Lp bounds of the Heat Kernel

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 $\mathbb R^d$: $$\mathcal H_t(x) = \frac{1}{(2 \pi t)^{d/2}} \exp{\left(-\frac{|x|^2}{4 \pi t}\right)}, \ \forall x \in \mathbb R^d.$$ It is classical that we have the following formula for the $L^p$ norm of this guy, $1 < p < \infty$: $$\| \mathcal H_t \|_{L^p} = C_p t^{-\left(1 - \frac{1}{p}\right)\frac{d}{2}}$$ 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 $$\| \mathcal H_t * u \|_{L^q} \leq c t^{-\left(\frac{1}{p} - \frac{1}{q}\right)\frac{d}{2}} \|u\|_{L^p}.$$ 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.
• @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? – Nate Eldredge Jul 30 '14 at 16:26