Is a sequence a frame iff it spans the space? Let $(v_k)_k\subseteq V$ be a sequence of vectors in some vector space $V$ which form a frame. This means that there are constants $0< A\le B<\infty$ such that, for all $v\in V$, we have
$$A \| v\|^2 \le \sum_k |\langle v_k,v\rangle|^2 \le B \|v\|^2.$$
In (Casazza 2013) (Link to pdf) the authors mention (Lemma 2 in page 14) that a family of vectors is a frame iff they are a spanning set for the vector space.
Very briefly, the reported proof is as follows:

*

*If $(v_k)_k$ does not span $V$, then there is some $x\in V$ such that $\langle v_k,x\rangle=0$ for all $k$, hence we don't have a lower frame bound and thus we don't have a frame.

*If $(v_k)_k$ is not a frame, then there is a sequence $(x_n)_n$ of normalised vectors such that $\sum_k |\langle v_k,x_n\rangle|^2<1/n$. Hence the limit $x$ of a convergent subsequence of $(x_n)_n$ satisfies $\langle x,v_k\rangle=0$ for all $k$, hence $(v_k)_k$ does not span the set.

While the first part is fine, I have some issues with the second side of the proof. More specifically, the authors seem to be saying that if $(v_k)_k$ is not a frame, then there's no lower frame bound, hence they show how to find a vector not in the span of the sequence.
But why can we make this implication? Couldn't it be possible that $(v_k)_k$ is not a frame for lack of an upper frame bound instead? That is, a situation where there is a lower frame bound $A>0$, but no finite upper $B$.
If this is in fact an actual imprecision in the proof, what's a good alternative way to prove this fact (assuming the statement itself does indeed hold)?
 A: Let $V$ be a finite dimensional vector space with complex inner product $\langle \cdot, \cdot \rangle$.
Remark: In the paper, a frame is a finite collection of vectors satisfying your property.
Let $(v_k)_{k=1}^m =: \mathcal{V} \subseteq V$ be a subset of vectors. Define the corresponding "frame map"
$$T_{\mathcal{V}} : V \rightarrow \mathbb{R}, \ \ T_{\mathcal{V}}(x) := \sum_{k=1}^m |\langle v_k, x\rangle|^2$$
I'll abbreviate this map as $T$.
We say that $\mathcal{V}$ is a frame if there are constants $0 < A \leq B$ so that
$$ A \|v\|^2 \leq T(v) \leq B \|v\|^2.$$
Goal: $\mathcal{V}$ is a frame iff $\text{Span}(\mathcal{V}) = V.$
Your argument works if we can show the following lemma.
Lemma: If $\mathcal{V}$ is not a frame, then $\inf \{ T(x) \ | \ \|x\| = 1\} = 0.$
Proof: Notice by Cauchy-Schwarz that if $\|x\| = 1$ then
$$ T(x) = \sum_{k=1}^m |\langle v_k, x \rangle|^2 \leq \sum_{k=1}^m \|v_k\|^2.$$
Set
$$B := \sum_{k=1}^m \|v_k\|^2 < \infty.$$
Then for any $v \in V$ we have
$$ B \geq T\left(\frac{v}{\|v\|}\right) = \frac{T(v)}{\|v\|^2} \implies T(v) \leq B \|v\|^2.$$
So the only way that $\mathcal{V}$ can fail to be a frame is if $\inf\{T(x) \ | \ \|x\| =1 \} = 0,$ i.e. we cannot find an $A > 0$. $\blacksquare$
Remark: Notice this lemma fails when dealing with infinite frames.
