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There are some functions which are invariant under Fourier transformation up to scaling factors, eg. sech(pi*x), Gaussian function etc.. Is there a set of basis functions, which form an invariant basis set? (eg like sine cosine functions, not being invariant though) If you think that such a basis cannot exist, explain why this is so, or even better add a proof.

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  • $\begingroup$ Does that mean you want to find an (orthonormal) basis $(f_i)$ with $\hat{f_i}=f_i$ for all $i$? Such a basis cannot exist. $\endgroup$ – PhoemueX Jul 14 '15 at 11:39
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    $\begingroup$ If you have a basis whose elements are invariant under the Fourier transform, then the Fourier transform would be the identity operator. You almost get that. The set of normalized Hermite functions $\{ h_{n}\}_{n=0}^{\infty}$ on $\mathbb{R}$ is an orthonormal basis of $L^{2}(\mathbb{R})$ for which $\hat{h_{n}}=(i)^{n}h_{n}$. So $L^{2}(\mathbb{R})$ decomposes in 4 similar and orthogonal subspaces $H_{0}\oplus H_{1}\oplus H_{2}\oplus H_{3}$ where $\mathcal{F}=(i)^{n}I$ on $H_{n}$. en.wikipedia.org/wiki/Hermite_polynomials#Hermite_functions $\endgroup$ – DisintegratingByParts Jul 14 '15 at 22:44
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    $\begingroup$ @TrialAndError To me too! I think your comment answers my question! $\endgroup$ – v217 Jul 21 '15 at 18:06
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    $\begingroup$ Good, glad I could help. $\endgroup$ – DisintegratingByParts Jul 21 '15 at 18:31
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    $\begingroup$ I don't know of anything that will give you what you want. $\endgroup$ – DisintegratingByParts Oct 9 '16 at 17:51

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