S_{-\xi}(t) f_-
\\
S_{+\japanese{\xi}}(t) g_+
\end{pmatrix},
\\
\mathbf N(\mathbf u) = \begin{pmatrix}
i \phi \psi_{-}
\\
i\phi \psi_{+}
\\
i \japanese{D_x}^{-1} \re \left( \overline{\psi_+} \psi_- \right)
\end{pmatrix},
\quad \mathbf L(\mathbf u) = \begin{pmatrix}
- iM\psi_{-} - M_{\mathfrak{K}_1}\psi_{+}
\\
- iM\psi_{+} - M_{\mathfrak{K}_1}\psi_{-}
\\
\end{pmatrix},
\quad
\mathbf M(\mathbf u) = \begin{pmatrix}
i \psi_{-} \mathfrak{K}_1
\\
i \psi_{+} \mathfrak{K}_1
\\
(i/2) \japanese{D_x}^{-1} \phi \mathfrak{K}_2
\end{pmatrix},
\end{gather*}

where $\phi$ stands for $\phi_+ + \overline{\phi_+}$. Let $s,r \in \R$ and $0 < b < 1/2$ be as in Lemma <ref>, and set $b'=-b$. Corresponding to $\mathbf s=(s,s,r)$ we then define the spaces $\mathbf H^{(s,s,r)}$ and $\mathbf X^{(s,s,r),b}$ as in (<ref>) and (<ref>). Define the Wiener process $W(t)$ using $K=L^2(\R,\R)$ with an orthonormal basis $\{e_j\}_{j \in \N}$. Assume that the convolution kernels satisfy $\mathfrak k_1 \in H^{\Abs{s}}(\R,\R)$ and $\mathfrak k_2 \in H^{\max(0,r-1)}(\R,\R)$.

The boundedness (<ref>) of $\mathbf M$ is now a consequence of Lemma <ref>. Concerning $\mathbf N$, the property (<ref>) is obvious, (<ref>), (<ref>) follow from Lemma <ref>, and (<ref>) holds by Lemma <ref>, if we take $\mathbf s' = (s',s',r')$ with $s'=r' > \max (r,s,1/2)$. For $\mathbf L$, the bound in (<ref>) holds by Lemma <ref>, and the bound in (<ref>) is trivial.

So with the above set-up, Theorem <ref>, with the exception of the charge conservation (considered below), follows from Theorems <ref> and <ref>.

§ CHARGE CONSERVATION

Let $s=0$ and $0 < r < 1/2$. We now prove the statement in Theorem <ref> about charge conservation of the local solution $(\psi_+,\psi_-,\phi)$, almost surely for $0 \le t < \tau$. Let $R > 0$. As explained in Section <ref>, the solution equals, up to the conditional stopping time $\tau_R$, the solution $(\psi_+^R,\psi_-^R,\phi^R)$ of the $R$-truncated problem, obtained in Theorem <ref> via the set-up in Section <ref>. Since $\tau_R \to \tau$ as $R \to \infty$, it clearly suffices to prove the charge conservation for $(\psi_+^R,\psi_-^R,\phi^R)$.

In the remainder of this section we fix $R > 0$, and to simplify the notation we drop the superscript $R$ on the solution. Thus, $(\psi_+,\psi_-,\phi)$ denotes the global solution of the truncated versions of (<ref>), (<ref>):

\begin{multline}
\label{mild_psi-truncated}
\psi_{\pm}(t)
S_{\pm\xi}(t) f_\pm
iM \int_0^t S_{\pm\xi}(t-s) \psi_{\mp}(s) \, ds
i \int_0^t  S_{\pm\xi}(t-s) ( \Theta \phi \Theta \psi_{\mp})(s) \, ds
\\
i \int_0^t S_{\pm\xi}(t-s) \psi_\mp(s) \mathfrak K_1 \, d W(s)
M_{\mathfrak K_1} \int_0^t S_{\pm\xi}(t-s) \psi_\pm(s) \, ds,
\end{multline}

\begin{multline}
\label{mild_phi-truncated}
\phi_+(t)
S_{+\japanese{\xi}}(t) g_+
i \int_0^t S_{+\japanese{\xi}}(t-s) \japanese{D_x}^{-1}
\re \left( \Theta\overline{\psi_+} \Theta\psi_- \right)(s) \, ds
\\
\frac{i}{2} \int_0^t S_{+\japanese{\xi}}(t-s) \japanese{D_x}^{-1}
\phi(s) \mathfrak K_2 \, dW(s)
\end{multline}

Here $\phi=2\re\phi_+$ and $\Theta(t)$ is defined as in (<ref>), with $s=0$. We assume that $\theta$ is even, so that $P_\mu f$ is real-valued if $f$ is.

Set $\psi=(\psi_+,\psi_-)$ and $\psi_0=(f_+,f_-)$. We will prove that the charge is almost surely conserved:

\[
\norm{\psi(t)}_{L^2}^2
\norm{\psi_0}_{L^2}^2 \quad \text{for $t \ge 0$.}
\]

To this end, we want to apply Itô's formula with the functional $\mathcal H \colon L^2(\R) \to \R$ given by

\[
\mathcal H(u) = \int_{\R} \Abs{u(x)}^2 \, dx.
\]

However, as discussed in Section <ref>, it is necessary to first regularise the problem. So for $\mu \ge 1$ we consider the solution $(\psi_+^\mu,\psi_-^\mu,\phi^\mu)$, obtained in Theorem <ref>, of the frequency-truncated equations

\begin{multline}
\label{mild_psi-truncated-regularised}
\psi_{\pm}^\mu(t)
S_{\pm\xi}(t) P_\mu f_\pm
iM \int_0^t S_{\pm\xi}(t-s) P_\mu^2 \psi_{\mp}^\mu(s) \, ds
i \int_0^t  S_{\pm\xi}(t-s) P_\mu \left( \Theta P_\mu \phi^\mu \cdot \Theta P_\mu \psi_{\mp}^\mu \right)(s) \, ds
\\
i \int_0^t S_{\pm\xi}(t-s) P_\mu \left(P_\mu \psi_\mp^\mu(s)\right) \mathfrak K_1 \, d W(s)
M_{\mathfrak K_1} \int_0^t S_{\pm\xi}(t-s) P_\mu^2 \psi_\pm^\mu(s) \, ds,
\end{multline}

\begin{multline}
\label{mild_phi-truncated-regularise}
\phi_+^\mu(t)
S_{+\japanese{\xi}}(t) P_\mu g_+
i \int_0^t S_{+\japanese{\xi}}(t-s) \japanese{D_x}^{-1}
P_\mu \re \left( \Theta P_\mu\overline{\psi_+^\mu} \cdot \Theta P_\mu \psi_-^\mu \right)(s) \, ds
\\
\frac i2 \int_0^t S_{+\japanese{\xi}}(t-s)  \japanese{D_x}^{-1}
P_\mu (P_\mu\phi^\mu)(s) \mathfrak K_2 \, dW(s),
\end{multline}

where $\phi^\mu=2\re \phi_+^\mu$. Set $\psi^\mu= \left(\psi_+^\mu,\psi_-^\mu \right)$.
Then by Theorem <ref> we have

\begin{equation}\label{DKG-convergence}
\mathbb E \left( \sup_{t \in [0,T]} \left( \norm{\psi^\mu(t) - \psi(t)}_{L^2}^2
+ \norm{\phi^\mu(t) - \phi(t)}_{H^r}^2
\right) \right) \to 0 \quad \text{as $\mu \to \infty$}
\end{equation}

for any $T > 0$. Moreover, the spatial Fourier transform of $\left(\psi_+^\mu,\psi_-^\mu,\phi^\mu \right)$ is supported in $[-2\mu,2\mu]$.

Notice that
\(
\mathcal H(\psi^\mu)
\mathcal H(\psi_+^\mu)
\mathcal H(\psi_-^\mu)
\)
and that the first and second derivatives of the functional $\mathcal H$ are given by the linear form
\(
\mathcal H'(u)
2 \re \langle \cdot, u \rangle _{L^2}
\)
and the bilinear form
\(
\mathcal H''(u)
2 \re \langle \cdot , \cdot \rangle _{L^2}
\)

In terms of $X_\pm(t) = S_{\pm\xi}(-t) \psi_\pm^\mu(t)$ we can rewrite (<ref>) as

\[
X_\pm(t) = X_\pm(0) + \int_0^t \Psi(s) \, ds + \int_0^t \Phi(s) \, dW(s),
\]

\[
\Psi(t) = -iM S_{\pm\xi}(-t) P_\mu^2 \psi_{\mp}^\mu(t)
i S_{\pm\xi}(-t) P_\mu ( \Theta P_\mu \phi^\mu \cdot \Theta P_\mu \psi_{\mp}^\mu)(t)
M_{\mathfrak K_1} S_{\pm\xi}(-t) P_\mu^2 \psi_\pm^\mu(t)
\]

\[
\Phi(t) = i S_{\pm\xi}(-t) P_\mu \left(P_\mu \psi_\mp^\mu(t)\right) \mathfrak K_1.
\]

Applying now Itô's formula, as stated in <cit.>, we get

\begin{multline*}
\mathcal H(X_\pm(t)) - \mathcal H(X_\pm(0))
\\
\int_0^t \mathcal H'(X_\pm(s)) \Psi(s) \, ds
\int_0^t \mathcal H'(X_\pm(s)) \Phi(s) \, dW(s)
+ \int_0^t \frac12 \trace \mathcal H''(X_\pm(s))\left(\Phi(s),\Phi(s)\right) \, ds.
\end{multline*}

Using the fact that the group $S_{\pm\xi}(t)$ is unitary on $L^2(\R)$, the above works out to be

\begin{multline*}
\mathcal H(\psi_\pm^\mu(t))
\mathcal H(\psi_\pm^\mu(0))
\\
\int_0^t 2\re \Innerprod{-iM P_\mu^2 \psi_{\mp}^\mu(s)
i P_\mu ( \Theta P_\mu \phi^\mu \cdot \Theta P_\mu \psi_{\mp}^\mu)(s)
M_{\mathfrak K_1} P_\mu^2 \psi_\pm^\mu(s)}
{\psi_\pm^\mu(s)}_{L^2} \, ds
\\
\int_0^t 2\re \Innerprod{i P_\mu \left(P_\mu \psi_\mp^\mu(s)\right) \mathfrak K_1 \cdot}
{\psi_\pm^\mu(s)}_{L^2} \, dW(s)
\\
+ \int_0^t
\sum _{j = 1}^{\infty}
\re
\left\langle
i P_{\mu} ( P_{\mu} \psi_\mp^\mu(s) ) \mathfrak K_1 e_j
i P_{\mu} ( P_{\mu} \psi_\mp^\mu(s) ) \mathfrak K_1 e_j
\right\rangle
\\
I_\pm + II_\pm + III_\pm.
\end{multline*}

Since $M, M_{\mathfrak K_1} \in \R$, $P_\mu \phi^\mu$ and $\Theta$ are real-valued, and $P_\mu$ is hermitian, it is clear that

\[
I_+ + I_-
M_{\mathfrak K_1} 2\re  \int_0^t \left(\Innerprod{ P_\mu \psi_+^\mu(s)}
\Innerprod{ P_\mu\psi_-^\mu(s)}
{P_\mu\psi_-^\mu(s)}_{L^2} \right)  \, ds.
\]

\[
I_+ + I_-
2 M_{\mathfrak K_1} \int_0^t
\norm{ P_{\mu} \psi^\mu(s) }_{L^2}^2
- \int_0^t
\norm{ \left( P_{\mu} \psi^\mu(s) \right) \mathfrak K_1}_{\mathcal L_2}^2
\]

where $\mathcal L_2 = \mathcal L_2(L^2,L^2)$ and the last equality holds by Remark <ref>, since $2M_{\mathfrak K_1} = \norm{\mathfrak k_1}_{L^2}^2$.
Next we notice that

\[
\sum _{j = 1}^{\infty}
\int_0^t
\im
\left\langle
( P_{\mu} \psi_\mp^\mu(s) ) \mathfrak K_1 e_j
P_{\mu} \psi_\pm^\mu(s)
\right\rangle
\, dB_j(s)
\]

\(
II_+ + II_- = 0
\)
since $\mathfrak K_1 e_j$ is a real-valued function.
Finally we notice that

\[
\int_0^t
\norm
P_{\mu} ( P_{\mu} \psi_\mp^\mu(s) ) \mathfrak K_1
_{\mathcal L_2}^2
\]

\[
III_+ + III_-
\int_0^t
\norm
P_{\mu} ( P_{\mu} \psi^\mu(s) ) \mathfrak K_1
_{\mathcal L_2}^2
\]

where the operator inside the norm is regarded as a composition
of three operators (first apply $\mathfrak K_1$, then multiplication by $P_{\mu} \psi^\mu$ and finally $P_\mu$).

Summing the contributions, we arrive at

\begin{equation}
\label{charge_control}
\mathcal H(\psi^\mu(t))
\mathcal H(\psi^\mu(0))
\int_0^t
\left(
\norm
P_{\mu} ( P_{\mu} \psi^\mu(s) ) \mathfrak K_1
_{\mathcal L_2}^2
\norm{ \left( P_{\mu} \psi^\mu(s) \right) \mathfrak K_1}_{\mathcal L_2}^2
\right)
\, ds,
\end{equation}

and letting $\mu \to \infty$ we get the charge conservation for $\psi$,

\[
\mathcal H(\psi(t))
\mathcal H(\psi(0))
0 \quad \text{for all $t \ge 0$}.
\]

Indeed, let $T > 0$. By (<ref>) we know that for some sequence $\mu_k \to \infty$ as $k \to \infty$, we have, almost surely,

\[
\sup_{t \in [0,T]} \norm{\psi^{\mu_k}(t) - \psi(t)}_{L^2} \to 0
\quad \text{as $k \to \infty$}.
\]

Thus, along $\mu=\mu_k$, the left hand side of (<ref>) converges to $\mathcal H(\psi(t)) - \mathcal H(\psi(0))$ for $0 \le t \le T$. Moreover, the right hand side converges to zero by the dominated convergence theorem. Indeed,
for $\mu$ large enough we have the bounds, uniformly in $s \in [0,T]$,

\[
\norm
P_{\mu} ( P_{\mu} \psi^\mu(s) ) \mathfrak K_1
_{\mathcal L_2}
\le
\norm
( P_{\mu} \psi^\mu(s) ) \mathfrak K_1
_{\mathcal L_2}
\le
\norm{\psi^\mu(s)}_{L^2} \norm{\mathfrak k_1}_{L^2}
\le
\left( \norm{\psi(s)}_{L^2}
+ 1 \right)
\norm{\mathfrak k_1}_{L^2},
\]

\[
0 \le
\norm{ (P_{\mu} \psi^\mu(s) ) \mathfrak K_1}_{\mathcal L_2}
\norm
P_{\mu} ( P_{\mu} \psi^\mu(s) ) \mathfrak K_1
_{\mathcal L_2}
\\
\le
C \left( \norm{\psi(s)}_{L^2} + 1 \right)
\norm{ (1-P_\mu) ( P_{\mu} \psi^\mu(s) ) \mathfrak K_1}_{\mathcal L_2}
\]

so it only remains to check that $\norm{ (1-P_\mu) ( P_{\mu} \psi^\mu(s) ) \mathfrak K_1}_{\mathcal L_2}$ tends to zero along $\mu=\mu_k$ as $k \to \infty$. But

\begin{multline*}
\norm{ (1-P_\mu) ( P_{\mu} \psi^\mu(s) ) \mathfrak K_1}_{\mathcal L_2}
\le
\norm{ (1-P_\mu) \psi(s) \mathfrak K_1}_{\mathcal L_2}
\norm{ (1-P_\mu) ( P_{\mu} \psi^\mu(s) - \psi(s) ) \mathfrak K_1}_{\mathcal L_2}
\\
\le
\norm{ (1-P_\mu) \psi(s) \mathfrak K_1}_{\mathcal L_2}
\norm{ (P_{\mu} \psi^\mu(s) -\psi(s) ) \mathfrak K_1}_{\mathcal L_2}
\end{multline*}

since $\norm{1 - P_{\mu}} \leq 1$.
Now the first term $\norm{ (1-P_\mu) \psi(s) \mathfrak K_1}_{\mathcal L_2} \to 0$ by (<ref>),
and the last term equals

\[
\norm{ P_{\mu} \psi^\mu(s) -\psi(s) }_{L^2}
\norm{\mathfrak k_1}_{L^2}
\le
\norm{ \psi^\mu(s) -\psi(s) }_{L^2}
\norm{\mathfrak k_1}_{L^2}
\norm{ (P_{\mu} - 1)\psi(s) }_{L^2}
\norm{\mathfrak k_1}_{L^2}
\]

that tends to zero along $\mu=\mu_k$ as $k \to \infty$.
This concludes the proof of charge conservation.

§ GLOBAL EXISTENCE

This section is devoted to the proof of Theorem <ref>.

So we suppose now

\begin{equation}\label{rb-cond}
s=0, \qquad \frac14 < r < \frac12, \qquad \max(r,1-2r) < b < 1/2.
\end{equation}

Fix some Lebesgue exponent

\[
p \ge \max\left( 4,  \frac{2b + 2r - 1}{b + 2r - 1} \right)
\]

and assume $f_\pm \in L^p \left( \Omega, L^2 \right)$, as well as $g_+ \in L^2(\Omega,H^r)$.

Consider the local solution $(\psi_+,\psi_-,\phi_+)$ from Theorem <ref>, existing up to the stopping time $\tau$. For $R \ge 1$ let $(\psi_+^R,\psi_-^R,\phi_+^R)$ be the solution of the truncated problem (<ref>), (<ref>), obtained in Theorem <ref>; it equals $(\psi_+,\psi_-,\phi_+)$ up to the stopping time $\tau_R$, which increases to $\tau$ as $R \to \infty$. As proved in the last section, we have almost surely the conservation of charge,

\begin{equation}\label{R-charge}
\norm{\psi_+^R(t)}_{L^2}^2 + \norm{\psi_-^R(t)}_{L^2}^2 = \norm{\psi_0}_{L^2}^2
\quad \text{for all $t \ge 0$}.
\end{equation}

And by (<ref>) we have

\begin{equation}\label{R-blowup}
\tau < \infty \implies
\norm{\psi_+^R}_{X^{0,b}_{+\xi}(0,\tau_R)}
\norm{\psi_-^R}_{X^{0,b}_{-\xi}(0,\tau_R)}
\norm{\phi_+^R}_{X^{r,b}_{+\japanese{\xi}}(0,\tau_R)}
\ge C \sqrt{R} \quad \text{for all $R$},
\end{equation}

where $C$ depends only on $b$.

We now claim that for all $R, T \ge 1$ we have bounds

\begin{equation}\label{phi-R-bound}
\norm{\phi_+^R}_{L^2 \left( \Omega, X^{r,b}_{+\japanese{\xi}}(0,T) \right)}
\le C(T) \left( \norm{g_+}_{L^2(\Omega,H^r)} + \norm{\psi_0}_{L^4(\Omega,L^2)}^2 \right)
\end{equation}

\begin{equation}\label{psi-R-bound}
\norm{\psi_\pm^R}_{L^1\left(\Omega,X^{0,b}_{\pm\xi}(0,T)\right)}
\le C\left(T, \norm{g_+}_{L^2(\Omega,H^r)},\norm{\psi_0}_{L^p(\Omega,L^2)} \right),
\end{equation}

which are uniform in $R$. That is, the right hand sides do not depend on $R$. Note that by Hölder's inequality, (<ref>) implies the $L^1(\Omega)$-bound

\begin{equation}\label{phi-R-bound-L1}
\norm{\phi_+^R}_{L^1 \left(\Omega,X^{r,b}_{+\japanese{\xi}}(0,T)\right)}
\le C(T) \left( \norm{g_+}_{L^2(\Omega,H^r)} + \norm{\psi_0}_{L^4(\Omega,L^2)}^2 \right).
\end{equation}

Granting the above claim for the moment, then we can almost surely exclude the scenario $\tau
< \infty$, as follows. Since $\{ \tau < \infty \} = \bigcup_{T=1}^\infty \{ \tau < T \}$, it suffices to check that $\mathbb P \left( \{ \tau < T \} \right) = 0$ for all $T \ge 0$. But (<ref>) implies that

\[
\norm{\psi_+^R}_{L^1\left(\Omega,X^{0,b}_{+\xi}(0,T)\right)}
\norm{\psi_-^R}_{L^1\left(\Omega,X^{0,b}_{-\xi}(0,T)\right)}
\norm{\phi_+^R}_{L^1\left(\Omega,X^{r,b}_{+\japanese{\xi}}(0,T)\right)}
\ge C\sqrt{R} \mathbb P \left( \{ \tau < T \} \right)
\]

for all $R \ge 1$. So if $\mathbb P \left( \{ \tau < T \} \right) > 0$, then letting $R \to \infty$ we get a contradiction to (<ref>) and (<ref>).

It remains to prove the claim. Fix $R,T \ge 1$. We first prove (<ref>). Here we follow the proof of the bound (<ref>) in Theorem <ref>, but with one crucial difference: to ensure that the bounds are uniform in $R$, we have to avoid using the cutoff bound (<ref>) at any point.

We cut $[0,T]$ into small subintervals $[0,\delta]$, $[\delta,2\delta]$ etc.
Fix now a subinterval $[S,S']$, $S'=S+\delta$. Recall that $\phi_+^R$ satisfies (<ref>), which we rewrite as

\[
\phi_+^R(t)
S_{+\japanese{\xi}}(t-S) \phi_+^R(S)
+ \Phi_S(t) + \Psi_S(t)
\]

\[
\Phi_S(t) = i\int_S^t S_{+\japanese{\xi}}(t-s) \japanese{D_x}^{-1}
\re \left( \Theta\overline{\psi_+^R} \Theta\psi_-^R \right)(s) \, ds,
\]

\[
\Theta(t) = \theta_R\left(
\norm{\psi_+^R}_{\widetilde X^{0,b}_{+\xi}(0,t)}^2
\norm{\psi_-^R}_{\widetilde X^{0,b}_{-\xi}(0,t)}^2
\norm{\phi_+^R}_{\widetilde X^{r,b}_{+\japanese{\xi}}(0,t)}^2
\right)
\]

\[
\Psi_S(t) = \frac{i}{2} \int_S^t S_{+\japanese{\xi}}(t-s) \japanese{D_x}^{-1}
\phi^R(s) \mathfrak K_2 \, dW(s).
\]

\[
\norm{\phi_+^R}_{Z(S,S')}
\norm{\phi_+^R}_{L^2\left(\Omega,X^{r,b}_{+\japanese{\xi}}(S,S')\right)}
\norm{\phi_+^R}_{L^2(\Omega,C([S,S'],H^r))}
\]

As in the proof of Theorem <ref> we obtain (using Corollary <ref>)

\[
\norm{\Psi_S}_{Z(S,S')} \le C \delta^b \norm{\phi_+^R}_{Z(S,S')}.
\]

So taking $\delta$ small enough, depending on $b$, we get, using also (<ref>) and (<ref>),

\[
\norm{\phi_+^R}_{Z(S,S')}
\le C\left( \norm{\phi_+^R(S)}_{L^2(\Omega,H^r)} + \norm{\Phi_S}_{L^2(\Omega,X^{r,1}_{+\japanese{\xi}}(S,S'))} \right).
\]

By (<ref>), and using $\abs{\Theta} \le 1$, we get, almost surely,

\[
\norm{\Phi_S}_{X^{r,1}_{+\japanese{\xi}}(S,S')}
\le
C \norm{\overline{\psi_+^R}\psi_-^R}_{L_t^2((S,S'),H^{r-1})}
\le
C \sqrt{\delta} \norm{\psi_0}_{L^2}^2,
\]

where we used the Sobolev product law (<ref>), (<ref>), and the conservation of charge (<ref>). So we conclude that on each subinterval $[S,S'] = [n\delta,(n+1)\delta]$ we have

\[
\norm{\phi_+^R}_{Z(n\delta,(n+1)\delta)}
\le C\left( \norm{\phi_+^R(n\delta)}_{L^2(\Omega,H^r)} +\sqrt{\delta} \norm{\psi_0}_{L^2}^2 \right).
\]

For $n=0$, $\phi_+^R(n\delta) = g_+$, while for $n \ge 1$,

\[
\norm{\phi_+^R(n\delta)}_{L^2(\Omega,H^r)}
\le \norm{\phi_+^R}_{Z((n-1)\delta,n\delta)}
\]

It follows that

\[
\norm{\phi_+^R}_{Z(n\delta,(n+1)\delta)}
\le C^{n+1} \norm{g_+}_{L^2(\Omega,H^r)} + \left( C + C^2 + \cdots + C^{n+1} \right) \sqrt{\delta} \norm{\psi_0}_{L^2}^2
\quad \text{for $n=0,\dots,T/\delta$}.
\]

Summing over the subintervals now yields (<ref>).

With (<ref>) in hand, we now prove (<ref>). The field $\psi_\pm^R$ satisfies (<ref>), which we write here as

\[
\psi_{\pm}^R(t)
S_{\pm\xi}(t) f_\pm
\sum_{j=1}^4 \Psi_{j,\pm}(t)
\]

\begin{align*}
\Psi_{1,\pm}(t)
- iM \int_0^t S_{\pm\xi}(t-s)  \psi_{\mp}^R(s) \, ds,
\\
\Psi_{2,\pm}(t)
- M_{\mathfrak K_1} \int_0^t S_{\pm\xi}(t-s) \psi_\pm^R(s) \, ds,
\\
\Psi_{3,\pm}(t)
i \int_0^t S_{\pm\xi}(t-s) \psi_\mp^R(s) \mathfrak K_1 \, d W(s),
\\
\Psi_{4,\pm}(t)
i \int_0^t  S_{\pm\xi}(t-s) \left( \Theta \phi^R \Theta \psi_{\mp}^R \right)(s) \, ds.
\end{align*}

In the estimates for these terms,
we use the bounds (<ref>) and (<ref>) on a time interval $[0,T]$ with $T \ge 1$,
hence the factors $\sqrt{T}$ and $T^{3/2}$ come up.
By (<ref>),

\begin{equation}\label{psi-R-bound-1}
\norm{S_{\pm\xi}(t) f_\pm}_{L^1\left(\Omega,X^{0,b}_{\pm \xi}(0,T)\right)}
\le C\sqrt{T} \norm{\psi_0}_{L^1(\Omega,L^2)}
\le C \sqrt{T} \norm{\psi_0}_{L^2(\Omega,L^2)}.
\end{equation}

By (<ref>) and the charge conservation (<ref>),

\begin{equation}\label{psi-R-bound-2}
\norm{\Psi_{1,\pm}}_{L^1\left(\Omega,X^{0,b}_{\pm \xi}(0,T)\right)}
\le
CM T^{3/2} \norm{\psi_{\mp}^R}_{L^1 \left(\Omega,X^{0,0}_{\pm \xi}(0,T) \right)}
CM T^{3/2} \norm{\psi_{\mp}^R}_{L^1 \left(\Omega,L^2((0,T) \times \R)\right)}
\le
CM T^2 \norm{\psi_0}_{L^2(\Omega,L^2)}.
\end{equation}

and similarly for $\Psi_{2,\pm}$.
Applying Lemma <ref> with the operator $M_1(f) = f\mathfrak K_1$, which by Remark <ref> satisfies $\norm{M_1(f)}_{\mathcal L_2(L^2,L^2)} = \norm{f}_{L^2} \norm{\mathfrak k_1}_{L^2}$,
we get

\begin{equation}\label{psi-R-bound-3}
\norm{\Psi_{3,\pm}}_{L^1 \left(\Omega,X^{0,b}_{\pm \xi}(0,T)\right)}
\le
\norm{\Psi_{3,\pm}}_{L^2 \left(\Omega,X^{0,b}_{\pm \xi}(0,T)\right)}
\le
C \norm{\psi_{\mp}^R}_{L^2(\Omega,L^2((0,T) \times \R))}
\le
C\sqrt{T} \norm{\psi_0}_{L^2(\Omega,L^2)}.
\end{equation}

By (<ref>) with the time regularity $1 - b > 1/2$ on the left hand side,

\[
\norm{\Psi_{4,\pm}}_{X^{0,b}_{\pm \xi}(0,T)}
\le
CT^{3/2} \norm{\Theta \phi^R \Theta \psi_{\mp}^R}_{X^{0,-b}_{\pm \xi}(0,T)}.
\]

Note that (<ref>) implies $1/2 - r < b/2 < b$.
So if one defines $\mu$ by

\[
\frac12 - r = \mu b
\]

then $0 < \mu < 1/2$, and we find ourselves in the assumption of Corollary <ref>.
Therefore, one can bound the norm on the right hand side as

\[
\norm{\Theta \phi^R \Theta \psi_{\mp}^R}_{X^{0,-b}_{\pm \xi}(0,T)}
\le
\norm{\Theta\phi_+^R}_{X^{r,b}_{+\japanese{\xi}}(0,T)}
\norm{\Theta\psi_{\mp}^R}_{X^{0,b}_{\mp \xi}(0,T)}^{\mu}
\norm{\Theta\psi_{\mp}^R}_{ L^2((0,T) \times \mathbb R) }^{ 1 - \mu }
\]

where the last norm

\[
\norm{\Theta\psi_{\mp}^R}_{L^2((0,T) \times \R)}
\le
\sqrt{T} \norm{\psi_0}_{L^2}
\]

by charge conservation and the bound $\Abs{\Theta} \le 1$.
The first two norms are estimated as

\[
\norm{\Theta\phi_+^R}_{X^{r,b}_{+\japanese{\xi}}(0,T)}
\le
\norm{\phi_+^R}_{X^{r,b}_{+\japanese{\xi}}(0,T)}
, \quad
\norm{\Theta\psi_{\mp}^R}_{X^{0,b}_{\mp \xi}(0,T)}
\le
\norm{\psi_{\mp}^R}_{X^{0,b}_{\mp \xi}(0,T)}
\]

where we used (<ref>) (with $\mathbf v=\mathbf 0$)
to dispose of the the cutoff in front of $\phi_+^R$ and $\psi_+^R$
(this is why there appears a constant $C(T)$ depending on $T$).
Combining the above estimates, we have

\[
\norm{\Psi_{4,\pm}}_{X^{0,b}_{\pm \xi}(0,T)}
\le
\norm{\phi_+^R}_{X^{r,b}_{+\japanese{\xi}}(0,T)}
\left( \sqrt{T} \norm{\psi_0}_{L^2} \right)^{1-\mu}
\norm{\psi_\mp^R}_{X^{0,b}_{\mp \xi}(0,T)}^{\mu}.
\]

Now write

\[
1 = \frac12 + \frac{1-\mu}{q} + \mu,
\quad \text{where} \quad
q = \frac{1-\mu}{1/2-\mu}
= \frac{2b + 2r - 1}{b + 2r - 1}.
\]

Then by Hölder's inequality,

\begin{equation}\label{psi-R-bound-4}
\norm{\Psi_{4,\pm}}_{L^1 \left(\Omega,X^{0,b}_{\pm \xi}(0,T) \right)}
\le
C(T) T^{(1-\mu)/2}
\norm{\phi_+^R}_{L^2 \left(\Omega,X^{r,b}_{+\japanese{\xi}}(0,T)\right)}
\norm{\psi_0}_{L^q(\Omega,L^2)}^{1-\mu}
\norm{\psi_{\mp}^R}_{L^1 \left(\Omega,X^{0,b}_{\mp \xi}(0,T) \right)}^{\mu}
\end{equation}

Note that $q \le p$, so that we can replace $L^q$ by $L^p$ in the last estimate.

So finally, combining (<ref>)–(<ref>), making use of (<ref>), and setting

\[
\norm{\psi_+^R}_{L^1 \left( \Omega,X^{0,b}_{+\xi}(0,T) \right)}
\norm{\psi_-^R}_{L^1 \left( \Omega,X^{0,b}_{-\xi}(0,T) \right)}
\]

we deduce that for all $T \ge 1$ holds

\[
f(T) \le CT \norm{\psi_0}_{L^2(\Omega,L^2)} + C(T)A(T) T^{(1-\mu)/2} \norm{\psi_0}_{L^p(\Omega,L^2)}^{1-\mu} \left[ f(T)\right]^\mu,
\]

where $A(T)$ stands for the expression on the right hand side of (<ref>). Recalling that $0 < \mu < 1/2$, we conclude that

\[
f(T) \le C T \left( 1 + C(T)A(T) \right) \left( 1 + \norm{\psi_0}_{L^p(\Omega,L^2)} \right)
\left( 1 +  \sqrt{f(T)} \right)
\]

for all $T \ge 1$. By the next lemma, we then get (<ref>), and this concludes the proof of global existence.

If $a,b \in [0,\infty)$ satisfy

\begin{equation}\label{boot-strap}
b \le a \left( 1 + \sqrt{b} \right),
\end{equation}

\[
b < 1 + 4a^2.
\]

To get a contradiction, assume that $b \ge 1 + 4a^2$.
Then $b \ge 1$, so (<ref>) implies $b \le a 2\sqrt{b}$, hence
$b \le 4a^2$, contradicting our assumption.

§ BOURGAIN ISOMETRY

This section is devoted to the proof of Lemma <ref>.
We start with a very general result related to Bochner spaces.
It may be difficult to find it in the existing literature,
so we provide here a complete proof.

Let $\mu$ be a $\sigma$-finite complete measure on $Y$
and $G$ be a separable Banach space.
Consider a closed subspace $G_0$ in $G$ with the quotient projection
\(
\pi_0 : G \to G / G_0
\)
The trivial case $G_0 = G$ is excluded, of course.
Then there exists a unique linear operator $\Phi$ making the following diagram commutative:

\begin{equation}
\label{Bochner_commutative_diagram}
\begin{tikzcd}
L^p(Y, \mu; G)
\arrow[r, "P"]
\arrow[d, " \pi : f \mapsto \bigl( y \mapsto \pi_0 ( f(y) ) \bigr) \;\; "']
L^p(Y, \mu; G) / L^p(Y, \mu; G_0)
\arrow[dl, "\Psi = \Phi^{-1}", bend left=10, dashed]
\\
L^p(Y, \mu; G / G_0)
\arrow[ur, "\Phi", dashed]
\end{tikzcd}
\end{equation}

where $P$ is the quotient projection and $p \in [1, \infty)$.
Moreover, $\Phi$ is invertible and
\(
\norm \Phi
\norm {\Phi^{-1}}
\)

We split the proof in several steps starting with the implied correctness of the diagram (<ref>).

\(
L^p(Y, \mu; G_0)
\)
is a closed subspace in
\(
L^p(Y, \mu; G)
\)
Hence $P$ is well defined as the corresponding quotient projection,
in particular,
\(
\norm P = 1
\)

For any
\(
f \in L^p(Y, \mu; G)
\)
the value $\pi (f)$ is the composition
\(
Y \xrightarrow{f} G \xrightarrow{\pi_0} G / G_0
\)
Here $G$ and $G / G_0$ are endowed with the Borel $\sigma$-algebras,
whereas $f$ is measurable and $\pi_0$ is continuous.
Therefore $\pi (f)$ is measurable.
Moreover, $\pi$ is linear with $\norm \pi \leqslant 1$.

\[
\norm{
\pi f
}_{ L^p(Y, \mu; G / G_0) }^p
\int_Y
\norm{
\pi_0 f(y)
}_{ G / G_0 }^p
d \mu(y)
\leqslant
\int_Y
\norm{
}_{ G }^p
d \mu(y)
\norm{
}_{ L^p(Y, \mu; G) }^p
\]

An important claim implied in the statement is surjectivity of $\pi$.
\(
h \in L^p(Y, \mu; G / G_0)
\)
We need to find an
\(
f \in L^p(Y, \mu; G)
\)
such that $\pi f = h$.
A direct pointwise construction may not necessarily lead even to a measurable function $f$ on $Y$.
So we approximate $h$ by a sequence $h_n, n \in \mathbb N,$ of simple functions in
\(
L^p(Y, \mu; G / G_0)
\)
Then for any $y \in Y$ and $n \in \mathbb N$ there is $f_n(y) \in G$ such that
\(
\pi_0( f_n(y) ) = h_n(y)
\)
and $f_n$ is simple.
Note that $f_n$ may not converge anywhere.
Let us choose a subsequence in $\{ h_n \}$, while keeping the same notation,
such that

\[
\sum_{n = 1}^{\infty}
\norm{
h_{n + 1} - h_n
}_{ L^p(Y, \mu; G / G_0) }
\infty
\]

Making use of the fact that $\mu$ is $\sigma$-finite,
that is
\(
Y = \bigcup \limits _{k = 1}^{\infty} Y_k
\)
with disjoint $Y_k$ having $\mu(Y_k) < \infty$,
we can introduce a measurable function
\(
\chi : Y \to (0, \infty)
\)
taking at most countably many values
(in fact we need below that it is constant on each $Y_k$)
and normalized by
\(
\int \chi d \mu = 1
\)
Pointwisely, we have

\[
\norm{
h_{n + 1}(y) - h_n(y)
}_{ G / G_0 }
\inf
\left \{
\norm F_G
\colon
F - ( f_{n + 1}(y) - f_n(y) ) \in G_0
\right \}
\]

Now let
\(
\left \{
\right \}
\)
be the partition associated with $h_n$.
Then for each $n \in \mathbb N$ we can approximate the infimum on the finest partition
\(
\left \{
Y_{l_{n + 1}}^{n + 1}
\bigcap
\bigcap
\right \}
\)
of $Y$ as follows

\[
\norm{
h_{n + 1}(y) - h_n(y)
}_{ G / G_0 }^p
\frac {\chi(y)}{ 2^{pn} }
\norm {F_n(y)}_G^p
\]

\(
F_n(y) - ( f_{n + 1}(y) - f_n(y) ) \in G_0
\)
and $F_n$ is constant on each
\(
Y_{l_{n + 1}}^{n + 1}
\bigcap
\bigcap
\)
One sets $F_0 = f_1$ and so $\pi F_0 = h_1$, in particular.
Integrating the above inequality one obtains the estimate

\[
\norm{
h_{n + 1} - h_n
}_{ L^p(Y, \mu; G / G_0) }
\frac 1{ 2^n }
\geqslant
\norm {F_n}_{ L^p(Y, \mu; G) }
\]

\[
\sum_{n = 1}^{\infty}
\norm {F_n}_{ L^p(Y, \mu; G) }
\leqslant
\sum_{n = 1}^{\infty}
\norm{
h_{n + 1} - h_n
}_{ L^p(Y, \mu; G / G_0) }
+ 1 < \infty
\]

\(
\sum_{n = 0}^{\infty}
\)
converges in
\(
L^p(Y, \mu; G)
\)
to an $F$ satisfying $\pi F = h$.

\[
\pi
\left(
\sum_{n = 0}^N
\right)
\sum_{n = 1}^N
( h_{n + 1} - h_n ) + h_1
h_{N + 1}
\to
\ \text{ as } \
N \to \infty
\]

Therefore $\pi$ is surjective.

Finally, we can define the linear mappings $\Phi$ and $\Psi$ in the diagram (<ref>).
Indeed, by the previous step for any
\(
h \in L^p(Y, \mu; G / G_0)
\)
there is an
\(
f_h \in L^p(Y, \mu; G)
\)
such that $\pi f_h = h$.
We set
\(
\Phi(h) = P f_h
\)
If we have two elements such that
\(
\pi f_h^1
\pi f_h^2
\)

\[
\int_Y
\norm{
\pi_0 f_h^1(y)
\pi_0 f_h^2(y)
}_{ G / G_0 }^p
d \mu(y)
\norm{
\pi f_h^1 - \pi f_h^2
}_{ L^p(Y, \mu; G / G_0) }^p
\]

Hence for a.e. $y$ the difference
\(
\in G_0
\)
and so
\(
\in L^p(Y, \mu; G_0)
\)
implying the equality
\(
\)
in $L^p(Y, \mu; G) / L^p(Y, \mu; G_0)$.
Thus $\Phi$ is a well defined linear operator.

Similarly, for any
\(
h \in L^p(Y, \mu; G) / L^p(Y, \mu; G_0)
\)
there is obviously an
\(
f_h \in L^p(Y, \mu; G)
\)
such that $P f_h = h$.
We can set
\(
\Psi(h) = \pi f_h
\)
If we have two elements satisfying
\(
P f_h^1
P f_h^2
\)
then their difference
\(
\in
L^p(Y, \mu; G_0)
\)
Hence for a.e. $y$ the difference
\(
\in G_0
\)
and so

\[
\left(
\pi f_h^1
\pi f_h^2
\right)
\pi_0
\left(
\right)
\ \text{ in } \
G / G_0
\]

implying the equality
\(
\pi f_h^1
\pi f_h^2
\)
in $L^p(Y, \mu; G / G_0)$.
Thus $\Psi$ is well defined as well.

Clearly, so defined $\Phi$ and $\Psi$ are the only linear operators
making the diagram (<ref>) commutative.
Moreover, it is straightforward to check that the compositions
$\Psi \Phi$ and $\Phi \Psi$ are identities in the spaces
\(
L^p(Y, \mu; G / G_0)
\)
\(
L^p(Y, \mu; G) / L^p(Y, \mu; G_0)
\)
Therefore $\Psi = \Phi^{-1}$.

In this step we prove the bound $\norm \Psi \leqslant 1$
which in turn will automatically imply that $\Phi$ is bounded as well and $\norm \Phi \geqslant 1$.
\(
h \in L^p(Y, \mu; G) / L^p(Y, \mu; G_0)
\)
then by steps <ref>, <ref>
we have

\[
\norm{
\Psi h
}_{ L^p(Y, \mu; G / G_0) }
\norm{
\pi f_h
}_{ L^p(Y, \mu; G / G_0) }
\leqslant
\norm{
}_{ L^p(Y, \mu; G) }
\]

In other words,
\(
\norm{
\Psi h
}_{ L^p(Y, \mu; G / G_0) }
\leqslant
\norm{
}_{ L^p(Y, \mu; G) }
\)
for every $f$ belonging to the preimage $P^{-1} \{ h \}$.
Therefore passing to infimum over this preimage one recovers
the quotient norm of $h$,
so that
\(
\norm{
\Psi h
}_{ L^p(Y, \mu; G / G_0) }
\leqslant
\norm{
}_{ L^p(Y, \mu; G) / L^p(Y, \mu; G_0) }
\)
which proves the claim $\norm \Psi \leqslant 1$ due to the arbitrary choice of $h$.

In order to complete the proof of the lemma it is only left to get the bound $\norm \Phi \leqslant 1$.
Here we need to be careful again about measurability of pointwise quotient norm approximations.
So we will check that the bound
\(
\norm{
\Phi h
}_{ L^p(Y, \mu; G) / L^p(Y, \mu; G_0) }
\leqslant
\norm{
}_{ L^p(Y, \mu; G / G_0) }
\)
holds for every simple $h$.
It is enough since $\Phi$ is already known to be bounded, by the previous step.

Similarly to the previous step, from the identity $\norm P = 1$ one deduces

\begin{equation}
\label{proof_step7_bound}
\norm{
\Phi h
}_{ L^p(Y, \mu; G) / L^p(Y, \mu; G_0) }
\leqslant
\norm{
}_{ L^p(Y, \mu; G) }
, \quad
f \in \pi^{-1} \{ h \}
\end{equation}

Let $\varepsilon > 0$ and function $\chi$ be as in step <ref>.

\[
\norm{
}_{ L^p(Y, \mu; G / G_0) }^p
\varepsilon
\int_Y
\left(
\inf_{ g \in \pi_0^{-1} \{ h(y) \} } \norm g _G^p
\varepsilon \chi(y)
\right)
d \mu(y)
\]

and so there exists $g_{\varepsilon} : Y \to G$ such that

\[
\inf_{ g \in \pi_0^{-1} \{ h(y) \} } \norm g _G^p
\varepsilon \chi(y)
\geqslant
\norm {g_{\varepsilon}(y)} _G^p
, \quad
\pi_0( g_{\varepsilon}(y) ) = h(y)
\]

and it takes at most countably many values,
its partition being the finest one for $h$ and $\chi$.
In particular,
\(
g_{\varepsilon} \in L^p(Y, \mu; G)
\)
\(
\pi ( g_{\varepsilon} )  = h
\)
which because of (<ref>) implies

\begin{equation*}
\norm{
\Phi h
}_{ L^p(Y, \mu; G) / L^p(Y, \mu; G_0) }^p
\leqslant
\norm{
}_{ L^p(Y, \mu; G) }^p
\leqslant
\norm{
}_{ L^p(Y, \mu; G / G_0) }^p
\varepsilon
\end{equation*}

Passing to the limit one obtains
\(
\norm{
\Phi h
}_{ L^p(Y, \mu; G) / L^p(Y, \mu; G_0) }
\leqslant
\norm{
}_{ L^p(Y, \mu; G / G_0) },
\)
completing the proof.

We formulate the following simple lemma, that can be either found in [22]
or easily proved with the help of an argument similar to one used in step <ref> of the previous proof.

Consider nontrivial closed subspaces $E_0, F_0$ in normed spaces $E, F$, respectively.
Let $\mathcal T : E \to F$ be a linear bounded operator satisfying $\mathcal T (E_0) \subset F_0$.
Then there exists a unique linear $\widetilde \mathcal T$ making the following diagram commutative:

\begin{equation}
\label{quotient_commutative_diagram}
\begin{tikzcd}
\arrow[r, "\mathcal T"]
\arrow[d, " P_1 \;\; "']
\arrow[d, " \;\; P_2 "]
\\
E / E_0
\arrow[r, "\widetilde \mathcal T", dashed]
F / F_0
\end{tikzcd}
\end{equation}

\(
\norm{ \widetilde \mathcal T }
\leqslant
\norm{ \mathcal T }
\)

We take
\(
E = X_{ h(\xi) }^{s, b}
Y = \mathbb R^d
d \mu(\xi) = \langle \xi \rangle^{2s} d \xi
G = H^b( \mathbb R )
\)
\(
F = L^2
\left(
\mathbb R^d
\langle \xi \rangle^{2s} d \xi
H^b( \mathbb R )
\right)
\)
We introduce $\mathcal T$ as the isometric extension of
\(
e^{ ith(\xi) } \mathcal F_x
\)
defined on the Schwartz space
$\mathcal S \left( \mathbb R^{d + 1} \right)$
with the property (<ref>),
so that

\begin{equation}
\label{abstract_XNorm_alternative}
\norm
_{ X_{h(\xi)}^{ s, b } }^2
\int
\langle \xi \rangle ^{2s}
\norm
\mathcal T u(\xi)
} _{ H^b(\R) }^2
\end{equation}

Now let us consider an interval $I = (S, T)$
and the closed subspaces
\(
\left \{
u \in H^b( \mathbb R )
\colon
u = 0 \text{ on } I
\right \}
\)

\(
\left \{
u \in X_{h(\xi)}^{ s, b }
\colon
u = 0 \text{ on } I \times \mathbb R^d
\right \}
\)
It is known that
\(
H^b( \mathbb R ) / G_0
\)
\(
X_{h(\xi)}^{ s, b }(I)
X_{h(\xi)}^{ s, b } / E_0
\)
endowed with the quotient norms, of course.
We claim that there exists a unique invertible isometry $\widetilde \mathcal T$ making the following diagram commutative

\begin{equation}
\label{Bourgain_isometry_diagram}
\begin{tikzcd}
%	\hspace{-5.5cm}
X_{h(\xi)}^{ s, b }
\arrow
r, " \mathcal T = e^{ ith(\xi) } \mathcal F_x "
yshift = 0.7ex
%        ,
%        start anchor={[xshift = -2.1cm]}
\arrow
d, " P_1 \;\; "'
%        ,
%        start anchor={[xshift = -2.9cm]}
%        ,
%        end anchor={[xshift = -2.9cm]}
\left(
\mathbb R^d
\langle \xi \rangle^{2s} d \xi
H^b( \mathbb R )
\right)
\arrow[
l, " \mathcal T^{-1} = \mathcal F_{\xi}^{-1} e^{ -ith(\xi) } "
yshift = -0.7ex
%        ,
%        end anchor={[xshift = -2.1cm]}
\arrow[r, " \pi "]
\arrow[d, " P_2 \;\; "']
\left(
\mathbb R^d
\langle \xi \rangle^{2s} d \xi
H^b( I )
\right)
\arrow
dl, "\Phi"
bend left=10
%        ,
%        shift right = -0.7ex
\\
%	\hspace{-5.5cm}
X_{h(\xi)}^{ s, b }(I)
\arrow[
r, "\widetilde \mathcal T", dashed
yshift = 0.7ex
%        ,
%        start anchor={[xshift = -2.1cm]}
\left(
\mathbb R^d
\langle \xi \rangle^{2s} d \xi
H^b( \mathbb R )
\right)
\left(
\mathbb R^d
\langle \xi \rangle^{2s} d \xi
\right)
\arrow[
ur, "\Phi^{-1}"
bend right = 2
%        ,
%        shift right = -0.7ex
\arrow[
l, "\widetilde \mathcal T^{-1}", dashed
yshift = -0.7ex
%        ,
%        end anchor={[xshift = -2.1cm]}
\end{tikzcd}
\end{equation}

This commutative diagram makes a complete sense of and proves the identity
In other words, the term
\(
e^{ it h(\xi) }
\mathcal F_x
\)
staying in (<ref>)
should be understood as the composition
\(
\Phi^{-1} \widetilde \mathcal T
\)
having operator norm equal to unity,

\begin{equation}
\label{abstract_XRestNorm_alternative}
\norm
_{ X_{h(\xi)}^{ s, b }(S,T) }
\left( \int
\langle \xi \rangle ^{2s}
\norm
\Phi^{-1} \widetilde \mathcal T
} _{ H^b (S,T)}^2
d\xi \right)^{1/2}
\end{equation}

In order to appeal to the previous two lemmas, and to prove the claim,
one needs only to show the inclusions

\[
\mathcal T(E_0)
\subset
\left(
\mathbb R^d
\langle \xi \rangle^{2s} d \xi
\right)
, \quad
\mathcal T^{-1}
\left(
\left(
\mathbb R^d
\langle \xi \rangle^{2s} d \xi
\right)
\right)
\subset
\]

Moreover, it is enough to check these on the smooth functions,
namely, that the following hold

\[
\mathcal T
\left(
\bigcap
\left \{
\phi \psi
\colon
\phi \in \mathcal S \left( \mathbb R^d \right)
\psi \in \mathcal S ( \mathbb R )
\right \}
\right)
\subset
\left(
\mathbb R^d
\langle \xi \rangle^{2s} d \xi
\right)
\]

\[
\mathcal T^{-1}
\left(
\left \{
\phi \psi
\colon
\phi \in \mathcal S \left( \mathbb R^d \right)
\psi \in \mathcal S ( \mathbb R ) \bigcap G_0
\right \}
\right)
\subset
\]

This is obvious and so (<ref>)
is commutative.
(<ref>), (<ref>)
(<ref>), (<ref>)
are fully justified.

§ A MODIFIED BOURGAIN NORM

In preparation for the proof of the cutoff estimates, we now investigate more closely the modified Bourgain norm defined in (<ref>), and derive some of its key properties.

Fix $s \in \R$, $b \in (0,1/2)$ and $h \in C(\R^d,\R)$. Write the norm (<ref>) as

\[
\norm{u}_{X^{s,b}}
= \left( \int_{\R^d} \norm{U(t,\xi)}_{H^b_t(\R)}^2 \, d\xi \right)^{1/2},
\]

where the transform $u(t,x) \mapsto U(t,\xi)$ is defined by

\[
U(t,\xi) = \japanese{\xi}^{s} e^{ith(\xi)} \widehat u(t,\xi).
\]

By (<ref>), the above norm is equivalent to, with constants depending only on $b$, the norm

\[
\norm{u}_{\widehat X^{s,b}}
:= \left( \int_{\R^d} \left( \norm{U(t,\xi)}_{L^2_t(\R)}^2
+ \norm{U(t,\xi)}_{S^b_t(\R)}^2 \right) \, d\xi \right)^{1/2}.
\]

Inserting here the characteristic function $\mathbb 1_{(S,T)}(t)$ of a time interval $(S,T)$, we compute

\begin{equation}
\label{CutoffNorm'}
\norm{\mathbb 1_{(S,T)} u}_{\widehat X^{s,b}}
\left( \int_{\R^d} \left( \int_S^T \Abs{U(t,\xi)}^2 \left( 1 + \frac{1}{b(t-S)^{2b}} + \frac{1}{b(T-t)^{2b}} \right) \, dt
+ \norm{U(t,\xi)}_{S^b_t(S,T)}^2 \right) \, d\xi \right)^{1/2}.
\end{equation}

By (<ref>), (<ref>)
the latter is equivalent to, with constants depending only on $b$,
the restriction norm

\begin{equation}\label{RestNorm}
\norm{u}_{X^{s,b}(S,T)}
= \left( \int_{\R^d} \norm{U(t,\xi)}_{H^b_t(S,T)}^2 \, d\xi \right)^{1/2}.
\end{equation}

The advantage of the norm (<ref>) is that it has an explicit expression; there are no restriction norms involved. It is still a bit tricky to work with, however. We will use instead the simpler, modified norm

\begin{equation}\label{CutoffNorm''}
\norm{u}_{\widetilde X^{s,b}(S,T)} =
\left( \int_{\R^d}
\left( \frac 1{ (T - S)^{2b} }
\int_S^T \Abs{ U(t, \xi) }^2 \, dt
\int_S^T \int_S^T
\frac
{ | U(t, \xi) - U(r, \xi) |^2 }
{ | t - r |^{1 + 2b} }
\, dr \, dt \right) \, d\xi \right)^{1/2},
\end{equation}

which is the same as (<ref>), and turns out to be equivalent to the two previous norms.

In fact, we have the following.

Let $T_0 > 0$ and $0 < b < 1/2$. Consider an interval $(S,T)$ with length at most $T_0$. Then the norms defined on $X^{s,b}(S,T)$ by (<ref>), (<ref>) and (<ref>) are pairwise equivalent, with constants depending only on $b$ and $T_0$.

Moreover, for all $T > 0$ we have the estimate

\begin{equation}\label{Large-T-estimate}
\norm{u}_{X^{s,b}(0,T)} \ge C \norm{u}_{\widetilde X^{s,b}(0,T)},
\end{equation}

where $C > 0$ depends only on $b$.

By translation invariance we may consider the interval $(0,T)$, where $0 < T \le T_0$. By (<ref>) we already know that (<ref>) and (<ref>) are equivalent, uniformly in $T$ (of any size). The equivalence of (<ref>) and (<ref>) reduces to proving the equivalence of the following norms on $H^b(0,T)$:

\[
M_T(\phi) = \left( \int_0^T \Abs{\phi(t)}^2 \left( 1 + \frac{1}{bt^{2b}} + \frac{1}{b(T-t)^{2b}} \right) \, dt
+ \int_0^T \int_0^T \frac{\Abs{\phi(t)-\phi(r)}^2}{\Abs{ t - r }^{1+2b} } \, dr \, dt \right)^{1/2}
\]

\[
N_T(\phi) = \left( T^{-2b} \int_0^T \Abs{\phi(t)}^2  \, dt
+ \int_0^T \int_0^T \frac{\Abs{\phi(t)-\phi(r)}^2}{\Abs{ t - r }^{1+2b} } \, dr \, dt \right)^{1/2}.
\]

First, since $0 < t < T$ implies $T^{-2b} < t^{-2b}$, it is clear that

\begin{equation} \label{MNineq}
\le b^{1/2} M_T(\phi) \quad \text{for all $T > 0$}.
\end{equation}

It remains to show

\[
M_T(\phi) \le C N_T(\phi).
\]

We claim that this holds for $T=T_0$. Granting this for the moment, it follows that the inequality holds also for $0 < T \le T_0$, since setting $g(s) = \phi\left(sT/T_0\right)$ and rescaling yields

\[
\le \left(\frac{T}{T_0}\right)^{1-2b} M_{T_0}(g)^2 \le C^2 \left(\frac{T}{T_0}\right)^{1-2b} N_{T_0}(g)^2 = C^2 N_T(\phi)^2.
\]

It remains to prove the claim, namely $M_{T_0}(g) \le C N_{T_0}(g)$ for $g \in H^b(0,T_0)$, with $C$ depending on $T_0$ and $b$. But on the one hand, (<ref>) implies $M_{T_0}(g) = \norm{\mathbb 1_{(0,T_0)} g}_{H^b(\R)}$. On the other hand, $N_{T_0}(g) \sim_{b,T_0}  \norm{g}_{H^b(0,T_0)}$ by (<ref>). By (<ref>) it now follows that $M_{T_0}(g) \sim_{b,T_0} N_{T_0}(g)$.

Finally, (<ref>) is immediate from (<ref>). This completes the proof of the lemma.

Before proceeding with the proof of the cutoff estimates, we mention some properties of the modified Bourgain norm (<ref>).

Let $T_0 > 0$ and $0 < b < 1/2$. Then for all $0 < T \le T_0$ and all $u \in X^{s,b}(0,T)$ we have the bounds

\begin{equation}\label{increase2}
\norm{u}_{\widetilde X^{s,b}(r,t)} \le C \norm{u}_{\widetilde X^{s,b}(0,T)} \quad \text{for} \quad 0 \le r < t \le T
\end{equation}

\begin{equation}\label{triangle}
\norm{u}_{\widetilde X^{s,b}(0,t)}
\le
C \left( \norm{u}_{\widetilde X^{s,b}(0,r)} + \norm{u}_{\widetilde X^{s,b}(r,t)} \right) \quad \text{for} \quad 0 \le r < t \le T,
\end{equation}

where the constants only depend on $b$ and $T_0$.

We apply Lemma <ref>. First, (<ref>) follows via the equivalence of (<ref>) with the restriction norm (<ref>), and the fact that the latter is increasing with respect to the interval to which we restrict. Second, (<ref>) follows via the equivalence of (<ref>) with the sharp cutoff norm (<ref>), writing $\mathbb 1_{(0,t)} = \mathbb 1_{(0,r)} + \mathbb 1_{(r,t)}$ (a.e.) and applying the triangle inequality.

Let $0 < b < 1/2$. Assume that $u \in X^{s,b}(0,T)$ and set

\[
f(t) = \norm{u}_{\widetilde X^{s,b}(0,t)}^2 \quad \text{for $0 < t \le T$}.
\]

Then $f$ is continuous on $(0,T]$. Moreover, if we additionally assume that $u \in C([0,T], H^s)$, then

\[
\lim_{t \searrow 0} f(t) = 0,
\]

so $f$ extends to a continuous function on $[0,T]$.

For $0 < t \le T$,

\begin{equation}\label{f-def}
t^{-2b} \int_0^t \norm{u(r)}_{H^s}^2 \, dr
\int_0^t\int_0^t \frac{\norm{U(r,\xi)-U(\sigma,\xi)}_{L^2_\xi}^2}{\Abs{r-\sigma}^{1+2b}} \, dr \, d\sigma.
\end{equation}

Continuity follows from the dominated convergence theorem. It remains to prove $\lim_{t \searrow 0} f(t) = 0$. For the second term in (<ref>), this is clear by dominated convergence, and the first term we bound by

\[
t^{-2b}  \int_0^t \norm{u(r)}_{H^s}^2  \, dr
\le
t^{1-2b} \sup_{0 \le r \le t} \norm{u(r)}_{H^s}^2,
\]

which tends to zero as $t \searrow 0$ if $u \in C([0,T], H^s)$.

Assume that $u \in X^{s,b}(0,T) \cap C([0,T], H^s)$ is an $H^s$-adapted random variable. Then the continuous function $f \colon [0,T] \to [0,\infty)$ from Lemma <ref> is adapted.

For both terms in (<ref>), $\mathcal F_t$-measurability follows by Tonelli's theorem, since $u(t)$ is progressively measurable.

§ CUTOFF ESTIMATES IN $H^B$ AND $X^{S,B}$

Here we prove Proposition <ref>, which we restate below for convenience. Let $0 < b < 1/2$. Fix a smooth, compactly supported function $\theta \colon \R \to \R$, and write $\theta_R(x) = \theta(x/R)$.

Via the transform $u(t,x) \mapsto U(t,\xi) = \japanese{\xi}^{s} e^{ith(\xi)} \widehat u(t,\xi)$, and recalling Lemma <ref>, we can identify $X^{s,b}_{h(\xi)}(S,T)$ with the space $L^2(\R^d,H^b(S,T))$ with norm (this corresponds to (<ref>))

\[
\norm{U}_{(S,T)}
\norm{
\left(
\frac 1{ (T - S)^{2b} }
\int_S^T \Abs{ U(t, \xi) }^2 \, dt
\int_S^T \int_S^T
\frac
{ | U(t, \xi) - U(r, \xi) |^2 }
{ | t - r |^{1 + 2b} }
\, dr \, dt
\right)^{1/2}
\]

If $S=0$, we simply write $\norm{U}_T$. This norm is associated to the inner product

\[
\innerprod{U}{V}_{T}
\int_{\R^d} \left( \frac{1}{T^{2b}} \int_0^T U(t,\xi) \overline{V(t,\xi)} \, dt
\int_0^T \int_0^T \frac{\left[U(t,\xi)-U(r,\xi)\right] \overline{\left[V(t,\xi)-V(r,\xi)\right]}}{\Abs{t-r}^{1+2b}} \, dr \, dt \right) \, d\xi.
\]

For a vector $\mathbf U=(U_1,\dots,U_n)$ we write

\[
\norm{\mathbf U}_{T} = \left( \sum_{i=1}^n \norm{U_i}_{T}^2 \right)^{1/2}.
\]

With this notation, and taking into account Lemma <ref>, we can now restate Proposition <ref> as follows.

Let $T_0 > 0$. Then for all $T \in (0, T_0]$ and $R > 0$ we have the estimates

\begin{equation}\label{MainCutoffLemma-2-1}
\norm{
\theta_R\left( \norm{\mathbf U}_{t} \right) \mathbf U
\leqslant
C \sqrt{R}
\end{equation}

\begin{equation}\label{MainCutoffLemma-2-2}
\norm{
\theta_R\left( \norm{\mathbf U}_{t}^2 \right) \mathbf U
\theta_R\left( \norm{\mathbf V}_{t}^2 \right) \mathbf V
\leqslant
C \norm{\mathbf U - \mathbf V}_{T},
\end{equation}

where the constant $C$ depends only on $b$, $T_0$ and $\theta$.

The remainder of this section is devoted to the proof of this result. For convenience, and without loss of generality, we assume that $\Abs{\theta} \le 1$ and $\theta$ is supported in $[-2,2]$. Throughout we assume that $T_0 > 0$, $b \in (0,1/2)$ and that functions $U,V$ etc.  are in $L^2(\R^d,H^b(S,T))$, and similarly for vectors $\mathbf U,\mathbf V$ etc.

We note the bound, for all $x,y \in \R$,

\begin{equation}\label{Trivial-theta-bound}
\Abs{\theta_R(x) - \theta_R(y)} \le \frac{C}{R} \Abs{x-y}.
\end{equation}

This holds with $C=\norm{\theta'}_{L^\infty}$.

We first prove some preliminary lemmas.

§.§ Preliminary estimates

For all $T \in (0,T_0]$ and $R > 0$ we have the estimate

\begin{equation}\label{KeyEst1}
\int_{\R^d} \int_0^T \Abs{U(t,\xi)}^2 \int_0^t \frac{\Abs{ \Innerprod{V}{W}_t - \innerprod{V}{W}_r  }^2}{(t-r)^{1+2b}} \, dr \, dt \, d\xi
\le C \norm{U}_T^2 \norm{V}_T^2 \norm{W}_T^2,
\end{equation}

where $C$ depends only on $b$ and $T_0$.

We write

\[
\innerprod{V}{W}_t = \mu(t) + \nu(t),
\]

\begin{align*}
\mu(t) &= t^{-2b} \int_{\R^d} \int_0^t V(s,\zeta) \overline{W(s,\zeta)} \, ds \, d\zeta,
\\
\nu(t) &= 2 \int_{\R^d} \int_0^t \int_0^s \frac{\left[V(s,\zeta)-V(\sigma,\zeta)\right] \overline{\left[W(s,\zeta)-W(\sigma,\zeta)\right]}}{(s-\sigma)^{1+2b}} \, d\sigma \, ds \, d\zeta.
\end{align*}

The left side of (<ref>) is therefore bounded by $2(I+J)$, where

\begin{align*}
I &= \int_{\R^d} \int_0^T \Abs{U(t,\xi)}^2 \int_0^t \frac{\Abs{\mu(t) - \mu(r)}^2}{(t-r)^{1+2b}} \, dr \, dt \, d\xi,
\\
J &= \int_{\R^d} \int_0^T \Abs{U(t,\xi)}^2 \int_0^t \frac{\Abs{\nu(t) - \nu(r)}^2}{(t-r)^{1+2b}} \, dr \, dt \, d\xi.
\end{align*}

To estimate $J$ we note that

\[
\nu(t) - \nu(r) = 2 \int_{\R^d} \int_r^t \int_0^s \frac{\left[V(s,\zeta)-V(\sigma,\zeta)\right] \overline{\left[W(s,\zeta)-W(\sigma,\zeta)\right]}}{(s-\sigma)^{1+2b}}  \, d\sigma \, ds \, d\zeta,
\]

so by Cauchy-Schwarz,

\begin{multline*}
\Abs{\nu(t) - \nu(r)}^2
\le
2 \int_{\R^d} \int_r^t \int_0^s \frac{\Abs{V(s,\zeta)-V(\sigma,\zeta)}^2}{(s-\sigma)^{1+2b}} \, d\sigma \, ds \, d\zeta
\\
\times 2 \int_{\R^d} \int_r^t \int_0^s \frac{\Abs{W(s,\zeta)-W(\sigma,\zeta)}^2}{(s-\sigma)^{1+2b}} \, d\sigma \, ds \, d\zeta,
\end{multline*}

where the last line is bounded by $\norm{W}_T^2$. Thus

\begin{multline}\label{I2bound}
J \le \norm{W}_T^2
2 \int_{\R^d} \int_0^T \Abs{U(t,\xi)}^2 \int_0^t \frac{1}{(t-r)^{1+2b}}
\\
\times \left( \int_{\R^d} \int_r^t \int_0^s \frac{\Abs{V(s,\zeta)-V(\sigma,\zeta)}^2}{(s-\sigma)^{1+2b}} \, d\sigma \, ds \, d\zeta \right) \, dr \, dt \, d\xi,
\end{multline}

and Lemma <ref> below yields the desired estimate $J \le C \norm{U}_T^2 \norm{V}_T^2 \norm{W}_T^2$.

In $I$ we split the innermost integral as $\int_0^{t/2} + \int_{t/2}^t$ and write $I = I_1 + I_2$
accordingly. The term $I_1$ is easy to handle since $t-r \sim t$. Applying Cauchy-Schwarz and (<ref>) to bound $\Abs{\mu(t)} \le \norm{V}_t \norm{W}_t \le C \norm{V}_T \norm{W}_T$, we then simply estimate

\[
\le
C \norm{V}_T^2 \norm{W}_T^2
\left( \int_{\R^d} \int_0^T \Abs{U(t,\xi)}^2 t^{-2b} \, dt \, d\xi \right)
\le
C' \norm{V}_T^2 \norm{W}_T^2 \norm{U}_T^2,
\]

where we used Lemma <ref> in the last step.

It remains to consider $I_2$. Writing

\[
\mu(t) = t^{1-2b} f(t), \quad \text{where} \quad f(t) = \frac{1}{t} \int_0^t g(s) \, ds,
\quad
g(s) = \int_{\R^d} V(s,\zeta) \overline{W(s,\zeta)} \, d\zeta,
\]

we expand

\begin{align*}
\mu(t) - \mu(r)
(t^{1-2b} - r^{1-2b}) f(t) + r^{1-2b} [f(t)-f(r)]
\\
(t^{1-2b} - r^{1-2b}) f(t) + r^{1-2b} \left[ \left( \frac{1}{t} - \frac{1}{r} \right) \int_0^r g(s) \, ds
+ \frac{1}{t} \int_r^t g(s) \, ds \right]
\\
\frac{t^{1-2b} - r^{1-2b}}{t^{1-2b}} \mu(t) + \frac{r-t}{t} \mu(r)
+ \frac{r^{1-2b}}{t} \int_r^t g(s) \, ds
\\
&=: \Delta_1 + \Delta_2 + \Delta_3.
\end{align*}

Thus $I_2 \le 3( K_1 + K_2 + K_3)$, where

\[
K_j = \int_{\R^d} \int_0^T \Abs{U(t,\xi)}^2
\int_{t/2}^t \frac{\Delta_j^2}{(t-r)^{1+2b}} \, dr \, dt \, d\xi \qquad (j=1,2,3).
\]

For $K_2$ we use once more the bound $\Abs{\mu(r)} \le C \norm{V}_T \norm{W}_T$ and get

\begin{multline*}
K_2 \le C \norm{V}_T^2 \norm{W}_T^2
\int_{\R^d} \int_0^T \Abs{U(t,\xi)}^2 t^{-2}
\left( \int_{t/2}^t (t-r)^{1-2b} \, dr \right) \, dt \, d\xi
\\
\le C' \norm{V}_T^2 \norm{W}_T^2
\int_{\R^d} \int_0^T \Abs{U(t,\xi)}^2 t^{-2b}
\, dt \, d\xi
\le C'' \norm{V}_T^2 \norm{W}_T^2 \norm{U}_T^2.
\end{multline*}

The same bound is obtained for $K_1$, since there $t^{1-2b} - r^{1-2b} \sim t^{-2b} (t-r)$.

For $K_3$ we bound, by Cauchy-Schwarz,

\[
(t-r)^{-2b} \int_r^t \Abs{g(s)} \, ds \le \norm{V}_{(r,t)} \norm{W}_{(r,t)}  \le C \norm{V}_T \norm{W}_T,
\]

where the last inequality follows from (<ref>). Thus

\begin{multline*}
\le
C \norm{V}_T^2 \norm{W}_T^2
\int_{\R^d} \int_0^T \Abs{U(t,\xi)}^2 t^{-4b}
\left( \int_{t/2}^t (t-r)^{2b-1} \, dr \right) \, dt \, d\xi
\\
\le C' \norm{V}_T^2 \norm{W}_T^2
\int_{\R^d} \int_0^T \Abs{U(t,\xi)}^2 t^{-2b}
\, dt \, d\xi
\le C'' \norm{V}_T^2 \norm{W}_T^2 \norm{U}_T^2.
\end{multline*}

This completes the proof of the lemma.

The following lemma was used in the above proof.

For all $T \in (0,T_0]$ and $R > 0$  we have the estimate

\begin{multline*}
\int_{\R^d} \int_0^T \Abs{U(t,\xi)}^2 \int_0^t \frac{1}{(t-r)^{1+2b}}
%  \\
%  \times
\left( \int_{\R^d} \int_r^t \int_0^s \frac{\Abs{V(s,\zeta)-V(\sigma,\zeta)}^2}{(s-\sigma)^{1+2b}} \, d\sigma \, ds \, d\zeta \right) \, dr \, dt \, d\xi
\\
\le C \norm{U}_T^2 \norm{V}_T^2,
\end{multline*}

with a constant $C$ depending only on $b$ and $T_0$.

Here $0 < r < s < t < T$ and $0 < \sigma < s$. Rearranging the order of the integrations, we rewrite the integral as

\begin{multline*}
\int_{\R^d} \int_0^T \int_0^s \left( \int_{\R^d} \int_s^T \Abs{U(t,\xi)}^2 \left( \int_0^s \frac{dr}{(t-r)^{1+2b}} \right) \, dt \, d\xi \right)
%  \\
%  \times
\frac{\Abs{V(s,\zeta)-V(\sigma,\zeta)}^2}{(s-\sigma)^{1+2b}} \, d\sigma \, ds \, d\zeta
\\
\le
\int_{\R^d} \int_0^T \int_0^s \left( \int_{\R^d} \int_s^T \Abs{U(t,\xi)}^2 \frac{1}{2b(t-s)^{2b}} \, dt \, d\xi \right) \frac{\Abs{V(s,\zeta)-V(\sigma,\zeta)}^2}{(s-\sigma)^{1+2b}} \, d\sigma \, ds \, d\zeta.
\end{multline*}

But by Lemma <ref> and (<ref>),

\[
\int_{\R^d} \int_s^T \Abs{U(t,\xi)}^2 \frac{1}{2b(t-s)^{2b}} \, dt \, d\xi
\le C \norm{U}_{(s,T)}^2 \le C \norm{U}_T^2,
\]

and the claimed inequality then follows.

We will also need the following double mean value theorem.

For all $x,y,X,Y \in \R$,

\begin{multline*}
\Abs{\theta(x) - \theta(y) - \theta(X) + \theta(Y)}
\\
\le \norm{\theta''}_{L^\infty}
\min\left(\Abs{x-y},\Abs{X-Y}\right)
\max\left(\Abs{x-X},\Abs{y-Y}\right)
%  \\
\norm{\theta'}_{L^\infty}\Abs{x-y-X+Y}.
\end{multline*}

Fix $x,y,X,Y$. By symmetry we may assume $\Abs{x-y} \le \Abs{X-Y}$. Defining $\kappa(t) = y + t(x-y)$ and $\rho(t) = Y + t(X-Y)$, we write

\[
\theta(x) - \theta(y)
= \left( \int_0^1 \theta'\left( \kappa(t) \right) \, dt \right) (x-y) =: I_1 (x-y)
\]

\[
\theta(X) - \theta(Y)
= \left( \int_0^1 \theta'\left( \rho(t) \right) \, dt \right) (X-Y) =: I_2 (X - Y).
\]

\[
\theta(x) - \theta(y) - \theta(X) + \theta(Y)
= (I_1-I_2)(x-y) + I_2 (x-y-X+Y).
\]

Clearly, $\Abs{I_2} \le \norm{\theta'}_{L^\infty}$, so it only remains to check that

\[
\Abs{I_1-I_2} \le \norm{\theta''}_{L^\infty}
\frac12 \left( \Abs{x-X} + \Abs{y-Y} \right).
\]

But this is clear since

\[
\int_0^1
\left ( \int_0^1
\theta''\left( \rho(t) + s \left[ \kappa(t) - \rho(t) \right] \right) \, ds
\right)
\left[ \kappa(t) - \rho(t) \right] \, dt
\]

\[
\Abs{\kappa(t) - \rho(t)} = \Abs{y - Y + t(x-y-X+Y)} \le (1-t)\Abs{y-Y} + t\Abs{x-X}.
\]

With these preliminary results in hand, we are now ready to start the proof of Proposition <ref>. We split the argument into several steps.

§.§ Cutoff estimate, version I

For all $T \in (0,T_0]$ and $R > 0$ we have the estimate

\begin{equation}\label{cutoff}
\norm{\theta_R\left( \norm{\mathbf U}_{t}^2 \right) V}_{T} \le C \left( 1 + R^{-1} \norm{\mathbf U}_{T}^2 \right) \norm{V}_{T},
\end{equation}

with a constant $C$ depending only on $b$ and $T_0$.

Setting $\psi(t) = \theta_R(\norm{\mathbf U}_{t}^2)$, the square of the left side of (<ref>) equals $A+2B$, where

\begin{align*}
A &= \int_{\R^d} T^{-2b} \int_0^T \Abs{\psi(t) V(t,\xi)}^2 \, dt \, d\xi,
\\
B &= \int_{\R^d} \int_0^T \int_0^t \frac{\Abs{\psi(t)V(t,\xi) - \psi(r)V(r,\xi)}^2}{(t-r)^{1+2b}} \, dr \, dt \, d\xi,
\end{align*}

and $\Abs{\psi(t)} \le 1$ implies $A \le \norm{V}_T^2$. Clearly, $B \le 2(B_1 + B_2)$,

\begin{align*}
B_1 &= \int_{\R^d} \int_0^T \int_0^t \Abs{\psi(r)}^2 \frac{\Abs{V(t,\xi) - V(r,\xi)}^2}{(t-r)^{1+2b}} \, dr \, dt \, d\xi,
\\
B_2 &= \int_{\R^d} \int_0^T \Abs{V(t,\xi)}^2 \int_0^t \frac{\Abs{\psi(t) - \psi(r)}^2}{(t-r)^{1+2b}} \, dr \, dt \, d\xi,
\end{align*}

where $\Abs{\psi(r)} \le 1$ implies $B_1 \le \norm{v_j}_T^2$. In $B_2$ we estimate, using (<ref>),

\begin{equation}\label{Trivial-cutoff-estimate}
\Abs{\psi(t) - \psi(r)}
\le \frac{C}{R} \Abs{ \norm{\mathbf U}_t^2 - \norm{\mathbf U}_r^2 },
\end{equation}

and obtain $B_2 \le R^{-2} \norm{\mathbf U}_T^4 \norm{V}_T^2$ as a consequence of Lemma <ref>.

The estimate (<ref>), using the equivalent norm (<ref>) instead of (<ref>), is claimed in [15], but there is a gap in the proof. To explain the problem, let us denote by $|\!|\!| \mathbf U |\!|\!|_T$ the norm used in [15], that is, the norm given by (<ref>):

\[
|\!|\!| \mathbf U |\!|\!|_T = \norm{\mathbb 1_{(0,T)} \mathbf u}_{\widehat X^{s,b}}, \quad \text{where $\mathbf U(t,\xi) = \japanese{\xi}^{s} e^{ith(\xi)} \widehat \mathbf u(t,\xi)$.}
\]

Then by the triangle inequality we have, for $0 < r < t$,

\begin{equation}\label{Too-rough}
\Abs{ |\!|\!| \mathbf U |\!|\!|_t - |\!|\!| \mathbf U |\!|\!|_r } \le \norm{\mathbb 1_{(r,t)} \mathbf u}_{\widehat X^{s,b}}
= |\!|\!| \mathbf U |\!|\!|_{(r,t)}.
\end{equation}

Combining (<ref>) with the analogue of (<ref>) for the norm $|\!|\!| \cdot |\!|\!|_t$ yields

\begin{equation}\label{Trivial-cutoff-estimate-2}
\Abs{\psi(t) - \psi(r)}
\le \frac{C}{R} |\!|\!| \mathbf U |\!|\!|_{(r,t)} |\!|\!| \mathbf U |\!|\!|_t,
\end{equation}

which is essentially what was used in [15], instead of (<ref>). But it is easy to see that (<ref>) is not enough to prove the estimate for $B_2$. Indeed, take $\mathbf U$ and $V$ both to be the function $\mathbb 1_{(0,T)}(t) f(\xi)$, where $0 < T < 1$ and $\norm{f}_{L^2} = 1$. Then by (<ref>) we have $\norm{\mathbf U}_{(r,t)} \sim (t-r)^{1/2-b}$ and $\norm{\mathbf U}_{t} \sim t^{1/2-b}$, so if we estimate $B_2$ by using (<ref>), then we get

\[
B_2 \le \frac{C}{R^2} \int_0^T t^{1-2b} \int_0^t (t-r)^{-4b} \, dr \, dt.
\]

But the right hand side equals $+\infty$ unless $b < 1/4$.

§.§ Cutoff estimate, version II

As a corollary to Lemma <ref>, we obtain the following variation on that result, proving the bound (<ref>) in Proposition <ref>, as well as (<ref>) in the case $\mathbf V=0$.

For all $T \in (0, T_0]$ and $R > 0$ we have the estimates

\begin{equation}
\label{cutoff2}
\norm{\theta_R\left( \norm{\mathbf U}_{t}^2\right) V}_{T} \le C \norm{V}_T,
\end{equation}

\begin{equation}
\label{particular_cutoff2}
\norm{\theta_R\left(\norm{\mathbf U}_{t}^2\right) \mathbf U}_{T}
\le C \sqrt R
\end{equation}

with a constant $C$ depending only on $b$ and $T_0$.

We first prove (<ref>). Recall the assumption $\supp\theta \subset [-2,2]$. So if $\norm{\mathbf U}_t^2 \ge 2R$ for all $t \in (0,T)$, then the left side of the inequality equals zero. It remains to consider the case where $\norm{\mathbf U}_t^2 < 2R$ for some $t \in (0,T)$. Define

\[
T_{\mathbf U} = \sup \left\{ t \in (0,T) \colon \norm{\mathbf U}_t^2 < 2R \right\}.
\]

\begin{equation}\label{Rbound}
\norm{\mathbf U}_{T_{\mathbf U}}^2 \le 2R,
\end{equation}

by the continuity of $\norm{\mathbf U}_t$ with respect to $t > 0$ (see Lemma <ref>). And if $T_{\mathbf U} < T$, then $\theta_R(\norm{\mathbf U}_{t}^2) = 0$ for $t \in [T_{\mathbf U},T]$. So (<ref>) yields (if $T_{\mathbf U} = T$, this holds trivially)

\[
\norm{\theta_R\left(\norm{\mathbf U}_{t}^2\right) V}_{T}
\le C \norm{\theta_R\left(\norm{\mathbf U}_{t}^2\right) V}_{T_{\mathbf U}},
\]

and by Lemma <ref> the right hand side is dominated by

\[
C\left( 1 + R^{-1} \norm{\mathbf U}_{T_{\mathbf U}}^2 \right) \norm{V}_{T_{\mathbf U}}
\le 3C \norm{V}_{T_{\mathbf U}}
\]

where we used (<ref>). By (<ref>),
$\norm{V}_{T_{\mathbf U}} \le C\norm{V}_{T}$, which proves (<ref>). Taking now $V=U_j$ and using the bound (<ref>), we get (<ref>).

§.§ Difference estimate, version I

For all $T \in (0, T_0]$ and $R > 0$ we have the estimates

\begin{equation}
\label{difference}
\norm{\theta_R\left(\norm{\mathbf U}_{t}^2\right) \mathbf U - \theta_R\left(\norm{\mathbf V}_{t}^2\right) \mathbf V}_{T}
%  \\
\le C \left( 1 + \frac{M^2}{R}
+ \frac{M^4}{R^2} \right)\norm{\mathbf U - \mathbf V}_T,
\end{equation}

\[
M = \norm{\mathbf U}_T + \norm{\mathbf V}_T
\]

and the constant $C$ depends only on $b$ and $T_0$

Setting $\psi(t) = \theta_R(\norm{\mathbf U}_{t}^2)$ and $\chi(t) = \theta_R(\norm{\mathbf V}_{t}^2)$, we reduce to proving, for $1 \le j \le n$,

\[
\norm{[\psi(t)-\chi(t)] U_j}_{T} \le \mathrm{r.h.s.}\eqref{difference}
\]

\[
\norm{\chi(t) (U_j - V_j)}_{T} \le C\norm{U_j - V_j}_{T}.
\]

The latter holds by (<ref>), so we concentrate on the former, whose left hand side squared equals $A+2B$, where

\begin{align*}
A &= \int_{\R^d} T^{-2b} \int_0^T \Abs{\psi(t)-\chi(t)}^2 \Abs{U_j(t,\xi)}^2 \, dt \, d\xi,
\\
B &= \int_{\R^d} \int_0^T \int_0^t \frac{\Abs{[\psi(t)-\chi(t)] U_j(t,\xi) - [\psi(r)-\chi(r)] U_j(r,\xi)}^2}{(t-r)^{1+2b}} \, dr \, dt \, d\xi.
\end{align*}

Using (<ref>) we estimate

\begin{equation}\label{alphabeta}
\Abs{\psi(t)-\chi(t)}
\le \frac{C}{R}
\Abs{\norm{\mathbf U}_t^2-\norm{\mathbf V}_t^2}.
\end{equation}

\begin{equation}\label{uvexp}
\norm{\mathbf U}_t^2-\norm{\mathbf V}_t^2 = \sum_{i=1}^n \left( \innerprod{U_i-V_i}{U_i}_t + \innerprod{V_i}{U_i-V_i}_t \right)
\end{equation}

and applying Cauchy-Schwarz yields

\begin{equation}\label{Diff1}
\left( \norm{\mathbf U}_t^2-\norm{\mathbf V}_t^2 \right)^2
\le
C M^2 \norm{\mathbf U - \mathbf V}_{T}^2 \quad \text{for $0 \le t \le T$},
\end{equation}

implying $A \le C R^{-2} M^4 \norm{\mathbf U - \mathbf V}_{T}^2$, as desired.

It remains to bound the term $B$. Defining

\[
\kappa = \psi-\chi
\]

we write $B \le 2(B_1 + B_2)$, where

\begin{align*}
B_1 &= \int_{\R^d} \int_0^T  \int_0^t \Abs{\kappa(r)}^2 \frac{\Abs{U_j(t,\xi)-U_j(r,\xi)}^2}{(t-r)^{1+2b}} \, dr \, dt \, d\xi,
\\
B_2 &= \int_{\R^d} \int_0^T \Abs{U_j(t,\xi)}^2 \int_0^t \frac{\Abs{\kappa(t)-\kappa(r)}^2}{(t-r)^{1+2b}} \, dr \, dt \, d\xi.
\end{align*}

By (<ref>) and (<ref>), $\Abs{\kappa(r)}^2 \le C R^{-2} M^2 \norm{\mathbf U - \mathbf V}_{T}^2$,

\[
B_1 \le C R^{-2} M^4 \norm{\mathbf U - \mathbf V}_{T}^2.
\]

In $B_2$ we write out

\[
\kappa(t)-\kappa(r)
= \theta_R(\norm{\mathbf U}_{t}^2) - \theta_R(\norm{\mathbf V}_{t}^2) - \left[\theta_R(\norm{\mathbf U}_{r}^2) - \theta_R(\norm{\mathbf V}_{r}^2) \right]
\]

and apply Lemma <ref> to get

\[
B_2 \le C \left( \frac{I}{R^4} + \frac{J}{R^2} \right),
\]

\begin{equation*}
I = \int_{\R^d} \int_0^T \Abs{U_j(t,\xi)}^2 \left( \norm{\mathbf U}_{t}^2 - \norm{\mathbf V}_{t}^2 \right)^2
%  \\
%  \times
\int_0^t \frac{\left( \norm{\mathbf U}_{t}^2 - \norm{\mathbf U}_{r}^2\right)^2
+ \left( \norm{\mathbf V}_{t}^2 - \norm{\mathbf V}_{r}^2 \right)^2}{(t-r)^{1+2b}} \, dr \, dt \, d\xi
\end{equation*}

\[
J = \int_{\R^d} \int_0^T \Abs{U_j(t,\xi))}^2
\int_0^t \frac{\left( \norm{\mathbf U}_{t}^2 - \norm{\mathbf V}_{t}^2 - \norm{\mathbf U}_{r}^2 + \norm{\mathbf V}_{r}^2 \right)^2}{(t-r)^{1+2b}} \, dr \, dt \, d\xi.
\]

By (<ref>) and Lemma <ref>,

\[
I \le C_b M^8 \norm{\mathbf U - \mathbf V}_{T}^2,
\]

which is acceptable. Finally, using (<ref>) and Lemma <ref>,

\[
J \le C_b M^4 \norm{\mathbf U - \mathbf V}_{T}^2,
\]

and this concludes the proof of the lemma.

§.§ Difference estimate, version II

We are now in a position to finish the proof of Proposition <ref>, by proving the bound (<ref>).

For all $T \in (0,T_0]$ and $R > 0$ we have the estimate

\begin{equation}\label{difference2}
\norm{\theta_R\left(\norm{\mathbf U}_{t}^2\right) \mathbf U - \theta_R\left(\norm{\mathbf V}_{t}^2\right) \mathbf V}_{T}
\le C \norm{\mathbf U-\mathbf V}_T,
\end{equation}

where the constant $C$ depends only on $b$ and $T_0$.

First, if $\norm{\mathbf U}_{t}^2 \ge 2R$ and $\norm{\mathbf V}_{t}^2 \ge 2R$ for all $t \in [0,T]$, then the left side equals zero.

Second, consider the hybrid case where $\norm{\mathbf U}_{t}^2 \ge 2R$ for all $t \in [0,T]$, whereas $\norm{\mathbf V}_{t}^2 < 2R$ for some $t \in [0,T]$. Then $\theta_R(\norm{\mathbf U}_{t}^2) = 0$ for $t \in [0,T]$. Defining $T_{\mathbf V}$ as in the proof of Lemma <ref>, we find

\[
\mathrm{l.h.s.}\eqref{difference2}
= \norm{\theta_R\left(\norm{\mathbf V}_{t}^2\right) \mathbf V}_{T}
\le
C \norm{\theta_R\left(\norm{\mathbf V}_{t}^2\right) \mathbf V}_{T_{\mathbf V}}
\le
C \sqrt{2R},
\]

by Lemma <ref>. So if $\norm{\mathbf U-\mathbf V}_{T_{\mathbf V}} \ge \sqrt{R}$, we are done. If, on the other hand, $\norm{\mathbf U-\mathbf V}_{T_{\mathbf v}} \le \sqrt{R}$, then by the triangle inequality, $\norm{\mathbf U}_{T_{\mathbf V}} \le (1+\sqrt{2})\sqrt{R}$, since $\norm{\mathbf V}_{T_{\mathbf V}}^2 \le 2R$. Then (<ref>) follows by applying Lemma <ref> to

\[
\norm{\theta_R(\norm{\mathbf V}_{t}^2) \mathbf V}_{T_{\mathbf v}} =
\norm{\theta_R(\norm{\mathbf U}_{t}^2) \mathbf U - \theta_R(\norm{\mathbf V}_{t}^2) \mathbf V}_{T_{\mathbf V}}.
\]

The other hybrid case is symmetric, so we are only left with the case where $\norm{\mathbf U}_t^2 < 2R$ for some $t \in [0,T]$ and $\norm{\mathbf V}_s^2 < 2R$ for some $s \in [0,T]$. Define $T_{\mathbf U}$ and $T_{\mathbf V}$ as in the proof of Lemma <ref>, so $\norm{\mathbf U}_{T_{\mathbf U}}^2 \le 2R$ and $\norm{\mathbf V}_{T_{\mathbf V}}^2 \le 2R$. By symmetry we may assume $T_{\mathbf U} \le T_{\mathbf V}$, and then (<ref>) yields

\[
\mathrm{l.h.s.}\eqref{difference2}
\le
C \norm{\theta_R(\norm{\mathbf U}_{t}^2) \mathbf U - \theta_R(\norm{\mathbf V}_{t}^2) \mathbf V}_{T_{\mathbf V}}.
\]

If $\norm{\mathbf U}_{T_{\mathbf V}}^2 \le 8R$, we now obtain (<ref>) by Lemma <ref>. If, on the other hand, $\norm{\mathbf U}_{T_{\mathbf V}}^2 > 8R$, then $\norm{\mathbf U - \mathbf V}_{T_{\mathbf V}} \ge \sqrt{2R}$, and using (<ref>) we obtain

\[
\mathrm{l.h.s.}\eqref{difference2}
\le
C \left( \norm{\theta_R(\norm{\mathbf U}_{t}^2) \mathbf U - \theta_R(\norm{\mathbf V}_{t}^2) \mathbf V}_{T_{\mathbf U}} +
\norm{ \theta_R(\norm{\mathbf V}_{t}^2) \mathbf V}_{(T_{\mathbf U},T_{\mathbf V})} \right).
\]

The first term on the right can be handled by Lemma <ref>, since $\norm{\mathbf U}_{T_{\mathbf U}}^2 \le 2R$ and  $\norm{\mathbf V}_{T_{\mathbf U}}^2 \le C\norm{\mathbf V}_{T_{\mathbf V}}^2 \le C2R$. For the second term, we get by (<ref>) and Lemma <ref>,

\[
\norm{ \theta_R(\norm{\mathbf V}_{t}^2) \mathbf V}_{(T_{\mathbf U},T_{\mathbf V})}
\le
C \norm{ \theta_R(\norm{\mathbf V}_{t}^2) \mathbf V}_{T_{\mathbf V}}
\le
C' \norm{ \mathbf V}_{T_{\mathbf V}}
\le
C' \norm{\mathbf U - \mathbf V}_{T_{\mathbf V}},
\]

where we used $\norm{\mathbf U - \mathbf V}_{T_{\mathbf V}} \ge \sqrt{2R} \ge \norm{\mathbf V}_{T_{\mathbf V}}$.
This concludes the proof of the lemma.

§.§ Sobolev–Slobodeckij norm on $H^b(0, T)$

All the properties discussed above in this section
are valid for usual Sobolev norms,
for functions depending only on the time variable $t$.
The proofs can be repeated directly without much of a difference.
However, we can get also those properties easily by

\[
u(t) = \phi(t) S_{h(\xi)}(t) f
\]

with  $\phi \in H^b(S,T)$ and $f \in H^s(\R^d)$.
Indeed, from (<ref>) and (<ref>) it is clear that

\[
\norm{u}_{X^{s,b}_{h(\xi)}(S,T)} = \norm{\phi}_{H^b(S,T)} \norm{f}_{H^s}
\quad \text{and} \quad
\norm{u}_{\widetilde X^{s,b}_{h(\xi)}(S,T)} = \norm{\phi}_{\widetilde H^b(S,T)} \norm{f}_{H^s},
\]

\[
\norm{\phi}_{\widetilde H^b(S,T)} =
\frac{1}{(T-S)^{2b}} \int_S^T \Abs{\phi(t)}^2 \, dt
+ \int_S^T \int_S^T \frac{\Abs{\phi(t)-\phi(r)}^2}{\Abs{t-r}^{1+2b}} \, dr \, dt.
\]

Normalising by taking $\norm{f}_{H^s} = 1$, we then get from Lemma <ref>, for $b \in (0,1/2)$,

\begin{equation}
\label{Sob-ModEquivalence}
C_{T_0,b}^{-1} \norm{\phi}_{H^b(S,T)}
\le
\norm{\phi}_{\widetilde H^b(S,T)}
\le C_{T_0,b}
\norm{\phi}_{H^b(S,T)} \qquad (0 \le S < T \le T_0),
\end{equation}

and Proposition <ref> has the following analogue.

Let $T_0 > 0$ and $b \in (0,1/2)$. Let $n \in \N$ and suppose that $\phi_i,\Phi_i \in H^b(0,T_0)$ for $1 \le i \le n$. Then for $T \in (0,T_0]$, $R > 0$ and $1 \le j \le n$ we have the estimates

\[
\norm{
\theta_R\left( \sum_{i=1}^n \norm{\phi_i}_{\widetilde H^{b}(0,t)}^2 \right)
\phi_j(t)
\leqslant
C \sqrt{R}
\]

\[
\norm{
\theta_R\left( \sum_{i=1}^n \norm{\phi_i}_{\widetilde H^{b}(0,t)}^2 \right)
\phi_j(t)
\theta_R\left( \sum_{i=1}^n \norm{\Phi_i}_{\widetilde H^{b}(0,t)}^2 \right)
\Phi_j(t)
\leqslant
C \sum_{i=1}^n \norm{\phi_i-\Phi_i}_{H^{b}(0,T)}
\]

where $C$ depends only on $b$, $T_0$ and $\theta$.

ED was partially supported by the ERC EU project 856408-STUOD
the Research Council of Norway through its Centres
of Excellence scheme, Hylleraas-senteret, project number 262695.
SS was partially supported by the Meltzer Research Fund.
The authors would like to thank the anonymous referees,
whose valuable comments helped us to improve the paper during the revision.

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-upcolorlinks=true, linkcolor=blue, anchorcolor=blue, citecolor=red, filecolor=blue, menucolor=blue, pagecolor=blue, urlcolor=blue

6.5 in

Stochastic Dirac-Klein-Gordon system

A conservative stochastic Dirac-Klein-Gordon system

Dinvay]Evgueni Dinvay
Selberg]Sigmund Selberg

Department of Chemistry

UiT The Arctic University of Norway

PO Box 6050 Langnes

N-9037 Tromsø

Department of Mathematics

University of Bergen

PO Box 7803

5020 Bergen

[2010]35Q53, 35Q60, 60H15

Considered herein is a particular nonlinear dispersive stochastic equation...

§ INTRODUCTION

Consideration is given to the following
Dirac-Klein-Gordon equations, one dimensional in the space variable,
containing multiplicative white noise of the Stratonovich type,

\begin{equation}
\label{Stratonovich_Dirac}
\left \{
\begin{aligned}
\left( -i \partial_t - i\alpha\partial_x + M\beta \right) \psi
\phi \beta \psi
\beta \psi \xi_1
\\
\left( \partial_t^2-\partial_x^2 + m^2 \right) \phi
\psi^* \beta \psi
\phi \xi_2
\end{aligned}
\right.
\end{equation}

with initial data

\begin{equation}\label{Data}
\psi(0,x,\omega) = \psi_0(x,\omega)
, \qquad
\phi(0,x,\omega) = \phi_0(x,\omega)
, \qquad
\partial_t\phi(0,x,\omega) = \phi_1(x,\omega).
\end{equation}

The unknowns are random processes $\psi(t, x, \omega) \in \C^2$ and $\phi(t, x, \omega) \in \R$, for $(t,x) \in \R^2$ and $\omega$ in a probability space $(\Omega,\mathcal F,\mathbb P)$. Here $m, M \geqslant 0$ are constants and
the $2 \times 2$ Dirac matrices $\alpha,\beta$ satisfy
\(
\alpha = \alpha^*,
\,
\beta = \beta^*,
\,
\alpha^2 = \beta^2 = I
\)
\(
\alpha \beta + \beta \alpha = 0.
\)
For simplicity we choose the particular representation

\[
\alpha
\begin{pmatrix}
\\
\end{pmatrix}
, \qquad
\beta
\begin{pmatrix}
\\
\end{pmatrix}
\]

Let $W$ be the cylindrical Wiener process defined by
a complete orthonormal sequence
\(
\{ e_j \} _{ j \in \mathbb N }
\)
\(
L^2( \R; \R )
\)
and a sequence
\(
\{ B_j \} _{ j \in \mathbb N }
\)
of independent real-valued Brownian motions on $(\Omega,\mathcal F,\mathbb P)$.
We assume that the noise is of the form

\[
\xi_j = \frac{d \mathcal W_j}{dt},
\qquad
\mathcal W_j = \mathfrak K_j W \qquad (j=1,2),
\]

where the $\mathfrak K_j$ are convolution operators

\begin{equation}
\label{convolution_operator}
\mathfrak K_jf(x)
\int_{ \R }
\mathfrak k_j(x - y) f(y) dy
\end{equation}

with real-valued kernels $\mathfrak k_j \in H^{\sigma_j}( \R )$, where $\sigma_j \ge 0$ will be chosen depending on the Sobolev regularity of the initial data.

Interpreting the stochastic integrals in the Stratonovich sense, we can then write (<ref>) as

\begin{equation}\label{StratonovichDKG}
\left\{
\begin{gathered}
d\psi = (-\alpha^j \partial_j - iM \beta) \psi \, dt + i\phi \beta \psi \, dt + i\beta \psi \mathfrak K_1 \circ dW,
\\
d\phi = \dot\phi \, dt,
\\
d\dot\phi = (\partial_x^2 - m^2) \phi \, dt + \psi^* \beta \psi \, dt + \phi \mathfrak K_2 \circ dW,
\end{gathered}
\right.
\end{equation}

where $\dot\phi = \partial \phi/\partial t$ and $\psi \mathfrak K_1$, $\phi \mathfrak K_2$ are understood as compositions of the convolution operators $\mathfrak K_1$, $\mathfrak K_2$ with the multiplication operators given by $\psi$, $\phi$. Thus

\[
(\psi\mathfrak K_1)f(x)
\psi(x)\int_{ \R }
\mathfrak k_1(x - y) f(y) dy
\]

and similarly for $\phi\mathfrak K_2$. Depending on the regularity of the initial data, the Sobolev regularity $\sigma_j$ of the kernels $\mathfrak k_j$ will be chosen so that the above compositions are Hilbert-Schmidt operators from $L^2(\R,\R)$ into suitable Sobolev spaces.

By introducing noise in the Stratonovich sense we respect two of the key physical
properties of the original deterministic Dirac-Klein-Gordon system:
the principle of least action and the conservation of the charge $\int \Abs{\psi(t,x)}^2 \, dx$.

From an analysis perspective it is more convenient to work
with the Itô stochastic integral. The Itô form of the above system is

\begin{equation}\label{ItoDKG}
\left\{
\begin{gathered}
d\psi =
(-\alpha^j \partial_j - iM \beta) \psi \, dt + i\phi \beta \psi \, dt
- M_{\mathfrak K_1} \psi \, dt + i\beta \psi \mathfrak K_1 \, dW,
\\
d\phi = \dot\phi \, dt,
\\
d\dot\phi = (\partial_x^2 - m^2) \phi \, dt + \psi^* \beta \psi \, dt + \phi \mathfrak K_2 \, dW,
\end{gathered}
\right.
\end{equation}

where $M_{\mathfrak K_1} = (1/2) \norm{\mathfrak k_1}_{L^2}^2$. To see this, write (<ref>) on the abstract form

\[
dX = AX dt + \mathcal N(X) dt + \mathcal M(X) \circ d W,
\]

\[
X = \begin{pmatrix} \psi \\ \phi \\ \dot\phi \end{pmatrix},
\qquad
A = \begin{pmatrix} -\alpha^j\partial_j - iM\beta & 0 & 0 \\ 0 & 0 & 1 \\ 0 & \Delta & 0 \end{pmatrix},
\qquad
\mathcal M(X) = \begin{pmatrix} i\beta\psi \mathfrak K_1 \\ 0 \\ \phi \mathfrak K_2 \end{pmatrix}.
\]

Then, at least formally, the corresponding Itô form is (see [19])

\[
dX = AX dt + \left( \mathcal N(X) + \frac12 \sum_k \mathcal M_k[ \mathcal M_k(X) ] \right) \, dt + \mathcal M(X) d W,
\]

where $\mathcal M_k(X) = \mathcal M(X)e_k$ and we calculate

\[
\mathcal M_k[ \mathcal M_k(X) ]
\begin{pmatrix} (i \mathfrak K_1 e_k \beta)^2\psi \\ 0 \\ 0\end{pmatrix}
\begin{pmatrix} - (\mathfrak K_1 e_k)^2 \psi \\ 0 \\ 0\end{pmatrix}
\]

and note that for all $x \in \R$,

\[
\sum_k \left( \mathfrak K_1e_k(x) \right)^2 =
\sum_k \innerprod{\mathfrak k_1(x - \cdot)}{e_j}_{L^2}^2 =
\norm{\mathfrak k_1}_{L^2}^2
\]

by Parseval's identity. Formally, this verifies the conversion from (<ref>) to (<ref>).

Our aim is to prove existence and uniqueness for (<ref>)
with initial data $(\psi,\phi,\dot\phi)(0) = (\psi_0,\phi_0,\phi_1)$ in the spaces

\begin{equation}
\label{DataRegularity}
\psi_0 \in L^2\left(\Omega,H^s( \R,\C^2 )\right)
, \qquad
\phi_0 \in L^2\left(\Omega,H^r( \R,\R )\right)
, \qquad
\phi_1 \in L^2\left(\Omega,H^{r-1}( \R, \R )\right)
\end{equation}

for a certain range of Sobolev indices $s,r \in \R$.
In particular, using the charge conservation we will prove global existence when $s = 0$ and $0 < r < 1/2$, under the additional assumption that $\psi_0 \in L^4(\Omega,L^2(\R))$.

Assuming for the moment that $m > 0$, then by a rescaling we may take $m=1$. Applying the linear transformation $(\psi,\phi,\dot\phi) \mapsto (\psi_+,\psi_-,\phi_+,\phi_-)$ given by

\[
\psi = \begin{pmatrix} \psi_+ \\ \psi_- \end{pmatrix},
\qquad
\phi=\phi_+ + \phi_-,
\qquad
\phi_\pm = \frac12 \left(\phi \pm \japanese{D_x}^{-1} i\dot\phi \right),
\]

where $D_x=-i\partial_x$ and $\japanese{\cdot} = (1+\Abs{\cdot}^2)^{1/2}$,
the Cauchy problem (<ref>), (<ref>) transforms to

\begin{equation}
\label{Ito-DKG-split}
\left\{
\begin{aligned}
&-i d \psi_+ + D_x \psi_+ \, dt
-  M \psi_- \, dt
\phi \psi_- \, dt
\psi_- \mathfrak K_1 \, d W
iM_{\mathfrak K_1} \psi_+ \, dt,
\\
&- i d \psi_- - D_x \psi_- \, dt
-  M \psi_+ \, dt
\phi \psi_+ \, dt
\psi_+  \mathfrak K_1 \, d W
iM_{\mathfrak K_1} \psi_- \, dt,
\\
&-id \phi_+ + \japanese{D_x} \phi_+ dt = - \japanese{D_x}^{-1} \re \left( \overline{\psi_+} \psi_- \right) dt
- \japanese{D_x}^{-1} \phi \mathfrak K_2 \, dW,
\\
&-id \phi_- - \japanese{D_x} \phi_- dt = + \japanese{D_x}^{-1} \re \left( \overline{\psi_+} \psi_- \right) dt
+ \japanese{D_x}^{-1} \phi \mathfrak K_2 \, dW,
\end{aligned}
\right.
\end{equation}

\begin{equation}
\label{Ito-DKG-split-data}
\psi_\pm(0) = f_\pm \in L^2\left(\Omega,H^a(\R,\C)\right),
\qquad
\phi_\pm(0) = g_\pm \in L^2\left(\Omega,H^r(\R,\C)\right),
\qquad
\overline{g_+} = g_-.
\end{equation}

Here $\japanese{D_x}^{-1} \phi \mathfrak K_2$ is understood as a composition of operators. We remark that $\overline{\phi_+} = \phi_-$. Thus $\phi=\phi_+ + \overline{\phi_+}$ and it suffices to solve for $\psi_+$, $\psi_-$ and $\phi_+$.

For an initial data
\(
\psi(0) = \psi_0
\)
it has the following mild form

\begin{equation}
\label{mild_Dirac}
\psi(t)
S(t) \psi_0
\int_0^t S(t - s) \ldots(s) ds
i \int_0^t S(t - s) \psi \Phi d W(s)
\frac 12 \int_0^t S(t - s) \psi F_{\Phi} ds
\end{equation}

where ...

The deterministic analogue of the Dirac-Klein-Gordon system was studied
in [11, 29, 30]
by iterating in Bourgain type spaces.
Important contributions to one space-dimensional case:
First global result 1d was obtained in [9],
global result 2d in [20],
global results in 3d for small data in [3, 8].
Expand this part.

In order to study well-posedness of the stochastic extension
(<ref>) we adopt the technique
based on analysis in Bourgain spaces and
developed in [15].
We are also taking advantage of the charge conservation,
\(
\norm{\psi(t)} _{L^2} = \mbox{const}
\)
and that is why we need a Stratonovich noise,
similar to one regarded in [14]
with a nonlinear Schrödinger equation.
We work in Bourgain spaces of low time regularity
and so we make use of product estimates proved in
<cit.>. Once these have been obtained, the local existence and uniqueness for (<ref>) follows from an abstract stochastic well-posedness theorem that we prove in section... Main ideas most clearly expressed in an abstract framework...

In the classical field theory one can determine the equations of motion
by the principle of least action.
As we restrict ourselves to the one dimensional space,
the action is an integral functional
\(
\mathcal S
%    (\psi, \phi)
\int
\mathcal L
%    (\psi, \phi, t)
\)
where the Lagrangian density, depending on the field and time,
is defined by the physical system under consideration.
The deterministic analogue of Equations (<ref>)
is related to a particular choice of
the density $\mathcal L(\psi, \phi, t)$,
as explained in
[5, 32].
We recall very briefly the corresponding physical background
and show how the noise can be naturally introduced here.
In particle physics, the Yukawa interaction [32]
explains how forces between nucleons are mediated
by massive particles called mesons.
Mathematically, this is described by the action integral
\(
\mathcal S(\psi,\phi)
\)
defined by the Lagrangian density

\[
\mathcal L(\psi,\phi)
\mathcal L_{\mathrm{Dirac}}(\psi)
\mathcal L_{\mathrm{meson}}(\phi)
\mathcal L_{\mathrm{Yukawa}}(\psi,\phi)
\]

Here $\psi$ is a spinor field (the fermion field),
$\phi$ is a real scalar field (the meson field),
whose free-field dynamics are determined by the Lagrangians

\[
\mathcal L_{\mathrm{Dirac}}(\psi)
\psi^*
\left(
i \partial_t + i \alpha \partial_x - M \beta
\right)
\psi
, \qquad
\mathcal L_{\mathrm{meson}}(\phi)
\frac 12 ( \partial_t \phi )^2
\frac 12 ( \partial_x \phi )^2
\frac12 m^2 \phi^2,
\]

corresponding to the free Dirac and Klein-Gordon equations;
$m,M \ge 0$ are masses,
$\psi^*$ the complex conjugate transpose of $\psi$.
The interaction is determined by the Yukawa coupling term

\[
\mathcal L_{\mathrm{Yukawa}}(\psi,\phi)
\phi \psi^* \beta \psi
\]

The corresponding system of Euler-Lagrange equations
is the deterministic Dirac-Klein-Gordon system.
We introduce the noise by adding

\[
\mathcal L_{\mathrm{noise}}(\psi, \phi, x, t)
\mathcal L_{\mathrm{Dirac}}^{\mathrm{noise}}(\psi, x,t)
\mathcal L_{\mathrm{meson}}^{\mathrm{noise}}(\phi, x,t)
\psi^* \beta \psi \xi_1
\frac 12 \phi^2 \xi_2
\]

One can think about $\mathcal L_{\mathrm{Dirac}}^{\mathrm{noise}}$
as stochastic fluctuations of the initial Dirac potential
\(
M \psi^* \beta \psi
\)
$\mathcal L_{\mathrm{meson}}^{\mathrm{noise}}$
represents a noisy extension of the potential
\(
m^2 \phi^2 / 2
\)
in the Klein-Gordon model.
Thus using the new Lagrangian density

\[
\mathcal L(\psi, \phi, x, t)
\mathcal L_{\mathrm{Dirac}}(\psi)
\mathcal L_{\mathrm{meson}}(\phi)
\mathcal L_{\mathrm{Yukawa}}(\psi,\phi)
\mathcal L_{\mathrm{noise}}(\psi, \phi, x, t)
\]

one arrives to a stochastic variational principle
leading to System (<ref>).
Note that the Stratonovich calculus obeys normal differentiation rules,
and so the derivation of the Euler-Lagrange equations
works out as in the deterministic case.

The paper is devoted to analysis of existence and uniqueness of
a mild solution to the Cauchy problem for
System (<ref>)
complemented with the initial data (<ref>).
It is organised as follows.
In the next section we introduce some preliminary notions
that will be used throughout the paper.
Section <ref> provides with the mild formulation
of the Cauchy problem and the main existence theorems.
Section <ref> is devoted to analysis
of the stochastic integrals we are dealing with.
Then in Section <ref> we prove bilinear estimates
necessary for treating nonlinear terms.
In Section <ref> we introduce and solve
a truncated system.
As a result we obtain a local mild solution to
(<ref>), (<ref>).
Section <ref> is devoted to the proof
of charge conservation.
Finally, in Section <ref>
we prove existence of a global solution.
Proofs of some very technical results are left for the last section,
where we prove in general terms the so-called cutoff estimates.
The idea of making use of a Slobodeckij norm for this
comes from [15].
However, it turns out that the treatment should be more delicate
than the argument given in [15].

§ PRELIMINARIES

First, we fix some general notational conventions.

As usual, $C$ will denote various positive constants, and its meaning can change from one instance to the next.

The characteristic function of a set $E$ will be denoted $\mathbb 1_E$. If $E$ is determined by some property $P$, say $E = \{ x \colon P(x) \}$, we will often use the convenient notation $\mathbb 1_{P(x)}$ for $\mathbb 1_E(x)$. If $E$ is a subset of a set $X$, and $f$ is a function defined on $E$, then by a slight abuse of notation we shall denote by $\mathbb 1_E f$ the extension of $f$ by zero outside $E$. We call this the trivial extension (of $f$).

We will adhere to the following convention regarding restrictions of $\sigma$-algebras. Suppose that $\mathcal M$ is a $\sigma$-algebra and that $E \in \mathcal M$. Let $\mathcal M \vert_E$ be the $\sigma$-algebra on $E$ consisting of all sets $A \cap E$, where $A \in \mathcal M$. Then if $f \colon E \to H$ is $\mathcal M \vert_E$-measurable, we will simply say that it is $\mathcal M$-measurable. This is of course equivalent to saying that the trivial extension is $\mathcal M$-measurable.

§.§ Random variables

We fix a filtered probability space $(\Omega,\mathcal F,\{\mathcal F_t\}_{t \ge 0},\mathbb P)$ admitting an independent sequence $\{B_k\}_{k \in \N}$ of one-dimensional Brownian motions. We write $\mathbb E(X) = \int_\Omega X(\omega) \, d\mathbb P(\omega)$ for $X \in L^1(\Omega)$.

A stochastic process $X(t)$, defined on a time interval $I = [S,T]$ or $I = [S,\infty)$, where $S \ge 0$, and taking values in a Hilbert space $H$, is said to be $H$-adapted (or just adapted if it is clear from the context which Hilbert space is meant) if $X(t)$ is $(\mathcal F_t,\mathcal B_H)$-measurable for all $t \in I$. In other words, $\innerprod{X(t)}{h}_H$ is $\mathcal F_t$-measurable for any $h \in H$. Here $\mathcal B_H$ denotes the Borel $\sigma$-algebra of $H$.

A process $Y(t)$ is a modification of $X(t)$ if for each $t \in I$ we have $X(t)=Y(t)$ a.s. We assume that $\mathcal F_0$ contains all sets in $\mathcal F$ with measure zero, so that any modification of an adapted process is itself adapted.

The process $X(t)$ is progressively measurable if for each $t \in I$ the map $(s,\omega) \mapsto X(s,\omega)$, from $[S,t] \times \Omega$ into $H$, is $(\mathcal B_{[S,t]} \otimes \mathcal F_t,\mathcal B_H)$-measurable. Progressive measurability implies adaptedness (see <cit.>), and the converse holds if the process has continuous paths (see <cit.> or Lemma <ref> below).

If $Z$ is some Banach space of functions from $[S,T]$ into $H$, we define

\begin{equation}\label{L2-prog}
\mathbb L^2(\Omega,Z) = \left\{ u \in L^2(\Omega,Z) \colon \text{$u$ is progressively measurable} \right\}.
\end{equation}

§.§ Stopping times

A stopping time is a random variable $\tau \colon \Omega \to [0,\infty]$ such that for all $t \ge 0$, the set
$\{ \tau \le t \} = \{ \omega \in \Omega \colon \tau(\omega) \le t \}$
is $\mathcal F_t$-measurable. Then also $\{ \tau < t \}$, $\{ \tau > t \}$ and $\{ \tau \ge t \}$ have this property, of course. Note that any constant $\tau \ge 0$ is a stopping time.

In the next three lemmas we establish some facts about stopping times. We only consider strictly positive times.

Let $\tau \colon \Omega \to (0,\infty]$ be a stopping time. Then the set

\[
E = \left\{ (s,\omega) \in [0,\infty) \times \Omega \colon 0 \le s < \tau(\omega) \right\}
\]

belongs to the product $\sigma$-algebra $\mathcal B_{[0,\infty)} \otimes \mathcal F$. For any $T > 0$, the set

\[
E_T = E \cap \left( [0,T] \times \Omega \right)
\]

belongs to $\mathcal B_{[0,T]} \otimes \mathcal F_T$.

Let $A_n$ be the set of numbers $iT/2^n$, $i=0,\dots,2^n$. Then
$E_T = \bigcup_{n \in \N} \bigcup_{t \in A_n} [0,t] \times \{t < \tau\}$,
and of course $E = \bigcup_{N \in \N} E_N$.

As a consequence of this lemma, if $u$ is a random variable defined on $E$ (so it is defined up to the time $\tau$), then it makes sense to ask whether $u$ is $\mathcal B_{[0,\infty)} \otimes \mathcal F$-measurable. That is, whether the trivial extension of $u$ has this property. Similarly, one can ask whether $u$ restricted to $E_t$ is $\mathcal B_{[0,t]} \otimes \mathcal F_t$-measurable for all $t \ge 0$, which amounts to progressive measurability of the trivial extension. The next lemma gives sufficient conditions for this to hold.

Let $\tau \colon \Omega \to (0,\infty]$ be a stopping time, and let $E$ and $E_t$, $t \ge 0$, be as in Lemma <ref>. Let $u \colon E \to H$, where $H$ is a separable Hilbert space. Assume that $u$ has continuous paths, in the sense that

\begin{equation}\label{Stopping-lemma-2-1}
\text{$t \mapsto u(t,\omega)$ is continuous on $[0,\tau(\omega))$, for each $\omega$},
\end{equation}

and assume that $u$ is adapted, in the sense that, for each $t \ge 0$ such that $\left\{ t < \tau \right\}$ is non-empty, we have

\begin{equation}\label{Stopping-lemma-2-2}
\text{$\omega \mapsto u(t,\omega)$ is $\mathcal F_t$-measurable.}
\end{equation}

Then $u \vert_{E_t}$ is $\mathcal B_{[0,t]} \otimes \mathcal F_t$-measurable for all $t \ge 0$. In other words, the trivial extension of $u$ is progressively measurable.

Extend $u$ trivially by setting $u(t,\omega) = 0$ for $t \ge \tau(\omega)$. Then (<ref>) holds for all $t \ge 0$. Now fix $t \ge 0$. For $n \in \N$ let $t_i = it/2^n$ for $i=0,\dots,2^n$, and define

\[
u_n(s,\omega) =  u(0,\omega) \mathbb 1_{\{0\}}(s)
+ \sum_{i=1}^{2^n} u(t_i,\omega) \mathbb 1_{(t_{i-1},t_i]}(s).
\]

Then $u_n$ is $\mathcal B_{[0,t]} \otimes \mathcal F_t$-measurable by (<ref>), and (<ref>) implies that $u_n$ converges pointwise to $u$ in $E_t$.

Let $\tau \colon \Omega \to (0,\infty]$ be a stopping time. Suppose that $f(t,\omega) \ge 0$ is defined for $0 \le t < \tau(\omega)$ and that for each $\omega$,

\begin{equation}\label{Stopping-lemma-3-1}
\text{$t \mapsto f(t,\omega)$ is continuous on $[0,\tau(\omega))$, and $f(0,\omega) =0$},
\end{equation}

and moreover that, for each $t \ge 0$ such that $\left\{ t < \tau \right\}$ is non-empty,

\begin{equation}\label{Stopping-lemma-3-2}
\text{$\omega \mapsto f(t,\omega)$ is $\mathcal F_t$-measurable.}
\end{equation}

For $R > 0$ define $\tau_R \colon \Omega \to (0,\infty]$ by

\[
\tau_R(\omega) = \sup \left\{ t \in [0,\tau(\omega)) \colon \text{$f(s,\omega) < R$ for $0 \le s \le t$} \right\}.
\]

Then $\tau_R$ is a stopping time. Moreover, for each $\omega$,

\begin{equation}\label{StoppingTime1}
\lim_{R \to \infty} \tau_R(\omega) = \tau(\omega),
\end{equation}

\begin{equation}\label{StoppingTime2}
0 \le t \le \tau_R(\omega) \implies f(t,\omega) \le R
\end{equation}

\begin{equation}\label{StoppingTime3}
\tau_R < \tau(\omega) \implies f(\tau_R(\omega),\omega) = R.
\end{equation}

For each $\omega$ there are two alternatives: (i) $f(s,\omega) < R$ for all $0 \le s < \tau(\omega)$,  or (ii) $f(s,\omega) \ge R$ for some $0 \le s < \tau(\omega)$. Equivalently, (i) $\tau_R(\omega) = \tau(\omega)$, or (ii) $\tau_R(\omega) < \tau(\omega)$. Moreover, in case (ii), $\tau_R(\omega)$ is the smallest $s \in (0,\tau(\omega))$ such that $f(s,\omega) \ge R$.

Now let $t \ge 0$ and write

\[
\left\{ \tau_R \le t \right\}
\left\{ \tau_R = \tau \le t \right\} \cup \left( \left\{ \tau_R < \tau\right\} \cap \left\{ \tau_R \le t \right\} \right).
\]

For $s \ge 0$ define $f_s \colon \{ s < \tau \} \to [0,\infty)$ by $f_s(\omega) = f(s,\omega)$. Using the above dichotomy, we see that

\[
\left\{ \tau_R < \tau\right\} \cap \left\{ \tau_R \le t \right\}
\bigcup_{0 \le s \le t} f_s^{-1}\left( [R,\infty) \right),
\]

where $f_s^{-1}\left( [R,\infty) \right)$ is $\mathcal F_s$-measurable, hence $\mathcal F_t$-measurable, by (<ref>). Now let $\{s_i\}$ be a dense, countable subset of $[0,\infty)$. Using (<ref>), and the compactness of the interval $[0,t]$, we now get

\[
\left\{ \tau_R < \tau\right\} \cap \left\{ \tau_R \le t \right\}
\bigcap_{k=1}^\infty \left( \bigcup_{0 \le s_i \le t} f_{s_i}^{-1}\left( \left[R-\frac{1}{k},\infty\right) \right) \right),
\]

which is evidently $\mathcal F_t$-measurable. Note that so far we did not use the assumption that $\tau$ is a stopping time.

Next, letting $P$ be any partition $t_0=0 < t_1 < \dots < t_N = t$ of the interval $[0,t]$, we write

\[
\left\{ \tau_R = \tau \le t \right\}
\bigcap_P \left( \bigcup_{\substack{t_k \in P\\ t_k < t}}  \bigcap_{0 \le s < t_k} f_{s}^{-1} \left(  \left[0,R\right] \right) \cap \{ t_k \le \tau \le t_{k+1} \} \right),
\]

where we used (<ref>). Moreover, we may restrict to partitions $P$ whose interior points are rational, and appealing once more to (<ref>), we see that

\[
\bigcap_{0 \le s < t_k} f_{s}^{-1} \left(  \left[0,R\right] \right)
\bigcap_{0 \le s_i < t_k} f_{s_i}^{-1} \left(  \left[0,R\right] \right).
\]

Finally, since $\{ t_k \le \tau \le t_{k+1} \}$ is $\mathcal F_t$-measurable, we conclude that $\left\{ \tau_R = \tau \le t \right\}$ is as well. So we have proved that $\left\{ \tau_R \le t \right\}$ is $\mathcal F_t$-measurable, hence $\tau_R$ is a stopping time.

If $0 < a < \tau(\omega)$, then taking $R > \sup_{0 \le s \le a} f(s,\omega)$ gives $\tau_R > a$, proving (<ref>). Finally, the properties (<ref>) and (<ref>) are immediate from the definition of $\tau_R$, and this concludes the proof of the lemma.

§.§ Stochastic integrals

In this section, let $K$ and $H$ be separable Hilbert spaces, with orthonormal bases $\{e_k\}$ and $\{f_j\}$, respectively.

We denote by $\mathcal L(K,H)$ the space of bounded linear operators from $K$ into $H$, with the operator norm, and by $\mathcal L_2(K,H)$ the class of Hilbert-Schmidt operators

\[
\mathcal L_2(K,H) = \left\{ T \in \mathcal L(K,H) \colon \trace(T^*T) < \infty \right\},
\]

which is a separable Hilbert space with the norm and inner product

\[
\norm{T}_{\mathcal L_2(K,H)} = \trace(T^*T)^{1/2} = \sum_k \norm{Te_k}_H^2,
\qquad
\innerprod{S}{T}_{\mathcal L_2(K,H)} = \trace(T^*S).
\]

One can think of Hilbert-Schmidt operators as infinite-dimensional matrices. Indeed, defining

\[
T_{jk} = \innerprod{Te_k}{f_j}_H,
\]

then $T \mapsto \{ T_{jk} \}$ is an isometry from $\mathcal L_2(K,H)$ onto $l^2(\N \times \N)$.

For later use we note the fact that if $S \in \mathcal L(H,H')$, where $H'$ is a Hilbert space, and $T \in \mathcal L_2(K,H)$ then the composition $ST$ belongs to $\mathcal L_2(K,H')$ and

\begin{equation}\label{HScomposition}
\norm{ST}_{\mathcal L_2(K,H')} \le \norm{S}_{\mathcal L(H,H')}  \norm{T}_{\mathcal L_2(K,H)}.
\end{equation}

Consider now the cylindrical Wiener process

\[
W(t) = \sum_{k=1}^\infty B_k(t) e_k,
\]

where the sum is formal. Given $T > 0$, the $H$-valued Itô integral of an adapted process

\[
F \in L^2\left([0,T] \times \Omega; \mathcal L_2(K,H) \right)
\]

is a natural generalisation of the $n$-dimensional Itô integral. It can be defined by

\[
\int_0^T F(t) \, dW(t)
\lim_{n \to \infty} \sum_{j,k=1}^n \left( \int_0^T F_{jk}(t) \, dB_k(t) \right) f_j,
\]

where the integrals on the right are ordinary Itô integrals and $F_{jk} = \innerprod{Fe_k}{f_j}_H$ are the matrix entries of $F$. The sum converges in $L^2(\Omega;H)$ and the Itô isometry holds:

\begin{equation}\label{Ito}
\mathbb E\left( \norm{\int_0^T F(t) \, dW(t)}_H^2 \right)
= \mathbb E\left( \int_0^T \norm{F(t)}_{\mathcal L_2(K,H)}^2  \, dt \right).
\end{equation}

Moreover, the $H$-valued random variable $I(t,\omega) = \int_0^t F(s,\omega) \, dW(s,\omega)$ is adapted and we can assume that it has continuous paths, since it has a modification with this property. Further, $I$ is a martingale, so by Doob's maximal inequality (see, e.g., [17]),

\begin{equation}\label{Maximal}
\mathbb E\left( \sup_{0 \le t \le T} \norm{\int_0^t F(s) \, dW(s)}_H^2 \right)
\le \mathbb E\left( \norm{\int_0^T F(s) \, dW(s)}_H^2 \right).
\end{equation}

§.§ Function spaces
Let $d \in \N$. First, $H^s(\R^d)$ denotes the usual Sobolev space with norm

\begin{equation}
\label{Sob-norm}
\norm{f}_{H^s(\R^d)} = \left( \int_{\R^d} \japanese{\xi}^{2s} \bigabs{\widehat f(\xi)}^2 \, d\xi \right)^{1/2},
\end{equation}

where $\japanese{\xi} = (1+\Abs{\xi}^2)^{1/2}$ and the Fourier transform is defined by

\[
\widehat f(\xi) = \mathcal F f(\xi) = \int_{\R^d} e^{-ix\xi} f(x) \, dx \qquad (\xi \in \R^d).
\]

Then the inverse transform is given by $\mathcal F^{-1} g(x) = (2\pi)^{-d} \int_{\R^d} e^{ix\xi} g(\xi) \, d\xi$, the Plancherel identity reads $\norm{\mathcal F f}_{L^2(\R)} = (2\pi)^{d/2} \norm{f}_{L^2(\R)}$, and we have $\widehat{fg} = (2\pi)^{-d} \widehat f * \widehat g$ and $\widehat{f * g} = \widehat f \, \widehat g$ for $f,g \in L^2(\R^d)$.

We recall the Sobolev product law

\begin{equation}\label{Sob-product-1}
\norm{fg}_{H^{-s_1}(\R^d)} \le C \norm{f}_{H^{s_2}(\R^d)}\norm{g}_{H^{s_3}(\R^d)},
\end{equation}

which holds for all Schwartz functions $f$ and $g$ on $\R^d$ if and only if $s_1,s_2,s_3 \in \R$ satisfy

\begin{equation}\label{Sob-product-2}
s_1+s_2+s_3 \ge \frac{d}{2}
\quad \text{and} \quad
\min_{i \neq j}(s_i+s_j) \ge 0,
\quad \text{which are not both equalities}.
\end{equation}

For $d=1$ and $0 < b < 1$ we will make use of the norm equivalence
(see <cit.>)

\begin{equation}\label{HbEquiv}
\norm{f}_{H^b(\R)}^2 = \norm{f}_{L^2(\R)}^2 + c_b \norm{f}_{S^b(\R)}^2,
\end{equation}

which holds for some $c_b > 0$.
Here $\norm{f}_{S^b(\Omega)}$ denotes the Slobodeckij seminorm on an open set $\Omega \subset \R$,

\[
\norm{f}_{S^b(\Omega)}^2 = \int_\Omega \int_\Omega \frac{\Abs{f(t)-f(r)}^2}{\Abs{t-r}^{1+2b}} \, dr \, dt.
\]

On any finite time interval $I=(S,T)$ there is a similar norm equivalence (see <cit.>)

\begin{equation}\label{HbEquivI}
\norm{f}_{H^b(I)} \sim_I \left( \norm{f}_{L^2(I)}^2 + \norm{f}_{S^b(I)}^2 \right)^{1/2},
\end{equation}

but with the caveat that the constants depend on the interval.
The norm on the left is the restriction norm, defined as the infimum
of $\norm{g}_{H^b(\R)}$ for $g \in H^b(\R)$ with $g=f$ on $I$.

Second, given a function $h \in C(\R^d;\R)$,
we denote by $X^{s,b}_{h(\xi)}(\R \times \R^d)$ the Bourgain space with norm

\begin{equation}
\label{Bourgain-norm}
\norm{u}_{X^{s,b}_{h(\xi)}}
\left( \int_{\R^d} \int_{\R} \japanese{\xi}^{2s} \japanese{\tau+h(\xi)}^{2b}
\Abs{\widehat u(\tau,\xi)}^2 \, d\tau \, d\xi \right) ^{1/2}
\end{equation}

where $\widehat u = \mathcal F_{(t,x)} u$ is the space-time Fourier transform.
The restriction to a time-slab $(S,T) \times \R$ is denoted $X^{a,b}_{h(\xi)}(S,T)$,
and is equipped with the norm

\begin{equation}
\label{XRestNorm}
\norm{u}_{X^{s,b}_{h(\xi)}(S,T)} = \inf \left\{ \norm{v}_{X^{s,b}_{h(\xi)}} \colon \text{$u(t)=v(t)$ for $t \in (S,T)$} \right\}
\end{equation}

The symbol $\tau+h(\xi)$ in (<ref>) is associated to the linear PDE
$-i\partial_t u + h(D_x) u = 0$ with the group

\[
S_{h(\xi)}(t) = e^{-ith(D_x)},
\]

whose action on $f(x)$ is given by $\mathcal F_x \left\{ S_{h(\xi)}(t) f \right\}(\xi) = e^{-ith(\xi)} \widehat f(\xi)$.
Then we can write

\begin{equation}
\label{XRestNorm_alternative}
\norm
_{ X_{h(\xi)}^{ s, b }(S, T) }^2
\int
\langle \xi \rangle ^{2s}
\norm
e^{ it h(\xi) }
\mathcal F_x u(t, \xi)
} _{ H_t^b(S, T) }^2
\end{equation}

Clearly, this coincides with the definition
given by (<ref>), (<ref>).

We now mention some well-known properties of Bourgain norms.
First, we note the conjugation property

\begin{equation}
\label{XRestNorm_conjugate}
\norm
\overline u
_{ X_{ h(\xi) }^{ s, b }(S, T) }
\norm
_{ X_{ -h(-\xi) }^{ s, b }(S, T) }
\end{equation}

By $L^2$ duality and Plancherel's theorem,

\begin{equation}\label{Xduality}
\norm{u}_{X^{s,b}_{h(\xi)}}
\sup_{\norm{v}_{X^{-s,-b}_{h(\xi)}} = 1} \Abs{ \int_{\R^d} \int_{\R} u(t,x) \overline{v(t,x)} \, dt \,dx}
\end{equation}

\begin{equation}\label{Xduality2}
\Abs{\int_{\R^d} \int_{\R} u(t,x) \overline{v(t,x)} \, dt \,dx}
\le
\norm{u}_{X^{s,b}_{h(\xi)}}\norm{v}_{X^{-s,-b}_{h(\xi)}}.
\end{equation}

Now let $\theta(t)$ be any smooth, compactly supported function.

\begin{equation}\label{Bourgain-1}
\norm{\theta(t) S_{h(\xi)}(t)f}_{X^{s,b}_{h(\xi)}} = \norm{\theta}_{H^b(\R)} \norm{f}_{H^s(\R^d)}
\quad \text{for $b \in \R$}
\end{equation}

\begin{equation}\label{Bourgain-2}
\norm{\theta(t) \int_0^t S_{h(\xi)}(t-\sigma)F(\sigma) \, d\sigma}_{X^{s,b}_{h(\xi)}} \le C \norm{\theta}_{H^1(\R)} \norm{F}_{X^{a,b-1}_{h(\xi)}(\R \times \R^d)}
\quad
\text{for $\frac12 < b \le 1$}.
\end{equation}

\begin{equation}\label{Bourgain-3}
\norm{u}_{X^{s,b}_{h(\xi)}(0,T)} \le C_{b,b'} T^{b'-b} \norm{u}_{X^{s,b'}_{h(\xi)}(0,T)}
\quad
\text{for $-\frac12 < b < b' < \frac12$ and $0 < T \le 1$}
\end{equation}

\begin{equation}
\label{Bourgain-4}
b > \frac12 \implies X^{s,b}_{h(\xi)}(0,T) \hookrightarrow C\left( [0,T], H^s \right)
\quad \text{with}
\quad
\sup_{0 \le t \le T} \norm{u(t)}_{H^s(\R^d)} \le C_b \norm{u}_{X^{s,b}_{h(\xi)}(0,T)}.
\end{equation}

We will also need the trivial fact that

\begin{equation}
\label{Bourgain-5}
b \ge 0 \implies X^{s,b}_{h(\xi)}(0,T) \hookrightarrow L^2\left( [0,T], H^s \right)
\quad \text{with}
\quad
\norm{u}_{L^2([0,T],H^s)} \le \norm{u}_{X^{s,b}_{h(\xi)}(0,T)}.
\end{equation}

In particular, this implies that the space $\mathbb L^2(\Omega, X^{s,b})$ is well-defined when $b \ge 0$, as in (<ref>).

Finally, we recall that for $0 \leqslant b < 1/2$
the restriction norm on $H^b(S, T)$ is equivalent to the
of the trivial extension.

More precisely, for all $\phi \in H^{b}(\R)$ we have

\begin{equation}
\label{H_characteristic_norm}
\norm{\phi}_{H^b(S, T)}
\leqslant
\norm{\mathbb 1_{(S, T)} \phi}_{H^b(\R)}
\leqslant
C_b \norm{\phi}_{H^b(S, T)}
\quad
\text{for $-\frac12 < b < \frac12$},
\end{equation}

where we emphasise that $C_b$ is independent of $(S,T)$. Similarly, for all $u \in X_{h(\xi)}^{a,b}(\R \times \R^d)$,

\begin{equation}
\label{X_characteristic_norm}
\norm{u}_{X^{s,b}_{h(\xi)}(S, T)}
\leqslant
\norm{\mathbb 1_{(S, T)} u}_{X^{s,b}_{h(\xi)}}
\leqslant
C_b \norm{u}_{X^{s,b}_{h(\xi)}(S, T)}
\quad
\text{for $-\frac12 < b < \frac12$.}
\end{equation}

Here $\mathbb 1_{(S, T)}(t)$ is the characteristic function
of the interval $(S, T)$.

An immediate consequence is that functions on adjacent time intervals can be glued together.

Let $-1/2 < b < 1/2$. Then there exists a constant $C_b$ such that if $u \in X^{s,b}_{h(\xi)}(t_0,t_1)$ and $v \in X^{s,b}_{h(\xi)}(t_1,t_2)$, where $t_0 < t_1 < t_2$, then the glued function

\[
[u,v](t) = \begin{cases}
u(t) &t_0 < t < t_1
\\
v(t) &t_1 < t < t_2
\end{cases}
\]

belongs to $X^{s,b}_{h(\xi)}(t_0,t_2)$ and

\[
\norm{[u,v]}_{X^{s,b}_{h(\xi)}(t_0,t_2)}
\le C_b \left(
\norm{u}_{X^{s,b}_{h(\xi)}(t_0,t_1)}
\norm{v}_{X^{s,b}_{h(\xi)}(t_1,t_2)}
\right).
\]

This follows from (<ref>) and the triangle inequality since $\mathbb 1_{(t_0,t_2)} = \mathbb 1_{(t_0,t_1)} + \mathbb 1_{(t_1,t_2)}$ a.e.

§.§ Cutoffs and a modified Bourgain norm

As usual with a multiplicative noise,
we have to truncate the nonlinearity in order to prove existence by iteration.
In the corresponding cutoffs we will use, for technical reasons,
not the restriction norm (<ref>),
but an equivalent norm, defined by

\begin{equation}
\label{Bourgain_Slobodeckij_norm}
\norm
_{ \widetilde X_{h(\xi)}^{ s, b }(S, T) }^2
\int_{\R^d}
\left(
\frac 1{ (T - S)^{2b} }
\int_S^T \Abs{ U(t, \xi) }^2 \, dt
\int_S^T \int_S^T
\frac
{ | U(t, \xi) - U(r, \xi) |^2 }
{ | t - r |^{1 + 2b} }
\, dr \, dt
\right)
\end{equation}

\(
U(t, \xi)
\langle \xi \rangle ^{s} e^{ it h(\xi) }
\mathcal F_x u(t, \xi)
\)

It is thoroughly studied below in Section <ref>.
Here it is crucial that the norm equivalence is uniform
with respect to the time interval to which we restrict.
To be precise, for any $T_0 > 0$ and $b \in (0,1/2)$ there exists a constant $C_{T_0,b}$
such that (see Lemma <ref>)

\begin{equation}
\label{ModEquivalence}
C_{T_0,b}^{-1} \norm{u}_{X^{s,b}_{h(\xi)}(S,T)}
\le
\norm{u}_{\widetilde X^{s,b}_{h(\xi)}(S,T)}
\le C_{T_0,b}
\norm{u}_{X^{s,b}_{h(\xi)}(S,T)}
\end{equation}

for all $u \in X_{h(\xi)}^{s,b}(S,T)$ with $0 < S < T \le T_0$.

The idea of exploiting the Slobodeckij seminorm,
the double integral over $(S, T)$ in (<ref>),
comes from [15].
However, we point out that the factor $1/(T - S)^{2b}$
in front of the $L^2$-norm
turns out to be important to claim the uniform equivalence (<ref>).
Moreover, the necessity of this factor becomes clear
when one calculates these norms on a concrete element,
\(
u = S_{ h(\xi) }(t) f,
\)
where $f \in H^s(\R^d)$.
Indeed, by (<ref>) it follows immediately that

\[
\norm
{ S_{ h(\xi) }(t) f }
{ X_{ h(\xi) }^{s,b}(0,T) }
\norm{1}_{H^b(0,T)}
\norm{f}_{H^s}
\]

for all $T > 0$ and $b \in \R$, and it is easy to see that

\[
\norm{1}_{H^b(0,T)}
\sim
T^{1/2 - b}
\quad
\text{for $0 \leqslant b < \frac12$ and $0 < T \leqslant 1$.}
\]

Indeed, recalling (<ref>), we can calculate the equivalent norm

\[
\norm{ \mathbb 1_{(0, T)} }_{H^b}^2
\int_{\R}
\Abs{
\int_0^T e^{-it\tau} dt
\japanese{\tau}^{2b} d\tau
\int_{\R}
\Abs{
\frac 2{\tau}
\sin \frac {\tau}2
\japanese{ \frac {\tau}T }^{2b} d\tau
\]

\[
\Abs{ \frac {\tau}T }^{2b}
\leqslant
\japanese{ \frac {\tau}T }^{2b}
\leqslant
c_b \left( 1 +
\Abs{ \frac {\tau}T }^{2b}
\right)
\quad
\text{and}
\quad
\Abs{
\frac 2{\tau}
\sin \frac {\tau}2
\lesssim \japanese{\tau}^{-2}
\]

On the other hand, it is easily seen from (<ref>), with $U(t,\xi) = \widehat f(\xi)$, that
\[
\norm{S_{h(\xi)}(t) f}_{\widetilde X_{ h(\xi) }^{s,b}(0,T) } = T^{1/2-b} \norm{f}_{H^s}.
\]

For the modified restriction norm (<ref>) we have the following key estimates,
proved in Section <ref>.

Let $T_0 > 0$ and $b \in (0,1/2)$. Let $\theta \colon \R \to \R$ be a smooth, compactly supported function and set $\theta_R(x) = \theta(x/R)$ for $R > 0$. Let $n \in \N$, and for $1 \le i \le n$ let $s_i \in \R$,  $h_i \in C(\R^d,\R)$ and $u_i,v_i \in X^{s_i,b}_{h_i(\xi)}(0,T_0)$. Then for $T \in (0,T_0]$, $R > 0$ and $1 \le j \le n$ we have the estimates

\[
\norm{
\theta_R\left( \sum_{i=1}^n \norm{u_i}_{\widetilde X^{s_i,b}_{h_i(\xi)}(0,t)}^2 \right)
\leqslant
C \sqrt{R}
\]

\[
\norm{
\theta_R\left( \sum_{i=1}^n \norm{u_i}_{\widetilde X^{s_i,b}_{h_i(\xi)}(0,t)}^2 \right)
\theta_R\left( \sum_{i=1}^n \norm{v_i}_{\widetilde X^{s_i,b}_{h_i(\xi)}(0,t)}^2 \right)
\leqslant
C \sum_{i=1}^n \norm{u_i-v_i}_{X^{s_i,b}_{h_i(\xi)}(0,T)}
\]

where $C$ depends only on $b$, $T_0$ and $\theta$.

We remark that the modified norm (<ref>) can be formulated also for functions $\phi(t)$ depending only on the time variable $t$, and that this gives cutoff estimates for functions in $H^b(\R)$, $0 < b < 1/2$. See Section <ref>.

§ MAIN RESULTS

We consider the mild form of (<ref>), which reads

\begin{multline}
\label{mild_psi}
\psi_{\pm}(t)
S_{\pm\xi}(t) f_\pm
iM \int_0^t S_{\pm\xi}(t-\sigma)  \psi_{\mp}(\sigma) \, d\sigma
i \int_0^t  S_{\pm\xi}(t-\sigma) ( \phi\psi_{\mp})(\sigma) \, d\sigma
\\
i \int_0^t S_{\pm\xi}(t-\sigma) \psi_\mp(\sigma) \mathfrak K_1 \, d W(\sigma)
M_{\mathfrak K_1} \int_0^t S_{\pm\xi}(t-\sigma) \psi_\pm(\sigma) \, d\sigma
\end{multline}

\begin{multline}
\label{mild_phi}
\phi_+(t)
S_{+\japanese{\xi}}(t) g_+
i \int_0^t S_{+\japanese{\xi}}(t-\sigma) \japanese{D_x}^{-1}
\re \left( \overline{\psi_+} \psi_- \right)(\sigma) \, d\sigma
\\
i \int_0^t S_{+\japanese{\xi}}(t-\sigma) \japanese{D_x}^{-1}
\phi(\sigma) \mathfrak K_2 \, dW(\sigma)
\end{multline}

\(
\phi
\phi_+ + \overline{\phi_+}
2 \re \phi_+
\)
We will look for solutions

\begin{equation}
\label{IterationSpace}
\psi_\pm \in X_{\pm\xi}^{s,b}(0,T)
, \qquad
\phi_+ \in X_{+\japanese{\xi}}^{r,b}(0,T)
\end{equation}

where $b < 1/2$ is taken sufficiently close to $1/2$, depending on $s$ and $r$.
Note that by the conjugation property (<ref>)
we have

\begin{equation}
\label{IterationSpace_conjugate}
\overline{\phi_+} \in X_{ -\japanese{\xi} }^{r,b}(0,T)
\end{equation}

It will be convenient to define

\[
\mathbf H^{(s,s,r)} = H^s(\R,\C) \times H^s(\R,\C) \times H^r(\R,\C)
\]

\[
\mathbf X^{(s,s,r),b}(0,T) = X_{+\xi}^{s,b}(0,T) \times X_{-\xi}^{s,b}(0,T) \times X_{+\japanese{\xi}}^{r,b}(0,T),
\]

with the product norms.

We now state our main results.

Assume that $s,r \in \R$ satisfy

\[
s > - \frac14, \qquad \Abs{s} \le r \le s+1, \qquad 0 < r < 1+2s.
\]

Then for any $b < 1/2$ sufficiently close to $1/2$, the following holds. Assume that

\[
(f_+,f_-,g_+) \in L^2\bigl( \Omega, \mathbf H^{(s,s,r)} \bigr)
\quad \text{is $\mathcal F_0$-measurable}.
\]

Then there exists a stopping time $\tau \colon \Omega \to (0,\infty]$ and a random process

\[
(\psi_+,\psi_-,\phi_+)(t) \in \mathbf H^{(s,s,r)} \quad \text{for $0 \le t < \tau$}
\]

such that for $0 < t < \tau$, (<ref>) and (<ref>) hold,

\[
(\psi_+,\psi_-,\phi_+)(t) \colon \{ t < \tau \} \to \mathbf H^{(s,s,r)} \quad \text{is $\mathcal F_t$-measurable}
\]

\[
(\psi_+,\psi_-,\phi_+) \in C\left( [0,t],\mathbf H^{(s,s,r)} \right) \cap \mathbf X^{(s,s,r),b}(0,t).
\]

Moreover, the solution is maximal in the sense that

\[
\tau < \infty \implies \limsup_{t \nearrow \tau} \norm{(\psi_+,\psi_-,\phi_+)(t)}_{\mathbf H^{(s,s,r)}} = \infty,
\]

and it is unique in the sense that if $(\Psi_+,\Psi_-,\Phi_+)$ is a solution with the same initial data, and satisfying the same assumptions but with a stopping time $\lambda$, then

\[
(\psi_+,\psi_-,\phi_+)(t) = (\Psi_+,\Psi_-,\Phi_+)(t)
\quad \text{for $0 \le t < \min(\tau,\lambda)$}.
\]

Further, if $s \ge 0$, then the charge is conserved:

\[
\int_{\R} \left( \Abs{\psi_+(t,x)}^2 + \Abs{\psi_-(t,x)}^2 \right) \, dx
= \int_{\R} \Abs{\psi_0(x)}^2 \, dx
\quad \text{for $0 \le t < \tau$},
\]

where $\psi_0 = (f_+,f_-)$.

This theorem is a consequence of the abstract well-posedness theory presented in Section <ref>.
The existence follows from Theorem <ref> and the uniqueness from Theorem <ref>. The necessary assumptions stated in Section <ref> are verified here on account of the bounds stated in Lemmas <ref>, <ref>, <ref> and <ref> below. The charge conservation is proved in Section <ref>.

Using the charge conservation, we will then deduce the following global result.

If $s=0$ and $0 < r < 1/2$, and if we impose the additional condition $\psi_0 \in L^4(\Omega,L^2(\R,\C))$ on the initial data, then the solution in Theorem <ref> extends globally in time. That is, $\tau=\infty$.

The proof is given in Section <ref>.

Although the local result will, as mentioned, be deduced from the abstract framework expounded in some detail in Section <ref>, we find it worthwhile to present here a broad outline of the key ideas behind the proof.

Existence for a short time interval $(0,T)$ will be proved by iteration in

\begin{equation}\label{iteration-space}
\mathbb L^2\left(\Omega,\mathbf X^{(s,s,r),b}(0,T) \right)
\cap
\mathbb L^2\left(\Omega,C([0,T],\mathbf H^{(s,s,r)}) \right),
\end{equation}

and the first thing to notice is that we cannot expect the stochastic integrals to be in this space unless $b$ is strictly less than $1/2$, the reason being that the paths of any one-dimensional Brownian motion belong to $H^b(\R)$ if and only if $b < 1/2$, as shown in [4].

For $b < 1/2$ the stochastic integrals can indeed be controlled in (<ref>), provided that we know that the linear operators

\begin{equation}\label{M1-M2}
M_1(f) = f \mathfrak K_1 \quad \text{and} \quad M_2(g) = \japanese{D_x}^{-1} g \mathfrak K_2
\end{equation}

map $H^s(\R)$ and $H^r(\R)$ into Hilbert-Schmidt operators from $L^2(\R,\R)$ into $H^s(\R)$ and $H^r(\R)$, respectively. Thus, we need the following.

Let $s,r$ be as in Theorem <ref>. Assume that $\mathfrak k_1 \in H^{\Abs{s}}(\R,\R)$ and $\mathfrak k_2 \in H^{\max(0,r-1)}(\R,\R)$. Then there exists a constant $C$ such that the linear operators $M_1$ and $M_2$ defined by (<ref>) satisfy

\[
\norm{M_1(f)}_{\mathcal L_2(L^2,H^s)} \le C \norm{f}_{H^s}
\quad \text{and} \quad
\norm{M_2(g)}_{\mathcal L_2(L^2,H^r)} \le C \norm{g}_{H^r}
\]

for all $f \in H^s(\R)$ and $g \in H^r(\R)$.

The proof of this lemma is given in Section <ref>, and in Section <ref> we show how it is applied to control the stochastic integrals.

Now let us turn our attention to the deterministic terms in (<ref>) and (<ref>). Here there is a difference from the purely deterministic case, where one works with $b > 1/2$, see e.g. [29]. Since now we are forced to take $b < 1/2$, the required bilinear estimates are a bit tighter. We will prove the following bilinear bounds, extending those obtained in [29] to the case where $b$ is less than, but close to, $1/2$.

Assume that $s,r \in \R$ satisfy

\[
s > - \frac14, \qquad \Abs{s} \le r \le s+1, \qquad 0 < r < 1+2s.
\]

Then for any $b < 1/2$ sufficiently close to $1/2$, there exists a constant $C$ such that

\begin{align}
\label{M+bound}
\norm{\phi \psi}_{X^{s,-b}_{+\xi}}
&\le C \norm{\phi}_{X^{r,b}_{\pm\japanese{\xi}}}
\norm{\psi}_{X^{s,b}_{-\xi}},
\\
\label{M-bound}
\norm{\phi \psi}_{X^{s,-b}_{-\xi}}
&\le C \norm{\phi}_{X^{r,b}_{\pm\japanese{\xi}}}
\norm{\psi}_{X^{s,b}_{+\xi}},
\\
\label{Nbound}
\norm{\overline{\psi} \psi'}_{X^{r-1,-b}_{\pm\japanese{\xi}}}
&\le C \norm{\psi}_{X^{s,b}_{+\xi}}
\norm{\psi'}_{X^{s,b}_{-\xi}},
\end{align}

for all Schwartz functions $\psi$, $\psi'$ and $\phi$ on $\R_t \times \R_x$.

This is proved in Section <ref>.

With $b$ as in the last lemma, and choosing $0 < \varepsilon < 1/2-b$, set

\[
B = -b+1-\varepsilon.
\]

Then $B > 1/2$, so we can apply (<ref>) and (<ref>) to control the deterministic integrals in $\mathbf X^{(s,s,r),b}(0,T)$. And then, crucially, by (<ref>) they are also controlled in $C([0,T];\mathbf H^{(s,s,r)})$ (and of course also in $\mathbf X^{(s,s,r),b}(0,T)$, since $b < B$). For example,

\[
\begin{alignedat}{2}
\norm{\int_0^t  S_{+\xi}(t-\sigma) ( \phi_+\psi_- )(\sigma) \, d\sigma}_{X^{s,B}_{+\xi}(0,T)}
C \norm{\phi_+\psi_-}_{X^{s,B-1}_{+\xi}(0,T)}&
\quad &\text{by \eqref{Bourgain-2}}
\\
C T^{\varepsilon} \norm{\phi_+\psi_-}_{X^{s,-b}_{+\xi}(0,T)}&
&\text{by \eqref{Bourgain-3}}
\\
C T^{\varepsilon} \norm{\phi_+}_{X^{r,b}_{+\japanese{\xi}}0,T)}
\norm{\psi_-}_{X^{s,b}_{-\xi}(0,T)}&
&\text{by Lemma \ref{DKG-bilinear-lemma}},
\end{alignedat}
\]

and similarly for the other bilinear terms, and also for the linear ones, for which we use the following.

Let $s \in \R$ and $b \ge 0$. Then

\[
\norm{u}_{X^{s,-b}_{h(\xi)}}
\le
\norm{u}_{L^2(\R_t,H^s)}
\le
\norm{u}_{X^{s,b}_{g(\xi)}}
\]

for all Schwartz functions $u$ on $\R_t \times \R_x$ and any choice of $g,h \in C(\R,\R)$.

This is obvious from the definitions (<ref>) and (<ref>), and Plancherel's theorem.

The bounds are pointwise in $\omega$, and for the iteration in the space (<ref>) one must now take the $L^2(\Omega)$ norm of them. So for example, one needs to control

\[
\mathbb E \left( \norm{\phi_+}_{X^{r,b}_{+\japanese{\xi}}0,T)}^2
\norm{\psi_- - \Psi_-}_{X^{s,b}_{-\xi}(0,T)}^2 \right),
\]

where $\psi_-$ and $\Psi_-$ represent different iterates. The only reasonable way to estimate this, seems to be

\[
\left( \sup_{\omega} \norm{\phi_+}_{X^{r,b}_{+\japanese{\xi}}0,T)}^2 \right) \mathbb E \left(
\norm{\psi_- - \Psi_-}_{X^{s,b}_{-\xi}(0,T)}^2 \right),
\]

so one needs to control the norms of the iterates uniformly in $\omega$. Here, a further difference from the deterministic case becomes apparent, since there one usually chooses $R > 0$ and considers initial data whose norm is at most $R$. Then for $T > 0$ small enough depending on $R$, the Bourgain norms in the iteration are all bounded by $R$ times some constant. This will not work in the stochastic case, since a bound in the space (<ref>) does not imply a pointwise bound in $\omega$. Instead, as is usual in stochastic problems with multiplicative noise, one must truncate the equations. Following the approach in [15], for a given $R > 0$ we consider a truncated version of (<ref>), (<ref>) where in the bilinear terms, each unknown is multiplied by the cutoff

\begin{equation}\label{DKG-cutoff}
\Theta(t) = \theta_R\left(
\norm{\psi_+}_{\widetilde X^{s,b}_{+\xi}(0,t)}^2
\norm{\psi_-}_{\widetilde X^{s,b}_{-\xi}(0,t)}^2
\norm{\phi_+}_{\widetilde X^{r,b}_{+\japanese{\xi}}(0,t)}^2
\right),
\end{equation}

where $\theta \colon \R \to \R$ is any smooth, compactly supported cutoff function with $\theta=1$ on $[0,1]$, and $\theta_R(x) = \theta(x/R)$. So for example, the integral term

\[
\int_0^t  S_{+\xi}(t-\sigma) ( \phi_+\psi_- )(\sigma) \, d\sigma,
\]

considered above, is replaced by

\[
\int_0^t  S_{+\xi}(t-\sigma) \left(
\Theta \phi_+
\Theta
\psi_- \right)(\sigma) \, d\sigma
\]

and similarly for the other bilinear terms. For technical reasons, inside the cutoffs we do not use the Bourgain restriction norm as defined in (<ref>), (<ref>), but rather the equivalent norm (<ref>), discussed in detail in Section <ref>.

With the truncation, and making use of Proposition <ref>, the bilinear terms can be controlled in the space (<ref>), and by iteration one can prove existence up to a small time $T > 0$ depending only on $R$. Repeating this argument one obtains existence on a time interval of any size. Letting $R$ tend to $\infty$, this implies the existence of a maximal solution of the original, non-truncated problem. The uniqueness requires a separate argument. The details are shown in Section <ref> in an abstract framework. As mentioned, the necessary assumptions in Section <ref> are verified on account of the bounds in Lemmas <ref>, <ref> and <ref>, as well as Lemma <ref>, which we now state.

The following estimates shows that the deterministic integrals make sense as Bochner integrals if the regularity is sufficiently high, and that this fails if  $s \le 0$.

Assume that

\[
s > 0, \qquad s \le r \le s+1, \qquad \frac12 < r < \frac12 + 2s.
\]

Then there exists a constant $C$ such that

\[
\norm{fg}_{H^{s}}
\le C \norm{f}_{H^{r}} \norm{g}_{H^{s}}
\quad \text{and} \quad
\norm{fg}_{H^{r-1}}
\le C \norm{f}_{H^{s}} \norm{g}_{H^{s}}
\]

for all Schwartz functions $f$ and $g$ on $\R$. Moreover, if $s \le 0$, the above estimates cannot both hold, for any $r \in \R$ and $C > 0$.

This follows from the Sobolev product law (<ref>), (<ref>).

The case $m = 0$ in (<ref>) does not bring
anything new to our analysis.
Indeed, if $m = 0$ then we can add $\phi$
to both sides of the second line in (<ref>),
which gives rise to Equation (<ref>) with an additional linear term

\[
-i \int_0^t S_{+\japanese{\xi}}(t-s) \japanese{D_x}^{-1}
\phi(s) ds
\]

on the right hand side.
It can be treated by Lemma <ref> and (<ref>)–(<ref>). NB! What about global existence?

§ BOUNDS FOR HILBERT-SCHMIDT OPERATORS

Our main aim in this section is to prove Lemma <ref>.

To this end, we require the following lemma. It corresponds to Lemma 2.6 in [15], but we remove an additional assumption made there, namely that the convolution kernel $\mathfrak k$ is in $L^1 \cap L^2$.

\(
\mathfrak k
\in
L^2( \R^d, \R )
\)

\(
v \in L^2( \R^d, \mathbb C )
\)

and let
\(
\mathfrak K
\)
be the convolution operator defined by

\begin{equation}\label{conv-op}
\mathfrak Kf(x)
\int_{ \R^d }
\mathfrak k(x - y) f(y) dy.
\end{equation}

Then for any orthonormal basis
\(
\{ e_j \} _{ j \in \mathbb N }
\)
\(
L^2( \R^d, \R )
\)
we have

\[
\sum_{j = 1}^{\infty}
\left|
\mathcal F ( v \mathfrak K e_j )(\xi)
\right| ^2
\frac{1}{(2\pi)^d} \int_{\R^d} \Abs{\widehat v(\xi-\eta) \widehat{\mathfrak k}(\eta)}^2 d\eta
< \infty
\]

for a.e. $\xi \in \R^d$.

Set $f_j(x) = e_j(-x)$. Then $\{f_j\}$ is also an orthonormal basis of $L^2(\R^d)$, and $\overline{\widehat{f_j}(\xi)} = \widehat{e_j}(\xi)$, by the assumption that $e_j$ is real valued.

\[
\int_{\R^d} \int_{\R^d} \Abs{\widehat v(\xi-\eta) \widehat{\mathfrak k}(\eta)}^2 d\xi \, d\eta
\norm{\widehat v\,}_{L^2}^2 \fixednorm{\widehat{\mathfrak k}}_{L^2}^2
(2\pi)^{2d} \norm{v}_{L^2}^2 \norm{\mathfrak k}_{L^2}^2,
\]

it follows that

\[
\norm{F_\xi}_{L^2}^2
\frac{1}{(2\pi)^d} \int_{\R^d} \Abs{\widehat v(\xi-\eta) \widehat{\mathfrak k}(\eta)}^2 d\eta
< \infty \quad \text{for a.e.~$\xi \in \R^d$},
\]

where $F_\xi(x) = \mathcal F^{-1}\left\{ \eta \mapsto \widehat v(\xi-\eta) \widehat{\mathfrak k}(\eta) \right\}$. Applying Parseval's identity we have, for a.e. $\xi$,

\[
\norm{F_\xi}_{L^2}^2
\sum_{j=1}^\infty \Abs{\innerprod{F_\xi}{f_j}_{L^2}}^2
\sum_{j=1}^\infty \Abs{\frac{1}{(2\pi)^d} \Innerprod{\widehat{F_\xi}}{\widehat{f_j}}_{L^2}}^2
\sum_{j=1}^\infty
\Abs{\frac{1}{(2\pi)^d} \int \widehat v(\xi-\eta) \widehat{\mathfrak k}(\eta) \widehat{e_j}(\eta) \, d\eta}^2
< \infty.
\]

To finish the proof we only have to notice that the Fourier transform of the $L^2$ function $v\mathfrak Ke_j$ is

\[
\mathcal F ( v \mathfrak K e_j )(\xi)
= \frac{1}{(2\pi)^d} \int \widehat v(\xi-\eta) \widehat{\mathfrak k * e_j}(\eta) \, d\eta
= \frac{1}{(2\pi)^d} \int \widehat v(\xi-\eta) \widehat{\mathfrak k}(\eta) \widehat{e_j}(\eta) \, d\eta.
\]

We claim that this equality holds in $L^2(\R^d)$, hence for a.e. $\xi$. To prove the claim, approximate the $L^2$ functions $v$, $\mathfrak k$ and $e_j$ by Schwartz functions, for which the equality clearly holds, and then pass to the limit using the fact that $(2\pi)^{-d/2} \mathcal F$ is an isometry on $L^2$, and the bound

\[
\norm{\int \widehat v(\xi-\eta) \widehat{\mathfrak k}(\eta) \widehat{e_j}(\eta) \, d\eta}_{L^2_\xi}
\le
\norm{\widehat v}_{L^2} \int \Abs{\widehat{\mathfrak k}(\eta) \widehat{e_j}(\eta)} \, d\eta
\le
\norm{\widehat v}_{L^2}
\fixednorm{\widehat{\mathfrak k}}_{L^2}
\norm{\widehat{e_j}}_{L^2},
\]

where we applied Minkowski's integral inequality and Hölder's inequality.

Integrating both sides of the equality in Lemma <ref>,
one recovers the well-known fact
that $v \mathfrak K$ is a Hilbert-Schmidt operator on $L^2(\R)$, with norm
\(
\norm{v} _{L^2}
\norm{\mathfrak k} _{L^2}
\)

Let $s \in \R$. Assume that $\mathfrak k \in H^{\Abs{s}}(\R^d,\R)$. Then the convolution operator $\mathfrak K$ defined by (<ref>) satisfies

\[
\norm{ v \mathfrak K }_{\mathcal L_2(L^2,H^{s})}
\le C \norm{v}_{H^{s}} \norm{\mathfrak k}_{H^{\Abs{s}}} \quad \text{for all $v \in \mathcal S(\R^d)$},
\]

where the constant depends only on $s$. Thus the map $v \mapsto v \mathfrak K$ extends to a bounded, linear map from $H^s(\R^d)$ into $\mathcal L_2(L^2(\R^d),H^s(\R^d))$.

Integrating both sides of the equality in Lemma
<ref> with respect to
\(
\japanese{\xi}^{ 2s } d\xi
\)
one obtains

\[
\norm{ v \mathfrak K }
_{ \mathcal L_2( L^2, H^s ) }^2
\frac 1{(2\pi)^d}
\int_{\R^d} \int_{\R^d}
\Abs{\widehat v(\xi-\eta) \widehat{\mathfrak k}(\eta)}^2
\japanese{\xi}^{ 2s }
d\xi \, d\eta
\]

Applying the inequality

\[
\japanese{\xi}^s \le C_s \japanese{\xi-\eta}^s \japanese{\eta}^{\Abs{s}}
\quad \text{for all $\xi,\eta \in \R^d$},
\]

the claimed bound follows immediately.

With this corollary in hand, we can now prove Lemma <ref>.

The bound on $M_1(f) = f \mathfrak K_1$ is immediate from Corollary <ref>. For $M_2(g) = \japanese{D_x}^{-1} g \mathfrak K_2$ we write

\[
\norm{M_2(g)}_{\mathcal L_2(L^2,H^r)} =  \norm{g \mathfrak K_2}_{\mathcal L_2(L^2,H^{r-1})}.
\]

If $0 \le r \le 1$, we can bound this by

\[
\norm{g \mathfrak K_2}_{\mathcal L_2(L^2,L^2)}
\le C\norm{\mathfrak k}_{L^2}\norm{g}_{L^2},
\]

using again the corollary, while if $r \ge 1$ we bound by

\[
C\norm{\mathfrak k}_{H^{r-1}}\norm{g}_{H^{r-1}}.
\]

This concludes the proof of the lemma.

§ BILINEAR BOUNDS

In this section we prove Lemma <ref>.

We first note the following basic product law for Bourgain norms.

Let $s_1,s_2,s_3 \in \R$ and $b_1,b_2,b_3 \ge 0$, and assume that

* $s_1 + s_2 + s_3 \ge 1/2$, and

* $\min_{i \neq j} (s_i + s_j) \ge 0$, and

* the two preceding inequalities are not both equalities, and

* $b_1 + b_2 + b_3 > 1/2$.

Then there is a constant $C$ such that the bound

\begin{equation}\label{XProd}
\norm{uv}_{X^{-s_1,-b_1}_{h_1(\xi)}} \le C \norm{u}_{X^{s_2,b_2}_{h_2(\xi)}} \norm{v}_{X^{s_3,b_3}_{h_3(\xi)}}
\end{equation}

holds for all Schwartz functions $u$ and $v$ on $\R_t \times \R_x$ and any choice of $h_1,h_2,h_3 \in C(\R,\R)$.

By Plancherel's theorem and $L^2$ duality, (<ref>) can be reformulated as

\begin{equation*}
\int_{\R^4}
\frac{
f_1(\tau,\xi) f_2(\tau-\lambda,\xi-\eta) f_3(\lambda,\eta)
\, d\lambda \, d\eta \, d\tau \, d\xi
\japanese{\xi}^{2s_1} \japanese{\xi-\eta}^{2s_2} \japanese{\eta}^{2s_3}
\japanese{\tau+h_1(\xi)}^{2b_1}
\japanese{\tau-\lambda+h_2(\xi-\eta)}^{2b_2}
\japanese{\lambda+h_3(\eta)}^{2b_3}
\le C \prod_{i=1}^3 \norm{f_i}_{L^2}^2,
\end{equation*}

where $f_i \ge 0$. By symmetry, it suffices to consider the region where $\japanese{\tau+h_3(\eta)}$ is the minimum among $\japanese{\tau+h_1(\xi)}$, $\japanese{\tau-\lambda+h_2(\xi-\eta)}$ and $\japanese{\lambda+h_3(\eta)}$, and then the left side is bounded by

\begin{equation*}
\int_{\R^4}
\frac{
f_1(\tau,\xi) f_2(\tau-\lambda,\xi-\eta) f_3(\lambda,\eta)
\, d\lambda \, d\eta \, d\tau \, d\xi
\japanese{\xi}^{2s_1} \japanese{\xi-\eta}^{2s_2} \japanese{\eta}^{2s_3}
\japanese{\lambda+h_3(\eta)}^{2(b_1+b_2+b_3)}
\end{equation*}

where we used the assumption $b_1,b_2,b_3 \ge 0$. This shows that it is enough to prove

\begin{equation}\label{XProd-2}
\norm{uv}_{X^{-s_1,0}_{h_1(\xi)}} \le C \norm{u}_{X^{s_2,0}_{h_2(\xi)}} \norm{v}_{X^{s_3,b_1+b_2+b_3}_{h_3(\xi)}}.
\end{equation}

But by the assumptions on $s_1,s_2,s_3$ we can apply the product law (<ref>) to get

\[
\norm{u(t)v(t)}_{H^{-s_1}} \le C_{s_1,s_2,s_3} \norm{u(t)}_{H^{s_2}} \norm{v(t)}_{H^{s_3}}
\]

for all $t$. Taking now the $L^2$ norm with respect to $t$ of both sides, and using (<ref>) to bound

\[
\sup_{t \in \R}\norm{v(t)}_{H^{s_3}} \le C_{b_1,b_2,b_3} \norm{v}_{X^{s_3,b_1+b_2+b_3}_{h_3(\xi)}},
\]

we then obtain (<ref>), and this concludes the proof.

Now we apply the above lemma to obtain estimates of the form

\begin{equation}\label{NullProd}
\norm{uv}_{X_{\pm\japanese{\xi}}^{-s_1,-b_1}} \le C \norm{u}_{X^{s_2,b_2}_{+\xi}} \norm{v}_{X^{s_3,b_3}_{-\xi}}.
\end{equation}

But here we can gain some regularity compared to the generic case, due to the opposite signs in the dispersion relations on the right hand side; the symbols $\tau+\xi$ and $\tau-\xi$ correspond to transport equations with propagation in transverse directions. Thus, (<ref>) is a null form estimate.

Let us denote by $b_\mathrm{min}, b_\mathrm{med}, b_\mathrm{max}$ the minimum, median and maximum, respectively, of the three numbers $b_1, b_2, b_3$.

We then have the following result.

Suppose $s_1,s_2,s_3 \in \R$ and $b_1,b_2,b_3 \ge 0$. Then the following conditions are sufficient for the null form estimate (<ref>) to hold for all Schwartz functions $u$ and $v$ on $\R_t \times \R_x$:

* $s_1 + s_2 + s_3 + b_\mathrm{min} \ge 1/2$, and

* $\min\left( s_2 + s_3 + b_\mathrm{min}, s_1 + s_2, s_1 + s_3 \right) \ge 0$, and

* the two preceding inequalities are not both equalities, and

* $b_\mathrm{med} + b_\mathrm{max} > 1/2$.

We reformulate (<ref>) as

\begin{equation*}
\int_{\R^4} \frac{f_1(\tau,\xi) f_2(\tau-\lambda,\xi-\eta) f_3(\lambda,\eta) \, d\lambda \, d\eta \, d\tau d\xi}
{\japanese{\xi}^{2s_1} \japanese{\xi-\eta}^{2s_2} \japanese{\eta}^{2s_3}
\japanese{\tau\pm\japanese{\xi}}^{2b_1} \japanese{\tau-\lambda+(\xi-\eta)}^{2b_2} \japanese{\lambda-\eta}^{2b_3} }
\le C \prod_{i=1}^3 \norm{f_i}_{L^2}^2,
\end{equation*}

where $f_i \ge 0$. By the triangle inequality,

\begin{align*}
2\Abs{\eta} &\le \Abs{\tau+\xi} + \Abs{\tau-\lambda+(\xi-\eta)} + \Abs{\lambda-\eta},
\\
2\Abs{\xi-\eta} &\le \Abs{\tau-\xi} + \Abs{\tau-\lambda+(\xi-\eta)} + \Abs{\lambda-\eta},
\end{align*}

\[
\min\left(\japanese{\eta},\japanese{\xi-\eta}\right)
\le C \max\left( \japanese{\tau\pm\japanese{\xi}}, \japanese{\tau-\lambda+(\xi-\eta)}, \japanese{\lambda-\eta} \right),
\]

since $\japanese{\tau\pm\Abs{\xi}}$ is comparable to $\japanese{\tau\pm\japanese{\xi}}$. Thus, we can reduce (<ref>) to estimates

\begin{align*}
\norm{uv}_{X_{\pm\japanese{\xi}}^{-s_1,-B_1}} &\le C \norm{u}_{X^{s_2+b_\mathrm{min},B_2}_{+\xi}} \norm{v}_{X^{s_3,B_3}_{-\xi}},
\\
\norm{uv}_{X_{\pm\japanese{\xi}}^{-s_1,-B_1}} &\le C \norm{u}_{X^{s_2,B_2}_{+\xi}} \norm{v}_{X^{s_3+b_\mathrm{min},B_3}_{-\xi}},
\end{align*}

where $B_1,B_2,B_3 \ge 0$ and $B_1+B_2+B_3=b_\mathrm{med} + b_\mathrm{max}$. Applying Lemma <ref> we therefore obtain the claimed result.

We are now ready to prove Lemma <ref>.

We start with (<ref>), which reads

\[
\norm{\overline{\psi} \psi'}_{X^{r-1,-b}_{\pm\japanese{\xi}}}
\le C \norm{\psi}_{X^{s,b}_{+\xi}}
\norm{\psi'}_{X^{s,b}_{-\xi}}.
\]

By (<ref>), $\norm{\psi}_{X^{s,b'}_{+\xi}} = \fixednorm{\overline\psi}_{X^{s,b'}_{+\xi}}$, so we can remove the conjugation on $\psi$. Now we apply Lemma <ref> with $s_1=1-r$, $s_1=s_2=s$ and $b_0=b_1=b_2=b$, and conclude that (<ref>) holds if

\begin{equation}\label{DKG-bilinear-lemma-1}
b > \frac14,
\quad
2s+b \ge 0,
\quad
1-r+s \ge 0,
\quad
1-r+2s+b > \frac12.
\end{equation}

Next, we consider (<ref>), which reads

\[
\norm{\phi \psi}_{X^{s,-b}_{+\xi}}
\le C \norm{\phi}_{X^{r,b}_{\pm\japanese{\xi}}}
\norm{\psi}_{X^{s,b}_{-\xi}}.
\]

By (<ref>),

\[
\norm{\phi \psi}_{X^{s,-b}_{+\xi}}
\sup_{\norm{\psi'}_{X^{-s,b}_{+\xi}} = 1}
\Abs{\int_{\R^2} \phi \psi \overline{\psi'} \, dt \,dx},
\]

where by (<ref>),

\[
\Abs{\int_{\R^2} \phi \psi \overline{\psi'} \, dt \,dx}
\le
\norm{\phi}_{X^{r,b}_{\pm\japanese{\xi}}}
\norm{\psi \overline{\psi'}}_{X^{-r,-b}_{+\japanese{\xi}}}.
\]

Thus we have reduced to obtaining a bound

\[
\norm{\overline{\psi'}\psi }_{X^{-r,-b}_{\pm\japanese{\xi}}}
\le C \norm{\psi'}_{X^{-s,b}_{+\xi}} \norm{\psi}_{X^{s,b}_{-\xi}}.
\]

Applying Lemma <ref> with $s_0=r$, $s_1=-s$, $s_2=s$ and $b_0=b_1=b_2=b$, we conclude that (<ref>) holds if

\begin{equation}\label{DKG-bilinear-lemma-2}
b > \frac14,
\quad
r+b > \frac12,
\quad
r - s \ge 0,
\quad
r + s \ge 0.
\end{equation}

If we take $b=1/2-\delta$ with $\delta > 0$ sufficiently small, then both (<ref>) and (<ref>) are satisfied if

\[
s > - \frac14, \qquad \Abs{s} \le r \le s+1, \qquad 0 < r < 1+2s,
\]

and this concludes the proof of Lemma <ref>.

§ ABSTRACT WELL-POSEDNESS FOR DISPERSIVE PDE SYSTEMS WITH NOISE

Let $W(t)$ be a cylindrical Wiener process, as in Section <ref>.