# 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.

My professor said this assertion is true but I haven't been able to solve this for some time. He also implied that this problem is highly non-trival (at least for me). Below are the some results I have derived:

1. Derivatives of any order of $a$ vanishes at $t = 1.$
2. 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 $n$.
3. If $a$ is analytic, then I can show that $a(t)\equiv 0.$
4. 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 $1$.

-

## migrated from mathoverflow.netMay 24 '14 at 7:29

This question came from our site for professional mathematicians.

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.
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.
The domain is given by $\lambda\ge 1$ and there is no assumption on the growth of $a(t)$. – dezdichado May 26 '14 at 7:57