We will proceed by induction. Assume $X$ is discrete and takes on $n$ values $x_1,\dots ,x_n$ with probabilities $p_1,\dots ,p_n.$

Base step, for $n=2$: Assume $X$ takes on two distinct values, $x_1$ and $x_2,$ with probabilities $p_1=\lambda ,p_2=1-\lambda .$ Then we want to show that

The last line is just the definition of a convex function, so it's obviously true.

Inductive step: Assume that (a) holds for any $n-1$ points. Take $n$ points with weights $p_i$ summing to $1.$ Let

$$w=\sum _{i=1}^{n-1}{p}_i=1-p_n.$$

Assuming let $q_i=p_i/w.$ Then, so the $q_i$ form an $(n-1)$-point distribution. Now, we'll rewrite $E[X]$ to bundle the first $n-1$ terms separately from the last term:

$$\begin{array}{cc}E[X]=\sum _{i=1}^n{p}_i{x}_i=w\sum _{i=1}^{n-1}{q}_i{x}_i+p_n{x}_n=w\bar{x}+p_n{x}_n,& \text{(b)}\end{array}$$

where $\bar{x}=\sum _{i=1}^{n-1}{q}_i{x}_i.$ Now, because $w+p_n=1,$ (b) is a two-point convex combination of $\bar{x}$ and $x_n.$ From our base case,

$$\phi (\sum _{i=1}^n{p}_i{x}_i)=\phi (w\bar{x}+p_n{x}_n)\le w\phi (\bar{x})+p_n\phi (x_n).$$

Now, from our inductive hypothesis we have that

$$\phi (\bar{x})=\phi (\sum _{i=1}^{n-1}{q}_i{x}_i)\le \sum _{i=1}^{n-1}{q}_i\phi (x_i).$$

Tying it all together we have

TODO: show the equality case only holds when the support is affine.

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