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Let $\phi(x)\in C_0^{\infty}(\mathbb{R})$ be an infinitely differentiable finite function with support $\operatorname{supp} \phi \subset [-c,c]$ and let $\mu_0$ be a zero of the function

$$\Phi(\lambda)=\int_{-c}^{c} \phi(x) \exp(-i \lambda x) dx \tag{1}$$

Then $$ \int_{-c}^{c}\sum_{j=0}^{n} c_j \phi^{(j)}(x)\exp(-i \mu_0 x) dx=0 \tag{2}$$

where $n$ is an arbitrary positive integer and $c_j$, $j=1,2,\dots,n$ are arbitrary complex numbers.

How to prove this proposition? Should I use integration by parts?

mike
  • 5,604

1 Answers1

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Yes, you should integrate by parts. Be careful to treat the case $\mu_0=0$ separately (this case only needs the fundamental theorem of calculus). For $\mu_0\ne 0$, $$\begin{split}0&=\int_{-c}^{c} \phi(x) \exp(-i \mu_0 x)\, dx \\&= -i\mu_0^{-1}\int_{-c}^{c} \phi'(x) \exp(-i \mu_0 x)\, dx \\&= (-i)^2\mu_0^{-2}\int_{-c}^{c} \phi''(x) \exp(-i \mu_0 x)\, dx \end{split}$$ and so on.