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:
$$\varphi\left(\int_a^b f\right) \leqslant \int_a^b \varphi(f).$$
Best Answer
First of all, Jensen's inequality requires a domain, $X$, where $$ \int_X\,\mathrm{d}x=1\tag{1} $$ 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 on $\mathcal{K}(f(X))\setminus\{t_0\}$. 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 $$ t_0=\int_Xf(x)\,\mathrm{d}x\tag{5} $$ 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 $$ \varphi\left(\int_Xf(x)\,\mathrm{d}x\right)\le\int_X\varphi(f(x))\,\mathrm{d}x\tag{8} $$