For a convex function $\phi$, not necessarily differentiable. If $\mu$ is a probability measure, and $f$ and $\phi(f)$ are integrable. We have:
$$
\phi \left( \int f \, d\mu \right) \leq \int \phi(f) \, d\mu
$$
If I recall correctly, the intuition of the proof is that $\phi$ has a subgradient at $\int f \, d\mu$, which in other words, the convex function is bounded below by a linear approximation.
Now for a strongly convex function, let's assume it to be differentiable. Is there any generalization of Jensen's inequality in this case? Strongly convex is defined as there exist $m > 0$ s.t. $f(y) \geq f(x) + \nabla f(x)^{\top}(y-x)+\frac{m}{2}\|y-x\|^{2}_{2}$ for all $x,y$.
Since this is a quadratic lower bound, I felt that there should be one. However, I am not aware of any.
Best Answer
If $\phi$ is a strongly convex function with parameter $m$ then function $$\phi(x) - \frac{m}{2}x^2 \triangleq \psi(x)$$ is also a convex function. For details see this article. If you apply Jensen's inequality to $\psi$ instead of $\phi$ you obtain $$ \phi\left ( \int f d\mu\right ) - \frac{m}{2}\left( \int f d\mu\right )^2 = \psi\left ( \int f d\mu \right ) \leq \int\psi(f) d\mu = \int\phi(f) - \frac{m}{2}f^2 d\mu $$ Re-arranging the first and the last you end up with \begin{align} \phi\left ( \int f d\mu\right ) &\leq \int\phi(f)d\mu - \frac{m}{2} \left[\int f^2d\mu - \left(\int fd\mu\right)^2\right ] \\ &= \int\phi(f)d\mu - \frac{m}{2}\text{Var}_{\mu}(f) \end{align} The variance term is quadratic like you have suggested. Also since the variance is non-negative, this approach can provide a tighter bound than Jensen's inequality.
Another interesting way of writing this inequality is $$ \frac{m}{2}\text{Var}_{\mu}(f) \leq \int\phi(f)d\mu - \phi\left (\int f d\mu\right ) $$ which actually lower bounds the gap (sometimes called the Jensen gap) by a typically non-negative constant.