{2}\leq|a_{12}|^{2},$

where the second inequality $\ll$ is in fact comparing for $\sigma$ fixed two
exponential decays. The numerical analysis of that inequality is certainly
quite difficult when $\sigma$ and $\tau$ are large since both sides are
converging to zero quite fast for $\sigma$ fixed and
$\tau\rightarrow+{\infty}$; of course taking the logarithm of both sides looks
quite reasonable, but in practice does not seem really easy numerically. When
$\sigma=0$, the situation is much better, since we had to compare (cf.
Subsection 5.3.1) an exponential decay $|1-a_{11}|\leq 2e^{-{\pi^{2}\tau}}$ to
a polynomial decay

$|\operatorname{Re}a_{12}|^{2}\sim\frac{1}{2^{6}\pi^{4}\tau^{2}},\quad\tau\rightarrow+{\infty},$

and this could be an a posteriori explanation for which our numerical argument
in [4] worked smoothly to disprove Flandrin’s conjecture. So to pick-up the
quarter-plane ((5.0.1) with $\sigma=0$) to produce a counterexample to that
conjecture was indeed a very wise choice: if you choose instead
$\mathcal{C}_{\sigma}$ for $\sigma$ large, our Theorem 5.20 shows that it is
also a counterexample to Flandrin’s conjecture151515As a convex subset of the
plane on which the integral of the Wigner distribution of some normalized
pulse is $>1$. , but we have a theoretical proof for that Theorem and if we
were depending on a numerical analysis, it is quite likely that checking
numerically the positivity of the function $\beta$ defined in (5.4.17) could
be rather difficult, even say for $\sigma=10$.

##  6\. Unboundedness is Baire generic

In this section we show that for plenty of subsets $E$ of the phase space
$\mathbb{R}^{2n}$, the operator ${\text{\rm Op}_{\text{\rm
w}}(\mathbf{1}_{E})}$ is not bounded on $L^{2}(\mathbb{R}^{n})$.

Acknowledgements. The author is grateful to H.G. Feichtinger and K. Gröchenig
for sharp comments on a first version of this section.

###  6.1. Preliminaries

####  6.1.1. Prolegomena

######  Lemma 6.1.

Let $u,v\in L^{2}(\mathbb{R}^{n})$ and let $\mathcal{W}(u,u),$
$\mathcal{W}(v,v),$ be their Wigner distributions. Then we have

$\|\mathcal{W}(u,u)-\mathcal{W}(v,v)\|_{L^{2}(\mathbb{R}^{2n})}\leq\|u-v\|_{L^{2}(\mathbb{R}^{n})}\bigl{(}\|u\|_{L^{2}(\mathbb{R}^{n})}+\|v\|_{L^{2}(\mathbb{R}^{n})}\bigr{)}.$

As a consequence if a sequence $(u_{k})$ is converging in
$L^{2}(\mathbb{R}^{n})$, then the sequence $(\mathcal{W}(u_{k},u_{k}))$
converges in $L^{2}(\mathbb{R}^{2n})$ towards $\mathcal{W}(u,u)$.

###### Proof.

We have by sesquilinearity

$\mathcal{W}(u,u)-\mathcal{W}(v,v)=\mathcal{W}(u-v,u)+\mathcal{W}(v,u-v),$

so that

$\|\mathcal{W}(u,u)-\mathcal{W}(v,v)\|_{L^{2}(\mathbb{R}^{2n})}\leq\|\mathcal{W}(u-v,u)\|_{L^{2}(\mathbb{R}^{2n})}+\|\mathcal{W}(v,u-v)\|_{L^{2}(\mathbb{R}^{2n})}\\\
\underbracket{\hskip 3.0pt=\hskip
3.0pt}_{\eqref{norm}}\|u-v\|_{L^{2}(\mathbb{R}^{n})}\bigl{(}\|u\|_{L^{2}(\mathbb{R}^{n})}+\|v\|_{L^{2}(\mathbb{R}^{n})}\bigr{)},$

proving the lemma. ∎

######  Lemma 6.2.

Let $(u_{k})$ be a converging sequence in $L^{2}(\mathbb{R}^{n})$ with limit
$u$. Let us assume that there exists $C_{0}\geq 0$ such that

(6.1.1) $\forall
k\in\mathbb{N},\quad\iint|\mathcal{W}(u_{k},u_{k})(x,\xi)|dxd\xi\leq C_{0}.$

Then we have $\iint|\mathcal{W}(u,u)(x,\xi)|dxd\xi\leq C_{0}.$

###### Proof.

Let $R>0$ be given. We check

$\iint_{|x|^{2}+|\xi|^{2}\leq
R^{2}}|\mathcal{W}(u,u)(x,\xi)-\mathcal{W}(u_{k},u_{k})(x,\xi)|dxd\xi\\\
\leq\iint_{|x|^{2}+|\xi|^{2}\leq
R^{2}}|\mathcal{W}(u-u_{k},u)(x,\xi)|dxd\xi+\iint_{|x|^{2}+|\xi|^{2}\leq
R^{2}}|\mathcal{W}(u_{k},u-u_{k})(x,\xi)|dxd\xi\\\
\leq\sqrt{|\mathbb{B}^{2n}|R^{2n}}\bigl{(}\|\mathcal{W}(u-u_{k},u)\|_{L^{2}(\mathbb{R}^{2n})}+\|\mathcal{W}(u_{k},u-u_{k})\|_{L^{2}(\mathbb{R}^{2n})}\bigr{)}\\\
=\sqrt{|\mathbb{B}^{2n}|R^{2n}}\|u-u_{k}\|_{L^{2}(\mathbb{R}^{n})}\bigl{(}\|u\|_{L^{2}(\mathbb{R}^{n})}+\|u_{k}\|_{L^{2}(\mathbb{R}^{n})}\bigr{)},$

and thus

$\displaystyle\iint_{|x|^{2}+|\xi|^{2}\leq
R^{2}}|\mathcal{W}(u,u)(x,\xi)|dxd\xi\leq\iint_{|x|^{2}+|\xi|^{2}\leq
R^{2}}|\mathcal{W}(u_{k},u_{k})(x,\xi)|dxd\xi$ $\displaystyle\hskip
155.0pt+\sqrt{|\mathbb{B}^{2n}|R^{2n}}\|u-u_{k}\|_{L^{2}(\mathbb{R}^{n})}\bigl{(}\|u\|_{L^{2}(\mathbb{R}^{n})}+\|u_{k}\|_{L^{2}(\mathbb{R}^{n})}\bigr{)}$
$\displaystyle\hskip 75.0pt\leq
C_{0}+\sqrt{|\mathbb{B}^{2n}|R^{2n}}\|u-u_{k}\|_{L^{2}(\mathbb{R}^{n})}\bigl{(}\|u\|_{L^{2}(\mathbb{R}^{n})}+\|u_{k}\|_{L^{2}(\mathbb{R}^{n})}\bigr{)},$

implying for all $R>0$,

$\iint_{|x|^{2}+|\xi|^{2}\leq R^{2}}|\mathcal{W}(u,u)(x,\xi)|dxd\xi\leq
C_{0},$

and thus the sought result. ∎

####  6.1.2. An explicit construction

We just calculate $\mathcal{W}(v_{0},v_{0})$ for

(6.1.2) $v_{0}=\mathbf{1}_{[-1/2,1/2]}.$

######  Remark 6.3.

When $u$ is supported in a closed convex set $J$, we have in the integral
(1.1.6) defining $\mathcal{W}$, $x\pm\frac{z}{2}\in J\Longrightarrow x\in J,$
so that $\operatorname{supp}\mathcal{W}(u,u)\subset J\times\mathbb{R}^{n}$.

We have

$\displaystyle\mathcal{W}(v_{0},v_{0})(x,\xi)=\int_{\begin{subarray}{c}-1/2\leq
x+z/2\leq 1/2\\\ -1/2\leq x-z/2\leq 1/2\end{subarray}}e^{2i\pi z\xi}dz,$

and the integration domain is

$-\min(1-2x,1+2x)=\max(-1-2x,2x-1)\leq z\leq\min(1-2x,1+2x),$

which is empty unless $1-2x,1+2x\geq 0$ i.e. $x\in[-1/2,+1/2]$, and moreover
we have the equivalence

$1-2x\leq 1+2x\Longleftrightarrow x\geq 0,$

so that

(6.1.3) $\mathcal{W}(v_{0},v_{0})(x,\xi)=H(x)\int_{-(1-2x)}^{1-2x}e^{2i\pi
z\xi}dz+H(-x)\int_{-(1+2x)}^{1+2x}e^{2i\pi z\xi}dz\\\
=H(x)\frac{e^{2i\pi\xi(1-2x)}-e^{-2i\pi\xi(1-2x)}}{2i\pi\xi}+H(-x)\frac{e^{2i\pi\xi(1+2x)}-e^{-2i\pi\xi(1+2x)}}{2i\pi\xi}\\\
=\mathbf{1}_{[0,1/2]}(x)\frac{\sin(2\pi\xi(1-2x))}{\pi\xi}+\mathbf{1}_{[-1/2,0]}\frac{\sin(2\pi\xi(1+2x))}{\pi\xi}.$

More generally for $a,b,\omega$ real numbers with $a<b$ and

(6.1.4) $u_{a,b,\omega}(x)=(b-a)^{-1/2}\mathbf{1}_{[a,b]}(x)e^{2i\pi\omega
x},$

we have

(6.1.5) $\mathcal{W}(u_{a,b,\omega},u_{a,b,\omega})(x,\xi)\\\