A simpler proof of Jensen's inequality

Solution 1:

Here is a proof without differentiation.

Since $\phi$ is convex, $\phi$ is the supremum of some affine functions $\alpha$, in the sense that $\phi(x)=\sup_\alpha \alpha(x)$ for every $x$, where each $\alpha$ is defined by $\alpha:x\mapsto a_\alpha x+b_\alpha$ for some $a_\alpha$ and $b_\alpha$.

Now, the integral is linear hence, for every $\alpha$, $\displaystyle\int\alpha(f)\mathrm d\mu=\int(a_\alpha f+b_\alpha)\mathrm d\mu=a_\alpha I+b_\alpha=\alpha(I)$, with $I=\displaystyle\int f\mathrm d\mu$. Since $\alpha(f)\leqslant \phi(f)$, $\displaystyle\int\alpha(f)\mathrm d\mu\leqslant\int\phi(f)\mathrm d\mu$ hence $\alpha(I)\leqslant\displaystyle\int\phi(f)\mathrm d\mu$.

This holds for every $\alpha$ hence $\sup_\alpha \alpha(I)\leqslant\displaystyle\int\phi(f)\mathrm d\mu$. Since $\phi(I)=\sup_\alpha \alpha(I)$, the proof is complete.