Is it impossible to prove Jensen's inequality by way of step functions?
Jensen's Inequality: Let $\varphi:\Bbb R\to \Bbb R$ be convex, and $f:[0,1]\to\Bbb R$ be integrable, and suppose $\varphi\circ f$ is integrable over [0,1]. Then $$ \varphi\left(\int_{[0,1]} f\right)\le \int_{[0,1]}\varphi\circ f $$ A proof from step functions:
I have seen a proof of this inequality for every step function f.
My question is, can we then extend this proof to all measurable f satisfying the assumptions of the proof?
I've encountered a number of other proofs of this theorem, but I'm particularly interested to see how one might do this from the result for step functions. If we knew that $\varphi$ were continuous then I would consider trying to use one of the several convergence theorems for the Lebesgue integral. But since we don't assume continuity then this seems like a bad route.
But if we are going to extend from the result for step functions, it seems like we must use $f = \lim_{n\to\infty} s_n$ for a sequence of step functions. But if the limit can't pass through the function $\varphi$ then it seems to me that there simply cannot be any way to infer the full result from the result for step functions.
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The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| whybecause | (no comment) | Feb 11, 2022 at 21:48 |
Right, due to the comment, we know the convex function is continuous and therefore the rest of the proof can be done using the MCT. In particular this is because every function is approached from below by a sequence of step functions.
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