Let $f$ be a realvalued 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 $[0,1]$?

1$\begingroup$ Since the terms of order $\leq 1$ in $s$ automaticallly satisfy the inequality, one finds by expanding in $s$ that the inequality is equivalent to $\int_0^1f(t) (e^{st}1st)/s^2 dt\leq C/s$ for all $s>1$. $\endgroup$ – user_1789 May 31 '17 at 8:24

2$\begingroup$ The integral is an entire function in $s$, so the inequality means that its Power series coefficients are zero from the second on. This is equivalent to $f$ being orthogonal to $x^k$ for $k=2,3,\dots$. $\endgroup$ – user1688 May 31 '17 at 9:08

2$\begingroup$ $\sin s$ is an entire function which is bounded by $s$ $\endgroup$ – Piero D'Ancona May 31 '17 at 10:35

4$\begingroup$ You can use PhragmenLindelof to show that the estimate actually holds for complex s. $\endgroup$ – Michael Renardy May 31 '17 at 21:13

1$\begingroup$ The power of $s$ is irrelevant, only the exponential type matters. See my answer below please. $\endgroup$ – Christian Remling Jun 1 '17 at 0:56
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 PhragmenLindelof 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.
The answer is, yes.
Suppose $f\not\equiv0$. By Stone–Weierstrass theorem, there is a sequence of nontrivial 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.

1$\begingroup$ The convergence $p_n e^{st}\to f e^{st}$ is not uniform in $s$, however, so I think you only obtain that for fixed $s$, you can find an $n=n(s)$ so that your inequality holds. $\endgroup$ – Christian Remling Jun 1 '17 at 1:56

$\begingroup$ I found the same problem stated by Christian Remling. If the $n$ depends on $s$, I doubt that the above argument can still work. $\endgroup$ – CooLee Jun 1 '17 at 2:14