Consider the equation: \begin{align*} u_t -\triangle u -a(t,x)u&= f(t,x),\\ u|_{\partial \Omega} &=0, \\ u(0)&=u_0. \end{align*} with $\Omega\subset \mathbb{R}^N$ open and $|\Omega|<\infty.$ Let $(T(t))_{t\geq0}$ the Heat semigroup. Suppose $a \in L^{\infty}((0,T), L^{\sigma }(\Omega))$ with $\sigma \geq 1$, $\sigma > \dfrac{N}{2}$, and $q \geq \sigma'$, where $\dfrac{1}{\sigma}+\dfrac{1}{\sigma'}=1$. My question is how to get the following estimate
$$\|T(t-s)a(s)u(s)\|_{L^q} \leq (t-s)^{\frac{-N}{2 \sigma}} \|a(s)\|_{L^\sigma}\|u(s)\|_{L^q},$$
What I tried to do was use this estimate: $$\|T(t)\phi\|_{L^q} \leq (4\pi t)^{\frac{-N}{2}\left( \frac{1}{p} - \frac{1}{q}\right) } \|\phi\|_{L^p},$$
but with her i get
$$\|T(t-s)a(s)u(s)\|_{L^q} \leq (4\pi (t-s))^{\frac{-N}{2}\left( \frac{1}{p} - \frac{1}{q}\right) } \|a(s)u(s)\|_{L^p},$$
Next I tried to use Holder inequality using the conjugate exponents $\dfrac{p}{\sigma}$ and $1-\dfrac{p}{\sigma}$ in $\|a(s)u(s)\|_{L^p}$ give you $\|a(s)\|_{L^\sigma}$, but I can't get $\|u(s)\|_{L^q}$. I don't see a way to put $\sigma$ in the exponent of $(t-s)^{\frac{-N}{2 \sigma}}$. This estimate is important to study the smoothing effect of heat equation. Does anyone know how to make this estimate? I also don't know if the way I thought of approaching it is correct, but the heat Kernel's estimate is the only one I know of that seems to have to do with the problem.
The quoted estimate is mentioned in the Book of Brezis-Cazenave Nonlinear Evolution Equations page 95.