Fractional Sobolev norm of a radial function 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.).
 A: 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.
A: A proof follows using interpolation:
First, notice that the claim can only hold if $\operatorname{supp} f\subset (0,1)$ (and not $[0,1]$ as written in the question). This is in order to avoid singularities at the origin after the transformation $f\mapsto F$.
As such, the operator $T[f] := F$ is a bounded linear operator $T:W_0^{k,p}((0,1))\to W_0^{k,p}(B_1(\mathbb{R}^n))$, where $B_1(\mathbb{R}^n)$ is the unit ball in $\mathbb{R}^n$, for any $k$, $p$ and $n$.
This follows by a direct calculation, using the fact that
$$D^k (Tf)(x) = \sum_{j=1}^k \frac{f^{(j)}(|x|)}{|x|^{k-j}}G_j^k\left(\frac{x}{|x|}\right), $$
where $G_j^k$ are smooth $k$-tensor valued functions (in fact, polynomials), which are independent of $f$.
The fact that $T:W_0^{k,p}((0,1))\to W_0^{k,p}(B_1(\mathbb{R}^n))$ is bounded follows by evaluating each term separately, using Jensen's inequality.
For a non-integer $s$, we then have that $T:W_0^{s,p}((0,1))\to W_0^{s,p}(B_1(\mathbb{R}^n))$ by interpolation, as the $W^{s,p}$ norm is equivalent to the interpolation norm of $(W^{k,p}_0(B_1(\mathbb{R}^n)), W^{k+1,p}_0(B_1(\mathbb{R}^n)))_{\sigma,p}$, where $s = k +\sigma$, $k$ an integer and $\sigma\in (0,1)$.
This interpolation (which slightly differs from the standard one because of the boundary conditions) follows from the one carried out in "A note on homogeneous Sobolev spaces of fractional order" by Brasco and Salort.
