Estimate of the norm of the radial part of a function

Question

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?

Answer 1

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 $$\|g \hat u\|_2^2=\sum_k \int_0^\infty |g(s)U_k(s)|^2 s^{N-1}ds \geq \int_0^\infty |g(s)U_0(s)|^2 s^{N-1}ds=\|g\hat{\tilde u}\|^2_2. $$