Proof that the eigenvalues of the frame operator $S$ are optimal frame bounds, and that $S = AI$ iff the frame is tight I'm studying frames, and I don't understant why the eigenvaule of S are the optimal frame bounds. I understand the proof that they are bounds, but in all the papers and book I've found they demonstrate that they're bounds and then say that they're optimals follows from taking $x$ as the largest/smallest eigenvector, but I can't see the reasoning: if we take $x = e_{min}$, where $e_{min}$ is the eigenvector corresponding to the smallest eigenvalue we have that $Sx = \lambda_{min}e_{min} $ and $\left<Se_{min}, e_{min}\right> = \lambda_{min} $, so $ \lambda_{min}\lVert x \rVert^2 \leq \sum_{j \in J}\left|\left<x, f_j\right>\right|^2$ becomes $ \lambda_{min}\lVert e_{min} \rVert^2 = \lambda_{min}$ which tells me that the norm squared of $e_{min}$ is 1, but I knew that already
Same thing for the proof that $S = AI$ if the frame is tight. The only proof that I found is in this book, page 26, but it just says that since $f = \frac{1}{A}\sum_{j \in J}\left<f, f_j\right>f_j$ the equation $A\lVert f\rVert^2 = \sum_{j \in J}\left|\left<f, f_j\right>\right|^2$ is equivalent to $S=AI$, and I can't see the connection
 A: Suppose that $A>\lambda_{min}$ is a larger frame bound for your frame $\{f_j\}_{j\in J}$, then we have
\begin{align}
A=A\|e_{min}\|^2&\leq \sum_{j\in J}|\langle e_{min},f_j\rangle|^2\\
&=\langle S e_{min},e_{min} \rangle\\
&=\lambda_{min}\langle e_{min},e_{min}\rangle\\
&=\lambda_{min},
\end{align}
which obviously contradicts $A>\lambda_{min}$. Since you already know that $\lambda_{min}$ is a frame bound and no larger frame bound can exist, we conclude that $\lambda_{min}$ is optimal. The argument for the upper frame bound works analogously.
If your frame is tight, then you have for all $f$
\begin{align}
A\|f\|^2&=\sum_{j\in J}|\langle f,f_j\rangle|^2\\
\Longleftrightarrow A\langle f,f\rangle&=\langle Sf,f\rangle\\
\Longleftrightarrow \langle (A\cdot I-S)f,f\rangle&=0.
\end{align}
Since $A\cdot I-S$ is self-adjoint, it follows from this post that $A\cdot I -S=0$.
Edit:
For the first equivalence, we use the definition of the norm, i.e., $\|f\|^2=\langle f,f\rangle$ and the definition of the frame operator $S$. For any $f$, we have that $Sf=\sum_{j\in J}\langle f,f_j\rangle f_j$ and therefore
\begin{align}
\langle Sf,f\rangle &= \bigg\langle \sum_{j\in J} \langle f,f_j\rangle f_j,f\bigg\rangle\\
&=\sum_{j\in J} \langle f,f_j\rangle\cdot \langle f_j,f\rangle\\
&=\sum_{j\in J} \langle f,f_j\rangle \cdot \overline{\langle f,f_j\rangle}\\
&=\sum_{j\in J} |\langle f,f_j\rangle|^2.
\end{align}
Edit 2:
\begin{align}
A\langle f,f\rangle&=\langle Sf,f\rangle\\
\Longleftrightarrow A\langle f,f\rangle-\langle Sf,f\rangle&=0\\
\Longleftrightarrow \langle Af-Sf,f\rangle&=0\\
\Longleftrightarrow \langle (A\cdot I-S)f,f\rangle&=0
\end{align}
A: Here's an argument working in finite dimensions.
Let $S$ be the frame operator associated with a frame $(v_k)_k\subseteq V$. By definition, this is a linear operator such that
$$\langle v,Sv\rangle = \sum_k |\langle v_k,v\rangle|^2, \qquad \forall v\in V. $$
We can furthermore easily see that $S$ is a Hermitian operator.
It thus has an orthonormal basis of eigenvectors with real eigenvalues, and from standard linear algebra we know that
$$\lambda_{\rm min}(S) \|v\|^2 \le \langle v, Sv\rangle \le \lambda_{\rm max}(S) \|v\|^2,$$
with the bounds saturated by the eigenvectors corresponding to smallest and largest eigenvalues, $\lambda_{\rm min}(S)$ and $\lambda_{\rm max}(S)$, respectively.
Given the definition of frame bounds, this observation is sufficient to conclude that if an Hermitian operator $S$ is the frame operator of a frame, then the optimal frame bounds must be smallest and largest eigenvalues.
It also follows immediately that if the frame is tight, which means that the frame bounds are equal, then $\lambda_{\rm min}(S)=\lambda_{\rm max}(S)$, hence $S=\lambda I$ for some $\lambda\in\mathbb{R}$.
