Convergence under a Hilbert space Let $\{\varphi_n\}_{n=1}^\infty$ be an orthonormal sequence (not necessarily a basis) in a Hilbert space. Let  $\{\lambda_n\}_{n=1}^\infty$ be a sequence of numbers
Define $T:H\to H$ by $Tx=
\sum_{n=1}^{\infty} \lambda_n\langle x, \varphi_n\rangle \varphi_n$
Show that if $sup_n|λ_n| < \infty$ then $\sum_{n=1}^{\infty} \lambda_n\langle x, \varphi_n\rangle \varphi_n$ converges for all
$x \in H$.
Well I have no clue for where to begin. Does it have something to do with Gram–Schmidt process?
 A: Hint:

Since $(\varphi_n)_{n=1}^{\infty}$ is an orthogonal sequence, the expression
  $$
\sum_{n=1}^{\infty}|\langle x, \varphi_n\rangle |^2
$$
  always converges. This follows from Bessel's inequality.

Using the fact that $\quad\sup_n|\lambda_n|<\infty\quad$, there's an $M>0$ such that $|\lambda_n|<M$ for all $n\in\Bbb N$.
It is not hard to see that 
$$
\sum_{n=1}^{\infty}M^2|\langle x, \varphi_n\rangle |^2
$$
is convergent for any $x\in H$. What does this say about the convergence of $\sum_{n=1}^{\infty}|\lambda_n\langle x, \varphi_n\rangle |^2$?

There's a theorem saying that for any orthonormal sequnce $(e_n)_{n=1}^{\infty}$,
  $$
\sum_{n=1}^{\infty}|\langle x, e_n\rangle |^2 \quad\text{converges iff}\quad \sum_{n=1}^{\infty}\langle x, e_n\rangle e_n\quad\text{converges.}
$$

Now all you have to do is put them all together.
A: Let $Tx=\sum_{n=1}^\infty \lambda_n (x, \varphi_n)_H \varphi_n$, then $T_N x \rightarrow T$ in $\mathcal{H}$, where $T_Nx=\sum_{n=1}^N \lambda_n (x,\varphi_n)_H \varphi_n$. By Bessel inequality we have
$\displaystyle \sum_{n=1}^N |(x,\varphi_n)_H|^2 \leq ||x||^{2}_H$
it's is valid for any orthonormal system. Therefore
$\displaystyle ||(T-T_N)x||_{H}^2 = \sum_{n > N} |\lambda_n(x, \varphi_n)_H|^2 \leq \sup_{N\in \mathbb{N}} |\lambda_N|^2 \sum_{n >N} |(x,\varphi_n)_H|^2 \leq \sup_{N \in \mathbb{N}}|\lambda_N|^2$
A: It seems you mean that $T$ is well-defined. Let $x\in H$,
$$\|\sum_m^n \lambda_i\langle x,\phi_i\rangle\phi_i\|^2=\sum_m^n |\lambda_i\langle x,\phi_i\rangle|^2\leq(\sup_{n\geq1} \lambda_n)\sum_m^n |\langle x,\phi_i\rangle|^2\to0$$ 
