I am working on the following exercise on Kai Lai Chung.
Let $F$ be a symmetric d.f. with ch.f. $f\ge0$, then $$\phi_F(h)=\int_{-\infty}^{\infty} \frac{h^2}{h^2+x^2} dF(x)=h\int_0^\infty e^{-ht}f(t)dt$$ is a sort of average concentration function. Prove that if $G$ is also a d.f. with ch.f. $g\ge0$, then we have $\forall h>0:$ $$\phi_{F*G}(h)\le\phi_F(h)\land\phi_G(h)$$ $$1-\phi_{F*G}(h)\le[1-\phi_F(h)][1-\phi_G(h)]$$
I feel this exercise simply wants me to use the fact that the characteristic function corresponding to $F*G$ is $fg$, then we can estimate on the last integral $$h\int_0^\infty e^{-ht}f(t)g(t)dt$$ but then I check the identity $$\int_{-\infty}^{\infty} \frac{h^2}{h^2+x^2} dF(x)=h\int_0^\infty e^{-ht}f(t)dt$$ By writing $f(t)=E(e^{itX})$ explicitly and changing the order of integration, I find that $F$ has to be symmetric for it to be valid. However, it doesn't tell whether $G$ is symmetric, so $F*G$ is not necessarily symmetric, and we can't use the formula directly. Then I don't know how to proceed.
Any help would be appreciated.
A random variable $X$ has a symmetric distribution iff its characteristic function is real valued. [This is because $X$ and $-X$ have the same characteristic function when the characteristic function is real valued]. In this case the characteristic functions of $F$ and $G$ are no non-negative so these distributions are symmetric.