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 $2^* = \frac{2n}{n-2}$.

  • $\begingroup$ I doubt about this. Take $M=\mathbb{R}$, and let $p=3$. then if my calculation is correct , the second derivative in $t$, at point $t=0$ is $C\int_{\mathbb{R}} (u'_{0}(x))^{3} dx$ where $u_{0}$ is your initial data, and $C>0$ is universal constant. Now it can have an arbitrary sign. For example, start with even $u_{0}$ and perturb it slightly. $\endgroup$ – Paata Ivanishvili Jun 1 '18 at 15:52
  • $\begingroup$ @Paata: Notice that $M$ is required to be compact. $\endgroup$ – Alex M. Jun 1 '18 at 15:55
  • $\begingroup$ Ops, this I missed. You are right. $\endgroup$ – Paata Ivanishvili Jun 1 '18 at 15:56
  • $\begingroup$ How is this function even defined if $u_t$ is negative and $p$ is not an integer? $\endgroup$ – Michael Renardy Jun 1 '18 at 16:07
  • 3
    $\begingroup$ @Paata $u''u'^2$ integrates to $0$ and, obviously, $\int u(u'')^2\ge 0$ $\endgroup$ – fedja Jun 2 '18 at 1:23

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 $$ \int_M \left(\frac{\partial}{\partial t}\right)^2v \geq \int_M 2 \frac{\partial}{\partial t}\Delta v-\Delta^2 v+\lambda \nabla\left( (p)_3 u^{p-3} (\nabla u)^3\right)=\int_M \Delta\left(2 \frac{\partial}{\partial t}v-\Delta v \right)+\lambda \int_M \nabla\left( (p)_3 u^{p-3} (\nabla u)^3\right)=0 $$

  • $\begingroup$ Great ! Thanks a lot ! I will check it next week and come back to you then. $\endgroup$ – Romain Gicquaud Jun 3 '18 at 5:41
  • $\begingroup$ I believe the second row should be $(1,2,1,1,0)$, not $(1,1,1,2,0)$ (you get the third derivative from the differentiation wrt $t$ of $|\nabla u|^2$ in $\Delta v$, so if there is a $2$ somewhere, it should be on this component). $\endgroup$ – fedja Jun 4 '18 at 16:03
  • $\begingroup$ @fedja: thx for pointing that out. I corrected my solution. Luckily, the method still seems to work and we even get nicer constants (less fractions). $\endgroup$ – Markus Sprecher Jun 4 '18 at 18:21
  • $\begingroup$ @RomainGicquaud Now, let's say we are on the 2D torus with conformal metric (multiply it by something of which the function doesn't depend to raise the dimension, if you insist on $n\ge 3$). Then if the expression is negative at a single point, we can enhance the importance of that point as much as we want by dropping the conformal factor there $A$ times (which will drop the volume form $A^2$ times but enhance the $4$-th order derivatives $A^4$ times. Thus Markus' range looks actually sharp. Am I talking nonsense? $\endgroup$ – fedja Jun 5 '18 at 0:00
  • $\begingroup$ @fedja: you are right. Set $\widehat{g} = f^2 g$ for some positive function. Then (in 2d) the conformal transformation law for the Laplacian is given by $\widehat{\Delta} u = f^{-2} \Delta u$ and for the measure it gives $d\widehat{v} = f^2 dv$. As a consequence, we have $$ \int u^{p-2} \left(\widehat{\Delta} u\right)^2 d\widehat{v} = \int u^{p-2} (\Delta u)^2 f^{-2}dv. $$ And the other two terms pick up the same $f^{-2}$ factor. So if you shrink the metric where the expression is negative, you can make it as negative as you want. $\endgroup$ – Romain Gicquaud Jun 5 '18 at 9:06

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