Differentiation under the ingtegral sign for a $W^{1, \infty}$ function

Question

I have the following question.

If $u\in L^1([0, T]; W^{1, \infty}(\Omega))$, then for every $t\in[0, T]$ can we conclude that the function $$ w^t:\Omega\longrightarrow\mathbb R,\qquad x\longmapsto\int_0^t u(s, x)ds $$ belongs to $W^{1, \infty}(\Omega)$ and $\frac{\partial w^t}{\partial x}=\int_0^t \frac{\partial}{\partial x}u(s, x)ds$?

I'm pretty sure that the answer is positive but I cannot find a book where to find it or some similar result which can confirm it. Could you help me?

Thank you

Answer 1

Let $\phi \in C_c^\infty(\Omega)$. Then by Fubini (twice) and the definition of weak derivative we have $$ \int_\Omega D_i\phi(x) \int_0^t u(s,x)ds\ dx = \int_0^t \int_\Omega D_i\phi(x) u(s,x)dx\ ds = -\int_0^t \int_\Omega \phi(x) D_i u(s,x)dx\ ds\\ = - \int_\Omega \phi(x) \int_0^t D_i u(s,x)ds\ dx $$ as claimed.