Jensen's inequality for integrals 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:
$$\varphi\left(\int_a^b f\right) \leqslant \int_a^b \varphi(f).$$
 A: 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.

A: 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.
A: 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.
A: First of all, Jensen's inequality requires a domain, $X$, where
$$
\int_X\,\mathrm{d}\mu=1\tag{1}
$$
Next, suppose that $\varphi$ is convex on the convex hull of the range of $f$, $\mathcal{K}(f(X))$; this means that for any $t_0\in \mathcal{K}(f(X))$,
$$
\frac{\varphi(t)-\varphi(t_0)}{t-t_0}\tag{2}
$$
is non-decreasing for $t\in\mathcal{K}(f(X))\setminus\{t_0\}$. 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
$$
t_0=\int_Xf(x)\,\mathrm{d}\mu\tag{5}
$$
and $(4)$ becomes
$$
\left(f(x)-\int_Xf(x)\,\mathrm{d}\mu\right)\Phi\le\varphi(f(x))-\varphi\left(\int_Xf(x)\,\mathrm{d}\mu\right)\tag{6}
$$
Integrating both sides of $(6)$ while remembering $(1)$ yields
$$
\left(\int_Xf(x)\,\mathrm{d}\mu-\int_Xf(x)\,\mathrm{d}\mu\right)\Phi\le\int_X\varphi(f(x))\,\mathrm{d}\mu-\varphi\left(\int_Xf(x)\,\mathrm{d}\mu\right)\tag{7}
$$
which upon rearranging, becomes
$$
\varphi\left(\int_Xf(x)\,\mathrm{d}\mu\right)\le\int_X\varphi(f(x))\,\mathrm{d}\mu\tag{8}
$$
A: Issue
This answer assumes that the integral belongs in to the domain of $\varphi$, i.e.,
$$
    \int_X\!f(x)\;dx \tag{1} \in {\rm dom}(\varphi)
$$
Even more, the statement we want to prove involves the evaluation of $\varphi$ at the value of the integral. So, a question remains: why does $(1)$ hold?
Let $a<b$ be the ends of interval $I$ where $\varphi$ is defined (they may or many not belong to $I$). In order to show $(1)$ we need a hypothesis, namely
$$
    \int_{\{f\notin I\}}dx = 0, \tag2
$$
i.e., $f\in I\rm\ a.e.$ (which does happen if $f(X)\subseteq{\rm dom}(\varphi)=I$)
Because of hypothesis $(2)$, to prove $(1)$ it is enough to show
$$
    \int_{\{f \in I\}}f(x)\;dx \in I.
$$
Since 
$$
    1 = \int_X dx = \int_{\{f\in I\}}dx + \int_{\{f\notin I\}}dx
         = \int_{\{f\in I\}}dx,
$$
we obtain
$$
    a = \int_{\{f\in I\}} a\;dx \le \int_{\{f\in I\}}f(x)\;dx
$$
Now, assume $a\notin I$ and
$$
    \int_{\{f\in I\}} f(x)\;dx = \int_Xf(x)\;dx = a.\tag3
$$
Then we would also have
$$
    \int_{\{f \ge a\}}f(x)\;dx = \int_{\{f\in I\}}f(x)\;dx = a
        = \int_{\{f\ge a\}}a\,dx.\tag4
$$
But $(4)$ means that $f = a,\rm\ a.e.$, which contradicts our hypothesis because $a\notin I$.
Thus $\int_X f(x)\,dx=a$ and $a\in I$ or $\int_X f(x)\,dx > a$. Similarly, $\int_X f(x)\,dx=b$ and $b\in I$ or $\int_X f(x)\,dx < b$.
In any case $(1)$ does hold and we can proceed as in the answer of reference.
