Suppose we have $u : [0,T] \times \Omega\to \mathbb{R}$ where $\Omega \subset \mathbb{R}$ is bounded and the following facts:
- $\partial_{x}u \in L^{\infty}(0,T; L^{\infty})$
- $\partial_{t}u \in L^{\infty}(0,T; L^{2})$
I am wondering if there is an easy way to deduce from either (or both) of these properties that $u \in L^{\infty}(0,T; L^{\infty})$ or $u \in L^{\infty}(0,T; L^{2})$?
It is not totally obvious to me. I would like to say from the first property that $u \in L^{\infty}(0,T; W^{1,\infty}) \hookrightarrow L^{\infty}(0,T; L^{\infty})$ but this is basically a circular argument since I am implicitly assuming $u \in L^{\infty}L^{\infty}$.
If $\Omega$ is unbounded, this is false, and $u(t,x) = x$ is a counterexample.
If $\Omega$ is a nice bounded domain, then $$\partial_x u\in L^\infty(\Omega) \implies \partial_x u \in L^1(\Omega)\\ \implies u\in L^\infty(\Omega)$$ by Sobolev's inequalities (uniformly in time), and so the result hold.