An integral transform and the Stone-Weierstrass theorem

Question

For a bounded function $\operatorname{F}: \mathbb{R}_{\,\ge\ 0} \to \mathbb{R}$ (not necessarily non-negative), if $$ \int_{0}^{\infty}\frac{x^{k}\,s}{(s^{2} + x^{2})^{\left(k + 3\right)/2}\,\,}\, \operatorname{F}\left(x\right)\,{\rm d}x = 0\quad \forall s > 0 $$ where $k \in \mathbb{N}$ is a positive constant, is it true that $$ \int_{0}^{\infty}\left(\frac{x^{k}\,s}{(s^{2} + x^{2})^{\left(k + 3\right)/2}\,\,}\right)^{h} \, \operatorname{F}\left(x\right)\,{\rm d}x = 0\quad \forall s > 0 $$ where $h \in \mathbb{N}$?

This question is inspired by a comment to the answer in https://math.stackexchange.com/questions/3996738/condition-for-an-integral-to-be-zero (which required to check the assumptions of the Stone-Weierstrass theorem).

More generally, ignoring the question above, my main concern is this:

• Use Stone-Weierstrass to prove that $$ \int_{0}^{\infty}\frac{x^{k}\,s}{(s^{2} + x^{2})^{\left(k + 3\right)/2}\,\,}\, \operatorname{F}\left(x\right)\,{\rm d}x = 0\quad \forall s > 0 $$ if and only if $F \equiv 0$.

Answer 1

Method 1: There is a clear harmonic-analytic interpretation: if \[ u(x, s) = C_k \int_{\mathbb R^{k+2}} \frac{s}{(s^2 + |x - y|^2)^{(k+3)/2}} \times |y|^{-1} F(|y|) dy , \] then $u$ a harmonic function in the half-space $x \in \mathbb R^{k+2}$, $s > 0$, and $u(0, s) = 0$ for all $s > 0$. We claim that $u$ is identically zero, and hence $F = 0$ almost everywhere. This follows by standard tools: we have $\partial_s^n u(0, 1) = 0$ for all $n$, and so $\Delta_x^n u(0, 1) = 0$ for all $n$, and consequently all terms in the Taylor expansion of $u$ about $(0, 1)$ vanish.

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Method 2 (Edit: This is the same as Georgio Metafune's answer, which I did not realise in the beginning. Sorry!) : Up to some powers of $s$, the integral is equal to the Mellin convolution of $F$ and the kernel \[ x \mapsto \frac{x^k}{(1 + x^2)^{(k+3)/2}} . \] The Mellin transform of that kernel is \[ t \mapsto \frac{\Gamma(\tfrac{k-t}{2}) \Gamma(\tfrac{k+t}{2})}{2 \Gamma(k)} , \] which has no zeros in the strip $0 \leqslant \Re t \leqslant k$. By the Mellin convolution theorem, we have $F = 0$.

The last step requires some care, since the Mellin transform of $F$ may fail to exist in the usual sense. One way around is to use distribution theory, as in Giorgio Metafune's answer. Another approach might be to split $F$ into two pieces ("small $x$" and "large $x$"), write the Mellin inversion formula for both of them, and deform the contour of integration.

Answer 2

Rewrite the assumption as $$\int_0^\infty \frac{y^k}{(1+y^2)^\frac{k+3}{2}}F(sy) dy=\int_{-\infty}^\infty \frac{e^{(k+1)x}}{(1+e^{2x})^{\frac{k+3}{2}}}F(e^{t+x})dx=0. $$ If $G(s)=F(e^s)$, this means that $G*v=0$, where $v(x)=\frac{e^{(k+1)x}}{(1+e^{2x})^{\frac{k+3}{2}}}$. Now $G$ is bounded and $v$ is in the Schwartz class and we can take the Fourier transform of $G$ and of $G*v$ in the sense of tempered distributions obtaining $0=\hat G \hat v$, that is $\langle \hat G, \hat v \phi \rangle=0$ for every $\phi$ in the Schwartz class. Now the question is reduced to the existence of real zeros of $\hat v$. If $\hat v(a)=0$, with $a$ real, then $G(s)=e^{ias}$ satisfies $G*v=0$. On the other hand, if $\hat v$ never vanishes on $\mathbb R$, then $\hat v \phi=\psi$ can be solved for every $\psi \in C_c^\infty (\mathbb R)$ and $\langle \hat G, \psi \rangle=0$ for every $\psi \in C_c^\infty (\mathbb R)$ imples that the same is true for every $\psi$ in the Schwartz class, hence $\hat G=0$ and $G=0$, too.

Concerning the zeros of $\hat v$, I checked in the book of Erdely (Tables of integral transforms, I, formula 20 pag 120) and it turns out that there are no real zeros when $k$ is odd (formula (20) gives the Fourier transform when $(k+3)/2$ is an integer). I do not know for even integers but some more thoughts should give the answer.