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.
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.
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.
