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What nice ways do you know in order to prove Jensen's inequality for integrals? I'm looking for some various approaching ways.
Supposing that $\varphi$ is a convex function on the real line and $g$ is an integrable real-valued function we have that:

$$\varphi\left(\int_a^b f\right) \leqslant \int_a^b \varphi(f).$$

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And by Jensen's inequality, do you mean something about convex functions? And not – GEdgar Jul 16 '12 at 17:44
@GEdgar: i know that in English is called Jensen's inequality for integrals (i hope i'm not wrong) and is related to convex functions. – user 1618033 Jul 16 '12 at 17:47
@Chris: That reply does not really clarify what you mean. Why not just edit your question such that it explicitly quotes the statement you want a proof of? – Henning Makholm Jul 16 '12 at 17:53
Do you have that $b-a=1$? – robjohn Jul 16 '12 at 19:27
@robjohn: i think you're right. b-a=1 – user 1618033 Jul 16 '12 at 19:38
up vote 10 down vote accepted

First of all, Jensen's inequality requires a domain, $X$, where $$ \int_X\,\mathrm{dx}=1\tag{1} $$ Next, suppose that $\varphi$ is convex on the range of $f$; this means that for any $t_0\in f(X)$, $$ \frac{\varphi(t)-\varphi(t_0)}{t-t_0}\tag{2} $$ is non-decreasing on $f(X)\setminus\{t_0\}$. This means that we can find a $\Phi$ so that $$ \sup_{t<t_0}\frac{\varphi(t)-\varphi(t_0)}{t-t_0}\le\Phi\le\inf_{t>t_0}\frac{\varphi(t)-\varphi(t_0)}{t-t_0}\tag{3} $$ and therefore, for all $t$, we have $$ (t-t_0)\Phi\le\varphi(t)-\varphi(t_0)\tag{4} $$ Now, let $t=f(x)$ and set $$ t_0=\int_Xf(x)\,\mathrm{d}x\tag{5} $$ and $(4)$ becomes $$ \left(f(x)-\int_Xf(x)\,\mathrm{d}x\right)\Phi\le\varphi(f(x))-\varphi\left(\int_Xf(x)\,\mathrm{d}x\right)\tag{6} $$ Integrating both sides of $(6)$ while remembering $(1)$ yields $$ \left(\int_Xf(x)\,\mathrm{d}x-\int_Xf(x)\,\mathrm{d}x\right)\Phi\le\int_X\varphi(f(x))\,\mathrm{d}x-\varphi\left(\int_Xf(x)\,\mathrm{d}x\right)\tag{7} $$ which upon rearranging, becomes $$ \varphi\left(\int_Xf(x)\,\mathrm{d}x\right)\le\int_X\varphi(f(x))\,\mathrm{d}x\tag{8} $$

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You are also taking $x=f(x)$, which is a confusing notation. – M Turgeon Jul 16 '12 at 18:57
@MTurgeon: I have changed to $t=f(x)$. – robjohn Jul 16 '12 at 19:13
@robjohn I corrected a small typo that had me scratching my head for a few moments. I hope you don't mind. – Potato Jan 27 '13 at 5:48
@Potato: thanks. That was a residual error from the previous edit. – robjohn Jan 27 '13 at 9:17
@robjohn would any steps in the proof change if $\varphi$ was convex on $\mathbb{R}^n$ instead of $\mathbb{R}$? – cap Nov 15 '15 at 8:26

I like this, maybe it is what you want ...

Let $E$ be a separable Banach space, let $\mu$ be a probability measure defined on $E$, let $f : E \to \mathbb R$ be convex and (lower semi-)continuous. Then $$ f\left(\int_E x d\mu(x)\right) \le \int_E f(x)\,d\mu(x) . $$ Of course we assume $\int_E x d\mu(x)$ exists, say for example $\mu$ has bounded support.

For the proof, use Hahn-Banach. Write $y = \int_E x d\mu(x)$. The super-graph $S=\{(x,t) : t \ge f(x)\}$ is closed convex. (Closed, because $f$ is lower semicontinuous; convex, because $f$ is convex.) So for any $\epsilon > 0$ by Hahn-Banach I can separate $(y,f(y)-\epsilon)$ from $S$. That is, there is a continuous linear functional $\phi$ on $E$ and a scalar $s$ so that $t \ge \phi(x)+s$ for all $(x,t) \in S$ and $\phi(y)+s > f(y)-\epsilon$. So: $$ f(y) -\epsilon < \phi(y)+s = \phi\left(\int_E x d\mu(x)\right)+s = \int_E (\phi(x)+s) d\mu(x) < \int_E f(x) d\mu(x) . $$ This is true for all $\epsilon > 0$, so we have the conclusion.


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If $\phi$ is a continuous linear functional on $E$, how do we define $\phi(y)$ (I assumed $y$ is in $\mathbb{R}$)? Thanks – Xiao Jul 12 at 15:41

One way would be to apply the finite Jensen's inequality $$\varphi\left(\frac{\sum a_i x_i}{\sum a_j}\right) \le \frac{\sum a_i \varphi (x_i)}{\sum a_j}$$ to each Riemann sum. The finite inequality is itself easily proved by induction on the number of points, using the definition of convexity.

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Here's a nice proof:

Step 1: Let $\varphi$ be a convex function on the interval $(a,b)$. For $t_0\in (a,b)$, prove that there exists $\beta\in\mathbb{R}$ such that $\varphi(t)-\varphi(t_0)\geq\beta(t-t_0)$ for all $t\in(a,b)$.

Step 2: Take $t_0=\int_a^bfdx$ and $t=f(x)$, and integrate with respect to $x$ to prove the desired inequality.

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There needs to be something about $b-a=1$ or else this doesn't work. – robjohn Jul 16 '12 at 19:16
@robjohn Of course. To be honest, I thought the OP took this as an "invisible assumption" – M Turgeon Jul 16 '12 at 19:48

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