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A sequence of distinct vectors $\{f_1,f_2,...\}$ belonging to a separable Hilbert space $H$ is said to be a Frame if there exist positive contants $A$, $B$ such that $$A\|f\|^2\leq\sum_{n=1}^\infty |(f,f_n)|^2\leq B \|f\|^2.$$ If we associate with a given frame $\{f_1,f_2,...\}$ a bounded operator $T$ on $H$ defined by $$Tf=\sum_{n=1}^\infty (f,f_n) f_n$$ it follows $$(Tf,f)=\sum_{n=1}^\infty |(f,f_n)|^2$$

Can we infer $$||Tf||^2\leq B^2 ||f||^2?$$

Thank you very much.

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The operator $T$ is called the Frame operator. A simple computation can prove that $T$ is self adjoint. Moreover, $T$ is positive and $AI\leq T\leq BI$, and hence $\|T\|\leq B$. Fix $f\in H$ and suppose $f\neq 0$. Consider $T(f/\|f\|)$. We have

$$\left\| T\left(\frac{f}{\|f\|}\right)\right\|\leq B.$$

Multiplying by $\|f\|$ we obtain $\|T(f)\|\leq B\|f\|$, and hence the result.

Edit:

Note that, as you wrote

$$(T(f),f)=\sum_{n=1}^\infty |(f,f_n)|^2,$$

and, by assumption

$$A\|f\|^2\leq\sum_{n=1}^\infty |(f,f_n)|^2\leq B \|f\|^2.$$

Hence

$$(AI(f),f)=A\|f\|^2\leq(T(f),f)\leq B \|f\|^2=(BI(f),f).$$

This proves that $AI\leq T\leq BI$.

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  • $\begingroup$ Added the proof of $AI\leq T\leq BI$. $\endgroup$ – MrSelberg Mar 25 '15 at 14:55
  • $\begingroup$ thank you! But, why T is self adjoint? $\endgroup$ – Mark Apr 8 '15 at 4:28
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    $\begingroup$ @Mark Here you can found a proof: math.stackexchange.com/questions/433006/…. $\endgroup$ – MrSelberg Apr 8 '15 at 4:32

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