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For a bounded function $\operatorname{F}: \mathbb{R}_{\,\ge\ 0} \to \mathbb{R}$ ( not necessarily non-negative ), is it true that $$ \int_{0}^{\infty}\frac{x^{k}\,s}{(s^{2} + x^{2})^{\left(k + 3\right)/2}\,\,}\, \operatorname{F}\left(x\right)\,{\rm d}x = 0\quad \forall s > 0 \iff \operatorname{F} \equiv 0 $$ where $k \in \mathbb{N}$ is a positive constant $?$ Of course, one implication ($\leftarrow$) is true. What about the other one $?$.

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  • $\begingroup$ Do you mean $F:\mathbb{R}^{\geq 0} \to \mathbb{R}^{\geq 0}$ above? $\endgroup$ – anomaly Jan 23 at 15:19
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    $\begingroup$ @anomaly No, F does not need to be positive $F: \mathbb{R}_{\ge 0} \to \mathbb R$ and is just bounded. $\endgroup$ – Jun Jan 23 at 15:21
  • $\begingroup$ Does $F$ need to be continuous? I'm pretty sure I could construct some sort of pathological function that would break this idea, but it wouldn't be continuous. EDIT: In response to your question edit, if we require $F\geq 0$ the statement is very easy to prove. $\endgroup$ – K.defaoite Jan 23 at 15:36
  • $\begingroup$ @K.defaoite No $F$ does not need to be continuous (although I would also be interested in seeing a positive result assuming continuity) $\endgroup$ – Jun Jan 23 at 15:47
  • $\begingroup$ The converse is always false. Think about it like this: if I tell you that some weighted average of $F$ is zero, can you say that $F$ is always equal to zero? $\endgroup$ – Thomas Bakx Jan 23 at 15:56
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Write

$$f_s(x) = \frac{x^ks}{(s^2+x^2)^{(k+3)/2}}$$

Notice that any such function vanishes at infinity. Let $A$ be the subalgebra of $\textbf{C}_0((0, \infty), \mathbb{R})$ generated by the family $f_s$. All these functions are positive on $(0, \infty)$ so the family doesn't vanish at any point, and it clearly separates points.

Thus by the locally compact version of the Stone-Weierstrass Theorem, $A$ is dense in $\textbf{C}_0((0, \infty), \mathbb{R})$, and $F$ satisfies your integral equation with $f_s$ replaced by any element of $A$. Thus as a functional, using the Riesz Representation Theorem (using the uniqueness part) you get that $F \equiv 0$ a.e.

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  • $\begingroup$ If you don't have the "locally compact version of Stone-Weierstrass" then just do your approximation on successively larger intervals $[\frac{1}{n}, n]$. $\endgroup$ – John Samples Jan 23 at 18:59
  • $\begingroup$ Thanks! Could you add some details about why the assumptions of the Stone-Weierstrass theorem are satisfied. In particular, why does the family $f_s$ generate a subalgebra of $C_0((0,\infty),\mathbb R)$? Why does it separate the points? $\endgroup$ – Jun Jan 27 at 10:26
  • $\begingroup$ You take the sub-algebra generated by the $f_n$, i.e. all finite linear combinations over $\mathbb{R}$. It's an algebra by definition. It separates any positive real (the case at $r = 0$ doesn't matter because a point is a null-set) because for any $r > 0$ you can find different $f_s$ that take different values on $r$. I mean, just write out the equation. If you're feeling very lazy, you can just add in functions $g_s = (1/2)f_s$ and take the algebra generated by both the $f$'s and $g$'s. Can you check the answer as correct? That's what you do on MSE when you get a correct answer. $\endgroup$ – John Samples Jan 29 at 15:33

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