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Let $\alpha_1,\dotsc,\alpha_n$ be complex numbers. Let $\xi_1,\dotsc,\xi_n$ be distinct real numbers.

Define a function $f:\mathbb{R}\rightarrow\mathbb{R}$ by $f(x)=\sum_{i=1}^n \alpha_ie^{2\pi i\xi_ix}$. Assume that $f(x)=0$ for every $x\in\mathbb{R}$.

Does it follow that $\alpha_i=0$ for every $i$?

Note: I'm tagging this "Fourier Analysis" because from the little I know about the Fourier transform, this seems to be related, but I'm new to this theory.

Edit: I think that if we can find integrable functions $g_1,\dotsc,g_n$ such that $\int g_i(x)e^{2\pi i \xi_j x} dx=\delta_{i,j}$, then I can prove that the answer is yes by exchanging a finite sum and integral. It is very likely that such functions exist (intuitively, $n$ "random" integrable functions should be enough to span such functions $g_1,\dotsc,g_n$).

Terry
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1 Answers1

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Let $D=\frac{d}{dx}$ be the operator of differentiation with respect to $x$. Then $$ (D-2\pi i\xi)e^{2\pi i\xi x}=0. $$ For any other $\xi'$, $$ (D-2\pi i\xi')e^{2\pi i\xi x}=2\pi i(\xi-\xi')e^{2\pi i\xi x} \ne 0. $$ Therefore, \begin{align} 0 &= (D-2\pi i\xi_2)(D-2\pi i\xi_3)(\cdots)(D-2\pi i\xi_n)\sum_{j=1}^{n}\alpha_je^{2\pi i\xi_j x} \\ & = (D-2\pi i\xi_2)(D-2\pi i\xi_3)(\cdots)(D-2\pi i\xi_n)\alpha_1 e^{2\pi i\xi_1 x} \\ & = (2\pi i)^{n-1}(\xi_1-\xi_2)(\xi_1-\xi_3)(\cdots)(\xi_1-\xi_n)\alpha_1 \\ & \implies \alpha_1 = 0. \end{align} The same argument applies to all of the other constants. The conclusion is that $\alpha_j=0$ for $1 \le j \le n$.

Disintegrating By Parts
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