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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 $1$.

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migrated from May 24 '14 at 7:29

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up vote 9 down vote accepted

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.

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I guess I need to study complex analysis a bit more now to completely grasp this proof. Thanks! – dezdichado May 24 '14 at 7:50
Sure. Complex analyis is useful, even for real analysis:-) If you already had an introductory (undergraduate) course, you can read Levin's books. – Alexandre Eremenko May 24 '14 at 12:12
Theorem 1 of Phragmen-Lindelof says that an entire function bounded on the boundaries of its domain D is bounded on D. So did you also mean to use Liouville's theorem to conclude that the function is constant? – dezdichado May 25 '14 at 20:48
Yes. You omitted in your statement the crucial condition about the growth of the function and the shape of the domain. – Alexandre Eremenko May 26 '14 at 6:59
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

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