# Is my proof of the conditionnal Jensen's inequality correct?

I want to prove the conditionnal Jensen's inequality. Let $$(\Omega, \mathcal H, \mathbb P)$$ be a probability space, $$\mathcal G \subset \mathcal H$$ a sub sigma algebra, $$\varphi : \mathbb R \longrightarrow \mathbb R$$ convex and $$X \in L^1(\mathcal H)$$. Then the Jensen inequality is $$\mathbb E [\varphi(X) | \mathcal G] \geq \varphi(\mathbb E [X | \mathcal G]).$$

For the proof the classical argument is that by basic properties of convex functions, forall $$x \in \mathbb R$$, $$\tag{1}\varphi(x) = \sup_{(a,b) \in E_\varphi} ax+b = \sup_{(a,b) \in E_\varphi \cap \mathbb Q^2} ax+b$$ where $$E_\varphi = \{ (a,b) \in \mathbb R^2 : \forall x \in \mathbb R ~~~\varphi(x) \geq ax+b \}.$$ The second equality of $$(1)$$ is false when $$\varphi$$ is affine with irrational growth ratio and I don't understand why we would need $$(a,b)$$ to vary in a countable set. This is how I proceed from the first equality of $$(1)$$ :

Fix $$(a,b) \in E_\varphi$$, forall $$x \in \mathbb R$$ we have $$\varphi(x) \geq ax+b$$ so $$\varphi(X) \geq aX+b$$. Taking the conditionnal expectation, which is non negative and linear, yields $$\mathbb E [\varphi(X) | \mathcal G] \geq a \mathbb E [X | \mathcal G] + b.$$ Now taking the supremum of the left hand side on $$(a,b)$$ and using the first equality of $$(1)$$ yields the result. Is my proof correct?

• Your proof makes sense to me. If you don't understand the second equality of $(1)$ I think you should remove it. Especially since you never use it. Commented Nov 18, 2021 at 13:13
• @jakobdt thank you for your answer. The second equality of $(1)$ is false in the case $\varphi$ is a line with irrational growth ratio. This is because in this case $E_\varphi$ is empty. In fact my proof is false as I detailed in my answer. Commented Nov 18, 2021 at 17:51
• That’s an interesting point. What you mean is that $E_\varphi\cap\mathbb{Q}^2$ is empty, right? Also, I finally understood why you need the countability. Commented Nov 18, 2021 at 20:19
• @jakobdt if $\varphi (x) = ax+b$ then $E_\varphi = \{a\} \times (-\infty,b]$ so asking $a$ irrationnal yields $E_\varphi \cap \mathbb Q^2$ empty. Commented Nov 18, 2021 at 20:34
• Perfect, then we agree. Commented Nov 18, 2021 at 21:08

In fact my proof is not correct and we should ask for a countable dense subset of $$E_\varphi$$. Indeed, fix a representant of $$X$$, it is true that for any $$(a,b) \in E_\varphi$$ we have everywhere on $$\Omega$$ $$\varphi(X) \geq aX+b.$$ But when taking the conditionnal expectation only get an inequality almost surely because the conditionnal expectation is only defined almost surely. Thus we have in fact $$\forall (a,b) \in E_\varphi,~~~\mathbb P -a.s., ~~~\mathbb E[\varphi(X) | \mathcal G] \geq a \mathbb E [X | \mathcal G] +b.$$ To swap the quantificators and since the application $$(a,b) \mapsto ax+b$$ is continuous we only need a countable dense subset of $$E_\varphi$$ which is possible because $$E_\varphi \subset \mathbb R^2$$ which is metric separable.