Let $f(x):\mathbb{R}\to \mathbb{R}$ be a smooth function, with $\operatorname{supp}(f) \subset [0,1]$. Consider the radial function $F:\mathbb{R}^d\to \mathbb{R}$, defined by $F(x) := f(|x|)$.

A direct calculation shows that for $k\in\mathbb{N}\cup \{0\}$ and $p\in[1,\infty]$, we have $$\|F\|_{W^{k,p}(\mathbb{R}^d)} \le C \|f\|_{W^{k,p}(\mathbb{R})},$$ For some $C=C(k,p,d)>0$, which is independent of $f$.

Is it true also for fractional Sobolev norms $W^{s,p}$, $s>0$?

For concreteness, I refer here to the norm $$ \| g \|_{W^{s,p}(\mathbb{R}^d)}^p = \| g\|_{W^{k,p}}^p + \int_{\mathbb{R}^d}\int_{\mathbb{R}^d} \frac{|D^kg(x)-D^kg(y)|^p}{|x-y|^{d+\sigma p}}\,dx\,dy, $$ where $s=k+\sigma$ such that $k\in\mathbb{N}\cup \{0\}$ and $\sigma\in(0,1)$, but I'll be also interested in results for other similar norms (e.g., using Fourier transform etc.).


I think that the answer is affirmative, at least for a certain range of exponents $p$, but I need to work a bit more to be sure. Meanwhile, I am posting here my partial progress.

The main problem is to estimate, for $\sigma \in (0, 1)$, $$ \iint_{\mathbb R^d\times \mathbb R^d} \frac{|F(x)-F(y)|^p}{|x-y|^{d+\sigma p}}\, dxdy \le C\iint_{[0, \infty)^2} \frac{|f(r)-f(s)|^p}{|r-s|^{1+\sigma p}}\, drds, $$ for some constant $C=C(d, \sigma, p)>0$. Integrating the left-hand side in polar coordinates, we see that this will follow from the estimate (to be proved) $$\tag{*} r^{d-1}s^{d-1}\iint_{\mathbb S^{d-1}\times \mathbb S^{d-1}} \frac{d\omega\, d\eta}{|r\omega - s\eta|^{d+\sigma p}} \le \frac{C}{|r-s|^{1+\sigma p}}.$$

I conjecture that (*) is true. To prove it, I think that we can use the Funk-Hecke theorem, like in this small paper of Han-Atkinson-Zheng (section 3), to compute $$ \int_{\mathbb S^{d-1}}\frac{d\omega}{|r\omega -s\eta|^{d+\sigma p}}.$$

Remark. I tried to solve the problem, in the $p=2$ case, using the Fourier transform, via the formula $$ \hat{F}(\xi)=C|\xi|^{-(d/2-1)}\int_0^\infty f(s)J_{\frac{d}2 -1}(|\xi|s)s^\frac{d}2\,ds, $$ (see, e.g., Stein-Weiss, Introduction to Fourier analysis on Euclidean spaces). The target would be to prove that $$ \int_{\mathbb R^d} |\hat{F}(\xi)|^2(1+|\xi|^2)^s\, d\xi \le C\int_0^\infty \left\lvert \int_0^\infty f(r) e^{-ir\rho}\,dr\right\rvert^2(1+\rho^2)^s\, d\rho.$$

But this seems to be more difficult.

  • $\begingroup$ Thanks! the Funk-Hecke theorem does seem to be helpful here, and I agree that (*) should suffice for the case $s\in (0,1)$. But when $s>1$ (say $s\in (1,2)$) there is an additional subtlety since $DF(x) = f'(r)\omega$, so $|DF(x) - DF(y)|$ is not independent of the polar factor. $\endgroup$ – C M May 16 at 15:25
  • $\begingroup$ @C M: right. This is weird. I thought it would be simpler. The result looks so obvious. But maybe there is a soft solution. If the result is true for integer values of s, then it should be true for nonintegers values too, by interpolation. $\endgroup$ – Giuseppe Negro May 16 at 21:35

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