About the use of Young and Hölder inequalities In the following Terraneo's paper Terr we have an idea of the proof for a certain Cauchy semilinear problem (theorem 3, page 5) but I'm stuck in one important detail in 1. How are they used the Young and Hölder inequalities in the third line of inequalities on page 6? I know the fact that $\lVert G_t\rVert_{L^r}=ct^{(\frac{N(1-r)}{2r})}$ and $e^{t\Delta}u=G_t\ast u$, where $G_t$ is the gaussian function associated with the heat kernel, but I worried about the next fact pointed in the paper
$$\lVert u^{\alpha+1}\rVert_{L^p}\leq \lVert u\rVert_{L^p}^{\alpha+1}$$
Any help will be very appreciated.
 A: I hope I am getting you right. You want to know why the inequality
$$\left\| \int_0^t e^{(t-s)\Delta} |u|^\alpha u(s) ~ds \right\|_p \leq C \int_0^t (t-s)^{-\tfrac{n \alpha}{2p}} s^{-\sigma(\alpha+1)} (s^\sigma \|u\|_p )^{\alpha+1} ~ds$$
holds?


*

*First notice that $\|e^{\tau \Delta} v \|_p \leq (4\pi \tau)^{-n/(2r)} \|v\|_q$ for $p > q$; where $\tfrac{1}{r}=\tfrac{1}{q}-\tfrac{1}{p}$. Hence we have
$$\|e^{(t-s) \Delta} |u|^\alpha u(s) \|_p \leq C (t-s)^{-n/(2r)} \||u|^\alpha u(s)\|_q$$
and setting $q=\tfrac{p}{\alpha+1}$ yields $r=\tfrac{p}{\alpha}$ and thus
$$\|e^{(t-s) \Delta} |u|^\alpha u(s) \|_p \leq C (t-s)^{-\tfrac{n\alpha}{2p}} \||u|^\alpha u(s)\|_{p/(\alpha+1)} \tag{1}.$$

*Second notice that we have $\tfrac{\alpha+1}{p}=\tfrac{\alpha}{p}+\tfrac{1}{p}$ and thus by Hölder's inequality
$$\||u|^\alpha u(s)\|_{p/(\alpha+1)} \leq \| |u|^\alpha\|_{p/\alpha} \|u(s)\|_p=\|u(s)\|_{p}^\alpha \|u(s)\|_p=\|u(s)\|_{p}^{\alpha+1} \tag{2} $$
Now, starting from the beginning, take the $L^p$-norm into the integral, do (1) and then (2).

EDIT: By applying Young's convolution inequality to $\|e^{\tau \Delta} v \|_p$ we have
$$\|e^{\tau \Delta} v \|_p=\|G_\tau *v \|_p \leq \|G_\tau\|_s \|v\|_q$$
for $\tfrac{1}{s}=1+\tfrac{1}{p}-\tfrac{1}{q}$. Notice that $\|G_{\tau/s}\|_1=1$ and hence
$$\| G_\tau \|_s = \left( (4\pi \tau)^{-sn/2}\int_{\mathbb{R}^n} e^{-s |x|^2/(4\tau)} ~dx \right)^{1/s}=(4\pi \tau)^{-n/2} (4\pi \tfrac{\tau}{s})^{n/(2s)}\leq (4\pi \tau)^{-n(s-1)/(2s)}$$
and setting yields $\tfrac{1}{r}:=-\tfrac{1}{s}+1=\tfrac{s-1}{s}$ yields finally for $\tfrac{1}{r}=\tfrac{1}{q}-\tfrac{1}{p}$
$$\| G_\tau \|_s \leq (4\pi \tau)^{-n(s-1)/(2s)}=(4\pi \tau)^{-n/(2r)}.$$
