Let be $(X,A,μ)$ a probability space and let $\textit{f}\in L^{1}(X)$ a real function such that $-\infty<a<f(x)<b<\inf$ for all $x\in X$.
Let $\varphi:(a,b)\rightarrow \mathbf{R}$ be a convex function. Prove the so called Jensen's inequality:
$$\varphi(\int_{X}\textit{f}d\mu)=\int_{X}\varphi(\textit{f})d\mu. $$
With the help of this inequality prove that if $h:X\rightarrow[0,\infty)$ is a measurable function, then
$$\sqrt{1+\left(\int_{X}h\,d\mu\right)^2}\le\int_{X}\sqrt{1+h^2}\,d\mu\le1+\int_{X}h\,d\mu$$
-I've just proved in general the Jensen's inequality $\textit{f}:X\rightarrow \mathbb{R}$ if $f $ is $\mu$-integrable and $\varphi$ convex and with $\operatorname {Dom}(\varphi)=\mathbb{R} but here I have different hypothesis. How could I do?
Thanks you very much!!
Best Answer
Since $\mu $ is a probability measure and $a <f <b $, you have $a <\int_Xf\,d\mu <b $. So $\varphi $ has the right domain.
If $f =\sum_j a_j\,1_{E_j}$ is simple, then $\int_Xf\,d\mu =\sum_j \mu (E_j)\,a_j $ is a convex combination and thus $$\varphi (\int_Xf\,d\mu)\leq\sum_j\mu (E_j)\,\varphi (a_j)=\int_X\varphi (f)\,d\mu. $$ Since $\varphi $ is continuous, the inequality extends to arbitrary measurable, bounded, $f $.
As $\sqrt {1+x^2} $ is convex, the first inequality follows directly from Jensen's. The second inequality is $\sqrt {1+h^2}\leq 1+|h| $ together with $h\geq0$ and $\mu (X)=1$.