Converse of Jensen's inequality

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Suppose $\varphi:\mathbb{R}\rightarrow\mathbb{R}$ and for all bounded measurable $f$, $$ \varphi\Big(\int_0^1fd\lambda\Big) \le \int_0^1\varphi(f)d\lambda $$ I'm asked to prove that $\varphi$ is a convex function. I have no idea how to even begin, only idea I've had is to try to suppose that $\varphi''(x)<0$ for some $x\in(0,1)$ but then I haven't got a clue.

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Given $x,y\in\mathbb R$ and $t\in(0,1)$, consider $$ f(s) = \begin{cases} x & s\leq t \\ y & s>t. \end{cases} $$ Then your inequality tells $$ \phi(tx+(1-t)y) \leq t\phi(x)+(1-t)\phi(y). $$