Assume for some finite positive constants $\tau,C$ and $\lambda$, the positive function $f$ satisfies $$\int_0^\tau f(t,s,x)ds\leq C \exp (-\lambda t),$$ where the constants are independent of $t$.
What kind of condition do we need on the function $f$ to obtain the following result? $$f(t,s,x)\leq C \exp (-\lambda t),$$ where $C$ is independent of time $t$. Is it enough that we just assume the function $f$ is continuous w.r.t $s$ in the interval [$0$,$\tau$]
Bounds on the integral of a function do not give bounds on the function itself. To illustrate this, let's ignore the variables $t$ and $x$ for now and just consider functions $f(s)$ on the interval $[0,1].$
For example the functions $f_\epsilon\colon[0,1]\rightarrow\mathbb{R}$ given by $f_\epsilon(s)=(s+\epsilon)^{-{1/2}}$ (for say $\epsilon < 1$) have a uniform integral bound: $$ \int_0^1 f_\epsilon(s) d s=\int_\epsilon^{1+\epsilon}r^{-1/2}dr=(1+\epsilon)^{1/2}-\epsilon^{1/2}\le3 $$ However, $\sup_{s\in[0,1]}f_\epsilon(s)=\epsilon^{-1/2}\rightarrow \infty$ for $\epsilon \rightarrow 0$.