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I refer to Wikipedia: https://en.wikipedia.org/wiki/Jensen%27s_inequality#Alternative_finite_form

"There is also an infinite discrete form."

There is no mention of how exactly the infinite discrete form looks like, but I guess that it is something like if: $\sum_{n=1}^\infty \lambda_n=1$, then $\varphi(\sum \lambda_n x_n)\leq \sum \lambda_n \phi(x_n)$. Is that correct?

How is the infinite discrete form proved?

Thanks a lot.

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    $\begingroup$ Caveat: This is true under the condition that $\sum_{n=1}^{\infty} \lambda_n x_n$ is finite. $\endgroup$ – Michael Nov 5 '15 at 15:53
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Define a probability measure $\mu$ on $\mathbb N$ by $\mu(\{n\}) = \lambda_n$ and extending to all subsets of $\mathbb N$ to make $\mu$ countably additive.

Let $f: \mathbb N \to \mathbb R$ be defined by $f(n) = x_n$.

Jensen's inequality states (under appropriate assumptions about $\phi$ and $f$) that $$\phi \left( \int_{\mathbb N} f \, d\mu \right) \le \int_{\mathbb N} \phi(f) \, d\mu.$$

Now just work out what these integrals are.

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The main point is that one has equality if $\varphi $ is an affine function.

Let $I$ be the interval of definition of your convex function, and $\Lambda$ the set of all affine functions $l(x)=ax+b$ such that $\forall y\in I, l(y) \leq \varphi (y)$. It is a basic property of convex function that $\varphi(x)= Sup_{l\in \Lambda} (l(x))$.

Now, let $l\in \Lambda$, $l(\Sigma _n \lambda _n x_n)=\Sigma _n \lambda _n l(x_n) $. Therefore $l(\Sigma _n \lambda _n x_n) \leq \Sigma _n \lambda _n \varphi (x_n)$. The right hand side do not depend on $l$, taking the supremum, we get $\varphi(\Sigma _n \lambda _n x_n) \leq \Sigma _n \lambda _n \varphi(x_n)$

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