Frame operator

Remarks

$S$ is bounded, self-adjoint and positive, since $\ip{Sf}{f} = \sum_i \abs{\ip{f}{f_i}}^2$. The frame inequality reads $A I \le S \le B I$, so $S$ is invertible with $\norm{S^{-1}} \le A^{-1}$. The family $(S^{-1} f_i)_{i \in I}$ is the canonical dual frame, and $f = \sum_i \ip{f}{S^{-1} f_i} f_i = \sum_i \ip{f}{f_i} S^{-1} f_i$ for all $f$. With the analysis operator $T f = (\ip{f}{f_i})_{i \in I} \in \ell^2(I)$, one has $S = T^* T$. A frame is tight with bound $A$ exactly when $S = A I$.

Example: for an orthonormal basis $(e_i)$, $S f = \sum_i \ip{f}{e_i} e_i = f$, so $S = I$.

Example: in $\C^2$ with the frame $(e_1, e_1, e_2)$, $S(x, y) = 2x e_1 + y e_2$, so $S = \mathrm{diag}(2, 1)$, with spectrum $\set{1, 2}$ equal to the optimal frame bounds.

Non-example: in $\ell^2(\N)$ the family $(e_n / n)_{n \ge 1}$ satisfies only the upper bound. The map $f \mapsto \sum_n \ip{f}{e_n/n} e_n/n$ is still defined, but it sends $e_n$ to $e_n/n^2$, so it has no bounded inverse; the family is not a frame and the reconstruction formula fails.