A hard problem on exponential integration Suppose $a : [0 , 1] \to \Bbb R$ is an infinitely smooth function. For $\lambda\ge1$, define $$F(\lambda) := \lambda \int_0^1 e^{\lambda t} a(t) \, dt.$$
If $\sup_{\lambda\ge1}|F(\lambda)|\lt\infty$, then $a$ is the identically zero function.
Below are the some results I have derived:


*

*Derivatives of any order of $a$ vanishes at $t = 1.$

*For any $\delta\lt1$, $a(t)$ has a zero in the open interval $(\delta , 1).$ 
The same is true for all the derivatives of $a(t)$. This tells us that the $n^\text{th}$ derivative of $a$ has infinitely many distinct zeros for all natural $n$.

*If $a$ is analytic, then I can show that $a(t)\equiv 0.$

*If $a(t)\ge0$ on $[0 , 1]$, then it is obvious that $a\equiv 0.$


I would appreciate any hint.
Actually $(1)$ and $(3)$ follows from $(2)$. For $(2)$, suppose that $a(1)\gt0$ (the case that $a(1)\lt0$ can be argued in the same way as the following), then there exists $\delta\lt1$ such that $a(t)$ is strictly positive on $I = [1-\delta , 1]$. Then write $F(\lambda) = \lambda\int_0^{1-\delta}e^{\lambda t}a(t)dt + \lambda\int_{1-\delta}^1e^{\lambda t}a(t)dt$. Now there exists $c\gt0$ such that $a(t)>c$ on $I$ since $I$ is compact, so that the second term is bounded below by $c (e^{\lambda}-e^{\lambda(1-\delta)})$. But the first term is only $O(e^{\lambda(1-\delta)})$, so this contradicts the fact that $F(\lambda)$ is bounded. Morever, using the same idea and integration by parts, one can see that $n^{th}$ derivative of $a$ must vanish at $1$. 
 A: This is a standard exercize on aplication of the Phragmén-Lindelöf Principle.
$F(\lambda)$ is an entire function (analytic in the whole complex plane).
On the imaginary axis we have:
$$|F(\lambda)|\leq c|\lambda|,$$
where $c$ is the $L^1$ norm of $a$.
On the real axis, it is bounded, by your assumption.
Moreover, this function is of exponential type $1$. It has at least one zero in the complex plane, say $\lambda_0$. (The only function of exponential type which has no zeros is
the exponential function, and it is clear that our function is not an exponential function).
Therefore $G(\lambda)/(\lambda-\lambda_0)$ is bounded on both coordinate axes,
and thus constant, by Phragmén-Lindelöf. Now it is easy to see that this is a contradiction.
A good reference to Phragmén-Lindelöf is any of the two books of Levin
(Lectures on entire functions, or Distribution of zeros of entire functions).
I believe that the first book is freely available on the web. Other books on entire
functions also contain this.
The assumption that $a$ is smooth was not used. The proof above only used that $a$
is integrable, but the fact is true even if we only assume that $a$ is a distribution, 
or even a hyprfunction.
