Suppose $f:I=(0,1)\to \mathbb R$ is a continuous function that satisfies
$$ \int\_I f(t) e^{at}\,dt \geq 0\quad \text{for all $a \in \mathbb R$}.$$


Does it follow that $f\geq 0$ on $I$?



A functional integral inequality
The answer is no. E.g., take
$$f(t):=1-\frac{3}{2} \max \left(0,1-2 \left| t-\frac{1}{2}\right| \right).$$
Then $f(1/2)=-1/2<0$, but
$$ \int\_0^1 f(t) e^{at}\,dt=
\frac{2 e^{a/2}}{a^2}\,\left(a \sinh\frac{a}{2}-3 \left(\cosh
 \frac{a}{2}-1\right)\right)\ge0$$
for all real $a$. (The inequality here follows from the inequality $u\sinh u\ge2(\cosh u-1)$ for $u\ge0$, which is easily checked by calculus.)


