integral depending on a parameter Let $f$ be a real-valued continuous function on the interval $[0,1]$ and satisfy the following estimate 
$$
\left|\int_0^1 f(t) e^{st}dt\right|\le Cs^{\frac12},\quad s>1,
$$
where the constant $C$ is independent of $s$. 
Can we assert that $f$ is identically zero on $[0,1]$?
 A: Michael has essentially answered this in his comment, but let me make this more explicit.
In fact, a stronger statement is true: If $F(z)=\int_0^1 f(t)e^{tz}\, dt$ satisfies $|F(s)|\lesssim e^{(a+\epsilon)s}$ for $s>1$ and all $\epsilon>0$ (but with possibly $\epsilon$ dependent implied constants), then $f=0$ on $[a,1]$. (This is of course extremely plausible right away, or how could there be cancellations between the various exponentials for large $s>1$?)
By splitting $0\le t\le 1$ into the two parts $[0,a+\epsilon]$ and $[a+\epsilon, 1]$, we see that the claim is equivalent to the following variant of it: If $G(z)=\int_0^b g(t)e^{tz}\, dt$ is bounded for $z=s\ge 0$, then $g\equiv 0$.
Since $G$ is of exponential type, the Phragmen-Lindelof principle applies to all sectors of opening $<\pi$, and in particular, it applies to quarter planes. Since $G$ is bounded on the imaginary axis and on $z=s\ge 0$, it is bounded on the right half plane. It is also, trivially, bounded on the left half plane. Thus $G$ is constant, and the constant is zero since $G$ is also square integrable on the imaginary axis.
A: The answer is, yes.
Suppose $f\not\equiv0$. By Stone–Weierstrass theorem, there is a sequence of non-trivial polynomials $P_n(t)$ converging uniformly to $f(t)$. That means, the inequality
$$
\left|\int_0^1 P_n(t) e^{st}dt\right|\le C's^{\frac12},\quad s>1,
$$
must persist for some polynomial $P_n(t)$ (perhaps $n$ large). However, it is easily checked that the integral against a polynomial grows exponentially with $s$. A contradiction. Thus, the claim follows.
