Let $u \in C^0(-T,T; L^2(B\_R))$ be a measurable function, then is the following true?


$$
\int\_0^R \sup\_{-T<t<T} \int\_{S\_r} |u(\sigma ,t)|^2 \ d \sigma \ dr = \sup\_{-T<t<T}\int\_0^R \int\_{S\_r} |u(\sigma ,t)|^2 \ d \sigma \ dr.
$$


Here $S\_r$ denotes the sphere of radius $r$ and $d\sigma$ is the standard measure on the sphere.



Interchange of integration and supremum
I don't think your left hand side is well defined for the class of $u$ you are considering, I can change each $u(.,t)$ to a large value on the zero-set $S\_{|t|}$, which will result in the supremum picking $t=r$ and changing the value of the left hand side.


But even for smooth functions, there is a counterexample in the same vein: Choose $u(.,t) = 1$ on $S\_{|t|}$ and going down to $0$ fast. Then the right hand side can be arbitrarily small, but the left hand side will always have $1$ as an integrand. In fact that is the general inequality you will have.


