For any measurable function $f$, Jensen inequality states that $\int \phi (f) d\mu \geq \phi ( \int f d\mu)$ for convex $\phi$.
Is there a proof that if the inequality holds with equality, that is $$\int \phi (f) d\mu = \phi ( \int f d\mu)$$ for every measurable $f$, then $\phi$ must be affine?
I was trying to assume, by contradiction, that (assuming differentiability and strict convexity of $\phi$ and strict convexity) $$\phi(f)>\phi(c)+\phi’(c)(f-c)$$ With $c= \int f d\mu$, and then I would integrate on both sides and reach a contradiction. However, the strict inequality above is not true in general, as it may exists an $x \in X$ such that the rhs is equal to the lhs, so my ‘proof’ would be invalid. Do you know if what I am trying to show is true? And what’s the proof?
The question implicitly used $\mu(X)=1$.
If $\phi$ is not affine, then there are numbers $a,b$ and $\lambda\in(0,1)$ such that $$ \phi(\lambda a + (1-\lambda)b)< \lambda \phi(a) + (1-\lambda)\phi(b). $$ Now let $A$ be a measurable set with measure equal to $\lambda$, then its complement $A^c$ has measure and $(1-\lambda)$. Define $$ f:=\chi_A \cdot a + \chi_{A^c}\cdot b, $$ which implies $$ \phi(\int f) = \phi(\lambda a + (1-\lambda)b) < \lambda \phi(a) + (1-\lambda)\phi(b) = \int \phi(f). $$ Thus, $f$ is the desired function, that yields strict inequality in Jensen.