Let $B_r$ be a ball of radius $r$ centered at $0$ and $u:B_r\times[t_1,t_2]\to(0,\infty)$ be an $L^p$ integrable function. Define the integral $$ \sup_{t_1<t<t_2}\int_{B_r}u^p(x,t)\,dx. $$ Then is it possible to choose $\tau\in[t_1,t_2]$ such that $$ \int_{B_r}u^p(x,\tau)\,dx=\sup_{t_1<t<t_2}\int_{B_r}u^p(x,t)\,dx? $$ Even for continuous function? Thanks.
Is it even possible for continuous function. I have seen such results are possible written in just one line. For example in the article: Equation (28) https://arxiv.org/pdf/1802.07649.pdf
Can somebody kindly help me. Thanks.
If $B_r$ is closed, then $B_r\times [t_1,t_2]$ is compact. So if $u: B_r\times [t_1,t_2]\to (0,\infty)$ is continuous, then $u$ is uniformly continuous there. Hence $u^p$ is uniformly continuous there. It follows that the function
$$t\to \int_{B_r} u^p(x,t)\,dx$$
is continuous on $[t_1,t_2].$ Hence it acheives a maximum at some $c\in [t_1,t_2].$ This gives the desired result for continuous $u.$
If $u$ is not continuous, there should be easy counterexamples.