Let $u\in W_0^{1,1}(B)$, where $B=\{x\in \mathbb{R}^N:\ |x|<1\}$. Assume that $u$ is radially symmetric, that is, $u(x)=u(y)$ if $|x|=|y|$.

Define $f:[0,1]\to \mathbb{R}$ by $f(r)=u(x)$ where $r=|x|$. Is it true that $f$ is an absolutely continuous function with $f(1)=0$?

It seems to me that the answer is true, because for a.e. $w\in \partial B$ the function $u$ is absolutely continuous on the set $\{tw:\ t\in [a,b]\}$, where $-1<a<b<1$.

Thus, by taking a direction $w$ where this is true, we see that $f$ is absolutely continuous in $[0,a]$ for every $0<a<1$, however, I fail to see why it would be that $f(1)=0$ and $f$ is absolutely continuous on $[0,1]$.


1 Answer 1


The answer is yes and no: First note that a radial $u\in W^{1,1}(\mathbb{R}^n)$ (extend your original $u$ as $0$ outside to obtain a $W^{1,1}(\mathbb{R}^n)$ radial function) implies that $$ f\in W^{1,1}((0,\infty), r^{n-1}dr), $$ where this last means that $$f, f' \in L^1((0,\infty), r^{n-1}dr).$$ To see this simply note that since $u$ is radial, we have $\nabla u(x)=f'(|x|)x/|x|$ so that $|\nabla u|=|f'|$ and use polar coordinates.

We see then that, by the usual Sobolev embedding in dimension 1, on any interval $(a,\infty)$, with $a>0$, $f$ is an aboslutely continuous function. In fact you can even get some pointwise decay at $\infty$ (see here for example).

On the other hand, singularities at $0$ can occur , for instance for the function $|x|^{-1}$ for $n\geq 2$.

  • $\begingroup$ @Tomás: $f(|x|)=u(x)$ so $f'$ is the derivative of $f$. $\endgroup$
    – Jose27
    Feb 25, 2015 at 0:15
  • $\begingroup$ @Tomás: A density argument as usual: Convolution of two radial functions is again radial. $\endgroup$
    – Jose27
    Feb 25, 2015 at 0:52

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