# An integral transform and the Stone-Weierstrass theorem

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 $$F \equiv 0$$.
• As argued in the thread you referred to, the first statement is true if and only $F$ is identically zero (almost everywhere). This means that the second statement is true as well. May 17, 2021 at 6:56
• @MatthiasLudewig In the thread, the answer uses Stone-Weierstrass to show that the first statement is true if and only if $F$ is zero. But the proof is missing a step: you need to verify that the subalgebra generated by the expression in brakets satisfies that the integral is equal to zero, that is you need to verify the second statement
– Jun
May 17, 2021 at 7:19
• Is the first statement only supposed to hold for some $k$ which is independent of $s$, but not necessarily for every $k$, right? May 17, 2021 at 9:40
• @user49822 Correct
– Jun
May 17, 2021 at 10:18

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.

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.

• I doubt this qualifies as "use Stone–Weierstrass to...", but might be of some interest anyway. May 18, 2021 at 9:23
• I prefer your method 1. By the way, the Mellin transform in method 2 should give the Fourier transform for $k$ even that I was unable to find. May 18, 2021 at 10:50
• @GiorgioMetafune: Ah, indeed — I only glanced at your answer and did not notice that in fact you were using the Mellin inversion formula in disguise. Sorry for that. Regarding the expression for the transform — I just used Mathamatica, but it is not the first time when I realise tables of Mellin transforms are more complete than tables of Fourier transforms. May 18, 2021 at 12:07
• No problem for me! Happy to discuss May 18, 2021 at 12:45
• @GiorgioMetafune and MateuszKwaśnicki: thanks to both of you for the interesting answers. Do you also have any idea about the approach using Stone-Weierstrass?
– Jun
May 18, 2021 at 19:26

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.

• Thanks! That's very interesting. What if we replace $\frac{x^{k}\,s}{(s^{2} + x^{2})^{\left(k + 3\right)/2}}$ by something more abstract but with the same scaling properties, like $s^{-\tfrac{k+2}{\alpha}} x^{k} g(|x|s^{-(k+2)/\alpha})$ for a continuous function $g$ and a fixed $\alpha \in \mathbb R$?
– Jun
May 18, 2021 at 19:33
• It depends on the vanishing of the Fourier tranform or of the Mellin transform. May 18, 2021 at 20:28