If $\{f_1,f_2,...\}$ a given frame, $T$ a bounded linear, prove that $(Tf,f)=\sum_{n=1}^\infty |(f,f_n)|^2$.

Question

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 linear operator $T$ on $H$ defined by $$Tf=\sum_{n=1}^\infty (f,f_n) f_n$$ can we infer that $$(Tf,f)=\sum_{n=1}^\infty |(f,f_n)|^2\,?$$

Thank you very much.

Answer 1

I don't see where the condition that $\{f_1,f_2,\ldots\}$ form a frame is needed. Just use that the inner product is continuous to obtain \begin{align*} (Tf,f)&=\left(\sum_{n=1}^\infty(f,f_n)f_n,f\right)= \sum_{n=1}^\infty\left((f,f_n)f_n,f\right)= \sum_{n=1}^\infty(f,f_n)\left(f_n,f\right) \\ &= \sum_{n=1}^\infty(f,f_n)\overline{\left(f,f_n\right)}= \sum_{n=1}^\infty|(f,f_n)|^2. \end{align*}