TITLE: Function defined by integrals convex?
QUESTION [5 upvotes]: Let $g$ be a positive integrable function in $[0,\infty)$, and $G$ its integral, that is $G(t) = \int_0^t g(u) \, du$.
Is the function f, defined as
$$
f(t) = \int_0^\infty g(u) e^{-(G(u+t) - G(u))} \, du
$$
convex?.
It is already decreasing and bounded by $1 - e^{-G(\infty)}$ and $G(\infty)$.
I have been testing numerically and has not failed. Also, I have reasons from the "physics" of the problem that I am working that this is the case.
Any answers, even with stronger hypotheses or counter examples are welcome.
Thanks in advance.

REPLY [2 votes]: I split the integrand
$$
f(t)=\int_0^{+\infty}g(u)e^{G(u)}e^{-G(u+t)}\,dt
$$
to stress that convexity of $f$ in $t$ will depend on the term $\phi(u,t)=e^{-G(u+t)}$. For convexity of $f$ it is sufficient to assume that the function $g(u)$ is decreasing (i.e. $u_2>u_1$ gives $g(u_2)\le g(u_1)$). It is not very restrictive I hope since $g$ must be integrable over $[0,+\infty)$.
If it is the case then the function $-G(u)$ is convex ($-G'(u)=-g(u)$ is increasing) and, hence, the function $-G(u+t)$ (superposition of convex and linear) as well as $\phi(u,t)=e^{-G(u+t)}$ (superposition of convex growing and convex) are convex. Finally, the function $f(t)$ is convex as a non-negative linear combination of convex functions (continuum many non-negative weights $g(u)e^{G(u)}$ , but it does not matter, the proof is identical - by definition of convexity).