Jensen's inequality for integrals

Remarks

Proof idea: let $m = \int g \, d\mu$. A convex function on an open interval has a supporting line at $m$: there is $c \in \R$ with $\varphi(y) \ge \varphi(m) + c (y - m)$ for all $y \in (a, b)$. Put $y = g(x)$ and integrate over $X$; the term $c(g - m)$ has integral $0$ because $\mu(X) = 1$.

Example: with $\varphi(y) = y^2$ it gives $(\int g \, d\mu)^2 \le \int g^2 \, d\mu$. In probability this is $\operatorname{Var}(Y) \ge 0$. With $\varphi = \exp$ it gives $\exp(\int g \, d\mu) \le \int e^{g} \, d\mu$.

The normalisation $\mu(X) = 1$ is needed. For $\mu = 2 \cdot$ Lebesgue measure on $[0, 1]$, $g = 1$ and $\varphi(y) = y^2$, the left side is $4$ and the right side is $2$. When $\mu$ is the probability measure with masses $w_i$ at finitely many points, the statement is the finite Jensen's inequality.

Going deeper

For a student

In probability the inequality reads $\varphi(\mathbb{E}[Y]) \le \mathbb{E}[\varphi(Y)]$ for an integrable random variable $Y$. It yields the comparison $\norm{f}_p \le \norm{f}_q$ for $p \le q$ on probability spaces, the AM–GM inequality, and the non-negativity of the Kullback–Leibler divergence. There is also a version for conditional expectations, used in martingale theory.