# Condition for an integral to be zero

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 $$?$$.

• Do you mean $F:\mathbb{R}^{\geq 0} \to \mathbb{R}^{\geq 0}$ above? – anomaly Jan 23 at 15:19
• @anomaly No, F does not need to be positive $F: \mathbb{R}_{\ge 0} \to \mathbb R$ and is just bounded. – Jun Jan 23 at 15:21
• 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. – K.defaoite Jan 23 at 15:36
• @K.defaoite No $F$ does not need to be continuous (although I would also be interested in seeing a positive result assuming continuity) – Jun Jan 23 at 15:47
• 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? – Thomas Bakx Jan 23 at 15:56

$$f_s(x) = \frac{x^ks}{(s^2+x^2)^{(k+3)/2}}$$
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.
• If you don't have the "locally compact version of Stone-Weierstrass" then just do your approximation on successively larger intervals $[\frac{1}{n}, n]$. – John Samples Jan 23 at 18:59
• 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? – Jun Jan 27 at 10:26
• 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. – John Samples Jan 29 at 15:33