we have $$\displaystyle\int_{E\times(0,T)}f(x,\tau)^{p-1}\partial_\tau g(x,\tau)\,dx\,d\tau$$
with $f\in C(0,T;L^p(E))$ for some domain $E$ in $\mathbb{R}^n$, $p>1$ and $g\in W^{1,p}(0,T;L^p(E))$. As continuous functions on compact sets (such as the interval $[0,T]$ in this case) are bounded, does the assumption $f\in C(0,T;L^p(E))$ suffice to guarantee boundedness of the above integral or am I misunderstanding Bochner spaces? I am unsure whether or not I need additionally $f\in L^p(0,T;L^p(E))$, i.e.: $f\in C(0,T;L^p(E))\cap L^p(0,T;L^p(E))$. As I said, unsure though if this really is needed but I am not that comfortable with Bochner spaces yet.
Thanks in advance and have a nice weekend!
$f\in C([0,T]; L^p(E))$ implies $f\in L^p(0,T;L^p(E))$ for all $p$, since $T$ is finite.
However this is not enough to guarantee the existence of the integral. Also the inner ($x$) integral has to exist: $f^{p-1}$ is in $L^\infty(0,T; L^{\frac{p}{p-1}})$, $\partial_\tau g$ belongs to $L^p(0,T;L^p(E))$. And the product is integrable. To see this, one can apply Hoelders inequality to show that the integral exists: $$ \begin{split} \int_{E\times(0,T)} |f^{p-1} \partial_\tau g| dx\ d\tau &\le \int_{(0,T)} \int_E |f^{p-1} \partial_\tau g| dx\ d\tau \\ &\le \int_{(0,T)} |f(\tau)^{p-1}|_{L^{\frac{p}{p-1}}(E)} |\partial_\tau g(\tau)|_{L^p(E)} d\tau \\ &\le \int_{(0,T)} |f(\tau)|_{L^p(E)}^{p-1} |\partial_\tau g(\tau)|_{L^p(E)} d\tau \\ &\le \big\|\ |f(\tau)|_{L^p(E)}^{p-1}\ \big\|_{L^\frac{p}{p-1}(0,T)}\|\partial_\tau g(\tau)\|_{L^p(0,T;L^p(E))} \\ &\le \| f\|_{L^p(0,T,L^p(E))}^{p-1}\|\partial_\tau g(\tau)\|_{L^p(0,T;L^p(E))} < +\infty. \end{split} $$