Let $z \colon \mathbb R^2 \to \mathbb{R}$ be $\mathcal C_B^{1,1}(\mathbb R^2)$. Fix $\epsilon>0$, choose $\rho \in \mathcal C^\infty $ a mollifier and consider the convolution of $z$: $$z_\epsilon(x_1,x_2)=\int_\mathbb{R}z(y_1,x_2) \frac 1 \epsilon \rho \left (\frac{x_1-y_1}{\epsilon} \right)dy_1$$ Now it is know that $z_\epsilon$ is differentiable, we denote by $Dz, Dz_\epsilon$ the gradients. How do I prove that $$|Dz_\epsilon(x_1,x_2)-Dz(x_1,x_2)| \leq C \epsilon$$ for some $C>0$ as claimed by P.L.Lions in [Viscosity solutions of fully nonlinear second-order equations and optimal stochastic control in infinite dimensions. Part I: the case of bounded stochastic evolutions] eq. (66) p.269 ?
The derivative wrt $x_2$ is simple since the integral is not involved, but how do you treat the derivative wrt $x_1$?
\begin{align} |\frac{\partial z}{\partial x_1}-\frac{\partial z_\epsilon}{\partial x_1}|=\left |\frac{\partial z}{\partial x_1} - \frac{\partial }{\partial x_1} \int_\mathbb{R}z(y_1,x_2) \frac 1 \epsilon \rho \left (\frac{x_1-y_1}{\epsilon} \right)dy_1 \right| \end{align} I should be able to pass the derivative of the convolution to $\rho$ but then how do I go on?
If your problem is about interchanging the integral and derivative, I think something like Dominated convergence theorem would be helpful.