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Jensen's inequality states that if $(X,\mu)$ is a measure space with $\mu(X) = 1$, $\phi$ is convex, and $f:X \rightarrow \mathbb R$ is integrable, then

$$\phi\left(\int fd\mu\right) \leq \int \phi \circ fd\mu.$$ I am trying to find an alternative proof which doesn't require the fact that convex functions are differentiable. I first prove the result for simple functions: if $f = \sum_{i=1} ^n a_i \chi_{A_i} $ and the $A_i$ to be disjoint with $X = \bigsqcup_{i=1} ^n A_i$. Then $\sum_{i=1} ^n \mu(A_i) = 1$, so

$$\phi\left(\int_X \sum_{i=1} ^n a_i \chi_{A_i} d\mu\right) = \phi\left(\sum_{i=1} ^n a_i \mu(A_i ) \right) \leq \sum_{i=1} ^n \phi(a_i) \mu(A_i ) = \int_X \phi \circ f d\mu .$$

I want to extend this to the general case by approximating $f$ with a sequence of simple functions $(f_n)$. However, I run into a problem showing that the integrals of $\phi \circ f_n$ converge to the integral of $\phi \circ f$. Any suggestions?

Davide Giraudo
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user15464
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  • I think the proof of Jensen's inequality in Rudin, Real and Complex Analysis, 3rd ed. (early in Chapter 3) doesn't use the differentiability of $\phi$, at least not overtly. That could be what you want. – Dimitrije Kostic Nov 22 '11 at 05:27

1 Answers1

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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.

Did
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