Consider a function $u\in L^2(\mathbb R^N)$, and another function $\varphi$ which is the unique solution to the Poisson equation $\Delta \varphi = u$ vanishing at $\infty.$ We know that the radial part of $u$, defined by
$$
\tilde u(r):=\frac{1}{S\_{N-1}}\int\_{\partial B(0,1)} u(ry)d\sigma\_y
$$
is radially symmetric and we have the inequality $\| \nabla\tilde u\|\_{L^2} \leq \|\nabla u\|\_{L^2}$, if $\nabla u$ is defined in the weak sense. Intuitively we have reduced the fluctuation by taking an average. In terms of Fourier transform, we have $\||\xi|\hat{\tilde u}\|\_{L^2} \leq \||\xi|\hat{u}\|\_{L^2}$.


Now, my question is, do we have a similar inequality for comparing other powers of $\xi,$ that is, and inequality for $s>0$ of the form
$$
\||\xi|^s\hat{\tilde u}\|\_{L^2} \leq \||\xi|^s\hat{u}\|\_{L^2} \space?
\label{1}
$$


And also, let $\Delta \tilde\varphi=u$ be solved by a function that vanish at $\infty$. For $s=-1,$ the above inequality becomes $\||\xi|^{-1}\hat{\tilde u}\|\_{L^2} \geq \||\xi|^{-1}\hat{u}\|\_{L^2}$ (I have reverse the direction of the inequality, because this makes more sense for an "antiderivative"). This corresponds to the inequality
$$
\|\nabla \tilde\varphi\|\_{L^2} \leq\|\nabla \varphi\|\_{L^2}.
$$


Is this inequality true?



Estimate of the norm of the radial part of a function
Yes, it is true. Take a function $u \in L^2(R^N)$ and expand for $x=r\omega$
$$
u(r\omega)=\sum\_{k=0}^\infty u\_k(r)P\_k(\omega)$$
where $(P\_k)$ an orthonormal basis of spherical harmonics in $L^2(S^{N-1})$ and $u\_0$ is your $\tilde u$. Then for $\xi=s\eta$
$$\hat u(s\eta)
=\sum\_{k=0}^\infty U\_k(s)P\_k(\eta),$$ see Stein-Weiss "Fourier Analysis....IV.4" and $U\_0=\hat{\tilde u}$. If you multiply by a radial weight $g(s)$, then
$$\|g \hat u\|\_2^2=\sum\_k \int\_0^\infty |g(s)U\_k(s)|^2 s^{N-1}ds \geq \int\_0^\infty |g(s)U\_0(s)|^2 s^{N-1}ds=\|g\hat{\tilde u}\|^2\_2.
$$


