These days, I am struggling with a problem which seems very straightforward (and I'm pretty sure it is straightforward) but it resists to my attempts to prove it. Here it is:

Let $\mathcal H_t$ be the heat kernel in $\mathbb R^d$: \begin{equation} \mathcal H_t(x) = \frac{1}{(2 \pi t)^{d/2}} \exp{\left(-\frac{|x|^2}{4 \pi t}\right)}, \ \forall x \in \mathbb R^d. \end{equation} It is classical that we have the following formula for the $L^p$ norm of this guy, $1 < p < \infty$: \begin{equation} \| \mathcal H_t \|_{L^p} = C_p t^{-\left(1 - \frac{1}{p}\right)\frac{d}{2}} \end{equation} where $C_p$ is an universal constant. In the paper Mathematische Zeitschrift, 1984, T. Kato claims that from this identity, it is easy to deduce that if $1 < p \leq q < \infty$, then \begin{equation} \| \mathcal H_t * u \|_{L^q} \leq c t^{-\left(\frac{1}{p} - \frac{1}{q}\right)\frac{d}{2}} \|u\|_{L^p}. \end{equation} Why is that last inequality true? Thanks for your explanations :)


This is nothing else than Youngs inequality on convolutions applied to the convolution $\mathcal H_t * u$. See Young's Inequality on Wikipedia.

  • $\begingroup$ The link looks broken. There are better ways to insert links, which avoid this problem. $\endgroup$ – user147263 Jul 30 '14 at 15:43
  • $\begingroup$ Yes I know that Young's inequality is one of my only tools, but I really can't find the good combination of parameters to obtain the result... $\endgroup$ – Dobby Jul 30 '14 at 16:22
  • $\begingroup$ I fixed the link. $\endgroup$ – Nate Eldredge Jul 30 '14 at 16:23
  • $\begingroup$ @Dobby: Renaming the variables appropriately, Young's inequality says $\|\mathcal{H}_t \ast u\|_q \le \|u\|_p \|\mathcal{H}_t\|_r$ where $\frac{1}{p} + \frac{1}{r} = \frac{1}{q}+1$. So what does $r$ have to be? $\endgroup$ – Nate Eldredge Jul 30 '14 at 16:26
  • $\begingroup$ Sorry i was away for too long: So @Nate Eldredge: Thanks for fixing the link and for clarifying the details. $\endgroup$ – frog Jul 31 '14 at 7:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.