Linear operator determined by its action on a complete orthonormal sequence Let $(e_n)$ be a complete orthonormal sequence in the Hilbert space H and let $(\lambda_n)$ be a sequence of scalars, not necessarily bounded. I'm trying to prove that the operator defined as
$$Tv = \sum \lambda_i \langle v, e_i \rangle e_i$$
is the unique operator taking $e_n$ to $\lambda_n e_n$. Proving uniqueness is a routine exercise if I assume the series converges, but I'm having trouble with that part. I know that the series
$$\sum \langle v, e_i\rangle e_i$$
converges for all $v\in H$, because $(e_n)$ is complete, so I know I have to use the fact that the partial sums of this series form a Cauchy sequence to prove the same for the modified series. So, I pick $\epsilon$ and $N$ such that
$$||S_n - S_m|| < \epsilon$$
for all $n, m > N$, where $S_n = \sum_1^n \langle v, e_i\rangle e_i$ so that the above can be written
$$||\sum_{i = m+1}^n \langle v, e_i \rangle e_i|| < \epsilon.$$
Then it's clear that 
$$||\sum_{m+1}^n \lambda_n \langle v, e_n \rangle e_n|| < \Lambda \epsilon, $$
where $$\Lambda = \max_{m < i \leq n} (\lambda_n).$$ From here, though, I'm blanking on how to complete the proof.
 A: Something more has to be said about the domain of $T$, even if you assume $T$ is closed. $T$ is unique if you assume the following:
$$
\begin{align}
   1.\;\; & T \mbox{ is linear with } Te_{n}=\lambda_{n}e_{n} \mbox{ for all } n; \\
   2.\;\; & T \mbox{ is closed};\\
   3.\;\; & \mathcal{D}(T)=\{ v \in H : \sum_{n}|\lambda_{n}(v,e_{n})|^{2} < \infty\}.
\end{align}
$$
If you replace (3) with $\{ v \in H : \sum_{n}|\lambda_{n}(v,e_{n})|^{2} < \infty\} \subseteq \mathcal{D}(T)$, then $T$ may not be unique because there are cases where the $T$ defined by (1)-(3) has a proper closed extension. That's why you're having trouble showing uniqueness.
For example, let $H=L^{2}[0,\pi]$, and $e_{n}(x)=\sqrt{2/\pi}\sin(nx)$. Suppose $\lambda_{n}=n^{2}$ for $n \ge 1$. Let $T$ be defined by (1)-(3) above. Then $\mathcal{D}(T)$ consists of all twice absolutely continuous functions $f$ on $[0,\pi]$ with $f(0)=f(\pi)=0$, and $Tf=-f''$. This operator $T$ is a densely-defined selfadjoint operator with a proper closed extension $T_{e}$ obtained by adding the constant function $1$ to the domain. $T_{e}$ is closed. So, what goes wrong? The constant function $1$ can be written as
$$
             1 = \sum_{n=1}^{\infty}(1,e_{n})e_{n}=\sum_{k=0}^{\infty}\sqrt{\frac{2}{\pi}}\frac{2}{(2k+1)}e_{2k+1}.
$$
Even though $T_{e}1=0$ makes sense in a simple and classical way, the following does not converge:
$$
                         \sum_{n=1}^{\infty}(1,e_{n})n^{2}e_{n}.
$$
And it can't because (1)-(3) results in a selfadjoint $T$ when $\lambda_{n}$ are real, while $T_{e}$ is not selfadjoint.
