# Jensen's Inequality in $L^1$.

Let be $$(X,A,μ)$$ a probability space and let $$\textit{f}\in L^{1}(X)$$ a real function such that $$-\infty for all $$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!!

• The title is wrong, sorry!!! Commented Jan 25, 2020 at 15:51

Since $$\mu$$ is a probability measure and $$a , you have $$a <\int_Xf\,d\mu . 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 $$\mu (X)=1$$.