\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. $\lvert u(x)\rvert\leq \lvert u^*(x)\rvert\,$?


1 Answer 1


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 $u(r) \geq (1+r^2)^{-1} \not \in L^2(\mathbb R^2)$ .


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