Let $f\in C^{\infty}([-1,0])$ be real-valued and suppose that


$$ \left| \int\_{-1}^{0} f(t)\,e^{\lambda t} \,{\rm d} t \right| \leq e^{-\sqrt{\lambda}},$$


for all $\lambda > 0$. Does it follow that $f(t)$ is zero on $-\epsilon \leq t\leq 0$, for some $\epsilon>0$?


My hunch is that this is actually not true but fail to come up with a counter example.



On an asymptotic integral decay
I'll change $[-1,0]$ to $[0,1]$ (so $t\to -t$), which seems easier on the brain.


$f(t)=e^{-1/t}$ is a counterexample: For $0<t<1/\sqrt{\lambda}$, we have $f\le e^{-\sqrt{\lambda}}$, so this part of the integral satisfies the desired bound, and if $t>1/\sqrt{\lambda}$, then $e^{-\lambda t}\le e^{-\sqrt{\lambda}}$.


