Suppose we have the time interval $[0,T]$, a domain $\Omega = [0,1]$. I am particularly interested in the case $p=1, q=2$. In other words if we have a function
$$u \in L^{\infty}(0,T; L^{1}(\Omega)) \cap L^{2}(0,T; L^{\infty}(\Omega)), $$ can we say that $$u \in L^{\infty}(0,T; L^{\infty}(\Omega)? $$ Intuitively, I can't decide if this statement should be true or not. I tried to construct a function in this space in order to look for a possible counterexample so I started with the simple case of $u = t^{\alpha}x^{\beta}$. I found that for $u$ to belong to the intersection of the above spaces means that $\alpha , \beta > 0$. Then this function would of course be in $L^{\infty}L^{\infty}$. This calculation gave me some feeling that the inclusion is true but on the other hand I'm not sure if one could construct an elaborate counterexample to show otherwise. I also tried to prove the statement using the $L^{p}$ embeddings $L^{p} \hookrightarrow L^{q}$ for $p > q$ but to no avail.
I would appreciate any help.
The answer is no, take for example \begin{equation} u(t, x) = t^{-1/3}\mathbf{1}_{0\le x\le t^{1/3}}\qquad (t,x)\in [0,1]^2 \end{equation} Then \begin{equation} \|u\|_{L^1_x} = 1 \in L^\infty_t\quad \text{and}\quad \|u\|_{L^\infty_x} = t^{-1/3}\in L^2_t \end{equation} but \begin{equation}\|u\|_{L^\infty_x}\not\in L^\infty_t\end{equation}