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Let $f$ be a continuous function on $\mathbb{R}$ with compact support with exactly one maximum. Form the functions $$ f_{m,k}(x)=f^m\left(x-\frac{k}{2^m}\right) $$ One can show that $\text{span}\{f_{m,k}(x)=f^m\left(x-\frac{k}{2^m}\right), k \in \mathbb{Z}, m>0\}$ is dense in $L_p(R)$.

Under which conditions does $\{f_{m,k}\}$ form a frame (or maybe a Riesz basis) for $L_p(R)$?

Thank you.

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    By $f^m$ you mean the pointwise power? Are you sure about your claim for2012-06-11
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    My objection about the lack of oscillation in $f_{m,k}$ applies to this construction as well. http://math.stackexchange.com/questions/155783/condition-for-frame-of-l-22012-06-11
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    @Leonid Kovalev: Thank you.2012-06-11
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    I just noticed tat my above comment is incomplete. I wanted to ask if you are sure about your claim $p=\infty$...2012-06-11
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    Yes, its for $p\ge 2$.2012-06-11
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    @Leonid Kovalev: How about if we fixed $m$. Then it seems we have translates of $f_{m,k}$ for the $m$ fixed and it seems a frame, isn't it?2012-06-20
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    @Michael No, translates of a fixed function can never form a frame. First insert $g=\chi_{[0,2\pi]}$ in the definition of the frame: upper bound must hold. Then replace $g$ with $g_N(x)=\sin(Nx)\chi_{[0,2\pi]}$: when $N$ is large, cancellation reduces the $\ell^2$ norm of coefficients, contradicting the lower bound... This is why the standard frame constructions include either scaling (as wavelets) or modulation (as Gabor frames).2012-06-20
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    @Leonid Kovalev: Thank you.2012-06-20

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