Support of function and support of its Fourier transform

Can anyone sort of give a proof for "if a function is concentrated in a cube then its Fourier transform is "mainly" concentrated in its dual cube"?

Also, i have seen similar arguments several times(like in the argument of wave packet decomposition). However, i never find formal arguments to prove relation of support of frequency function and support of its oscillatory integrals. I would really appreciate it if someone can suggest some reference about such relation.

• The statement is similar to those given by the Paley-Wiener theorems, you should check the precise formulations of those. – Desiderius Severus Apr 5 '17 at 15:47
• But this statement is wrong: it contradicts the indeterminacy principle. – Alexandre Eremenko Apr 6 '17 at 3:40
• Surely it's the other way: if u is supported in a cube then û has a positive fraction of its mass outside the dual cube. (A form of the uncertainty [a.k.a. indeterminacy] principle mentioned by Alexandre Eremenko.) – Noam D. Elkies Apr 6 '17 at 3:54
• @NoamD.Elkies: Without the motivation in terms of wavelets, I would agree with you and cite Hardy's version of the uncertainty principle as a prime example. However one does frequently see arguments in harmonic analysis where something resembling what the OP is describing is used; the main difference being that the principle is applied for one fixed, well-chosen, function across multiple scalings, rather than arbitrary functions at one scaling. (One example is the Knapp counterexample for Tomas-Stein restriction theorem.) – Willie Wong Apr 7 '17 at 14:20
• @OwenKING : Can you please state the definition of dual cube. The crux of the question seems to be in Fourier transform and I am lost just because I have never came across "dual cube". I guess what you are talking about is the product of second order moments of the function and its FT has an lower bound, which is termed as principle of uncertainity for reasons in Quantum physics. If you are interested in reaching the limit, then you may want to look at Prolate spheroidal wave functions. – Rajesh Dachiraju Apr 9 '17 at 14:38

This statement is wrong. Fourier transform of the characteristic function of the unit cube in dimention 1 equals $$\int_{-1}^1e^{-itx}dx=2(\sin t)/t.$$ Where is it "concentrated"?

Can anyone sort of give a proof for "if a function is concentrated in a cube then its Fourier transform is "mainly" concentrated in its dual cube"?

No one can, because it is false.

Take $f$ to be the Gaussian function, and take $g_y(x) = e^{ix\cdot y}f(x)$ for some $y \in \mathbb{R}^d$. Then the functions $f + g_y$ are all concentrated in the same cube (of size $1$), but there is no uniform control on their Fourier supports (for large $y$, the Fourier transform is supported on the disjoint union of two cubes separated a distance $y$ apart).

Since you mentioned wavelet theory, maybe what you meant is the statement

If both $f(x)$ and $\hat{f}(x)$ is "concentrated in a cube of size 1", and if $g(x) = f(\lambda x)$, then $g$ is concentrated in a cube of size $\lambda^{-1}$ and $\hat{g}$ concentrated in a cube of size $\lambda$.

This follows directly from the scaling properties of the Fourier transform. Since in wavelet decomposition the mother wavelet is frequently a well-chosen function which is concentrated in both physical and frequency space, and the daughters are all generated by rescaling, translations, and modulations, you can use something resembling what you quoted as a guiding principle for the analysis.

• What is a dual cube? – Christian Remling Apr 5 '17 at 20:04
• @ChristianRemling: I assume the OP means that, if a function is concentrated in a cube of side length $\approx \lambda$ then its Fourier transform is concentrated in a cube of side length $\approx \lambda^{-1}$. Of course, the OP has also used the phrase "the dual cube", and I admit I wouldn't know how to identify a preferred cube of side length $\lambda^{-1}$ from a given cube of side length $\lambda$: I can't see any reason to prefer one center over another. – Willie Wong Apr 5 '17 at 21:24
• (I meant to write its instead of the in the emphasized word in my previous comment.) – Willie Wong Apr 5 '17 at 21:42
• I suspect that the dual polytope is meant, in which case it is an octahedron (in higher dimensions, the cross-polytope) – მამუკა ჯიბლაძე Jan 4 '18 at 5:31
• @მამუკაჯიბლაძე Up to constant factors (with constant depending on the dimension), isn't concentrating on the dual polytope the same as concentrating on the dual cube? – Willie Wong Jan 4 '18 at 17:14

This is a delayed answer, but it seems good to clarify what the OP is asking. This kind of statements: "since $f$ is concentrated in such a ball then its Fourier transform is essentially concentrated in such a ball" is quite common in certain fields, but it doesn't mean that the function and its Fourier transform are compactly supported, because this is wrong, as noticed already. People usually omit the details of what is meant, however there are several ways of formalizing this heuristic.

Fix a smooth function $\zeta$ supported in the unit cube $Q$ centered at the origin in $\mathbb{R}^n$. Then, by standard methods, we see that $|\hat{\zeta}(\xi)|\le C_N\frac{1}{(1+|\xi|^2)^{N/2}}$, where $C_N$ is a constant depending on $N$ and $\zeta$; but as $\zeta$ is fixed, we forget about it. Whatever the reason is, you want to use a cut-off function for some parallelepiped $P$, so you take the affine transformation $A$ transforming $Q$ into $P$, hence $\zeta_A(x)=\zeta(A^{-1}x)$ works as a cut-off and its Fourier transform is $\widehat{\zeta_A}(\xi)=|\det A|\hat{\zeta}(A^t\xi)$, and $|\widehat{\zeta_A}(\xi)|\le C_N|\det A|\frac{1}{(1+|A^t\xi|)^{N/2}}$ , hence we see that $|\widehat{\zeta_A}(\xi)|$ decays strongly outside the "dual parallelepiped" $A^{-t}Q$ or is "concentrated" in $A^{-t}Q$; it is in general unimportant the position of $A^{-t}Q$, but its dimensions and orientation in space. This had been basically noted by Willie Wong.

People usually replace $|\widehat{\zeta_A}|$ by $\chi_{A^{-t}Q}$ when they try to get upper bounds, because $|\widehat{\zeta_A}(\xi)|\le C_N\sum_{\nu\in A^{-t}\mathbb{Z}^n}\frac{1}{(1+|\nu|^2)^{N/2}}\chi_{A^{-t}Q}(\xi-\nu)$. As in every field, there is a toolkit you acquire after some time, so it's hard to provide a single reference of the many ways this heuristic is applied.

By the way, to say that if $f$ is supported in a cube then its Fourier transform is concentrated in the dual cube is not quite precise and depends on the context. What is true in general, is that if $f$ is supported in a cube, then $|\hat{f}|$ is "essentially" constant in translations of the dual cube.