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:
- 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 $1$.