Why $\overline {\operatorname{span}} \left\{ f_k \right\}_{k = 1}^\infty = \cal H$?

Question

If ${\left\{ f_k \right\}_{k = 1}^\infty }$ be a frame for Hilbert space ${\cal H}$, then

$$\overline {\operatorname{span}} \left\{ {{f_k}} \right\}_{k = 1}^\infty =\cal H.$$

Why?

Answer 1

Let $\mathcal{V} := \{f_k\}_{k = 1}^{\infty}$ be a frame for a Hilbert space $\mathcal{H}$. Then there exists, in particular, an $A>0$ such that, for all $ f \in \mathcal{H}$, $$ A \| f \|^2 \leq \sum_{k=1}^{\infty} |\langle f, f_k \rangle |^2. $$

Let $f \in \mathcal{H}$ be such that $ f \perp f_k$ for all $k$, i.e., let $f \in \mathcal{V}^{\perp}$. Then $\sum_{k=1}^{\infty} |\langle f, f_k \rangle |^2 = 0$, and hence $ f = 0$. Thus $ \mathcal{V}^{\perp} = \{0 \}$, which implies that $ \mathcal{V} $ is total in $\mathcal{H}$.

Answer 2

The lower bound in the frame condition implies that every vector has at least one nonzero orthogonal projection onto an element of your frame. This implies that the orthogonal complement of the span is $\{ 0\} $, and so the span is dense.