Theorem 3.3 from W. Rudin, Real and complex analysis, says:
Let $\mu$ be a probabilistic measure on a $\sigma$-algebra of subsets of a given set $\Omega$.
If a function $f:X \rightarrow \mathbb R$ is in $L^1(\mu)$, $a<f(x)<b$ for $x\in \Omega$ and if a function $\phi:(a,b)\rightarrow \mathbb R$ is a convex then
$$
\phi\left(\int_a^b fd \mu\right) \leq \int_a^b (\phi \circ f) d \mu.
$$
Is it known when in this inequality holds equality?
Maybe, it is iff $\phi$ is affine a.e. ?
Best Answer
We can look at the proof to see when equality occurs.
The convexity of $\phi$ gives us
$$\lambda := \sup_{a < s < c} \frac{\phi(c)-\phi(s)}{c-s} \leqslant \rho := \inf_{c < t < b} \frac{\phi(t)-\phi(c)}{t-c}$$
for every $c\in (a,b)$. Letting $c := \int_X f\,d\mu$, it follows that for every $\kappa \in [\lambda,\rho]$ and $t\in (a,b)$ we have
$$\phi(t) \geqslant \phi(c) + \kappa\cdot (t-c)\tag{1}$$
and hence
$$\phi(f(x)) \geqslant \phi(c) + \kappa\cdot \bigl(f(x) - c\bigr)\tag{2}$$
for every $x\in X$. Integrating $(2)$ gives Jensen's inequality, and it follows that we have the equality
$$\int_X \phi\circ f\,d\mu = \phi\left(\int_X f\,d\mu\right)$$
if and only if we have equality a.e. in $(2)$.
That we have equality a.e. in $(2)$ means that $\phi$ coincides with an affine function on [the convex hull of] the essential range of $f$, but $\phi$ need not be affine globally on $(a,b)$.
If $f$ is essentially constant, that is no restriction on $\phi$, then equality holds in Jensen's inequality for all $\phi$. If $\phi$ is affine, we have equality for all $f$. If $\phi$ is strictly convex, Jensen's inequality is strict for all $f$ except the essentially constant ones, but if $\phi$ is not strictly convex, equality also holds for some (essentially) non-constant $f$.