How to prove the stated statement in the title ?
For any two elements $\{ a_j \} , \{ b_j \} \in \mathbb{C}^n$ the associated inner product is $$\langle\{ a_j \} , \{ b_j \}\rangle=\sum_{j=1}^n a_j \overline{b_j}.$$
For any $\{ f_j \}_{j=1}^{n} \in \mathbb{C}^n,$ there exists constants $A,B >0$ such that $$A \|f\|^2 \leq \sum_{j=1}^n |\langle f,f_j\rangle|^2 \leq B \|f\|^2.$$
The "frame operator" $S:\mathbb{C}^n \rightarrow \mathbb{C}^n$ is $Sf=TT^*f=\sum_{j=1}^n \langle f,f_j\rangle f_j$ and $\langle Sf,f\rangle =\sum_{j=1}^n |\langle f,f_j\rangle|^2.$
May be this is obvious. But I'don't seem to be able to prove this. Any help is much appreciated.
In general, it's not true - letting all $f_j$ be $0$ is clearly a counterexample.
I imagine $f_1,...,f_n$ is supposed to be linearly independent. In this case, it's true and follows by observing that the function $$\mathbb{C}^n \backslash \{0\} \rightarrow \mathbb{R}, \; f \mapsto \frac{\sum_{j=1}^n | \langle f,f_j \rangle|^2}{\|f\|^2}$$ is invariant under scaling $f \mapsto \lambda f$, so it is actually a function on the (compact) unit sphere that is always strictly positive; therefore, it takes a positive minimum and positive maximum.