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Let $\{ f_k \}_{k=1}^m$ be a frame for an $n$-dimensional vector space $V,$ and let $B$ denote the optimal upper bound. Prove that $$B \leq \sum_{k=1}^m \| f_k \|^2 \leq n B .$$

My approach: From the upper frame condition or even as a consequence of Cauchy-Schawarz we have $$\sum_{k=1}^m \left\vert\langle f, f_k \rangle\right\vert^2 \leq B \cdot \| f \|^2, ~ \forall f \in V.$$ For $f \neq 0,$ dividing by $\| f \|^2,$ $$\sum_{k=1}^m \left\vert\left\langle \frac{f}{\| f \|}, f_k \right\rangle\right\vert^2 \leq B < n B.$$ I need to some way other set $f=f_k$ to get the sum in what was asked to prove. But I don't know how to continue from here onwards. Any help is much appreciated.

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By Cauchy-Schwarz we have $$ \sum_{k=1}^m|\langle f,f_k\rangle|^2\leq \|f\|^2\sum_{k=1}^m \|f_k\|^2. $$ Since $B$ is the optimal upper bound for $(f_k)_{1\leq k\leq m}$, we have $B\leq \sum_{k=1}^m \|f_k\|^2$.

On the other hand, let $(e_j)_{1\leq j\leq n}$ be an orthonormal basis of $V$. By the definition of $B$ we have $$ \sum_{k=1}^m |\langle e_j,f_k\rangle|^2\leq B. $$ Summation over $j$ yields $$ nB\geq \sum_{j=1}^n\sum_{k=1}^m|\langle e_j,f_k\rangle|^2=\sum_{k=1}^m \sum_{j=1}^n |\langle e_j,f_k\rangle|^2=\sum_{k=1}^m \|f_k\|^2. $$ The last equality follows from Parseval's identity.

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  • $\begingroup$ @ MaoWao, thank you very much. $\endgroup$ – user358174 Oct 17 '16 at 14:22

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