Let $(X,\Sigma,\mu)$ be a probability space; in particular, $\mu(X)=1$. The integral form of Jensen's inequality can be phrased in terms of permuting a convex function $\varphi$ (say, with the property that $\varphi(0)=0$) and the $L^1(d\mu)$ norm:
$$\varphi\Big(\|f\|_{L^1(d\mu)}\Big)\leq \|\varphi\circ f\|_{L^1(d\mu)}.$$
My question is: does a similar statement hold for $\|\cdot\|_{L^p(d\mu)}$ norms with $p>1$? More precisely, given a convex function $\varphi$ such that $\varphi(0)=0$ and an exponent $1\leq p\leq \infty$, does there exist a constant $C_p<\infty$ such that
$$\varphi\Big(\|f\|_{L^p(d\mu)}\Big)\leq C_p\|\varphi\circ f\|_{L^p(d\mu)}$$
for every $f\in L^p(d\mu)$?
Thank you.
Best Answer
Let $g=|f|^p$ and $\psi(t)=\varphi(t^{1/p})^p$. Then the inequality $$\varphi\Big(\|f\|_{L^p(d\mu)}\Big)\leq C_p\|\varphi\circ |f|\|_{L^p(d\mu)} \tag1$$ becomes $$\psi\left(\int g\,d\mu\right)\le C_p^p\int (\psi\circ g)\,d\mu \tag2$$ For (2) to be true for arbitrary $g$, we need $\psi$ to be convex to $[0,\infty)$. This can be phrased as a differential inequality for $\varphi$ which is neither stronger nor weaker than $\varphi''\ge 0$. For example, with $p=2$ my computations show that $\psi''\ge 0$ is equivalent to $$t(\varphi\varphi''+(\varphi')^2)-\varphi\varphi'\ge 0 \tag3$$