Let $(X,\Sigma,\mu)$ be a probability space, and let $g:X \to \mathbb [0,\infty)$ be measurable.
Let $\phi:\mathbb [0,\infty) \to [0,\infty)$ be a convex $C^1$ function.
Furthermore, assume that there exist an $a \in (0,\infty)$ such that
- $\, \phi$ is strictly convex on $[a,\infty)$.
- $c:=\int_X g d\mu \in [a,\infty)$.
Finally, we are given the fact that $\int_X \phi\circ g=\phi(\int_X g)$.
I am trying to prove that $g$ must be constant a.e.
In fact, it suffices to prove that $g(x) \ge a$ a.e, since then the strict convexity of $\phi$ can be applied directly. However, I don't see how to use that to obtain a simpler proof.
I think I found a proof, but I am not sure, and I wonder if there is an easier approach or lemma lying behind here:
First, as is proved here for instance, we deduce from the Jensen equality that
$$ \phi(g(x))-\phi(c)=\phi'(c)\big( g(x)-c\big) \tag{1}$$
for almost every $x \in X$.
Now let $x \in X$ be a generic point, where equation $(1)$ holds. We claim that $g(x) \le c$. Assume otherwise, that $g(x) > c$. Then the mean value theorem implies that $\phi'(c)=\phi'(m(x))$ for some $g(x) >m(x) > c$.
This contradicts the fact that $g$ is strictly convex on $[a,\infty)$, so its first derivative must be strictly increasing on that interval, and in particular $\phi'(c)<\phi'(m(x))$.
Thus , for a.e. $x \in X$, we have $g(x) \le c$. Integrating, we see that $g(x)=c$ a.e. as required.
Best Answer
First, note that the strict convexity of $\phi$ in $[a,\infty)$ implies that for $x>a\ge y$, $$\phi(t x + (1-t) y) = t\phi(x) + (1-t)\phi(y)$$ implies $t\in \{0,1\}$.
Now, denote $A = g^{-1}([a,\infty))$. Obviously, $\mu(A)>0$. If $\mu(A) = 1$, we are done. Otherwise, denote $x=(\mu(A))^{-1}\int_A g$, and $y=\mu(A^c)^{-1}\int_{A^c} g$. We obviously have $y< a$ and $c= \mu(A)x + \mu(A^c)y$, hence $x>a$.
Thus, using the equality assumption and Jensen's inequality, $$\phi(c) = \int_X \phi\circ g = \int_A \phi\circ g + \int_{A^c} \phi\circ g \ge \mu(A)\phi(x) + \mu(A^c)\phi(y)\ge \phi(c).$$ Therefore equality must hold, hence either $\mu(A)=1$ or $\mu(A^c)=1$, and both options result in a contradiction.