0
$\begingroup$

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.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks for the answer, is there some additional constraint that can be imposed on $u$ such that the equality holds? I also have that $u$ is a weak solution to some linear parabolic equation, will that help? $\endgroup$
    – Adi
    Commented May 5, 2021 at 20:48
  • $\begingroup$ @Adi The only condition that I can see is that the $t$ on the left hand side is the same for almost all $r$. $\endgroup$
    – mlk
    Commented May 5, 2021 at 20:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .