\; x\in\mathcal{H}.$$ Such an operator always exists [@Renardy95] and we refer to [@CL4] for various examples. Using the operator $B$ we define for $\gamma>0$ the space $\mathcal{H}_{-\gamma}$ to be the completion of $\mathcal{H}$ under the norm $$\|x\|_{-\gamma}=\|B^{\frac{\gamma}{2}}x\|.$$ We need to impose another set of assumptions on $b$ and $L$.

\[hp:section4\] There exist a constant $K>0$ and a local modulus of continuity $\omega(\cdot,\cdot)$ such that: $$\|b(t,x,u)-b(s,y,u)\| \leq K \|x-y \|_{-1} + \omega(|t-s|, \|x\|
\vee \|y\|)$$ and $$|L(t,x,u)-L(s,y,u)| \leq \omega( \|x-y \|_{-1} + |t-s|, \|x\| \vee
\|y\|)$$

Let $m\geq 2$. Modifying slightly the functions introduced in [@CL5] we define for a function $w:(0,T)\times \mathcal{H}\to
\mathbb{R}$ and $\epsilon,\beta,\lambda>0$ its sup- and inf-convolutions by $$w^{\lambda,\epsilon,\beta}(t,x)=\sup_{(s,y)\in(0,T)\times
\mathcal{H}} \left\{w(s,y)-\frac{\|x-y\|_{-1}^2}{2\epsilon}
-\frac{(t-s)^2}{2\beta}-\lambda e^{2mK(T-s)}\|y\|^m\right\},$$ $$w_{\lambda,\epsilon,\beta}(t,x)=\inf_{(s,y)\in(0,T)\times
\mathcal{H}} \left\{w(s,y)+\frac{\|x-y\|_{-1}^2}{2\epsilon}
+\frac{(t-s)^2}{2\beta}+\lambda e^{2mK(T-s)}\|y\|^m\right\}.$$

\[lem2\] Let $w$ be such that $$\label{aaa2} w(t,x)\leq C(1+\|x\|^k)\quad(\hbox{respectively,}\,\,\,
w(t,x)\geq -C(1+\|x\|^k))$$ on $(0,T)\times \mathcal{H}$ for some $k\geq 0$. Let $m>k$. Then:

-   For every $R>0$ there exists $M_{R,\epsilon,\beta}$ such that if $v=w^{\lambda,\epsilon,\beta}$ (respectively, $v=w_{\lambda,\epsilon,\beta}$) then $$\label{aaa6} |v(t,x)-v(s,y)|\leq
    M_{R,\epsilon,\beta}(|t-s|+\|x-y\|_{-2})\quad
    \hbox{on}\,\,\,(0,T)\times B_R$$

-   The function $$w^{\lambda,\epsilon,\beta}(t,x)+\frac{\|x\|_{-1}^2}{2\epsilon}
    +\frac{t^2}{2\beta}$$ is convex (respectively, $$w_{\lambda,\epsilon,\beta}(t,x)-\frac{\|x\|_{-1}^2}{2\epsilon}
    -\frac{t^2}{2\beta}$$ is concave).

-   If $v=w^{\lambda,\epsilon,\beta}$ (respectively, $v=w_{\lambda,\epsilon,\beta}$) and $v$ is differentiable at $(t,x)\in (0,T)\times B_R$ then $|v_t(t,x)|\leq
    M_{R,\epsilon,\beta}$, and $Dv(t,x)=Bq$, where $\|q\|\leq
    M_{R,\epsilon,\beta}$

**(i)** Consider the case $v=w^{\lambda, \epsilon, \beta}$. 