2
$\begingroup$

\begin{align*} L^2 (\mathbb{R}^3)& {}=\{ u : \int_{\mathbb{R}^3} \lvert u\rvert^2 dx<+\infty \}. \\ H^1(\mathbb{R}^3) & {}=\{ u\in L^2 (\mathbb{R}^3):\, \lvert\nabla u\rvert\in L^2(\mathbb{R}^3) \}. \\ H_r(\mathbb{R}^3) & {}=\{ u\in H^1 (\mathbb{R}^3):\, \text{$u$ is radial} \}. \end{align*}

How to connect the functions in spaces $H^1$ and $H_r$?

I saw a lemma like that, for every $u\in H^1(\mathbb{R}^N)$, $u\geq 0$, there results $u^*\in H_r$, $u^*\geq 0$,

\begin{align*} \int_{\mathbb{R}^N}|\nabla u^*|^2dx\leq \int_{\mathbb{R}^N}|\nabla u|^2dx\quad and \quad\quad\quad\\ \int_{\mathbb{R}^N}|u^*|^pdx = \int_{\mathbb{R}^N}|u|^pdx,\quad for\;all\;\, p>1. \end{align*}

So the functions in these two Spaces can be related by integration.

Is there some other relationship (e.g. inequality) between the functions in these two spaces $H^1$ and $H_r$?

e.g. (I guess) $\,\forall\, u\in H^1$, is there an $u^*\in H_r$ s.t. $\lvert u(x)\rvert\leq \lvert u^*(x)\rvert\,$?

$\endgroup$

1 Answer 1

1
$\begingroup$

This is not true since a radial majorant might not be in $L^2$. A counterexample is $u(x,y)=(1+x^2+y^4)^{-\frac 12} \in H^1(\mathbb R^2)$. If $u^*$ is radial and majorizes $u$, and $r=\sqrt {x^2+y^2}$, then $u(r) \geq (1+r^2)^{-1} \not \in L^2(\mathbb R^2)$ .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.