Inequalities that need only be checked on a dense subspace 
Let $\left\{x_n\right\}$ be a sequence in a Hilbert space $H$ and
  suppose there exist $0<A\leq B$ such that  $$A\|x\|^2\leq
 \sum_{n=1}^{\infty}\left|\left\langle x,x_n\right\rangle\right|^2\leq
 B\|x\|^2,\qquad \forall x\in D $$ where $D$ is a dense subspace of
  $H$. Prove that the above inequalities are satisfied for all $x\in H$,
  i.e. $\left\{x_n\right\}$ is a frame for $H$.

It seems like this statement is completely obvious to everyone except for me. I hope someone can enlighten me or I will not be able to sleep.
Attempt (following the given hint). The operator $C:D\to \ell^2$, $x\mapsto (\left\langle x,x_n\right\rangle)_{n\in \mathbb{N}}$ is such that $\|Cx\|^2=\sum_{n=1}^{\infty}\left|\left\langle x,x_n\right\rangle\right|^2\leq B\|x\|^2$, so $C$ is bounded in $D$ with $\|C\|\leq \sqrt{B}$. Hence it can be extended to a unique bounded linear operator $\tilde{C}$ on $H$. However, can I deduce that this extension $\tilde{C}$ is such that $\tilde{C}x=(\left\langle x,x_n\right\rangle)_{n\in \mathbb{N}}$ for all $x\in H$? I cannot use uniqueness of the bounded extension since a priori the map $x\mapsto (\left\langle x,x_n\right\rangle)$ is not bounded on $H$. How do I exclude the possibility that the map $x\mapsto (\left \langle x,x_n\right\rangle)_{n\in \mathbb{N}}$ is bounded on $D$ yet unbounded on $H$?
 A: It is sufficient to assume that $D$ is a dense subset of $H$, not neccesarily a subspace.
Suppose $\exists x_0 \in H$ such that $\sum_{n=1}^\infty \left|\langle x_0, x_n\rangle\right|^2 > B\|x_0\|^2$. Pick $\varepsilon > 0$ such that $\sum_{n=1}^\infty \left|\langle x_0, x_n\rangle\right|^2 > B\|x_0\|^2 + \varepsilon$.
There exists $n_0 \in \mathbb{N}$ such that $\sum_{n=1}^{n_0} \left|\langle x_0, x_n\rangle\right|^2 > B\|x_0\|^2$. The map $$y \mapsto \sum_{n=1}^{n_0} \left|\langle y, x_n\rangle\right|^2 - B\|y\|^2$$
is continuous on $H$ so there exists $\delta > 0$ such that $\|x_0 - y\| < \delta$ implies $$\left|\left(\sum_{n=1}^{n_0}\left|\langle x_0, x_n\rangle\right|^2 - B\|x_0\|^2\right) - \left(\sum_{n=1}^{n_0}\left|\langle y, x_n\rangle\right|^2 - B\|y\|^2\right)\right| < \varepsilon$$
In particular, $D$ intesects the ball $B(x_0, \delta)$ so $\exists y \in D$ such that $$\sum_{n=1}^\infty \left|\langle y, x\rangle\right|^2 - B\|y\|^2 \ge  \sum_{n=1}^{n_0}\left|\langle y, x\rangle\right|^2 - B\|y\|^2 > \sum_{n=1}^{n_0}\left|\langle x_0, x_n\rangle\right|^2 - B\|x_0\|^2 - \varepsilon > 0$$
which is a contradiction.
Therefore $\sum_{n=1}^\infty \left|\langle x, x_n\rangle\right|^2 \le B\|x\|^2, \forall x \in H$.
Hence, the map $U : H \to \ell^2$ given by $Ux = (\left \langle x,x_n\right\rangle)_{n\in \mathbb{N}}$ is bounded on $H$.
The map $x \mapsto \|Ux\|_2^2 - A\|x\|^2 = \sum_{n=1}^\infty \left|\langle x, x_n\rangle\right|^2 - A\|x\|^2$ is then surely continuous on $H$.
Furthermore, it is $\ge 0$ on a dense subset $D$, so it is also $\ge 0$ on all of $H$.
Therefore $\sum_{n=1}^\infty \left|\langle x, x_n\rangle\right|^2 \ge A\|x\|^2, \forall x \in H$.
