Let $f\in L^1(0,\infty)$ and $F(x):=\int_0^\infty xe^{-xy}\int_0^yf(t)dtdy$ on$(0,\infty)$ Suppose there exists $c\in \mathbb{R}$, for any $n\in \mathbb{N}$, $F(nc)=0$. Then $F=0$
I checked $F$ is well-defined on $(0,\infty)$ I tried to use uniqueness of Fourier's transformation, but I couldn't. Someone know how to prove $F=0$?
Observe that $xe^{-xy}=-\frac{\partial}{\partial y}e^{-xy}$, so that we can apply integration by parts: $$ F(x)=\lim_{Y\to\infty}-e^{-xY}\int_0^Yf(t)dt+\int_0^\infty e^{-xy}f(y)dy. $$ The first term is zero since $|\int_0^Yf(t)dt|\le \|f\|_{L^1}$, so in fact $$ F(x)=\int_0^\infty e^{-xy}f(y)dy $$ is just the Laplace transform of $f$. Let us try to avoid using complex analysis to invert this transform.
The condition that $F(nc)=0$ says that if $E_{n}(x)=e^{-ncx}$, then $\int_0^\infty E_{n}(x)f(x)dx=0$ for all $n$, which means this problem is analogous to proving $\phi(\theta)=0$ if the Fourier coefficients $\int_0^{2\pi}e^{in\theta}\phi(\theta)d\theta=0$ for all $n$. One way to do this is the Stone-Weierstrass theorem: since the algebra generated by $e^{i\theta}$, which we write as $<e^{i\theta}>$, is an algebra of continuous functions on $C(\mathbb S^1)$ which separates points of $\mathbb S^1$, it is dense in $C(\mathbb S^1)$. This means $\int_0^{2\pi}g(\theta)\phi(\theta)d\theta=0$ for all $g\in C$, which obviously implies $\phi(\theta)\equiv 0$.
There is a version of the Stone-Weierstrass theorem for locally compact Hausdorff spaces, such as $[0,\infty)$. It is clear that the algebra $<e^{-cx}>$ separates points of $[0,\infty)$ and vanishes nowhere, so it is dense in $C_0([0,\infty))$. As before, this means $\int_0^\infty g(x)f(x)dx=0$ for all $g\in C_0$, which gives $f\equiv 0$.
If I'm not mistaken, one might prove the locally compact version of Stone-Weierstrass by successively "multiplying the algebra" by a bump function $\psi(x/k)$, applying compact SW, and letting $k\to\infty$.