Preliminaries: Let $u$ be the solution of the Cauchy problem for the heat equation with initial datum $u_0 \in L^1 \cap L^p$. Then I know that the following estimates hold: $$\Vert u(t,\cdot)\Vert_{L^p} \le Ct^{-\frac{N}{2}\left(1-\frac{1}{p}\right)}\Vert u_0\Vert_{L^1}$$ and $$\Vert u(t,\cdot)\Vert_{L^p} \le \Vert u_0 \Vert_{L^p}$$ where $N$ is the space dimension, $p \in [0,\infty]$ and $C$ a constant.

Question 1: Now let $q \in [1,2]$. and $u_0 \in L^q$ I'm not able to prove the following estimate (which should be classical) $$\Vert \partial^k_t D^\alpha u(t,\cdot)\Vert_{L^p} \le Ct^{-\frac{N}{2}\left(\frac{1}{p}-\frac{1}{q}\right)-\frac{|\alpha|}{2} - k}\Vert u_0\Vert_{L^q},$$ where $p \in [q,\infty]$, $C$ is a constant, and $|\alpha|$ is the lenght of the multi-index $\alpha$. How does one do it?

I've tried to compute $$\partial_{x_k} K(t,x) = \frac{-x_k}{t}K(t,x)$$ $$\partial_{x_kx_k} K(t,x) = \left(\frac{x^2_k}{t^2}- \frac{1}{t}\right)K(t,x)$$ $$\partial_{t} K(t,x) = \frac{1}{2}\left(\frac{|x|^2}{t^2}- \frac{N}{t}\right)K(t,x)$$

(where $K$ is the heat kernel), but I don't see where this leads.

Question 2: To obtain the first estimates that I stated, I used the explicit form of the heat kernel. I guess one should use it to derive the last one too. Can we also obtain these estimates without making use of the form of the heat kernel K? That is, can we also obtain these estimates using the fact that for the heat kernel we have $$\mathcal{F} K_t(\xi) = (\pi/t)^{\frac{N}{2}}K_{(4t)^{-1}}(\xi),$$ where $\mathcal{F}$ is the Fourier transform.

  • $\begingroup$ @WillieWong How does Q1 follow from interpolation? I've problems with the actual computation and estimates of the derivatives (I've added some computations to the question, but I don't know how iterating them leads to the desired estimate) and also with the more theoretical considerations of Q2 $\endgroup$ – Jun Jun 23 '17 at 20:37
  • $\begingroup$ that's what I was not clear about: I was not sure whether your problem was going between $L^p$ spaces or that you just don't have the higher derivative estimates to start with. $\endgroup$ – Willie Wong Jun 23 '17 at 20:44
  • $\begingroup$ @WillieWong The problem is mainly the derivatives' estimates. $\endgroup$ – Jun Jun 23 '17 at 20:53

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