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 (?)](https://mathoverflow.net/questions/301673/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 $2^\* = \frac{2n}{n-2}$.



$L^p$-norm under the heat flow
If I didn't do any miscalculations I believe I have proven the case $1\leq p\leq 2$. I will write $u$ instead of $u\_t$. Let $(p)\_k=p(p-1)\ldots(p-k+1)$ and
$$
w=\begin{pmatrix}
pu^{p-1}\Delta^2 u\\ (p)\_2u^{p-2}\nabla\Delta u\cdot \nabla u\\ (p)\_2u^{p-2}(\Delta u)^2\\ (p)\_3u^{p-3}(\Delta u)|\nabla u|^2\\ (p)\_4u^{p-4}|\nabla u|^4\end{pmatrix}
$$
For $v=u^p$ we have
$$
\begin{pmatrix} \Delta^2v\\ \frac{\partial}{\partial t}\Delta v\\ \left(\frac{\partial}{\partial t}\right)^2v\end{pmatrix}=
 \begin{pmatrix}1 & 4 & 3 & 6 & 1\\ 1 & 2 & 1 & 1 & 0\\1 & 0 & 1 & 0 & 0 \end{pmatrix}w\\
$$
Hence
$$
\left(\frac{\partial}{\partial t}\right)^2v = 2 \frac{\partial}{\partial t}\Delta v-\Delta^2 v+\begin{pmatrix}2 & 4 & 1 \end{pmatrix}\begin{pmatrix}(p)\_2u^{p-2}(\Delta u)^2\\ (p)\_3u^{p-3}(\Delta u)|\nabla u|^2\\ (p)\_4u^{p-4}|\nabla u|^4\end{pmatrix}.\\
$$
For $1\leq p\leq 2$ we have $(p)\_2\geq 0, (p)\_4\geq 0$. We try to prove that the last term is positive. Note that $\sqrt{2(2-p)(3-p)}\geq 2(2-p)$. Hence by AM-GM
$$
\begin{eqnarray}
2(p)\_2u^{p-2}(\Delta u)^2+ (p)\_4u^{p-4}|\nabla u|^4&\geq& \sqrt{2(p)\_2u^{p-2}(\Delta u)^2(p)\_4u^{p-4}|\nabla u|^4}\\
&=&2(p)\_2\sqrt{2(p-2)(p-3)}u^{p-3} |\Delta u||\nabla u|^2 \\
&\geq&2(p)\_22(2-p)u^{p-3} (\Delta u)||\nabla u|^2\\
&\geq&4(p)\_3u^{p-3}(\Delta u)|\nabla u|^2.
\end{eqnarray}.
$$
Hence we have for $1\leq p \leq 2$
$$
\left(\frac{\partial}{\partial t}\right)^2v \geq 2 \frac{\partial}{\partial t}\Delta v-\Delta^2 v
$$
and therefore
$$
\int\_M \left(\frac{\partial}{\partial t}\right)^2v \geq \int\_M 2 \frac{\partial}{\partial t}\Delta v-\Delta^2 v=\int\_M \Delta\left(2 \frac{\partial}{\partial t}v-\Delta v \right)=0
$$




---


Prove for $0\leq p\leq 3$ (added 6.6.2018):


Note that
$$
0=\int\_M \nabla\left( (p)\_3 u^{p-3} (\nabla u)^3\right)=3\int\_M (p)\_3 u^{p-3} |\nabla u|^2 (\Delta u)+\int\_M (p)\_4 u^{p-4} (\nabla u)^4.
$$
Hence we have
$$
\int\_M 2(p)\_2 u^{p-2} (\Delta u)^2+4 (p)\_3 u^{p-3} |\nabla u|^2 (\Delta u)+(p)\_4 u^{p-4} (\nabla u)^4\\
=\int\_M 2(p)\_2 u^{p-2} (\Delta u)^2+(4+3\lambda) (p)\_3 u^{p-3} |\nabla u|^2 (\Delta u)+(1+\lambda)(p)\_4 u^{p-4} (\nabla u)^4
$$
for any $\lambda \in \mathbb{R}$. We try to prove that there exist $\lambda$ such that the integrand is positive everywhere. A sufficient condition is
$$
2(p-2)(p-3)(\lambda+1)\geq \left(\frac{4+3\lambda}{2}\right)^2(p-2)^2\\
\Leftrightarrow 0\geq\frac{9}{4}(p-2)^2\lambda^2+(p-2)(4p-6)\lambda+2(p-2)(p-1),
$$
for which a $\lambda$ exist if and only if
$$
(p-2)^2(4p-6)^2\geq 4\frac{9}{4}(p-2)^2 2(p-2)(p-1)\\
\Leftrightarrow 0\geq p(p-3) \Leftrightarrow 0\leq p \leq 3.
$$
Hence for $0\leq p \leq 3$ we have
$$
\left(\frac{\partial}{\partial t}\right)^2v \geq 2 \frac{\partial}{\partial t}\Delta v-\Delta^2 v+\lambda \nabla\left( (p)\_3 u^{p-3} (\nabla u)^3\right)$$
and therefore
$$
\int\_M \left(\frac{\partial}{\partial t}\right)^2v \geq \int\_M 2 \frac{\partial}{\partial t}\Delta v-\Delta^2 v+\lambda \nabla\left( (p)\_3 u^{p-3} (\nabla u)^3\right)=\int\_M \Delta\left(2 \frac{\partial}{\partial t}v-\Delta v \right)+\lambda \int\_M \nabla\left( (p)\_3 u^{p-3} (\nabla u)^3\right)=0
$$


