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let $f\in L^2$ and 2pi periodic

let $f_x(y)=f(x-y)$ and $V = span\{f_x: x\in [0, 2\pi)\}$

show V is dense in $L^2([0,2\pi))$ iff all of the fourier coefficients of f are non zero

i would be grateful for any tips

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1 Answer 1

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First, suppose that all the Fourier coefficients of $f$ do not vanish. Suppose $g$ is orthogonal to all the translates. I claim that $g =0$. (This suffices by Hahn-Banach.)

Since $g$ is orthogonal to any translate of $f$, we get that $$\sum e^{i n t}\hat{f}(n) \overline{\hat{g}(n) }= 0$$ by taking the dot product of $g$ with a translate of $f$ by $t$ and using Parseval's theorem. This means that the Fourier series $\sum e^{i n t}\hat{f}(n)\overline{ \hat{g}(n)}$ is trivial; that means that all the coefficients vanish, which gives what you want.

For the other direction, if $f$ has a vanishing Fourier coefficient, a suitable exponential will be orthogonal to all the translates of $f$.

Note that there is a harder theorem of Wiener (the Tauberian theorem): if $f$ is an $L^1$ function whose Fourier transform vanishes nowhere, then the sums of its translates form a dense subset.

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I like this answer. Hahn-Banach is one way to see that a proper closed subspace is orthogonal to some nonzero vector, and that version generalizes to other spaces. However, one doesn't need Hahn-Banach to show that closed subspaces of Hilbert space are orthogonally complemented. –  Jonas Meyer Mar 30 '11 at 1:22

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