This question is about the concavity of a particular function. Consider a probability space $(\Omega, \mathcal{A}, \mu)$ and let $\|f\|_p=\left(\int_\Omega |f|^p\mathrm{d}\mu\right)^{\frac1p}$ for a function $f\in L^p(\Omega, \mu)$, with $p>0$.
I would like to know whether the map $r\mapsto \log\|f\|_{r}$ with $r>0$ is concave, where $f\in L^p(\Omega, \mu)$ with $\int_\Omega f\mathrm{d}\mu=1$. A setting in which it seems to be true is the following: take $\Omega=\{-1, 1\}$ with $\mu$ uniform on $\Omega$. In this space, functions with $\int_\Omega f\mathrm{d}\mu=1$ can be represented by $f(x)=1+ax$ with $|a|\leq 1$, and the map of interest can be plotted.
Some related statements about the convexity of similar-looking functions:
- The map $r\mapsto \log\|f\|_{\frac1r}$ with $r>0$ is convex (see for example Lemma 9 in Terry Tao's notes for a proof). The proof follows from an application of Hölder's inequality, which I attempted to adapt using the reverse form of Hölder's inequality, but to no avail.
- The map $r\mapsto \log\|f\|_r^r$ is with $r>0$ is convex. This can be seen from the fact that for a convex function $\varphi(x)$, its so-called perspective function $x\varphi(1/x)$ is also convex. Since the function $\log\|f\|_r^r$ is precisely the perspective functionof the function in the previous point, it is also convex.
Are there any existing results about concavity involving logarithms of norms? Any help would be appreciated, thank you!
