Let $\mathcal{H}$ be an infinite dimensional Hilbert space and $\{e_i\}_{i=1}^{N}\subseteq \mathcal{H}$ be an arbitrary finite elements in $\mathcal{H}$ such that $||e_i||=1$ for all $i$. Define the operator $C:\mathcal{H} \rightarrow \ell^2(\mathbb{N})$ via $f \mapsto \{ \langle f, e_i \rangle \}_{i=1}^N$; in frame theory, this is called the synthesis operator. If we compute the operator norm of $C$, we get $$||C||^2=\sup_{||f||=1}||Cf||^2=\sum_{i=1}^{N}\sup_{||f||=1}|\langle f,e_i \rangle|^2 =\sum_{i=1}^{N}||e_i||^2 = N.$$ Therefore we have $||C|| = \sqrt{N}$. I can do a similar computation to show that the frame operator $S=:C^*C = \sum_{i=1}^{N}\langle f,e_i \rangle e_i$ has the same operator norm using the self-adjointness of $S$. Therefore, we every $f\in \mathcal{H}$
\begin{align} ||f|| = &||f-Sf+Sf|| \\ \leq & ||I-S|| \cdot ||f|| + ||Sf|| \\ \end{align}
\begin{align} \implies (1-||I-S||)^2||f||^2 \leq ||Sf||^2 = \sum_{i=1}^{N}|\langle f, e_i \rangle |^2. \end{align} Thus if $||I-S|| \neq1$, we have a lower frame bound for the infinite dimensional Hilbert space $\mathcal{H}.$ We can compute this norm (note that $S$ is a positive operator):
\begin{align} ||I-S||^2 = & \sup_{||f||=1}||f-Sf||^2 \\ =& \sup_{||f||=1}(||f||^2 -2\langle f, Sf \rangle + ||Sf||^2) \\ =& 1-2||S||+||S||^2 \\ =&(1-||S||)^2 \\ =& (1-\sqrt{N})^2 \end{align} Therefore $||I-S|| = |1-\sqrt{N}|$, so if we just choose $N$ such that $N \neq 4$, we have a lower frame bound for the Hilbert space $\mathcal{H}$, but this would imply that we have a finite span for $\mathcal{H}$, and since we assumed that $\mathcal{H}$ is infinite dimensional, then we have a contradiction. Therefore I must have made a wrong step along the way with this 'proof'. Can anybody help me here?
$\sup_{\|f\|=1}\displaystyle\sum_{i=1}^{n}|\left<f,e_{i}\right>|^{2}\ne\sum_{i=1}^{n}\sup_{\|f\|=1}|\left<f,e_{i}\right>|^{2}$.
This is something like, $\sup_{a\in A}[f(a)+g(a)]\ne\sup_{a\in A}f(a)+\sup_{a\in A}g(a)$. In general, we only have $\sup_{a\in A}[f(a)+g(a)]\leq\sup_{a\in A}f(a)+\sup_{a\in A}g(a)$.