Let $(M, g)$ be a compact Riemannian manifold.

Assume that $u_0$ is a positive smooth function on $M$ and let $u_t = e^{t \Delta} u_0$ be the solution to the heat equation on $(M, g)$ with initial data $u_0$.

Given $2a > 1$, is it true that the function $$ f: t \mapsto \int_M (u_t)^{2a} $$ is a convex function?

This question gives an answer to Functional decaying under the heat flow (?) in the case $p=2$. Indeed,

\begin{eqnarray*} \frac{d}{dt} \int_M u_t^{2a} dv &=& 2 a \int_M u_t^{2a-1} \Delta u_t dv\\ &=& -2 a \int_M \left\langle \nabla\left(u_t^{2a-1}\right), \nabla u_t\right\rangle dv\\ &=& -2 a (2a-1) \int_M u_t^{2a-2} \left|\nabla u_t\right|^2 dv\\ &=& -2 a (2a-1) \int_M \left|u_t^{a-1}\nabla u_t\right|^2 dv\\ &=& -\frac{4a-2}{a} \int_M \left|\nabla (u_t^a)\right|^2 dv. \end{eqnarray*} Hence, showing that $f$ is convex means that $f'' \geq 0$ or, equivalently, from the previous calculation $$ \frac{d}{dt} \int_M \left|\nabla (u_t^a)\right|^2 dv \leq 0. $$

The value of $a$ for interest in my problem is $$ 2a = 2^* + 2 = 4 \frac{n-1}{n-2}, $$ where $n$ is the dimension of the manifold and $2^* = \frac{2n}{n-2}$.