other hand, going backward of the above equalities one easily finds that if $u$ is given as in (\[right term\]), then it satisfies (\[time-space\]). The proof of the lemma is now complete.

Recall $\gamma_0=(2\gamma-1)_+/{\beta}<2$. Fix $\varepsilon_0\in (0,1)$ and define $$\label{e:3.23}
\sigma_0:=\gamma_0+ (\varepsilon_0 1_{\gamma=1/2})=
\begin{cases}
 (2\gamma-1) /\beta \quad &\hbox{if } \gamma >1/2\\
 \varepsilon_0 &\hbox{if } \gamma =1/2\\
 0 &\hbox{if } \gamma <1/2.
 \end{cases}$$

\[thm laplace\] For any $f\in {\mathbb{H}}^{\sigma}_2(T)$, $g\in {\mathbb{H}}^{\sigma+\sigma_0}_2(T,\ell_2)$ and $u_0\in U^{\sigma+1}_2$, equation (\[time-space\]) has a unique solution $u\in {\mathcal{H}}^{\sigma+2}_2(T)$, and for this solution we have $$\label{main estimate1}
\|u\|_{{\mathbb{H}}^{\sigma+2}_2(T)}\leq N\left(\|u_0\|_{U^{\sigma+1}_2}+\|f\|_{{\mathbb{H}}^{\sigma}_2(T)}+\|g\|_{{\mathbb{H}}^{\sigma+\sigma_0}_2(T,\ell_2)}\right),$$ where $N$ depends only on $d$ and $T$.

Without loss of generality we only need to prove the case $\sigma=0$.

For the deterministic equation, the theorem including the estimate is proved in [@Za]. Thus the uniqueness for equation (\[time-space\]) easily follows.

Define $w$ as in (\[eqn deterministic\]). Then by considering $u-w$, we may assume without loss of generality that $u_0=0$ and $f=0$.

First, assume $g\in {\mathbb{H}}^{\infty}_0(T,\ell_2)$. Then by Lemmas \[lemma main estimate\] and \[lemma formula\], equation (\[time-space\]) has a unique solution $u\in {\mathbb{H}}^2_2(T)$ and estimate holds.

For general $g\in  {\mathbb{H}}^{\sigma_0}_2(T,\ell_2)$, take a sequence of $g_n\in {\mathbb{H}}^{\infty}_0(T,\ell_2)$ so that $g_n \to g$ in ${\mathbb{H}}^{\sigma_0}_2(T,\ell_2)$. Define $u_n$ as the solution of 