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Let $f$ be continuous, real valued and compactly supported with exactly one maximum function in $L_2$. Form the functions $ f_{m,k}=f^m(x-2^k) $

Under which conditions $\{f_{m,k}\}$ would be a frame?

(A function $f\in L_2(R)$ is said to generate a frame $\{f_{m,k}\}$ of $L_2(\mathbb{R})$ if for some $A$,$B>0$ we have $A\|f\|^2_2\le \sum_{j, k \in Z}|\langle f, f_{j,k}\rangle|^2\le B\|f\|^2_2$)

Thank you.

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    Crossposted to MO: http://mathoverflow.net/questions/99147/frame-of-l-2-space (At a first glance, I am not sure which site it belongs on.)2012-06-08

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My impression is: never. A frame sequence must be bounded in $L^2$, so we must have $|f|\le 1$ a.e. Also, the set where $|f|=1$ must have measure zero, otherwise the characteristic function of this set will violate the lower bound. Since $f$ is continuous and compactly supported, $\sup |f|<1$... But the main issue is, neither $k$ nor $m$ increases the frequency of oscillation of $f$, which is necessary to handle highly oscillatory inputs such as $g_N(x)=\sin(Nx)\chi_{[0,2\pi]}$. Indeed, the family $f_{m,k}$ is equicontinuous, and as $N\to\infty$, this will make $\langle g_N,f^{j,k}\rangle$ small, breaking the lower frame bound.