Assume that $G$ is a finite vector space over a finite field with order $|G|$. (For example, $G=Z_p^k$). Assume that $\{f_n\}_n$ is a Parseval frame for $l^2(G)$. Can we say that the sequence $\{f_n\}_n$ is independent? In other words, can we say that the sequence is of length $|G|$? If the sequence is an orthonormal basis, then the answer is yes.
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2 Ok. I have the answer to my own question. It is No. Here is how we prove it: Let $G=\Bbb Z_p^d$. Let $F=\{e_1, e_2, \cdots, e_d\}$ denote the standard orthonormal basis for $\Bbb Z_p^d$. For any $e_i\in F$ define the dirac function $\delta_{e_i}$ which is defined by $\delta_{e_i}(x) = 1$ for $x=e_i$ and zero elsewhere. Then the set $\{\frac{1}{\sqrt{2}}\delta_{e_1}, \frac{1}{\sqrt{2}}\delta_{e_1}, \delta_{e_2}, \delta_{e_3}, \cdots \delta_{e_d} \}$ is a Parseval frame for $L^2(\Bbb Z_p^d)$ with the frame bounds $A=B=1$. And, the set is not independent.
- 1$\begingroup$ Does $A=B$ mean this is a tight frame? As an outsider, I find some of the definitions confusing. $\endgroup$kodlu– kodlu2016-06-17 04:04:22 +00:00Commented Jun 17, 2016 at 4:04
- $\begingroup$ Yes. if the bounds (upper and lower bounds) of a frame equal, then the frame is called tight. $\endgroup$Melody– Melody2016-10-22 15:22:16 +00:00Commented Oct 22, 2016 at 15:22