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I found this as an exercise in Kolmogorov-Fomin, let me state the problem precisely:

Let $S_{\infty}$ be the space of functions on the real line for which there exists a colelction of constants $C_{pq}$ for all $p, q$ such that $$ |x^pf^{(q)}(x)|\leq C_{pq}. $$ Is it true that if for all $p\geq 0$ $\int_{-\infty}^{\infty} x^pf(x)=0$, then $f=0$?

The exercise follows a section on the Fourier transform of functions of $S_{\infty}$. Now, I guess one should prove this exercise (I am fairly sure it is true) using some properties of the Fourier transform, but I cannot see how.

So my question is, how is this to be proved with the fundamental properties of the Fourier transform in $S_{\infty}$?

RandomGuy
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  • Yes, it is actually a duplicate but I don't find the answer there particularly clear or useful. – RandomGuy Jul 29 '17 at 11:55
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    The linked answer tells that (1) the given condition is equivalent to $\hat{f}^{(q)}(0) = 0$ for all $q \geq 0$, and that (2) you can indeed find a non-trivial function $f \in S_{\infty}$ for which this is satisfied by taking inverse Fourier-transform to some $g \in S_{\infty}$, for instance, $g(x) = \exp(-x^2-x^{-2})$. – Sangchul Lee Jul 29 '17 at 12:31

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