$p\to 1+$. The limit function coincides with $B_{2}$ described in Subsection
4.4.2.

First, we note that the domain
$\Omega_{\varepsilon}^{3}=\Omega_{\varepsilon;p}^{3}$ for $p=1$ is the limit
of the corresponding domains in a natural sense: for any point
$(x_{1},x_{2})\in\Omega^{2}_{\varepsilon}$ the values
$\boldsymbol{B}_{1;\varepsilon}^{\pm}(x_{1},x_{2})$ that define the upper and
the lower boundaries of $\Omega_{\varepsilon;1}^{3}$ are the limits of
$\boldsymbol{B}_{p;\varepsilon}^{\pm}(x_{1},x_{2})$, when $p\to 1+$, see
Subsection 2.1. For $p=1$ we have $m_{1}(u)=1$ for $u\geqslant 0$, therefore,
the function $A_{m_{1}}(x_{1},x_{2})$ has a very simple description:

$A_{m_{1}}(x_{1},x_{2})=\begin{cases}|x_{1}|,&(x_{1},x_{2})\in\Omega_{\varepsilon}^{2}\setminus\big{(}\omega_{0}\cup\omega_{\pm
1}\big{)};\\\
\frac{x_{2}}{2\varepsilon},&(x_{1},x_{2})\in\omega_{0}\cup\omega_{\pm
1}.\end{cases}$ (11.1)

Second, the domains $\Xi_{{}_{\mathrm{L}+}}$, $\Xi_{{}_{\mathrm{ch}+}}$, and
$F(0)$ do not exist for $p=1$ (see (3.12), (3.14), and (3.5)). This happens
because all the chords of the form $[U(a),U(b)]$, $0\leqslant a\leqslant
b\leqslant a+2\varepsilon$ lie on the lower boundary of
$\Omega_{\varepsilon;1}^{3}$. We conclude that the whole domain
$\Omega_{\varepsilon;1}^{3}$ is the union of $\Xi_{{}_{\mathrm{R}+}}$, its
symmetric domain $\Xi_{{}_{\mathrm{L}-}}$, and the domain $R$ foliated by the
two-dimensional leaves $R(v)$, $v\in[0,2\varepsilon]$. We state that the
function $B_{2}$ defined on these domains in Subsection 4.4.2 for $p=1$
coincides with $\boldsymbol{B}_{1,r;\varepsilon}^{+}$.

Note that $w_{2}(\,\cdot\,,p,\varepsilon,x_{1},x_{2})$ defined in (4.90)
converges to $w_{2}(\,\cdot\,,1,\varepsilon,x_{1},x_{2})$ uniformly in
$\xi\in[0,u]$ as $p\to 1+$ when $\varepsilon$ and $(x_{1},x_{2})$ are fixed.
The limit function $w_{2}(\,\cdot\,,1,\varepsilon,x_{1},x_{2})$ is continuous
and strictly increasing. Therefore, for every
$x_{3}\in\big{(}w_{2}(0,1,\varepsilon,x_{1},x_{2}),w_{2}(u,1,\varepsilon,x_{1},x_{2}))$
we have the convergence

$w_{2}\big{(}w_{2}^{-1}(x_{3};p,\varepsilon,x_{1},x_{2});r;\varepsilon,x_{1},x_{2}\big{)}\to
w_{2}\big{(}w_{2}^{-1}(x_{3};1,\varepsilon,x_{1},x_{2});r;\varepsilon,x_{1},x_{2}\big{)},\qquad
p\to 1+.$

Thus, in the interior of $\Xi_{{}_{\mathrm{R}+}}$ we have the pointwise
convergence of the functions $B_{2}$ to the limit function when $p\to 1+$.
Convergence on $R$ is obvious. As a result, we get convergence on the interior
of $\Omega_{\varepsilon;1}^{3}$. All these functions $B_{2}$ are locally
concave for $p>1$, therefore, the limit function is also locally concave on
the interior of $\Omega_{\varepsilon;1}^{3}$. Since it is also continuous on
$\Omega_{\varepsilon;1}^{3}$, it is locally concave on the entire
$\Omega_{\varepsilon;1}^{3}$, therefore, due to item 3) of Proposition 2.3

$B_{2}(x)\geqslant\boldsymbol{B}_{1,r;\varepsilon}^{+}(x),\qquad
x\in\Omega_{\varepsilon;1}^{3}.$

For each point $x\in\Xi_{{}_{\mathrm{R}+}}$ one can construct an optimizer in
the same way as it was done in (7.2). The only difference is that for $p=1$
the variable $\xi$ runs from $0$ to $\min(u,\varepsilon)$, and by this reason
we only use optimizers given by (7.4). Therefore,

$B_{2}(x)\leqslant\boldsymbol{B}_{1,r;\varepsilon}^{+}(x),\qquad
x\in\Xi_{{}_{\mathrm{R}+}}.$

In the domain $R$ we use the same reasoning as before: the function $B_{2}$ is
linear on $R(v)$ and coincides with the locally concave function
$\boldsymbol{B}^{+}_{1,r;\varepsilon}$ at the vertices of $R(v)$, therefore,
$B_{2}\leqslant\boldsymbol{B}_{1,r;\varepsilon}^{+}$ on $R(v)$.

Thus, we have proved that $B_{2}=\boldsymbol{B}_{1,r;\varepsilon}^{+}$
everywhere on $\Omega_{\varepsilon;1}^{3}$. As a consequence, we have that the
statements of Theorems 1.1, 1.3, 1.4, 1.5, and 1.7 and Corollaries 1.6, 1.8
hold for $p=1$.

Finally, we compute the sharp constant in the inequality (1.1) for $p=1$.
Repeating the argument from Section 9, we arrive at

$C^{r}(1,r)=\sup_{x_{2},x_{3}}\frac{\boldsymbol{B}_{1,r;1}^{+}(0,x_{2},x_{3})}{x_{3}}\,.$
(11.2)

Since the domain $F(0)$ does not exist for the case $p=1$, it suffices to find
the maximum of the function $\frac{\boldsymbol{B}^{+}_{1,r;1}}{x_{3}}$ in the
domain $R$. Recall that in this domain our function is defined by a simple
formula (4.98), whence

$\frac{\boldsymbol{B}_{1,r;1}^{+}(0,x_{2},x_{3})}{x_{3}}=\Big{(}\frac{x_{2}}{x_{3}}\Big{)}^{r-1}\\!\\!.$
(11.3)

From (3.8) we have $x_{3}\geqslant\tfrac{1}{2}x_{2}$ and the maximum value is
attained on the lower boundary, i. e. , $x_{3}=\tfrac{1}{2}x_{2}$, which
implies

$C(1,r)=2^{1-\frac{1}{r}}.$

Thus, we have proved the formula stated in Remark 1.2 for the case
$1<r\leqslant 2$. For $r\geqslant 2$ it was proved in [6], see (1.2).

##  12 Comments on multidimensional case

The proof of Theorem 1.7 follows the reasoning from [7]. Following the
arguments of Lemma 5.2 there it is possible to prove the following assertion.
If $G$ is a non-negative locally concave function on
$\Omega_{\varepsilon}^{3}$ and $\phi\in\mathrm{BMO}(\mathbb{R}^{d})$,
$\|\phi\|_{\scriptscriptstyle K}<\varepsilon$, then

$G\Big{(}\phi_{\scriptscriptstyle
K}(y,t),\big{(}\phi^{2}\big{)}_{\scriptscriptstyle
K}(y,t),\big{(}|\phi|^{p}\big{)}_{\scriptscriptstyle
K}(y,t)\Big{)}\geqslant\big{(}G(\phi,\phi^{2},|\phi|^{p})\big{)}_{\scriptscriptstyle
K}(y,t),\qquad y\in\mathbb{R}^{d},\quad t>0.$

Fix a point $(y,t)\in\mathbb{R}^{d}\times\mathbb{R}_{+}$ and use this
inequality for the function $G=\boldsymbol{B}_{p,r;\varepsilon}^{+}$ and for
$\phi=\varphi-\varphi_{\scriptscriptstyle K}(y,t)$. Then we obtain

$\boldsymbol{B}_{p,r;\varepsilon}^{+}\Big{(}0,\big{(}(\varphi-\varphi_{\scriptscriptstyle
K}(y,t))^{2}\big{)}_{\scriptscriptstyle
K}(y,t),\big{(}|\varphi-\varphi_{\scriptscriptstyle
K}(y,t)|^{p}\big{)}_{\scriptscriptstyle
K}(y,t)\Big{)}\geqslant\big{(}|\varphi-\varphi_{\scriptscriptstyle
K}(y,t)|^{r}\big{)}_{\scriptscriptstyle K}(y,t).$ (12.1)

Now, we use the definitions of the Bellman function
$\boldsymbol{B}_{p,r;\varepsilon}^{+}$ and the constant $C(p,r)$ in the
corresponding inequality. We estimate the left-hand side of (12.1):

$\boldsymbol{B}_{p,r;\varepsilon}^{+}\Big{(}0,\big{(}(\varphi-\varphi_{\scriptscriptstyle
K}(y,t))^{2}\big{)}_{\scriptscriptstyle
K}(y,t),\big{(}|\varphi-\varphi_{\scriptscriptstyle
K}(y,t)|^{p}\big{)}_{\scriptscriptstyle K}(y,t)\Big{)}\leqslant
C(p,r)^{r}\varepsilon^{r-p}\big{(}|\varphi-\varphi_{\scriptscriptstyle
K}(y,t)|^{p}\big{)}_{\scriptscriptstyle K}(y,t).$

Therefore, we obtain

$\varepsilon^{r-p}C(p,r)^{r}\big{(}|\varphi-\varphi_{\scriptscriptstyle
K}(y,t)|^{p}\big{)}_{\scriptscriptstyle
K}(y,t)\geqslant\big{(}|\varphi-\varphi_{\scriptscriptstyle
K}(y,t)|^{r}\big{)}_{\scriptscriptstyle K}(y,t).$

Rewrite the last inequality in the form

$\varepsilon^{r-p}C(p,r)^{r}\int\limits_{\mathbb{R}^{d}}|\varphi(\tilde{y})-\varphi_{\scriptscriptstyle
K}(y,t)|^{p}K_{t}(y-\tilde{y})d\tilde{y}\geqslant\int\limits_{\mathbb{R}^{d}}|\varphi(\tilde{y})-\varphi_{\scriptscriptstyle
K}(y,t)|^{r}K_{t}(y-\tilde{y})d\tilde{y}.$ (12.2)

At this moment we recall that $K_{t}$ coincides either with $P$ or $H$. For
each of these kernels there exists the function

$Q(t)=\begin{cases}\frac{\pi^{\frac{d+1}{2}}}{\Gamma(\frac{d+1}{2})}t^{d},&K=P,\\\
(4\pi t)^{\frac{d}{2}},&K=H,\end{cases}$

such that $Q(t)K_{t}(y)\to 1$ for any $y\in\mathbb{R}^{d}$ when $t\to+\infty$.
Moreover, this convergence is monotone and uniform on compact subsets of
$\mathbb{R}^{d}$. We also note that since the function $\varphi$ lies in
$L^{p}(\mathbb{R}^{d})$, we have $\varphi_{\scriptscriptstyle K}(y,t)\to 0$
for any $y\in\mathbb{R}^{d}$ when $t\to+\infty$. Multiplying both sides of
(12.2) by $Q(t)$ and passing $t\to+\infty$, we finally get

$\varepsilon^{r-p}C(p,r)^{r}\int_{\mathbb{R}^{d}}|\varphi(\tilde{y})|^{p}d\tilde{y}\geqslant\int_{\mathbb{R}^{d}}|\varphi(\tilde{y})|^{r}d\tilde{y}.$
(12.3)

Theorem 1.7 is proved.

##  Appendix A Smoothness of the Bellman function $B_{2}$

In this appendix we prove $C^{1}$-smoothness of $B_{2}$. Since we will deal
only with this candidate we omit index $2$. Then it will be convenient to
denote the partial derivatives of $B$ placing the corresponding variable in
index: $B_{x_{i}}=\frac{\partial B}{\partial x_{i}}$.

######  Lemma A.1.

The function $B$ is $C^{1}$-smooth.

##### Proof.

By the definition, it is clear that $B$ is $C^{2}$-smooth on each subdomain.
Therefore, we need to check that this function is $C^{1}$-smooth on the
borders between subdomains. Since all borders consists of extremals, $B$ is
linear in corresponding direction. Therefore, it suffices to verify continuity
of the derivatives in any two transversal directions. We will check continuity
of $B_{x_{3}}$ and $B_{x_{2}}$. The only exception is the boundary between
$\Xi_{{}_{\mathrm{L}+}}$ and $F(0)$ that goes along $x_{3}$ (it belongs to the
plane $x_{2}=2\varepsilon x_{1}$). In this case we check continuity of
$B_{x_{1}}$ instead of $B_{x_{3}}$.

Let us start with the domain $\Xi_{{}_{\mathrm{L}+}}$. On the boundary
$x_{2}=2\varepsilon x_{1}$ (or more exactly $x_{1}=\Delta_{-}$) the function
$B$ is continuous by definition: we have defined $B$ on $F(0)$ using this
continuity.

Boundary $\mathbf{\Xi_{{}_{\mathrm{L}+}}\Big{|}F(0)}$. To check
$C^{1}$-smoothness, we calculate the derivative $B_{x_{1}}$ and show that it
is zero on the boundary $x_{1}=\Delta_{-}$. It is what we need because the
function $B$ does not depend on $x_{1}$ on the domain $F(0)$. We use the
representation of $B$ in it initial form (4.7):

$B_{x_{1}}=rv^{r-1}\frac{\partial v}{\partial x_{1}}+\Big{(}\frac{\partial
K_{{}_{\mathrm{L}}}}{\partial u}\cdot\frac{\partial u}{\partial
x_{1}}+\frac{\partial K_{{}_{\mathrm{L}}}}{\partial h}\cdot\frac{\partial
h}{\partial x_{1}}\Big{)}(x_{1}-v)+K_{{}_{\mathrm{L}}}\Big{(}1-\frac{\partial
v}{\partial x_{1}}\Big{)},$ (A.1)

where $v=u-\varepsilon=x_{1}-\Delta_{-}$, therefore,

$\frac{\partial v}{\partial x_{1}}=\frac{\partial u}{\partial
x_{1}}=\frac{d+x_{1}}{d}\,,$

and $v=0$ on the boundary. To find the derivative $\frac{\partial h}{\partial
x_{1}}$ we use formula (4.10) and the relation $\varepsilon
m_{p}^{\prime}(v)=m_{p}(v)-pv^{p-1}$ (see (2.15)). As a result, we get

$\frac{\partial h}{\partial x_{1}}\Big{|}_{x_{1}=\Delta_{-}}\\!\\!\\!=\
\frac{h}{d}\,.$

From (4.15) we see that

$h\frac{\partial K_{{}_{\mathrm{L}}}}{\partial
h}\big{|}_{x_{1}=\Delta_{-}}=K_{{}_{\mathrm{L}}}-\varepsilon\frac{\partial
K_{{}_{\mathrm{L}}}}{\partial u}.$

Gathering all these formulas, we obtain

$B_{x_{1}}\big{|}_{x_{1}=\Delta_{-}}=\frac{\partial
K_{{}_{\mathrm{L}}}}{\partial u}\cdot\frac{d+x_{1}}{d}\cdot
x_{1}+\Big{(}K_{{}_{\mathrm{L}}}-\varepsilon\frac{\partial
K_{{}_{\mathrm{L}}}}{\partial
u}\Big{)}\cdot\frac{x_{1}}{d}-\frac{x_{1}}{d}K_{{}_{\mathrm{L}}}=\frac{\partial
K_{{}_{\mathrm{L}}}}{\partial
u}\cdot\frac{x_{1}}{d}\cdot(x_{1}+d-\varepsilon)=0\,,$ (A.2)

since $\varepsilon-d=\Delta_{-}=x_{1}$ on this boundary.

To prove continuity of $B_{x_{2}}$, note that from (4.92) for the points in
the intersection $F(0)\cap\Xi_{{}_{\mathrm{L}+}}=F_{L}(\varepsilon)$ we have

$B_{x_{2}}\Big{|}_{{}_{F(0)}}\\!\\!\\!=\
\frac{1}{2\varepsilon}B_{x_{1}}\Big{|}_{{}_{\Xi_{{}_{\mathrm{L}+}}}}\\!\\!\\!+\
B_{x_{2}}\Big{|}_{{}_{\Xi_{{}_{\mathrm{L}+}}}},$

and the first summand at the right-hand vanishes due to (A.2).

Boundary $\mathbf{F(0)|R}$. Now, we check $C^{1}$-smoothness of $B$ on the
boundary $x_{3}=(2\varepsilon)^{p-2}x_{2}$ between $F(0)$ and $R$ (see (5.6)).
According to (4.98) the value of $B$ from $R$ is $(2\varepsilon)^{r-2}x_{2}$.
The value from $F(0)$ looks much more difficult (see (4.93)):

$\Big{[}e\Psi_{{}_{\mathrm{L}}}\Big{(}\frac{(2\varepsilon)^{p}}{2e}-\frac{\varepsilon^{p}}{e}\Gamma(p+1)\Big{)}+\varepsilon^{r-1}\Gamma(r+1)\Big{]}\frac{x_{2}}{2\varepsilon}\,.$

However, the fact that these two expressions are equal is known: this is just
formula (4.19) for $u=\varepsilon$.

The calculations will be simpler if we use representation (4.94)–(4.95) for
the function on $F(0)$. Then the boundary $x_{3}=(2\varepsilon)^{p-2}x_{2}$
corresponds to $\xi=\varepsilon$ and we have

$w_{{}_{\mathrm{L}}}(\varepsilon;s,\varepsilon)=\frac{(2\varepsilon)^{s}}{2e}-\frac{\varepsilon}{e}m_{s}(0)\qquad\text{and}\qquad
B(x)=(2\varepsilon)^{r-2}x_{2}\,.$ (A.3)

Since this boundary is a piece of two-dimensional plane where $B$ is linear,
for continuity of the gradient it is sufficient to check continuity of
$B_{x_{3}}$ only. On $R$ we have (see (4.98)):

$B_{x_{3}}=\frac{r-2}{p-2}\Big{(}\frac{x_{3}}{x_{2}}\Big{)}^{\frac{r-p}{p-2}}\qquad\text{and}\qquad
B_{x_{3}}\big{|}_{x_{3}=(2\varepsilon)^{p-2}x_{2}}=\frac{r-2}{p-2}(2\varepsilon)^{r-p}\,.$

On $F(0)$ we use the same representation (4.94)–(4.95) and get

$B_{x_{3}}=\frac{w_{{}_{\mathrm{L}}}^{\prime}(\xi;r,\varepsilon)}{w_{{}_{\mathrm{L}}}^{\prime}(\xi;p,\varepsilon)}\,.$
(A.4)

Formula (4.63) yields
$w_{{}_{\mathrm{L}}}^{\prime}(\varepsilon;s,\varepsilon)=\tfrac{1}{2e}(s-2)(2\varepsilon)^{s-1}$,
whence

$B_{x_{3}}\big{|}_{\xi=\varepsilon}=\frac{r-2}{p-2}(2\varepsilon)^{r-p}\,.$
(A.5)

We have considered all boundaries of $F(0)$ and now we consider boundaries of
$\Xi_{{}_{\mathrm{L}+}}$. It has two neighbours: already considered $F(0)$ and
$\Xi_{{}_{\mathrm{ch}+}}$, which we have to check.

Boundary $\mathbf{\Xi_{{}_{\mathrm{L}+}}\Big{|}\Xi_{{}_{\mathrm{ch}+}}}$. This
is the boundary

$x_{3}=\frac{\Delta_{-}(x_{1}+\Delta_{+})^{p}+\Delta_{+}(x_{1}-\Delta_{-})^{p}}{2\varepsilon}$

(see (5.8)–(5.10)). This boundary is foliated by the left halves of the chords
$[U(u-\varepsilon),U(u+\varepsilon)]$, where $B$ is linear with the prescribed
values at the ends. Therefore, it takes the same values not depending on what
side of the boundary we consider our function. However, we can verify this
formally plugging the values $\xi=u=v+\varepsilon=x_{1}+d$ into (4.82):

$B(x)=(u-\varepsilon)^{r}+\Big{[}m_{r}(u-\varepsilon)+\frac{1}{2\varepsilon}\big{(}(u+\varepsilon)^{r}-(u-\varepsilon)^{r}\big{)}-m_{r}(u-\varepsilon)\Big{]}\Delta_{-}=\frac{\Delta_{-}(x_{1}+\Delta_{+})^{r}+\Delta_{+}(x_{1}-\Delta_{-})^{r}}{2\varepsilon}\,.$

This expression coincides with (4.101), since
$a=u-\varepsilon=x_{1}-\Delta_{-}$ and $b=u+\varepsilon=x_{1}+\Delta_{+}$.

Now, we have to check continuity of $B_{x_{3}}$ and $B_{x_{2}}$ on this
boundary. In $\Xi_{{}_{\mathrm{L}+}}$ we have:

$B_{x_{3}}{\buildrel\eqref{050801}\over{=}}\varepsilon\Psi_{{}_{\mathrm{L}}}^{\prime}(\xi){\buildrel\eqref{Psi_L_def}\over{=}}\frac{w_{{}_{\mathrm{L}}}^{\prime}(\xi;r,\varepsilon)}{w_{{}_{\mathrm{L}}}^{\prime}(\xi;p,\varepsilon)}{\buildrel\eqref{280401}\over{=}}\frac{A(\xi,\varepsilon,r)}{A(\xi,\varepsilon,p)}\,.$
(A.6)

On $\Xi_{{}_{\mathrm{ch}+}}$ formula (5.3) yields

$\displaystyle 1{\buildrel\eqref{231002}\over{=}}w_{x_{3}}(a,b;p)$
$\displaystyle=\frac{a_{x_{3}}(b-x_{1})}{(b-a)^{2}}A(\tfrac{1}{2}(b+a),\tfrac{1}{2}(b-a),p),$
(A.7) $\displaystyle
B_{x_{3}}{\buildrel\eqref{090601}\over{=}}w_{x_{3}}(a,b;r)$
$\displaystyle=\frac{a_{x_{3}}(b-x_{1})}{(b-a)^{2}}A(\tfrac{1}{2}(b+a),\tfrac{1}{2}(b-a),r),$
(A.8)

whence

$B_{x_{3}}=\frac{A(\tfrac{1}{2}(b+a),\tfrac{1}{2}(b-a),r)}{A(\tfrac{1}{2}(b+a),\tfrac{1}{2}(b-a),p)}\,.$
(A.9)

Therefore, on the line $\xi=u$, $b=u+\varepsilon$, $a=u-\varepsilon$, the
expressions from (A.6) and (A.9) are equal to

$\frac{A(u,\varepsilon,r)}{A(u,\varepsilon,p)}\,.$

On this boundary, it remains to check continuity of $B_{x_{2}}$. In
$\Xi_{{}_{\mathrm{L}+}}$ we use formula (4.29) rewritten in terms of
$w_{{}_{\mathrm{L}}}$ with the help of (4.23) and (4.24):

$B_{x_{2}}(x)=\frac{1}{2\varepsilon}\Big{[}\frac{1}{\varepsilon}e^{\frac{u}{\varepsilon}}w_{{}_{\mathrm{L}}}(\xi;r,\varepsilon)-\frac{1}{\varepsilon}e^{\frac{u}{\varepsilon}}w_{{}_{\mathrm{L}}}(\xi;p,\varepsilon)\frac{w_{{}_{\mathrm{L}}}^{\prime}(\xi;r,\varepsilon)}{w_{{}_{\mathrm{L}}}^{\prime}(\xi;p,\varepsilon)}+\varepsilon
m_{r}^{\prime}(v)-\varepsilon
m_{p}^{\prime}(v)\frac{w_{{}_{\mathrm{L}}}^{\prime}(\xi;r,\varepsilon)}{w_{{}_{\mathrm{L}}}^{\prime}(\xi;p,\varepsilon)}\Big{]}\,.$

On the boundary $\xi=u=v+\varepsilon=x_{1}+d$ we use relations (4.20), (4.66),
(2.15) and rewrite this formula as follows:

$B_{x_{2}}(x)\big{|}_{\xi=u}=\frac{1}{2\varepsilon}\Big{[}\Big{(}\frac{(u+\varepsilon)^{r}-(u-\varepsilon)^{r}}{2\varepsilon}-rv^{r-1}\Big{)}-\Big{(}\frac{(u+\varepsilon)^{p}-(u-\varepsilon)^{p}}{2\varepsilon}-pv^{p-1}\Big{)}\frac{A(u,\varepsilon,r)}{A(u,\varepsilon,p)}\Big{]}\,.$
(A.10)

In $\Xi_{{}_{\mathrm{ch}+}}$ we differentiate formula (5.2) with respect to
$x_{2}$ and as in (5.3) we get

$w_{x_{2}}=\frac{a_{x_{2}}(b-x_{1})\big{[}sa^{s-1}(b-a)+a^{s}-b^{s}\big{]}+b_{x_{2}}(x_{1}-a)\big{[}sb^{s-1}(b-a)+a^{s}-b^{s}\big{]}}{(b-a)^{2}}\,.$
(A.11)

To obtain relation between $a_{x_{2}}$ and $b_{x_{2}}$ we differentiate
(4.102):

$1=b_{x_{2}}(x_{1}-a)-a_{x_{2}}(b-x_{1})\,.$ (A.12)

Now we replace the expression $a_{x_{2}}(b-x_{1})$ in (A.11) using (A.12) and
get

$\displaystyle w_{x_{2}}=$
$\displaystyle\frac{b_{x_{2}}(x_{1}-a)\big{[}s(b-a)(a^{s-1}+b^{s-1})+2a^{s}-2b^{s}\big{]}-\big{[}sa^{s-1}(b-a)+a^{s}-b^{s}\big{]}}{(b-a)^{2}}$
$\displaystyle=$
$\displaystyle\frac{b_{x_{2}}(x_{1}-a)}{(b-a)^{2}}A\big{(}\tfrac{1}{2}(b+a),\tfrac{1}{2}(b-a),s\big{)}-\frac{sa^{s-1}(b-a)+a^{s}-b^{s}}{(b-a)^{2}}\,.$

Whence,

$\displaystyle B_{x_{2}}{\buildrel\eqref{090601}\over{=}}w_{x_{2}}(a,b;r)$
$\displaystyle=\frac{b_{x_{2}}(x_{1}-a)}{(b-a)^{2}}A\big{(}\tfrac{1}{2}(b+a),\tfrac{1}{2}(b-a),r\big{)}-\frac{ra^{r-1}(b-a)+a^{r}-b^{r}}{(b-a)^{2}},$
(A.13) $\displaystyle 0{\buildrel\eqref{231002}\over{=}}w_{x_{2}}(a,b;p)$
$\displaystyle=\frac{b_{x_{2}}(x_{1}-a)}{(b-a)^{2}}A\big{(}\tfrac{1}{2}(b+a),\tfrac{1}{2}(b-a),p\big{)}-\frac{pa^{p-1}(b-a)+a^{p}-b^{p}}{(b-a)^{2}}.$
(A.14)

Therefore,

$B_{x_{2}}=\frac{b^{r}-a^{r}-ra^{r-1}(b-a)}{(b-a)^{2}}-\frac{b^{p}-a^{p}-pa^{p-1}(b-a)}{(b-a)^{2}}\cdot\frac{A\big{(}\tfrac{1}{2}(b+a),\tfrac{1}{2}(b-a),r\big{)}}{A\big{(}\tfrac{1}{2}(b+a),\tfrac{1}{2}(b-a),p\big{)}}\,.$
(A.15)

Since on the boundary we have $b=u+\varepsilon$, $a=u-\varepsilon=v$, we see
that the expressions in (A.10) and in (A.15) coincide.

Now we consider two remaining boundaries of $\Xi_{{}_{\mathrm{ch}+}}$. This is
the boundary $x_{3}=x_{2}^{p-1}x_{1}^{2-p}$ with $R$ (if
$(x_{1},x_{2})\in\omega_{2}\cup\omega_{3}$, see (5.8) and (5.9)) and the
boundary
$x_{3}=\frac{\Delta_{-}(x_{1}-\Delta_{+})^{p}+\Delta_{+}(x_{1}+\Delta_{-})^{p}}{2\varepsilon}$
with $\Xi_{{}_{\mathrm{R}+}}$ (if $(x_{1},x_{2})\in\omega_{4}$, see (5.10)).

Boundary $\mathbf{\Xi_{{}_{\mathrm{ch}+}}\Big{|}R}$. This is the boundary
$x_{3}=x_{2}^{p-1}x_{1}^{2-p}$. It consists of the chords with $[U(0),U(b)]$,
$b\in[0,2\varepsilon]$: $x_{2}=bx_{1}$, $x_{3}=b^{p-1}x_{1}$. The value of our
function on this chord is $B=b^{r-1}x_{1}$ in both domains.

The derivatives on $R$ (see (4.98)) are:

$\displaystyle
B_{x_{3}}=\frac{r-2}{p-2}\Big{(}\frac{x_{3}}{x_{2}}\Big{)}^{\frac{r-p}{p-2}}=\frac{r-2}{p-2}\cdot
b^{r-p},$ (A.16) $\displaystyle
B_{x_{2}}=\frac{p-r}{p-2}\Big{(}\frac{x_{3}}{x_{2}}\Big{)}^{\frac{r-2}{p-2}}=\frac{p-r}{p-2}\cdot
b^{r-2}.$ (A.17)

Since
$A(\tfrac{1}{2}b,\tfrac{1}{2}b,s){\buildrel\eqref{051003}\over{=}}(s-2)b^{s}$,
on this boundary the value of (A.9) coincides with (A.16) and the value of
(A.15) coincides with (A.17).

Boundary $\mathbf{\Xi_{{}_{\mathrm{ch}+}}\Big{|}\Xi_{{}_{\mathrm{R}+}}}$. This
boundary is foliated by the right half of the chords
$[U(u-\varepsilon),U(u+\varepsilon)]$ ($u\geqslant\varepsilon$), and $B$ is
linear on such chord with the prescribed values at the ends. Therefore, the
boundary values from both sides of the boundary are the same.

Now, we check continuity of $B_{x_{3}}$. On $\Xi_{{}_{\mathrm{ch}+}}$ these
derivatives are already calculated, we need only to plug the boundary values
$b=u+\varepsilon$ and $a=u-\varepsilon$ into (A.9):

$B_{x_{3}}=\frac{A(u,\varepsilon,r)}{A(u,\varepsilon,p)}\,.$ (A.18)

On $\Xi_{{}_{\mathrm{R}+}}$ we get the same expression if we substitute
$\xi=u$ to the following formula:

$B_{x_{3}}\Big{|}_{\Xi_{{}_{\mathrm{R}+}}}{\buildrel\eqref{250601}\over{=}}-\varepsilon\Psi_{{}_{\mathrm{R}}}^{\prime}{\buildrel\eqref{Psi_R_def}\over{=}}\frac{w_{{}_{\mathrm{R}}}^{\prime}(\xi;r,\varepsilon)}{w_{{}_{\mathrm{R}}}^{\prime}(\xi;p,\varepsilon)}{\buildrel\eqref{w_ww_relation}\over{=}}\frac{w_{{}_{\mathrm{L}}}^{\prime}(\xi;r,\varepsilon)}{w_{{}_{\mathrm{L}}}^{\prime}(\xi;p,\varepsilon)}{\buildrel\eqref{280401}\over{=}}\frac{A(\xi,\varepsilon,r)}{A(\xi,\varepsilon,p)},\qquad\xi\geqslant\varepsilon.$
(A.19)

Verifying continuity of $B_{x_{2}}$, we use (4.53) and (4.54) to rewrite the
second formula of (4.58) in terms of $w_{{}_{\mathrm{R}}}$:

$B_{x_{2}}(x)=\frac{1}{2\varepsilon}\Big{[}\Big{(}\varepsilon
k_{r}^{\prime}(v)+\frac{1}{\varepsilon}e^{-\frac{u}{\varepsilon}}w_{{}_{\mathrm{R}}}(\xi;r,\varepsilon)\Big{)}-\Big{(}\varepsilon
k_{p}^{\prime}(v)+\frac{1}{\varepsilon}e^{-\frac{u}{\varepsilon}}w_{{}_{\mathrm{R}}}(\xi;p,\varepsilon)\Big{)}\frac{w_{{}_{\mathrm{R}}}^{\prime}(\xi;r,\varepsilon)}{w_{{}_{\mathrm{R}}}^{\prime}(\xi;p,\varepsilon)}\Big{]}\,.$
(A.20)

On the boundary $\xi=u=v-\varepsilon=x_{1}-d$ we apply already used relations
(4.50), (4.68), (4.66), (2.15), and rewrite this formula as follows:

$B_{x_{2}}(x)\big{|}_{\xi=u}=\frac{1}{2\varepsilon}\Big{[}\Big{(}\frac{(u-\varepsilon)^{r}-(u+\varepsilon)^{r}}{2\varepsilon}+rv^{r-1}\Big{)}-\Big{(}\frac{(u-\varepsilon)^{p}-(u+\varepsilon)^{p}}{2\varepsilon}+pv^{p-1}\Big{)}\frac{A(u,\varepsilon,r)}{A(u,\varepsilon,p)}\Big{]}\,.$
(A.21)

On the face of it, this expression differs from (A.15) with
$b=u+\varepsilon=v$, $a=u-\varepsilon$. However, in fact they are equal. It
can be verified by the direct calculation using the definition of the function
$A$ (see (4.64)). But we check this in other way, deducing an alternative
formula for $B_{x_{2}}$ on $\Xi_{{}_{\mathrm{ch}+}}$. Namely, in (A.11) we
remove not $a_{x_{2}}$ but $b_{x_{2}}$ using the same relation (A.12):

$\displaystyle w_{x_{2}}=$
$\displaystyle\frac{a_{x_{2}}(b-x_{1})\big{[}s(b-a)(a^{s-1}+b^{s-1})+2a^{s}-2b^{s}\big{]}+\big{[}sb^{s-1}(b-a)+a^{s}-b^{s}\big{]}}{(b-a)^{2}}$
$\displaystyle=$
$\displaystyle\frac{a_{x_{2}}(b-x_{1})}{(b-a)^{2}}A\big{(}\tfrac{1}{2}(b+a),\tfrac{1}{2}(b-a),s\big{)}+\frac{ba^{s-1}(b-a)+a^{s}-b^{s}}{(b-a)^{2}}\,.$

Then instead of (A.13) and (A.14) we get

$\displaystyle B_{x_{2}}{\buildrel\eqref{090601}\over{=}}w_{x_{2}}(a,b;r)$
$\displaystyle=\frac{a_{x_{2}}(b-x_{1})}{(b-a)^{2}}A\big{(}\tfrac{1}{2}(b+a),\tfrac{1}{2}(b-a),r\big{)}+\frac{rb^{r-1}(b-a)+a^{r}-b^{r}}{(b-a)^{2}},$
(A.22) $\displaystyle 0{\buildrel\eqref{231002}\over{=}}w_{x_{2}}(a,b;p)$
$\displaystyle=\frac{a_{x_{2}}(b-x_{1})}{(b-a)^{2}}A\big{(}\tfrac{1}{2}(b+a),\tfrac{1}{2}(b-a),p\big{)}+\frac{pb^{p-1}(b-a)+a^{p}-b^{p}}{(b-a)^{2}}.$
(A.23)

Therefore,

$B_{x_{2}}=-\frac{b^{r}-a^{r}-rb^{r-1}(b-a)}{(b-a)^{2}}+\frac{b^{p}-a^{p}-pb^{p-1}(b-a)}{(b-a)^{2}}\cdot\frac{A\big{(}\tfrac{1}{2}(b+a),\tfrac{1}{2}(b-a),r\big{)}}{A\big{(}\tfrac{1}{2}(b+a),\tfrac{1}{2}(b-a),p\big{)}}\,,$
(A.24)

and this expression evidently coincides with (A.21) on the boundary
$b=u+\varepsilon=v$, $a=u-\varepsilon$.

It remains to check one boundary, namely, between $\Xi_{{}_{\mathrm{R}+}}$ and
$R$.

Boundary $\mathbf{\Xi_{{}_{\mathrm{R}+}}\Big{|}R}$. This is the boundary
$x_{3}=(x_{1}+\Delta_{-})^{p-2}x_{2}$ or $\xi=u=v-\varepsilon=x_{1}-d$ and it
is foliated by the extremal lines $x_{3}=v^{p-2}x_{2}$,
$x_{2}=2(v-\varepsilon)x_{1}-v^{2}+2v\varepsilon$, $\varepsilon\leqslant
v\leqslant 2\varepsilon$. On this boundary we have to use the first line of
definition of $w_{{}_{\mathrm{R}}}$ (4.50) for $\xi\leqslant\varepsilon$.
Therefore, (see (4.87)), on any such extremal line we have

$B=v^{r}-\big{(}k_{r}(v)-k_{r}(v)+2uv^{r-2}\big{)}\Delta_{-}=v^{r-2}(v^{2}-2u\Delta_{-})=v^{r-2}x_{2}\,,$

that coincides with the value on $R$ (see (4.99)).

Using as before (4.54) and the first formula in (4.58) for $B_{x_{3}}$ on
$\Xi_{{}_{\mathrm{R}+}}$, we get

$B_{x_{3}}=\frac{w_{{}_{\mathrm{R}}}^{\prime}(\xi;r,\varepsilon)}{w_{{}_{\mathrm{R}}}^{\prime}(\xi;p,\varepsilon)}{\buildrel\eqref{ww_dif}\over{=}}\frac{r-2}{p-2}(\xi+\varepsilon)^{r-p},$

and for $\xi=u=v-\varepsilon$ we get the same expression as in (A.16) for
$x_{3}=v^{p-2}x_{2}$. The second formula in (4.58) for $B_{x_{2}}$ yields
(A.20), where we have use another expression for $w_{{}_{\mathrm{R}}}$ (see
the first line of (4.50)). As a result, we get not (A.21), but

$\displaystyle B_{x_{2}}\big{|}_{\xi=u}=$
$\displaystyle\frac{1}{2\varepsilon}\Big{[}\big{(}rv^{r-1}-2uv^{r-2}\big{)}-\big{(}pv^{p-1}-2uv^{p-2}\big{)}\frac{(r-2)v^{r-3}}{(p-2)v^{p-3}}\Big{]}$
$\displaystyle=$
$\displaystyle\frac{1}{2\varepsilon}\Big{[}\big{(}(r-2)v^{r-1}+2\varepsilon
v^{r-2}\big{)}-\big{(}(p-2)v^{p-1}+2\varepsilon
v^{p-2}\big{)}\frac{r-2}{p-2}v^{r-p}\Big{]}=\frac{p-r}{p-2}v^{r-2}\,,$

therefore, we get the same expression as in (A.17) for $x_{3}=v^{p-2}x_{2}$.

The only boundary intersecting the plane $x_{1}=0$ is the boundary between
$F(0)$ and $R$ considered above. $C^{1}$-smoothness of $B$ in the domain
$x_{1}<0$ follows from the symmetry. This completes the proof of
$C^{1}$-smoothness of the function $B$. $\square$

##  Appendix B Convexity/concavity of the Bellman function $B_{2}$

In this appendix we prove convexity/concavity of $B_{2}$. Since we will deal
only with this candidate we omit index $2$, as it was done in Appendix A.

######  Lemma B.1.

The equality $\operatorname{sign}B_{x_{3}x_{3}}=\operatorname{sign}(r-2)(r-p)$
holds on subdomains $F(0),$ $R,$ $\Xi_{{}_{\mathrm{L}\pm}},$
$\Xi_{{}_{\mathrm{R}\pm}},$ $\Xi_{{}_{\mathrm{c}h\pm}}$.

##### Proof.

We start with $F(0)$, where by (A.4) we have

$B_{x_{3}}=\frac{w_{{}_{\mathrm{L}}}^{\prime}(\xi;r,\varepsilon)}{w_{{}_{\mathrm{L}}}^{\prime}(\xi;p,\varepsilon)}\buildrel\eqref{280401}\over{=}\frac{A(\xi,\varepsilon,r)}{A(\xi,\varepsilon,p)}.$
(B.1)

The same expression we have in $\Xi_{{}_{\mathrm{L}+}}$, as we have seen in
(A.6). In $\Xi_{{}_{\mathrm{R}+}}$ we have the same formula for
$\xi\geqslant\varepsilon$ due to (A.19). Therefore, in all three cases we have

$B_{x_{3}x_{3}}=\left(\frac{A(\xi,\varepsilon,r)}{A(\xi,\varepsilon,p)}\right)^{\prime}\\!\cdot\frac{\partial\xi}{\partial
x_{3}}\,.$

Recall that we use symbol ′ to denote differentiation with respect to $\xi$.
In the cases under consideration we have

$\displaystyle F(0):\qquad x_{3}^{\prime}$
$\displaystyle\buildrel\eqref{220601}\over{=}\frac{ex_{2}}{2\varepsilon^{2}}w_{{}_{\mathrm{L}}}^{\prime}(\xi;p,\varepsilon)\,;$
$\displaystyle\Xi_{{}_{\mathrm{R}+}}:\qquad x_{3}^{\prime}$
$\displaystyle\buildrel\eqref{141003}\over{=}\frac{\Delta_{-}}{\varepsilon}e^{-\frac{u}{\varepsilon}}w_{{}_{\mathrm{R}}}^{\prime}(\xi;p,\varepsilon)\buildrel\eqref{w_ww_relation}\over{=}\frac{\Delta_{-}}{\varepsilon}e^{\frac{2\xi-u}{\varepsilon}}w_{{}_{\mathrm{L}}}^{\prime}(\xi;p,\varepsilon),\qquad\xi\geqslant\varepsilon\,;$
$\displaystyle\Xi_{{}_{\mathrm{L}+}}:\qquad x_{3}^{\prime}$
$\displaystyle\buildrel\eqref{141001}\over{=}\frac{\Delta_{-}}{\varepsilon}e^{\frac{u}{\varepsilon}}w_{{}_{\mathrm{L}}}^{\prime}(\xi;p,\varepsilon)\,.$

Therefore, in all cases Lemma 4.1 implies

$\operatorname{sign}\frac{\partial\xi}{\partial
x_{3}}=\operatorname{sign}x_{3}^{\prime}=\operatorname{sign}w_{{}_{\mathrm{L}}}^{\prime}(\xi;p,\varepsilon)=\operatorname{sign}(p-2)\,,$

and we need to check that

$\operatorname{sign}\left(\frac{A(\xi,\varepsilon,r)}{A(\xi,\varepsilon,p)}\right)^{\prime}=\operatorname{sign}\big{(}A^{\prime}(\xi,\varepsilon,r)A(\xi,\varepsilon,p)-A^{\prime}(\xi,\varepsilon,p)A(\xi,\varepsilon,r)\Big{)}=\operatorname{sign}(p-2)(r-2)(r-p)\,.$
(B.2)

Differentiating (4.65) with respect to $\alpha$ and then integrating by parts,
we get

$\frac{\partial}{\partial\alpha}A(\alpha,\beta,s)=s(s-1)(s-2)\\!\\!\int\limits^{\;\beta}_{-\beta\
\ }\\!\\!2\lambda(\lambda+\alpha)^{s-3}d\lambda\,.$ (B.3)

After dividing the expression in (B.2) over $r(r-1)(r-2)p(p-1)(p-2)$, we need
to verify that $\operatorname{sign}(r-p)$ is the sign of the following
expression:

$\displaystyle\int\limits^{\;\beta}_{-\beta\ \
}\\!\\!2\lambda(\lambda+\alpha)^{r-3}\,d\lambda\\!\\!\int\limits^{\;\beta}_{-\beta\
\
}\\!\\!(\beta^{2}-\lambda^{2})(\lambda+\alpha)^{p-3}d\lambda-\\!\\!\int\limits^{\;\beta}_{-\beta\
\
}\\!\\!2\lambda(\lambda+\alpha)^{p-3}d\lambda\\!\\!\int\limits^{\;\beta}_{-\beta\
\ }\\!\\!(\beta^{2}-\lambda^{2})(\lambda+\alpha)^{r-3}\,d\lambda$
$\displaystyle=2\\!\\!\int\limits^{\;\beta}_{-\beta\ \
}\\!\\!\\!\\!\int\limits^{\;\beta}_{-\beta\ \
}\\!\\!\big{[}\lambda(\beta^{2}-\mu^{2})-\mu(\beta^{2}-\lambda^{2})\big{]}(\mu+\alpha)^{p-3}(\lambda+\alpha)^{r-3}\,d\lambda\,d\mu$
$\displaystyle=2\\!\\!\int\limits^{\;\beta}_{-\beta\ \
}\\!\\!\\!\\!\int\limits^{\;\beta}_{-\beta\ \
}\\!\\!(\lambda-\mu)(\beta^{2}+\lambda\mu)(\mu+\alpha)^{p-3}(\lambda+\alpha)^{r-3}\,d\lambda\,d\mu\,.$

After symmetrization (interchanging $\mu$ and $\lambda$) we get:

$\int\limits^{\;\beta}_{-\beta\ \ }\\!\\!\\!\\!\int\limits^{\;\beta}_{-\beta\
\
}\\!\\!(\beta^{2}+\lambda\mu)(\mu+\alpha)^{p-3}(\lambda+\alpha)^{p-3}\Big{[}(\lambda-\mu)\big{(}(\lambda+\alpha)^{r-p}-(\mu+\alpha)^{r-p}\big{)}\Big{]}\,d\lambda\,d\mu\,.$

Since the function $t\mapsto t^{s}$ is increasing for $s>0$ and decreasing for
$s<0$, the sign of expression in the square brackets coincides with
$\operatorname{sign}(r-p)$, and this is just what we need, because all other
terms are positive.

To complete the proof for these three domains, it remains to consider the
subdomain of $\Xi_{{}_{\mathrm{R}+}}$ with $\xi<\varepsilon$:

$B_{x_{3}}\Big{|}_{\Xi_{{}_{\mathrm{R}+}}}{\buildrel\eqref{250601}\over{=}}-\varepsilon\Psi_{{}_{\mathrm{R}}}^{\prime}{\buildrel\eqref{Psi_R_def}\over{=}}\frac{w_{{}_{\mathrm{R}}}^{\prime}(\xi;r,\varepsilon)}{w_{{}_{\mathrm{R}}}^{\prime}(\xi;p,\varepsilon)}{\buildrel\eqref{ww_dif}\over{=}}\frac{r-2}{p-2}(\xi+\varepsilon)^{r-p},\qquad\
\xi<\varepsilon\,,$

whence $\operatorname{sign}B_{x_{3}\xi}=\operatorname{sign}(p-2)(r-2)(r-p)$.
Since $\operatorname{sign}\frac{\partial\xi}{\partial
x_{3}}=\operatorname{sign}w_{{}_{\mathrm{R}}}^{\prime}(\xi;p,\varepsilon)=\operatorname{sign}(p-2)$
by Lemma 4.2, we obtain the required property:
$\operatorname{sign}B_{x_{3}x_{3}}=\operatorname{sign}(r-2)(r-p)$.

Now, we start to calculate the sign of $B_{x_{3}x_{3}}$ on the domain
$\Xi_{{}_{\mathrm{ch}+}}$. On this domain we will use notation
$\alpha=\tfrac{1}{2}(b+a)$ and $\beta=\tfrac{1}{2}(b-a)$. Due to formula (A.9)
we have

$B_{x_{3}x_{3}}=\frac{\partial}{\partial
x_{3}}\left(\frac{A(\alpha,\beta,r)}{A(\alpha,\beta,p)}\right)=\frac{A_{x_{3}}(r)A(p)-A(r)A_{x_{3}}(p)}{A(p)^{2}}\,.$

For brevity, we omit arguments $\alpha$ and $\beta$ if this does not lead to
misunderstanding. The principal step will be the same as before, but we need
some auxiliary calculations here. First of all, we will use another integral
form of the function $A$. Changing the variable of integration in (4.65), we
get the following formula

$A(s)=s(s-1)(s-2)\int\limits_{a}^{b}(b-\lambda)(\lambda-a)\lambda^{s-3}d\lambda\,,$
(B.4)

whence

$A_{x_{3}}(s)=s(s-1)(s-2)\int\limits_{a}^{b}\big{[}b_{x_{3}}(\lambda-a)-(b-\lambda)a_{x_{3}}\big{]}\lambda^{s-3}d\lambda\,.$
(B.5)

From (A.7) and (5.1) we have

$(x_{1}-a)b_{x_{3}}=(b-x_{1})a_{x_{3}}=\frac{(b-a)^{2}}{A(p)},$ (B.6)

therefore, (B.5) can be rewritten as follows:

$\displaystyle A_{x_{3}}(s)$
$\displaystyle=\frac{s(s-1)(s-2)(b-a)^{2}}{(x_{1}-a)(b-x_{1})A(p)}\int\limits_{a}^{b}\big{[}(b-x_{1})(\lambda-a)-(b-\lambda)(x_{1}-a)\big{]}\lambda^{s-3}d\lambda$
(B.7)
$\displaystyle=\frac{s(s-1)(s-2)(b-a)^{3}}{(x_{1}-a)(b-x_{1})A(p)}\int\limits_{a}^{b}(\lambda-
x_{1})\lambda^{s-3}d\lambda\,,$

whence

$\displaystyle B_{x_{3}x_{3}}$
$\displaystyle=\frac{r(r-1)(r-2)p(p-1)(p-2)(b-a)^{3}}{(x_{1}-a)(b-x_{1})A(p)^{3}}\times$
$\displaystyle\quad\times\Bigg{\\{}\int\limits_{a}^{b}(\lambda-
x_{1})\lambda^{r-3}d\lambda\int\limits_{a}^{b}(b\\!-\\!\mu)(\mu\\!-\\!a)\mu^{p-3}d\mu\\!-\\!\int\limits_{a}^{b}(b\\!-\\!\lambda)(\lambda\\!-\\!a)\lambda^{r-3}d\lambda\int\limits_{a}^{b}(\mu\\!-\\!x_{1})\mu^{p-3}d\mu\Bigg{\\}}$
$\displaystyle=\frac{r(r-1)(r-2)p(p-1)(p-2)(b-a)^{3}}{(x_{1}-a)(b-x_{1})A(p)^{3}}\int\limits_{a}^{b}\\!\\!\int\limits_{a}^{b}\big{[}\lambda\mu\\!-\\!ab\\!+\\!(a\\!+\\!b\\!-\\!\lambda\\!-\\!\mu)x_{1}\big{]}(\lambda\\!-\\!\mu)\lambda^{r-3}\mu^{p-3}d\lambda\,d\mu$
$\displaystyle=\frac{r(r-1)(r-2)p(p-1)(p-2)(b-a)^{3}}{2(x_{1}-a)(b-x_{1})A(p)^{3}}\times$
$\displaystyle\quad\times\int\limits_{a}^{b}\\!\\!\int\limits_{a}^{b}\big{[}\lambda\mu-
ab+(a+b-\lambda-\mu)x_{1}\big{]}(\lambda-\mu)(\lambda^{r-p}-\mu^{r-p})\lambda^{p-3}\mu^{p-3}d\lambda\,d\mu\,.$
(B.8)

In the latter equality, we use symmetrization with respect $\lambda$ and
$\mu$, as it has been made before. Now it is easy to determine the sign of
this expression. Since $a<x_{1}<b$ and
$\operatorname{sign}A(p)=\operatorname{sign}(p-2)$, the sign of the factor in
front of the integral is $\operatorname{sign}(r-2)$. The expression in the
square brackets is positive. To check this it suffices to note that this is a
linear function with respect to $x_{1}$ and it is positive at both endpoints
of the interval $[a,b]$:

  * •

at $x_{1}=a$ we have $(\lambda-a)(\mu-a)$;

  * •

at $x_{1}=b$ we have $(b-\lambda)(b-\mu)$.

Clearly,

$\operatorname{sign}(\lambda^{r-p}-\mu^{r-p})(\lambda-\mu)=\operatorname{sign}(r-p)\,.$

Therefore, we get what is needed:
$\operatorname{sign}B_{x_{3}x_{3}}=\operatorname{sign}(r-2)(r-p)$.

It remains to calculate $\operatorname{sign}B_{x_{3}x_{3}}$ on $R$. Since the
function $B$ has a simple explicit expression there (see (4.98))

$B=x_{2}^{\frac{p-r}{p-2}}x_{3}^{\frac{r-2}{p-2}},$

it is easy to calculate $B_{x_{3}x_{3}}$ directly:

$\operatorname{sign}B_{x_{3}x_{3}}=\operatorname{sign}\frac{r-2}{p-2}\Big{(}\frac{r-2}{p-2}-1\Big{)}x_{2}^{\frac{p-r}{p-2}}x_{3}^{\frac{r-2}{p-2}-2}=\operatorname{sign}(r-2)(r-p)\,.$

The consideration of this last case completes the proof of the lemma.
$\square$

Since in the domains $F(0)$ and $R$ the function $B$ is linear on two-
dimensional planes transversal to the direction of $x_{3}$, just proved Lemma
B.1 ensures the required in Theorem 6.1 concavity/convexity of $B$ inside
these domains. The other domains are foliated by one-dimensional extremal
lines transversal to $x_{2}x_{3}$-plane, therefore, to prove
concavity/convexity there we need to check positivity of the minor
$B_{x_{2}x_{2}}B_{x_{3}x_{3}}-B_{x_{2}x_{3}}^{2}$.

We start with domain $\Xi_{{}_{\mathrm{L}+}}$.

######  Lemma B.2.

$\det\\{B_{x_{i},x_{j}}\\}_{2\leqslant i,j\leqslant
3}=B_{x_{2}x_{2}}B_{x_{3}x_{3}}-B_{x_{2}x_{3}}^{2}\geqslant 0$ on
$\Xi_{{}_{\mathrm{L}+}}$.

##### Proof.

To calculate the sign of this determinant we use formula (4.32):

$B_{x_{2}x_{2}}B_{x_{3}x_{3}}-B_{x_{2}x_{3}}^{2}=\frac{e^{-\frac{u}{\varepsilon}}\Psi_{{}_{\mathrm{L}}}^{\prime\prime}(e^{-\frac{u}{\varepsilon}}h)}{4(x_{1}-u)(x_{1}-v)}\Big{(}e^{\frac{u}{\varepsilon}}\Psi_{{}_{\mathrm{L}}}(e^{-\frac{u}{\varepsilon}}h)-\big{(}h+\varepsilon^{3}m_{p}^{\prime\prime}(v)\big{)}\Psi_{{}_{\mathrm{L}}}^{\prime}(e^{-\frac{u}{\varepsilon}}h)+\varepsilon^{2}m_{r}^{\prime\prime}(v)\Big{)}.$

Since $v<x_{1}<u=v+\varepsilon$ and
$\operatorname{sign}\Psi_{{}_{\mathrm{L}}}^{\prime\prime}=\operatorname{sign}(r-2)(r-p)$
(see (4.28) and Lemma B.1) we deduce that
$\operatorname{sign}\det\\{B_{x_{i},x_{j}}\\}_{2\leqslant i,j\leqslant 3}$
coincides with

$-\operatorname{sign}(r-2)(r-p)\times\operatorname{sign}\Big{(}e^{\frac{u}{\varepsilon}}\Psi_{{}_{\mathrm{L}}}(e^{-\frac{u}{\varepsilon}}h)-\big{(}h+\varepsilon^{3}m_{p}^{\prime\prime}(v)\big{)}\Psi_{{}_{\mathrm{L}}}^{\prime}(e^{-\frac{u}{\varepsilon}}h)+\varepsilon^{2}m_{r}^{\prime\prime}(v)\Big{)}.$

So, we need to check that

$\operatorname{sign}\Big{(}e^{\frac{u}{\varepsilon}}\Psi_{{}_{\mathrm{L}}}(e^{-\frac{u}{\varepsilon}}h)-\big{(}h+\varepsilon^{3}m_{p}^{\prime\prime}(v)\big{)}\Psi_{{}_{\mathrm{L}}}^{\prime}(e^{-\frac{u}{\varepsilon}}h)+\varepsilon^{2}m_{r}^{\prime\prime}(v)\Big{)}\buildrel?\over{=}\operatorname{sign}(p-r)(r-2)\,.$
(B.9)

We return to variable $\xi$ (see (4.22)) and use the definition of
$\Psi_{{}_{\mathrm{L}}}$ in terms of $w_{{}_{\mathrm{L}}}$ (see (4.24)) and
rewrite (B.9) as follows:

$\operatorname{sign}\left(\frac{e^{\frac{u}{\varepsilon}}}{\varepsilon}w_{{}_{\mathrm{L}}}(\xi;r,\varepsilon)-\Big{(}e^{\frac{u}{\varepsilon}}w_{{}_{\mathrm{L}}}(\xi;p,\varepsilon)+\varepsilon^{3}m_{p}^{\prime\prime}(v)\Big{)}\frac{1}{\varepsilon}\frac{w_{{}_{\mathrm{L}}}^{\prime}(\xi;r,\varepsilon)}{w_{{}_{\mathrm{L}}}^{\prime}(\xi;p,\varepsilon)}+\varepsilon^{2}m_{r}^{\prime\prime}(v)\right)\buildrel?\over{=}\operatorname{sign}(p-r)(r-2),$

which is equivalent to

$\operatorname{sign}\left(\frac{w_{{}_{\mathrm{L}}}(\xi;r,\varepsilon)+e^{-\frac{u}{\varepsilon}}\varepsilon^{3}m_{r}^{\prime\prime}(v)}{w_{{}_{\mathrm{L}}}^{\prime}(\xi;r,\varepsilon)}-\frac{w_{{}_{\mathrm{L}}}(\xi;p,\varepsilon)+e^{-\frac{u}{\varepsilon}}\varepsilon^{3}m_{p}^{\prime\prime}(v)}{w_{{}_{\mathrm{L}}}^{\prime}(\xi;p,\varepsilon)}\right)\buildrel?\over{=}\operatorname{sign}(p-r)\,,$
(B.10)

because
$\operatorname{sign}w_{{}_{\mathrm{L}}}^{\prime}(\xi;r,\varepsilon)=\operatorname{sign}(r-2)$
by Lemma 4.1. In other words, we need to prove that the function

$s\mapsto\frac{w_{{}_{\mathrm{L}}}(\xi;s,\varepsilon)+e^{-\frac{u}{\varepsilon}}\varepsilon^{3}m_{s}^{\prime\prime}(v)}{w_{{}_{\mathrm{L}}}^{\prime}(\xi;s,\varepsilon)}$

decreases for all $\xi\geqslant u$.

We split this function into the sum of two

$\frac{w_{{}_{\mathrm{L}}}(\xi;s,\varepsilon)+e^{-\frac{u}{\varepsilon}}\varepsilon^{3}m_{s}^{\prime\prime}(v)}{w_{{}_{\mathrm{L}}}^{\prime}(\xi;s,\varepsilon)}=\frac{w_{{}_{\mathrm{L}}}(\xi;s,\varepsilon)+e^{-\frac{\xi}{\varepsilon}}\varepsilon^{3}m_{s}^{\prime\prime}(\xi-\varepsilon)}{w_{{}_{\mathrm{L}}}^{\prime}(\xi;s,\varepsilon)}+\frac{e^{-\frac{u}{\varepsilon}}\varepsilon^{3}m_{s}^{\prime\prime}(v)-e^{-\frac{\xi}{\varepsilon}}\varepsilon^{3}m_{s}^{\prime\prime}(\xi-\varepsilon)}{w_{{}_{\mathrm{L}}}^{\prime}(\xi;s,\varepsilon)}$
(B.11)

and prove that both of them are decreasing.

We begin with a simpler second function rewriting the expression in the
numerator. By (2.13) we have

$e^{-\frac{u}{\varepsilon}}\varepsilon^{3}m_{s}^{\prime\prime}(v)-e^{-\frac{\xi}{\varepsilon}}\varepsilon^{3}m_{s}^{\prime\prime}(\xi-\varepsilon)=\varepsilon^{2}s(s-1)(s-2)\int\limits_{u}^{\xi}e^{-\frac{t}{\varepsilon}}(t-\varepsilon)^{s-3}dt\,,$

and due to (4.66) and (4.65) the question is reduced to verification that the
function

$s\mapsto\frac{\int\limits_{u}^{\xi}e^{-\frac{t}{\varepsilon}}(t-\varepsilon)^{s-3}dt}{\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon^{2}-\lambda^{2})(\lambda+\xi)^{s-3}d\lambda}$
(B.12)

is decreasing. Indeed, for $\varepsilon<t<\xi$ and $\lambda>-\varepsilon$ we
have $\frac{\lambda+\xi}{t-\varepsilon}>1$, and therefore, the function

$s\mapsto\Big{(}\frac{\lambda+\xi}{t-\varepsilon}\Big{)}^{s-3}$

is increasing. After integration of this family of increasing functions with
positive weight we get the following increasing function:

$s\mapsto\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon^{2}-\lambda^{2})\Big{(}\frac{\lambda+\xi}{t-\varepsilon}\Big{)}^{s-3}d\lambda\,,$

or the following decreasing function:

$s\mapsto\frac{(t-\varepsilon)^{s-3}}{\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon^{2}-\lambda^{2})(\lambda+\xi)^{s-3}d\lambda}\,.$
(B.13)

We integrate once more family of functions (B.13) with the positive weight
$e^{-\frac{t}{\varepsilon}}$ and finally obtain the decreasing function
(B.12).

Now, we consider the first summand in (B.11). Again, we start with simplifying
the expression in the numerator. Using twice (2.15), we get

$\varepsilon^{2}m_{s}^{\prime\prime}(v)=\varepsilon\big{(}m_{s}(v)-sv^{s-1}\big{)}^{\prime}=m_{s}(v)-sv^{s-1}-\varepsilon
s(s-1)v^{s-2}.$

Then, we use the definition (4.20) of $w_{{}_{\mathrm{L}}}$:

$\displaystyle w_{{}_{\mathrm{L}}}(\xi;$ $\displaystyle
s,\varepsilon)+e^{-\frac{\xi}{\varepsilon}}\varepsilon^{3}m_{s}^{\prime\prime}(\xi-\varepsilon)$
$\displaystyle=e^{-\frac{\xi}{\varepsilon}}\Big{(}\frac{(\xi+\varepsilon)^{s}-(\xi-\varepsilon)^{s}}{2}-\varepsilon
m_{s}(\xi-\varepsilon)\Big{)}+\varepsilon
e^{-\frac{\xi}{\varepsilon}}\Big{(}m_{s}(\xi-\varepsilon)-s(\xi-\varepsilon)^{s-1}-\varepsilon
s(s-1)(\xi-\varepsilon)^{s-2}\Big{)}$
$\displaystyle=e^{-\frac{\xi}{\varepsilon}}\Big{(}\frac{(\xi+\varepsilon)^{s}-(\xi-\varepsilon)^{s}}{2}-\varepsilon
s(\xi-\varepsilon)^{s-1}-\varepsilon^{2}s(s-1)(\xi-\varepsilon)^{s-2}\Big{)}$
$\displaystyle=\frac{1}{4}s(s-1)(s-2)e^{-\frac{\xi}{\varepsilon}}\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon-\mu)^{2}(\mu+\xi)^{s-3}d\mu\,.$

Therefore, we need to check that the function

$s\mapsto\frac{\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon-\mu)^{2}(\mu+\xi)^{s-3}d\mu}{\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon^{2}-\lambda^{2})(\lambda+\xi)^{s-3}d\lambda}$
(B.14)

is decreasing. After differentiating this function, we get in numerator the
following expression:

$\displaystyle\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon-\mu)^{2}$
$\displaystyle(\mu+\xi)^{s-3}\log(\mu+\xi)\,d\mu\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon^{2}-\lambda^{2})(\lambda+\xi)^{s-3}d\lambda$
$\displaystyle\qquad\qquad-\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon-\mu)^{2}(\mu+\xi)^{s-3}d\mu\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon^{2}-\lambda^{2})(\lambda+\xi)^{s-3}\log(\lambda+\xi)\,d\lambda$
$\displaystyle=\int\limits_{\\!-\varepsilon\ \
}^{\;\varepsilon}\\!\\!\\!\int\limits_{\\!-\varepsilon\
}^{\;\varepsilon}(\varepsilon-\mu)^{2}(\varepsilon^{2}-\lambda^{2})(\mu+\xi)^{s-3}(\lambda+\xi)^{s-3}\big{(}\log(\mu+\xi)-\log(\lambda+\xi)\big{)}\,d\mu\,d\lambda\,.$

Now, we symmetrize this expression (interchanging $\mu$ and $\lambda$):

$\displaystyle\frac{1}{2}\\!\\!\int\limits_{\\!-\varepsilon\ \
}^{\;\varepsilon}\\!\\!\\!\int\limits_{\\!-\varepsilon\ }^{\;\varepsilon}$
$\displaystyle\big{[}(\varepsilon-\mu)^{2}(\varepsilon^{2}-\lambda^{2})-(\varepsilon-\lambda)^{2}(\varepsilon^{2}-\mu^{2})\big{]}(\mu+\xi)^{s-3}(\lambda+\xi)^{s-3}\big{(}\log(\mu+\xi)-\log(\lambda+\xi)\big{)}\,d\mu\,d\lambda$
$\displaystyle=\varepsilon\\!\\!\int\limits_{\\!-\varepsilon\ \
}^{\;\varepsilon}\\!\\!\\!\int\limits_{\\!-\varepsilon\
}^{\;\varepsilon}(\lambda-\mu)(\varepsilon-\mu)(\varepsilon-\lambda)(\mu+\xi)^{s-3}(\lambda+\xi)^{s-3}\big{(}\log(\mu+\xi)-\log(\lambda+\xi)\big{)}\,d\mu\,d\lambda\,.$

Since $\log$ is an increasing function, we have

$(\lambda-\mu)\big{(}\log(\mu+\xi)-\log(\lambda+\xi)\big{)}<0\,.$

This means that the derivative we have calculated is negative and the function
(B.14) decreases. This completes the proof of the lemma. $\square$

######  Lemma B.3.

$B_{x_{2}x_{2}}B_{x_{3}x_{3}}-B_{x_{2}x_{3}}^{2}\geqslant 0$ on
$\Xi_{{}_{\mathrm{R}+}}$.

##### Proof.

To calculate the sign of this determinant, we use formula (4.62):

$B_{x_{2}x_{2}}B_{x_{3}x_{3}}-B_{x_{2}x_{3}}^{2}=\frac{e^{-\frac{u}{\varepsilon}}\Psi_{{}_{\mathrm{R}}}^{\prime\prime}(e^{\frac{u}{\varepsilon}}h)}{4(x_{1}-u)(x_{1}-v)}\Big{(}e^{-\frac{u}{\varepsilon}}\Psi_{{}_{\mathrm{R}}}(e^{\frac{u}{\varepsilon}}h)-\big{(}h-\varepsilon^{3}k_{p}^{\prime\prime}(v)\big{)}\Psi_{{}_{\mathrm{R}}}^{\prime}(e^{\frac{u}{\varepsilon}}h)+\varepsilon^{2}k_{r}^{\prime\prime}(v)\Big{)}.$

Since $u<x_{1}<v=u+\varepsilon$ and
$\operatorname{sign}\Psi_{{}_{\mathrm{R}}}^{\prime\prime}=-\operatorname{sign}(r-2)(r-p)$
(see (4.59) and Lemma B.1), we deduce that
$\operatorname{sign}\det\\{B_{x_{i},x_{j}}\\}_{2\leqslant i,j\leqslant 3}$
coincides with

$\operatorname{sign}(r-2)(r-p)\times\operatorname{sign}\Big{(}e^{-\frac{u}{\varepsilon}}\Psi_{{}_{\mathrm{R}}}(e^{\frac{u}{\varepsilon}}h)-\big{(}h-\varepsilon^{3}k_{p}^{\prime\prime}(v)\big{)}\Psi_{{}_{\mathrm{R}}}^{\prime}(e^{\frac{u}{\varepsilon}}h)+\varepsilon^{2}k_{r}^{\prime\prime}(v)\Big{)}.$

So, we need to check that

$\operatorname{sign}\Big{(}e^{-\frac{u}{\varepsilon}}\Psi_{{}_{\mathrm{R}}}(e^{\frac{u}{\varepsilon}}h)-\big{(}h-\varepsilon^{3}k_{p}^{\prime\prime}(v)\big{)}\Psi_{{}_{\mathrm{R}}}^{\prime}(e^{\frac{u}{\varepsilon}}h)+\varepsilon^{2}k_{r}^{\prime\prime}(v)\Big{)}\buildrel?\over{=}\operatorname{sign}(r-p)(r-2)\,.$
(B.15)

As in the proof of the preceding lemma, we return to variable $\xi$ (see
(4.52)) and use the definition of $\Psi_{{}_{\mathrm{R}}}$ in terms of
$w_{{}_{\mathrm{R}}}$ (see (4.54)) to rewrite (B.15) as follows:

$\operatorname{sign}\left(-\frac{e^{-\frac{u}{\varepsilon}}}{\varepsilon}w_{{}_{\mathrm{R}}}(\xi;r,\varepsilon)+\Big{(}e^{-\frac{u}{\varepsilon}}w_{{}_{\mathrm{R}}}(\xi;p,\varepsilon)-\varepsilon^{3}k_{p}^{\prime\prime}(v)\Big{)}\frac{1}{\varepsilon}\frac{w_{{}_{\mathrm{R}}}^{\prime}(\xi;r,\varepsilon)}{w_{{}_{\mathrm{R}}}^{\prime}(\xi;p,\varepsilon)}+\varepsilon^{2}k_{r}^{\prime\prime}(v)\right)\buildrel?\over{=}\operatorname{sign}(r-p)(r-2),$

which is equivalent to

$\operatorname{sign}\left(\frac{w_{{}_{\mathrm{R}}}(\xi;r,\varepsilon)-e^{\frac{u}{\varepsilon}}\varepsilon^{3}k_{r}^{\prime\prime}(v)}{w_{{}_{\mathrm{R}}}^{\prime}(\xi;r,\varepsilon)}-\frac{w_{{}_{\mathrm{R}}}(\xi;p,\varepsilon)-e^{\frac{u}{\varepsilon}}\varepsilon^{3}k_{p}^{\prime\prime}(v)}{w_{{}_{\mathrm{R}}}^{\prime}(\xi;p,\varepsilon)}\right)\buildrel?\over{=}\operatorname{sign}(p-r)\,,$
(B.16)

because
$\operatorname{sign}w_{{}_{\mathrm{R}}}^{\prime}(\xi;r,\varepsilon)=\operatorname{sign}(r-2)$
by Lemma 4.2. In other words, we need to prove that the function

$s\mapsto\frac{w_{{}_{\mathrm{R}}}(\xi;s,\varepsilon)-e^{\frac{u}{\varepsilon}}\varepsilon^{3}k_{s}^{\prime\prime}(v)}{w_{{}_{\mathrm{R}}}^{\prime}(\xi;s,\varepsilon)}$

decreases for all $\xi\leqslant u$.

We split this function into the sum of two

$\frac{w_{{}_{\mathrm{R}}}(\xi;s,\varepsilon)-e^{\frac{u}{\varepsilon}}\varepsilon^{3}k_{s}^{\prime\prime}(v)}{w_{{}_{\mathrm{R}}}^{\prime}(\xi;s,\varepsilon)}=\frac{w_{{}_{\mathrm{R}}}(\xi;s,\varepsilon)-e^{\frac{\xi}{\varepsilon}}\varepsilon^{3}k_{s}^{\prime\prime}(\xi+\varepsilon)}{w_{{}_{\mathrm{R}}}^{\prime}(\xi;s,\varepsilon)}+\frac{e^{\frac{\xi}{\varepsilon}}\varepsilon^{3}k_{s}^{\prime\prime}(\xi+\varepsilon)-e^{\frac{u}{\varepsilon}}\varepsilon^{3}k_{s}^{\prime\prime}(v)}{w_{{}_{\mathrm{R}}}^{\prime}(\xi;s,\varepsilon)}$
(B.17)

and prove that both of them are decreasing.

Again, we begin with a simpler second function rewriting the expression in the
numerator. From (2.14) we have

$e^{\frac{\xi}{\varepsilon}}\varepsilon^{3}k_{s}^{\prime\prime}(\xi+\varepsilon)=s(s-2)\varepsilon^{s}+\varepsilon^{2}s(s-1)(s-2)\int\limits_{0}^{\xi}e^{\frac{t}{\varepsilon}}(t+\varepsilon)^{s-3}dt\,,$

and, therefore, since $v=u+\varepsilon$, we have

$e^{\frac{\xi}{\varepsilon}}\varepsilon^{3}k_{s}^{\prime\prime}(\xi+\varepsilon)-e^{\frac{u}{\varepsilon}}\varepsilon^{3}k_{s}^{\prime\prime}(v)=-\varepsilon^{2}s(s-1)(s-2)\int\limits_{\xi}^{u}e^{\frac{t}{\varepsilon}}(t+\varepsilon)^{s-3}dt\,.$

If $\xi\geqslant\varepsilon$ then due to (4.68), (4.66), and (4.65) the
question is reduced to verification that the function

$s\mapsto\frac{\int\limits_{\xi}^{u}e^{\frac{t}{\varepsilon}}(t+\varepsilon)^{s-3}dt}{\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon^{2}-\lambda^{2})(\lambda+\xi)^{s-3}d\lambda}\,$
(B.18)

is increasing. If $\xi\leqslant\varepsilon$ we refer to formula (4.67) and
check that the function

$s\mapsto\frac{s(s-1)\int\limits_{\xi}^{u}e^{\frac{t}{\varepsilon}}(t+\varepsilon)^{s-3}dt}{(\xi+\varepsilon)^{s-3}}$
(B.19)

increases. Since $s\mapsto s(s-1)a^{s}$ increases on $(1,\infty)$ for $a>1$,
the function (B.19) is increasing as well, because $t>\xi$. To prove that
(B.18) increases, we repeat the chain of arguments we already used for proving
that (B.12) is decreasing. Since $t>\xi$ and $\lambda<\varepsilon$, we have
$\frac{\lambda+\xi}{t+\varepsilon}<1$, and therefore, the function

$s\mapsto\Big{(}\frac{\lambda+\xi}{t+\varepsilon}\Big{)}^{s-3}$

is decreasing. After integrating this family of decreasing functions with a
positive weight, we get the following decreasing function:

$s\mapsto\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon^{2}-\lambda^{2})\Big{(}\frac{\lambda+\xi}{t+\varepsilon}\Big{)}^{s-3}d\lambda\,,$

or the following increasing function:

$s\mapsto\frac{(t+\varepsilon)^{s-3}}{\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon^{2}-\lambda^{2})(\lambda+\xi)^{s-3}d\lambda}\,.$
(B.20)

We integrate once more family of functions (B.20) with the positive weight
$e^{\frac{t}{\varepsilon}}$ and finally obtain the increasing function (B.18).

Now, we consider the first summand in (B.17). First, we simplify the
expression in the numerator. Using twice (2.15), we get

$\varepsilon^{2}k_{s}^{\prime\prime}(v)=\varepsilon\big{(}sv^{s-1}-k_{s}(v)\big{)}^{\prime}=\varepsilon
s(s-1)v^{s-2}-sv^{s-1}+k_{s}(v).$

Then we use definition of $w_{{}_{\mathrm{R}}}$ (see (4.50)) first for
$\xi\leqslant\varepsilon$:

$\displaystyle w_{{}_{\mathrm{R}}}(\xi;$ $\displaystyle
s,\varepsilon)-e^{\frac{\xi}{\varepsilon}}\varepsilon^{3}k_{s}^{\prime\prime}(\xi+\varepsilon)$
$\displaystyle=\varepsilon
e^{\frac{\xi}{\varepsilon}}\Big{(}k_{s}(\xi+\varepsilon)-2\xi(\xi+\varepsilon)^{s-2}\Big{)}-\varepsilon
e^{\frac{\xi}{\varepsilon}}\Big{(}\varepsilon
s(s-1)(\xi+\varepsilon)^{s-2}-s(\xi+\varepsilon)^{s-1}+k_{s}(\xi+\varepsilon)\Big{)}$
$\displaystyle=\varepsilon e^{\frac{\xi}{\varepsilon}}(s-2)(\xi-\varepsilon
s)(\xi+\varepsilon)^{s-2}\,,$

and the function

$\frac{w_{{}_{\mathrm{R}}}(\xi;s,\varepsilon)-e^{\frac{\xi}{\varepsilon}}\varepsilon^{3}k_{s}^{\prime\prime}(\xi+\varepsilon)}{w_{{}_{\mathrm{R}}}^{\prime}(\xi;s,\varepsilon)}\buildrel\eqref{ww_dif}\over{=}\frac{\varepsilon(\xi-\varepsilon
s)(\xi+\varepsilon)}{\xi^{2}+\varepsilon^{2}}$

decreases in $s$.

For $\xi\geqslant\varepsilon$ we have:

$\displaystyle w_{{}_{\mathrm{R}}}(\xi;$ $\displaystyle
s,\varepsilon)-e^{\frac{\xi}{\varepsilon}}\varepsilon^{3}k_{s}^{\prime\prime}(\xi+\varepsilon)$
$\displaystyle=e^{\frac{\xi}{\varepsilon}}\Big{(}\frac{(\xi-\varepsilon)^{s}-(\xi+\varepsilon)^{s}}{2}+\varepsilon
k_{s}(\xi+\varepsilon)\Big{)}-\varepsilon
e^{\frac{\xi}{\varepsilon}}\Big{(}\varepsilon
s(s-1)(\xi+\varepsilon)^{s-2}-s(\xi+\varepsilon)^{s-1}+k_{s}(\xi+\varepsilon)\Big{)}$
$\displaystyle=e^{\frac{\xi}{\varepsilon}}\Big{(}\frac{(\xi-\varepsilon)^{s}-(\xi+\varepsilon)^{s}}{2}+\varepsilon
s(\xi+\varepsilon)^{s-1}-\varepsilon^{2}s(s-1)(\xi+\varepsilon)^{s-2}\Big{)}$
$\displaystyle=-\frac{1}{4}e^{\frac{\xi}{\varepsilon}}s(s-1)(s-2)\\!\\!\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon+\mu)^{2}(\mu+\xi)^{s-3}d\mu\,.$

Therefore, we need to check that the function

$s\mapsto\frac{\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon+\mu)^{2}(\mu+\xi)^{s-3}d\mu}{\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon^{2}-\lambda^{2})(\lambda+\xi)^{s-3}d\lambda}$
(B.21)

is increasing. After differentiating this function, we get the following
expression in the numerator:

$\displaystyle\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon+\mu)^{2}$
$\displaystyle(\mu+\xi)^{s-3}\log(\mu+\xi)\,d\mu\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon^{2}-\lambda^{2})(\lambda+\xi)^{s-3}d\lambda$
$\displaystyle\qquad\qquad-\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon+\mu)^{2}(\mu+\xi)^{s-3}d\mu\int\limits_{-\varepsilon\
}^{\;\varepsilon}(\varepsilon^{2}-\lambda^{2})(\lambda+\xi)^{s-3}\log(\lambda+\xi)\,d\lambda$
$\displaystyle=\int\limits_{\\!-\varepsilon\ \
}^{\;\varepsilon}\\!\\!\\!\int\limits_{\\!-\varepsilon\
}^{\;\varepsilon}(\varepsilon+\mu)^{2}(\varepsilon^{2}-\lambda^{2})(\mu+\xi)^{s-3}(\lambda+\xi)^{s-3}\big{(}\log(\mu+\xi)-\log(\lambda+\xi)\big{)}\,d\mu\,d\lambda\,.$

Now, we symmetrize this expression (interchanging $\mu$ and $\lambda$):

$\displaystyle\frac{1}{2}\\!\\!\int\limits_{\\!-\varepsilon\ \
}^{\;\varepsilon}\\!\\!\\!\int\limits_{\\!-\varepsilon\ }^{\;\varepsilon}$
$\displaystyle\big{[}(\varepsilon+\mu)^{2}(\varepsilon^{2}-\lambda^{2})-(\varepsilon+\lambda)^{2}(\varepsilon^{2}-\mu^{2})\big{]}(\mu+\xi)^{s-3}(\lambda+\xi)^{s-3}\big{(}\log(\mu+\xi)-\log(\lambda+\xi)\big{)}\,d\mu\,d\lambda$
$\displaystyle=\varepsilon\\!\\!\int\limits_{\\!-\varepsilon\ \
}^{\;\varepsilon}\\!\\!\\!\int\limits_{\\!-\varepsilon\
}^{\;\varepsilon}(\mu-\lambda)(\varepsilon+\mu)(\varepsilon+\lambda)(\mu+\xi)^{s-3}(\lambda+\xi)^{s-3}\big{(}\log(\mu+\xi)-\log(\lambda+\xi)\big{)}\,d\mu\,d\lambda\,.$

Since $\log$ is an increasing function, we have

$(\mu-\lambda)\big{(}\log(\mu+\xi)-\log(\lambda+\xi)\big{)}>0\,.$

This means that the derivative we have calculated is positive and the function
(B.21) increases. This completes the proof of the lemma. $\square$

######  Lemma B.4.

$B_{x_{2}x_{2}}B_{x_{3}x_{3}}-B_{x_{2}x_{3}}^{2}\geqslant 0$ on
$\Xi_{{}_{\mathrm{ch}+}}$.

##### Proof.

Using (A.9), we take a half sum of expressions (A.15) and (A.24) to obtain:

$B_{x_{2}}=D(r)-D(p)B_{x_{3}}\,,$ (B.22)

where

$D(s)=D(s;a,b)=\frac{s}{2}\cdot\frac{b^{s-1}-a^{s-1}}{b-a}\,.$

Using this representation, we express the derivatives of $B$ with respect to
$x_{2}$ via derivatives with respect to $x_{3}$:

$\displaystyle B_{x_{2}x_{3}}$
$\displaystyle=D_{x_{3}}(r)-D_{x_{3}}(p)B_{x_{3}}-D(p)B_{x_{3}x_{3}};$
$\displaystyle B_{x_{2}x_{2}}$
$\displaystyle=D_{x_{2}}(r)-D_{x_{2}}(p)B_{x_{3}}-D(p)B_{x_{2}x_{3}}$
$\displaystyle=\big{[}D_{x_{2}}(r)-D(p)D_{x_{3}}(r)\big{]}-\big{[}D_{x_{2}}(p)-D(p)D_{x_{3}}(p)\big{]}B_{x_{3}}+D(p)^{2}B_{x_{3}x_{3}}\,.$

Using these formulas, after some simplifications, we rewrite the minor under
consideration in the following form:

$B_{x_{2}x_{2}}B_{x_{3}x_{3}}-B_{x_{2}x_{3}}^{2}=\big{[}E(r)-E(p)B_{x_{3}}\big{]}B_{x_{3}x_{3}}-\big{[}D_{x_{3}}(r)-D_{x_{3}}(p)B_{x_{3}}\big{]}^{2}\,,$
(B.23)

where $E(s)=D_{x_{2}}(s)+D(p)D_{x_{3}}(s)$.

Now, we plan to find some integral representations for each factor in (B.23),
from where it will be clear that this expression is positive. We already have
an integral representation for $B_{x_{3}x_{3}}$, this is formula (B.8). Let us
show that the factor $E(r)-E(p)B_{x_{3}}$ has almost the same representation,
the difference consists in the factor in front of the integral.

We start with an integral representation for $D$:

$D(s)=\frac{s(s-1)}{2(b-a)}\int\limits_{a}^{b}\lambda^{s-2}d\lambda\,.$ (B.24)

Whence

$\displaystyle D_{a}(s)$
$\displaystyle=\frac{s(s-1)(s-2)}{2(b-a)^{2}}\int\limits_{a}^{b}(b-\lambda)\lambda^{s-3}d\lambda\,;$
$\displaystyle D_{b}(s)$
$\displaystyle=\frac{s(s-1)(s-2)}{2(b-a)^{2}}\int\limits_{a}^{b}(\lambda-a)\lambda^{s-3}d\lambda\,,$

and

$D_{x_{i}}(s)=\frac{s(s-1)(s-2)}{2(b-a)^{2}}\int\limits_{a}^{b}\big{[}(b-\lambda)a_{x_{i}}+(\lambda-a)b_{x_{i}}\big{]}\lambda^{s-3}d\lambda\,.$
(B.25)

We use this expression to get a representation for $E$:

$\displaystyle E(s)$ $\displaystyle=D_{x_{2}}(s)+D(p)D_{x_{3}}(s)$ (B.26)
$\displaystyle=\frac{s(s-1)(s-2)}{2(b-a)^{2}}\int\limits_{a}^{b}\big{[}(b-\lambda)(a_{x_{2}}+D(p)a_{x_{3}})+(\lambda-a)(b_{x_{2}}+D(p)b_{x_{3}})\big{]}\lambda^{s-2}d\lambda\,.$

We have formula (B.6) for $a_{x_{3}}$ and $b_{x_{3}}$:

$(b-x_{1})a_{x_{3}}=(x_{1}-a)b_{x_{3}}=\frac{(b-a)^{2}}{A(p)}\,,$ (B.27)

and formulas (A.23) and (A.14) for $a_{x_{2}}$ and $b_{x_{2}}$:

$(b-x_{1})a_{x_{2}}=\frac{b^{p}-a^{p}-pb^{p-1}(b-a)}{A(p)}\qquad\text{and}\qquad(x_{1}-a)b_{x_{2}}=-\frac{b^{p}-a^{p}-pa^{p-1}(b-a)}{A(p)}\,.$

If we plug these expressions into the parts of (B.26), we get

$\displaystyle a_{x_{2}}+D(p)a_{x_{3}}$
$\displaystyle=\frac{1}{b-x_{1}}\Big{[}\frac{b^{p}-a^{p}-pb^{p-1}(b-a)}{A(p)}+\frac{p}{2}\cdot\frac{b^{p-1}-a^{p-1}}{b-a}\cdot\frac{(b-a)^{2}}{A(p)}\Big{]}$
$\displaystyle=\frac{1}{2A(p)(b-x_{1})}\big{[}2b^{p}-2a^{p}-p(b^{p-1}+a^{p-1})(b-a)\big{]}=-\frac{1}{2(b-x_{1})}\,,$

and similarly

$b_{x_{2}}+D(p)b_{x_{3}}=\frac{1}{2(x_{1}-a)}\,.$

As a result, we can rewrite (B.26) as follows

$\displaystyle E(s)$
$\displaystyle=\frac{s(s-1)(s-2)}{4(b-a)^{2}}\int\limits_{a}^{b}\Big{[}\frac{\lambda-a}{x_{1}-a}-\frac{b-\lambda}{b-x_{1}}\Big{]}\lambda^{s-3}d\lambda$
(B.28)
$\displaystyle=\frac{s(s-1)(s-2)}{4(b-a)(x_{1}-a)(b-x_{1})}\int\limits_{a}^{b}(\lambda-
x_{1})\lambda^{s-3}d\lambda=\frac{A(p)}{4(b-a)^{4}}A_{x_{3}}(s)\,.$

In the last equality we use representation (B.7).

Finally, we calculate the first factor in (B.23):

$\displaystyle E(r)-E(p)B_{x_{3}}$
$\displaystyle\buildrel\eqref{230603}\over{=}\frac{E(r)A(p)-E(p)A(r)}{A(p)}=\frac{1}{4(b-a)^{4}}\big{(}A_{x_{3}}(r)A(p)-A_{x_{3}}(p)A(r)\big{)}$
$\displaystyle=\frac{A(p)^{2}}{4(b-a)^{4}}\frac{\partial}{\partial
x_{3}}\Big{(}\frac{A(r)}{A(p)}\Big{)}\buildrel\eqref{230603}\over{=}\frac{A(p)^{2}}{4(b-a)^{4}}B_{x_{3}x_{3}}\,.$

Therefore,

$\displaystyle\big{[}E(r)$
$\displaystyle-E(p)B_{x_{3}}\big{]}B_{x_{3}x_{3}}=\Big{(}\frac{A(p)}{2(b-a)^{2}}B_{x_{3}x_{3}}\Big{)}^{2}$
(B.29)
$\displaystyle=\Bigg{(}\frac{r(r-1)(r-2)p(p-1)(p-2)(b-a)}{4(x_{1}-a)(b-x_{1})A(p)^{2}}\times$
$\displaystyle\qquad\times\int\limits_{a}^{b}\\!\\!\int\limits_{a}^{b}\big{[}\lambda\mu-
ab+(a+b-\lambda-\mu)x_{1}\big{]}(\lambda-\mu)(\lambda^{r-p}-\mu^{r-p})\lambda^{p-3}\mu^{p-3}d\lambda\,d\mu\Bigg{)}^{2}.$

Now, we find an integral representation for another term in (B.23):

$\displaystyle D_{x_{3}}$
$\displaystyle(r)-D_{x_{3}}(p)B_{x_{3}}=\frac{D_{x_{3}}(r)A(p)-D_{x_{3}}(p)A(r)}{A(p)}$
(B.30)
$\displaystyle\buildrel\eqref{270801},\eqref{030902}\over{=}\frac{r(r-1)(r-2)p(p-1)(p-2)(b-a)}{2A(p)(b-a)^{2}}\times$
$\displaystyle\quad\times\int\limits_{a}^{b}\\!\\!\int\limits_{a}^{b}\big{[}(\lambda-a)(\mu-a)b_{x_{3}}-(b-\lambda)(b-\mu)a_{x_{3}}\big{]}(\lambda-\mu)\lambda^{r-3}\mu^{p-3}d\lambda\,d\mu$
$\displaystyle\buildrel\eqref{030904}\over{=}\frac{r(r-1)(r-2)p(p-1)(p-2)}{4(x_{1}-a)(b-x_{1})A(p)^{2}}\times$
$\displaystyle\quad\times\int\limits_{a}^{b}\\!\\!\int\limits_{a}^{b}\big{[}(\lambda-a)(\mu-a)(b-x_{1})-(b-\lambda)(b-\mu)(x_{1}-a)\big{]}(\lambda-\mu)(\lambda^{r-p}-\mu^{r-p})\lambda^{p-3}\mu^{p-3}d\lambda\,d\mu\,,\hskip-20.0pt$

where we, as usual, used symmetrization to get the last formula in the chain.

It remains to compare the different parts of expression in (B.29) and (B.30).
We see that

$(b-a)\big{[}\lambda\mu-
ab+(a+b-\lambda-\mu)x_{1}\big{]}\geqslant\big{|}(\lambda-a)(\mu-a)(b-x_{1})-(b-\lambda)(b-\mu)(x_{1}-a)\big{|}$

for $\lambda,\mu,x_{1}\in[a,b]$. Indeed, on the left-hand side we have a
linear function in $x_{1}$ and we have a convex piecewise linear function on
the right-hand side. The values of both functions coincide at the endpoints of
the interval $a\leqslant x_{1}\leqslant b$. Therefore, inside the interval the
left function is strictly larger than the right one. Since the remaining terms
are the same and they do not change its sign inside the interval, we have the
required inequalities for the squares of integrals. $\square$

## References

  * [1] A. Burchard, G. Dafni, R. Gibara, _Mean oscillation bounds on rearrangements_ , https://arxiv.org/pdf/2011.09111.pdf.
  * [2] J. B. Garnett, _Bounded analytic functions_. Graduate Texts in Mathematics, 236. Springer, New York, 2007. 459 pp. ISBN: 978-0-387-33621-3; 0-387-33621-4.
  * [3] P. Ivanisvili, D. M. Stolyarov, V. I. Vasyunin, P. B. Zatitskiy, _Bellman function for extremal problems in $\mathrm{BMO}$ $\mathrm{II}$: evolution_, Memoirs of the AMS, 255:1220 (2018); https://doi.org/10.1090/memo/1220, preprint: http://arxiv.org/abs/1510.01010.
  * [4] A. Osȩkowski, _Sharp Martingale and Semimartingale Inequalities_ , Springer Basel, (2012), 464pp, ISBN: 978-3-0348-0369-4.
  * [5] L. Slavin, V. Vasyunin, _Sharp results in the integral form John–Nirenberg inequality_ , Trans. Amer. Math. Soc. 363: 8 (2011), 4135–4169.
  * [6] L. Slavin and V. Vasyunin, _Sharp $L^{p}$ estimates on $\mathrm{BMO}$_, Indiana Univ. Math. J. 61: 3 (2012), 1051–1110.
  * [7] L. Slavin, P. B. Zatitskiy, _Dimension-free estimates for semigroup $\mathrm{BMO}$ and $A_{p}$_, to appear in Indiana University Mathematics Journal, preprint: https://arxiv.org/abs/1908.02602.
  * [8] Dmitriy M. Stolyarov, Vasily I. Vasyunin, Pavel B. Zatitskiy, _Monotonic rearrangements of functions with small mean oscillation_ , Studia Mathematica 231 (2015), 257–267, https://doi.org/10.4064/sm8326-2-2016, preprint: https://arxiv.org/pdf/1506.00502.pdf.
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Vasily Vasyunin
Department of Mathematics and Computer Science,
St. Petersburg State University, 14-th Line Vasilyevsky Island, 29,
199178, St. Petersburg, Russia
<EMAIL_ADDRESS>

Pavel Zatitskiy
Department of Mathematics and Computer Science,
St. Petersburg State University, 14-th Line Vasilyevsky Island, 29,
199178, St. Petersburg, Russia
<EMAIL_ADDRESS>

Ilya Zlotnikov
Department of Mathematics and Computer Science,
St. Petersburg State University, 14-th Line Vasilyevsky Island, 29,
199178, St. Petersburg, Russia
<EMAIL_ADDRESS>


# Correlation effects in the electronic structure of the Ni-based
superconducting $KNi_{2}S_{2}$

Feng Lu1, Wei-Hua Wang2, Xinjian Xie3, and Fu-Chun Zhang1,4 1 Department of
Physics, and Center of Theoretical and Computational Physics, The University
of Hong Kong, Hong Kong, China
2 Department of Electronics and Tianjin Key laboratory of Photo-Electronic
Thin film Device and Technology, Nankai University, Tianjin, China
3 School of materials and engineering, Hebei University of Technology,Tianjin
300130, People s Republic of China 4 Department of Physics, Zhejiang
University, Hangzhou, China

(today)

###### Abstract

The LDA plus Gutzwiller variational method is used to investigate the ground-
state physical properties of the newly discovered superconducting
$KNi_{2}S_{2}$. Five Ni-3d Wannier-orbital basis are constructed by the
density-functional theory, to combine with local Coulomb interaction to
describe the normal state electronic structure of Ni-based superconductor. The
band structure and the mass enhanced are studied based on a multiorbital
Hubbard model by using Gutzwiller approximation method. Our results indicate
that the correlation effects lead to the mass enhancement of $KNi_{2}S_{2}$.
Different from the band structure calculated from the LDA results, there are
three energy bands across the Fermi level along the X-Z line due to the
existence of the correlation effects, which induces a very complicated Fermi
surface along the X-Z line. We have also investigated the variation of the
quasi-particle weight factor with the hole or electron doping and found that
the mass enhancement character has been maintained with the doping.

###### pacs:

74.20.-z, 74.20.Pq, 71.15.Mb

##  I Introduction

The discovery of the second class of high-transition temperature
superconductors in LaFeAsO1-xFxKamihara-1  has stimulated great interest in
the study of iron pnictides. Over the past five years, many iron-based
compounds were reported to show superconductivity after doping or under high
pressuresKamihara-1 ; Chen-G-F-1 ; Lei-H-1 ; C-1 ; and the superconducting
(SC) transition temperature exceeds 50K in the $RFeAsO_{1-x}F_{x}$($R$: Rare
earth atoms)Wug-1 . Recently, one family $AFe_{2}Se_{2}(A=K,Rb$, or $Tl)$
compounds with the $ThCr_{2}Si_{2}$-type crystal structure, such as
$K_{y}Fe_{2}Se_{2}$Guo-Jin-1  and $(Tl,K)_{y}Fe_{2-x}Se_{2}$ Fang-Wang-1 ;
Wang-Dong-1 , have been found to become superconductors at about 30 K.
Different from other crystal structures of the iron-based superconductor,
these compounds $AFe_{2}Se_{2}(A=K,Rb$, or $Tl)$ have many unique physical
properties, such as absence of hole-like Fermi surface (FS)in
$K_{0.8}Fe_{1.7}Se_{2}$Qian-Wang-1  and the intrinsic $\sqrt{5}\times\sqrt{5}$
Fe vacancies superstructure in $K_{0.8}Fe_{1.6}Se_{2}$ Wei-Bao-1 ; Wei-Bao-2 ;
Wei-Bao-3 . Thus, searching for superconductors with the $ThCr_{2}Si_{2}$-type
crystal structure has become one of the interesting topics in condensed matter
physicsLee-P-A-1 ; C-2 .

Figure 1:  (Color online) $KNi_{2}S_{2}$ with the $ThCr_{2}Si_{2}$-type
structure.

Very recently, a novel group of superconductors with $ThCr_{2}Si_{2}$-type
crystal structure, $KNi_{2}A_{2}(A=S,Se)$, have been discoveredNeilson-1 ;
Neilson-2 . The experiments show that the Ni-based superconductor exhibits a
rich and unusual electronic and structural phase diagramNeilson-1 ; Neilson-2
. For example, based on high-resolution synchrotron X-ray diffraction and
time-of-flight neutron scattering experiments, James et al. have found that
the mixed valence compound $KNi_{2}S_{2}$ displays a number of highly unusual
structural transitions and presents the charge density wave
fluctuationsNeilson-2 . The specific heat measurement reveals that
$KNi_{2}S_{2}$ exhibits the correlated behavior at low-temperature with an
enhanced effective electron mass, $m^{*}=11m_{b}$ to $24m_{b}$. Below
$T_{c}=0.46K$, $KNi_{2}S_{2}$ becomes superconducting without the partial
substitution of K or Ni deficientNeilson-2 . As discussed above, the Ni-based
superconductor $KNi_{2}S_{2}$ has a similar crystal structure as the iron-
based superconductor $AFe_{2}S_{2}$ compound. However, there are also many
differences between these two kinds of compounds. For example, the localized
magnetism is absent in the Ni-based superconductor $KNi_{2}S_{2}$Neilson-2 .
Thus, it is critical to understand the electronic structures of the parent
$ANi_{2}S_{2}$ compounds at first in order to investigate the mechanism of
superconductivity in this material.

In this paper, we have performed the first-principle electronic structure
calculations on $KNi_{2}S_{2}$. To overcome the problem of dealing with
Coulomb interaction in the local density approximation (LDA) calculation, we
use LDA plus the Gutzwiller variational approach to investigate this typical
correlated material. This method has proved to be effective in dealing with
Coulomb interaction and has been widely applied to strongly correlated
systemsDeng_Wang_1 ; Wang_Qian-2 . The calculation shows that $KNi_{2}S_{2}$
is a kind of multi-orbital system with five Ni-3d orbital near the Fermi
level. Our results indicate that the correlation effects of the 3d local
electrons lead to mass enhancement of $KNi_{2}S_{2}$. This behavior is always
present with the hole or electron doping, which is well agreement with the
experimental resultsNeilson-2 . The correlation effect is not only presented
by the narrowing of the bandwidth, but also by the change of the FS near the
high symmetry line. In the LDA calculation, there are two bands across the X-Z
line. Actually, three electron bands across the Fermi level along the X-Z
line, which induce a very complicated FS structure along the X-Z line. The
relationship among the quasi-particle weight factors of the different orbital
is also discussed in this paper.

##  II Model and method

The first-principles calculations were performed within LDA using the plane
wave basis methodP-Giannozzi-1 . The generalized gradient approximation (GGA)
was adopted for the exchange-correlation potential of the density function
theory (DFT)Perdew-1 . We used the ultrasoft pseudopotentials to model the
electron-ion interactionsD-Vanderbilt-1 . The kinetic energy cutoff and the
charge density cutoff were taken as 4800 and 6400 eV, respectively. A mesh of
$18\times 18\times 9$ k points were used to perform the Brillouin zone
integration. $KNi_{2}S_{2}$ has $ThCr_{2}Si_{2}$-type structure (space group
I4$/$mmm, ($\\#139$, Z=1)) and each unit cell contains two formula units, as
shown in Fig.1. In our calculation, the structure parameters optimized by the
energy minimization, a=3.7792Å, c=12.7139Å, and $z(S)$=0.35, were adopted,
which are in good agreement with the experimental values a= 3.80899Å,
c=12.67609Å, and $z(S)$=0.34642 determined by synchrotron X-ray powder
diffractionNeilson-2 . Since the LDA can not describe the Coulomb interaction
adequately, in order to take into account more explicitly the correlated
effects of the d electrons, we have performed Gutzwiller variational approach
to deal with the Coulomb interaction.

Figure 2: (Color online) (a) Energy band structure of $KNi_{2}S_{2}$ by
LDA(black line) and Wannier function(dotted line) ; (b) The Fermi surface of
$KNi_{2}S_{2}$ by LDA calculation.

##  III Results and discussions

We first present the electronic structure of $KNi_{2}S_{2}$ by LDA
calculation. The corresponding band structures and FS are given in Fig. 2a and
Fig. 2b, respectively. Similar to the iron-based superconductorMazin-I-I-1 ;
Singh-D-1 ; Boeri-L-1 ; C-Cao-1 ; F-Ma-1 ; lufeng-1 , there are several bands
across the Fermi level, which indicates that the $KNi_{2}S_{2}$ is a kind of
multi-orbital superconducting system. The multi-orbital character can also be
seen from the complicated FS sheets. As shown in Fig.2 b, there is a large
hole-like FS around the $\Gamma$ point, two 2D-cylinder-like ones around the
corners of the Brillouin zone and a very complicated pocket at the middle of
the four boundaries of the Brillouin zone. As mentioned above, the structure
of the FS is very complicated due to the multi-orbital characters, which
implies that the $KNi_{2}S_{2}$ should be a kind of multiple-gap
superconductors similar to the iron based superconductor, in contrast to the
single-orbital copper-based superconductors.

Figure 3:  (Color online) (a) DOS and PDOS of $KNi_{2}S_{2}$ . (b) PDOS of
$KNi_{2}S_{2}$ for the five Fe-3d orbitals.

To investigate the orbital characters near the FS, the corresponding density
of states (DOS) and the projecting the density of states (PDOS) of
$KNi_{2}S_{2}$ are plotted in Fig.3. The basic shape of the DOS and PDOS is
very similar to that of $AFe_{2}Se_{2}$Yan_Miao_3 . However, the position of
the Fermi energy is higher due to the more electron count. As shown by the DOS
in Fig.3, the total DOS near the $E_{f}$ is manly contributed by the Ni-3d
orbitals, which mix with the S-3p orbitals near the $E_{f}$. In detail, the
PDOS of the Ni-3d orbitals consist of the DOS from -3 eV to 1 eV around the
$E_{f}$, and at the same time there is some contribution from S-3p orbitals.
Evidently, the contribution of S-3p orbitals is smaller than that from the
Ni-3d orbitals. The value of DOS at the Fermi energy N($E_{f}$) is about 6.9
states per eV per formula unit both spins.

Figure 4: (Color online) The quasi-particle weight factor Z as a function of
Coulomb interaction U at J=0.8eV. (a) The quasi-particle weight factor Z for
$d_{xy}$ orbital; (b) The quasi-particle weight factor Z for $d_{yz}$ orbital;
(c) The quasi-particle weight factor Z for $d_{xz}$ orbital; (d) The quasi-
particle weight factor Z for $d_{x^{2}-y^{2}}$ orbital; (e) The quasi-particle
weight factor Z for $d_{z^{2}}$ orbital; (f) The quasi-particle weight factor
Z as a function of doping $x$ for $K_{x}Ni_{2}S_{2}$ at U=8 eV and J=0.8eV.

As we know, both LDA and GGA fail to depict the strongly correlated electron
systems due to the insufficient treatment of electron correlation. In the last
20 years, a lot of new methods and new computational tools have been developed
to improve the calculation of strongly correlated electron materials, such as
the LDA+UAnisimov_Zaanen-1 , LDA+DMFT(dynamical mean-field theory)
Kotliar_Savrasov-1  and LDA+G (Gutzwiller method)Deng_Wang_1 . Among all these
methods, LDA+G is thought as one that is practically efficient and can capture
the key feature of the correlation effect as well in the ground-state studies.
The Gutzwiller variational approachGutzwiller-1  was firstly introduced by
Gutzwiller and has been proved to be quite efficient and accurate for the
ground-state physical properties studies, such as the the Mott transition,
ferromagnetism, and superconductivityDeng_Wang_1 ; Gutzwiller-1 ; Vollhardt-1
; Brinkman-1 ; Zhan_Gros_1 . In our calculation, we also used the LDA+G
framework to treat the correlation effects between the localized d electrons.
In other words, we firstly construct the multi-orbital Hubbard model and then
solve this model by the Gutzwiller variational approach (GW). Firstly, the
Wannier function is constructed by the maximally localized Wannier functions
(MLWFs) approachNicola_Arash-1 , as shown by the dotted line in Fig.2 a.
Secondly, combining with the Coulomb interaction, the multi-orbital model can
effectively describe the multi-orbital characters of $KNi_{2}S_{2}$. Then the
GW method is used to deal with this many-body model and the physical
properties in this system Deng_Wang_1 ; Gutzwiller-1 ; Vollhardt-1 ;
Brinkman-1 ; Zhan_Gros_1 .

We start from this five-orbital Hubbard model,

$\displaystyle H_{Tot}$ $\displaystyle=$ $\displaystyle H_{wannier}+H_{int},$
(1)

where $H_{wannier}$ describes the on-site energies and hopping integrals of
electrons between the Ni-3d Wannier orbitals; and $H_{int}$ is the local
interaction term for the Coulomb interaction between the d electron described
by the parameters $U$, $U^{\prime}$, $J$. The parameters $U$, $U^{\prime}$,
and $J$ denote the intra-orbital Coulomb interaction, inter-orbital Coulomb
interaction and Hund’s coupling. In what follows, considering the realistic
wavefunctions of 3d-orbitals Castellani  and the spin rotational symmetry, we
adopt the relationships $U=U^{\prime}+2J$.

The true ground state of $H_{Tot}$ is described by the Gutzwiller wave
function $|\Psi_{G}>$, which is constructed by a many-particle projection
operator $\hat{P}_{G}$ on the uncorrelated wave function $|\Psi_{0}>$

$\displaystyle|\Psi>$ $\displaystyle=$
$\displaystyle\hat{P}_{G}|\Psi_{0}>=\prod_{i}\hat{P}_{i}|\Psi_{0}>$ (2)

with Gutzwiller projector operator

$\displaystyle\hat{P}_{i}$ $\displaystyle=$
$\displaystyle\sum_{\Gamma}\lambda_{\Gamma}|\Gamma><\Gamma|$ (3)

where $\lambda_{\Gamma}$ is the variational parameter for the i-th site with
the atomic eigenstates $|\Gamma>$, which is determined by minimizing the
ground state total energy. And the renormalization of kinetic energy can be
described by the bandwidth renormalization or quasi-particle weight factor Z,
as defined in Ref. 17.

Generally speaking, the essence of the correlation effects under Gutzwiller
approach is to suppress the double occupancy and to renormalize the kinetic
energy. Specifically, the Coulomb interaction U will enhance the orbital
polarization; however, the interorbital Hund s coupling J favors even
distribution of electrons among five orbitals. To understand the role of
Coumlomb interaction and the change of the band structure in the presence of
interactions, the value of the quasi-particle weight factor Z of the kinetic
energy as a function of U at fixed J=0.8 eV is shown in Fig. 4. As shown by
the solid line in Figs. 4(a), the Z factor decreases with the increase of the
Coulomb interaction U, due to the suppression of the double occupancy by U.
Because of the valence state of $Ni^{1.5+}$, each five Ni-3d orbitals have 8.5
electrons and are non-integer filling. Thus, the metal insulator transition
does not occur in this system and $KNi_{2}S_{2}$ is a multi-orbital metal
system, which is agreement with the experimentNeilson-2 . In the iron-based or
the copper-based superconductors, different doping with holes ($p$ type) or
electrons ($n$ type) has important influence on the superconducting transition
temperature, which lead to a rich phase diagram of these high temperature
superconductors. For example, superconductivity only occurs in a doping range,
which indicates that electron doping or hole doping may affect the electronic
properties in these materials. To study the doping effect on the physical
properties of $KNi_{2}S_{2}$ , we investigate the quasi-particle weight factor
Z at different doping situations, as shown in Fig. 4(a-e). The dotted line
shows Z at the $20\%$ hole doping, the black solid line shows Z without doping
and the dashed line shows Z at the $20\%$ electron doping, respectively. When
the Coulomb interaction U is not too large, the electron or hole doping almost
do not affect the quasi-particle weight factor Z. But, the difference of the
quasi-particle weight factor Z with different doping become more and more
obvious with the increase of the U in the large range; and the value of Z is
also dependent on the orbital degree of freedom due to the orbital anisotropy.
For example, the doping effect is very weak for the $d_{xy}$ orbital but
important for the $d_{z^{2}}$ orbital in the large U range, as shown in Fig.
4(a-e). At fixed Coulomb interaction value U=7 eV,J=0.8 eV, the variation of
the quasi-particle weight factor Z with the different doping is also
investigated and plotted in Fig. 4f. The results show that the correlated
electron character of the $KNi_{2}S_{2}$ has been maintained with the
electron-like or hole-like doping for all the Ni 3-$d$ orbitals, which is well
agreement with the experimental measurementslei-1 .

Figure 5:  (Color online) Electronic band structure of $KNi_{2}S_{2}$
calculated by the Gutzwiller method.

Figure 6:  (Color online) DOS and PDOS of $KNi_{2}S_{2}$ calculated by the
Gutzwiller method.

The corresponding band structures and PDOS of the Ni-based superconducting
compound $KNi_{2}S_{2}$ calculated by the GW method are presented in Fig. 5
and Fig. 6. Since the electron occupation number is 8.5 for each Ni-3d
orbitals and the five Ni-3d orbitals are almost occupied by the electrons, the
added Coulomb interactions can not make $KNi_{2}S_{2}$ become a Mott insulator
state, which is agreement with the experimentNeilson-2 . At a typical value of
Coulomb interaction U=8 eV and J=0.8 eV Lu_Zhao-1 ; Stewart_Liu-1 ; Han_Wang-2
, the quasi-particle weight factor is about 0.5. In other words, the overall
bandwidth calculated by the GW method is smaller than that by the LDA
calculation due to the strong correlation effect among the Ni-3d orbitals
electrons. In the LDA results, the band structure near Fermi level along the
X-Z line is very simple, which causes a two FS structure along the X-Z line.
Actually, our GW calculation shows that there are three Bloch bands across the
Fermi level along the X-Z line due to the existence of the correlation effect,
which would induce a very complicated FS along the X-Z line. The corresponding
electron FS will dominate the low energy properties along the X-Z line, which
is easier to be observed by the Angle-resolved photoemission spectroscopy
experiments (ARPES). As shown above, comparing with the LDA calculation, the
GW calculation can improve the correct band narrowing and mass
renormalization, which can also be seen from the DOS calculated by the LDA+G
method, as shown in the Fig.6. The corresponding orbital character is also
plotted in Fig.6. Comparing the orbital character of Ni-based superconductor
with that of the iron based superconductorWang_Qian-2 , we can clearly find
that all the Ni-3d orbitals are renormalized by the correlation effect. We
obtain the value of DOS at the Fermi energy N($E_{f}$) is about 11.5 states
per eV per formula unit both spins, which corresponds to a bare specific heat
coefficient, $\gamma=27$ mJ mol f.u.-1 K-2 based on the formula
$\gamma=\pi^{2}k^{2}N(E_{f})/3$Singh_2 ; Beck_Claus_1 . This $\gamma$ value is
2.5 times smaller than the experimental value $\gamma=68$ mJ mol f.u.-1
K-2Neilson-2 , but it is better than that of the LDA calculation,
$\gamma=16.2$ mJ mol f.u.-1.

##  IV Conclusions

In conclusion, we have presented the results of the electronic structure of
$KNi_{2}S_{2}$ based on LDA plus Gutzwiller variational calculations. The
results show that the ground state of $KNi_{2}S_{2}$ is a strongly correlated
metal and the energy bands near Fermi level are dominated by the Ni-3d
electron states. The multi-orbital correlation effect between the Ni-3d
electrons determines the main physical properties of newly discovered Ni-based
superconductor $KNi_{2}S_{2}$, such as the mass enhancement behavior and the
multi-orbital characters. Different from the LDA understanding, the band
structure has a lot of changes under correlation effect, especially for the
band structure along the X-Z high symmetry line. In particular, the bands are
narrowed by a factor of about 2, and the five Ni-3d orbital are all almost
filled, which determine the low energy physical properties of $KNi_{2}S_{2}$.
The calculation about the electronic structure provides a reasonable starting
point for the investigations of the Ni-based superconductor. The variation of
the quasi-particle weight factor with the doping is also investigated and
compared to the experiments. Our results confirm that $KNi_{2}S_{2}$ is a
multi-orbital strongly correlated system with a reduced renormalized bandwidth
by the correlation effect.

###### Acknowledgements.

The authors thank X. Dai for helpful discussions. FCZ thanks the support of
RGC HKU707211 and AOE/P-04/08.

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