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Prove Jensen's inequality, that is: if $f$ is convex, then $f(E[X|\mathcal{A}])\leq E([f(X)|\mathcal{A}])$

My attempt: because $f$ is convex, we can define $$f(x)=\sup_{n\in\mathbb{N}}a_nx+b_n$$ $$\{a_n\},\{b_n\}\in\mathbb{R}$$ Hence, $\forall n\in\mathbb{N} : f(x)\geq a_nx+b_n =\bar f(x)$ for a fixed $n\in\mathbb{N}$. So we obtain: $$f(E[X|\mathcal{A}])\geq a_nE[X|\mathcal{A}]+b_n=E[a_nX+b_n|\mathcal{A}]=E[\bar{f}(X)|\mathcal{A}]$$ So this must also hold for the supreme of $\bar{f}(x)$, "proving" the opposite result. Could anyone point out the mistake I am making?

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    What are $a_n$ and $b_n$? – 2017-02-15
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    Some real numbers, I'll edit it – 2017-02-15
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    You omitted the last step, which is to explain what you do with $E(\bar f(X)\mid\mathcal A)$. If you add this step, you should see what happened. – 2017-02-15
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    $E[\bar{f}(X)|\mathcal{A}]\leq E[\sup \bar{f}(X)|\mathcal{A}]$, so we can't conclude anything useful for this theorem this way? – 2017-02-15
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    I do not understand "conclude anything useful for this theorem this way". If you mean: "Can we prove conditional Jensen's inequality using similar approaches than in this post?" then the answer is irrevocably "Yes", but not starting from the useless double inequality $$f(E(X\mid\mathcal A))\geqslant E(\bar f(X)\mid\mathcal A)\leqslant E(\sup\bar f(X)\mid\mathcal A)$$ – 2017-02-15

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