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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 :)

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

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  • $\begingroup$ 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

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