Proof on sequence of finite-rank operators I had difficulties figuring out the solutions to the following problem:

Let $H$ be a separable infinite dimensional Hilbert space. Let
{${L_n}$} be a sequence of linear finite rank operators from $H$ to
into itself such that {${L_n}x$} converges in $H$ for any $x \in H$.
Prove that:
i) there exist $L: H \rightarrow H$ linear and bounded such that $L_nx
 \rightarrow Lx$ for any $x \in H$.
ii) $L$ can be either compact or not by showing and discussing two
examples.

As an idea (maybe wrong idea) I propose my solution concerning point (i):
Fixed $x$, we have that {${L_n}x$} is convergent; it follows that $||{L_n}x|| < \infty$ for any $n$.  Consider now n being fixed. We have that  $||{L_n}x|| < \infty$ for any $x \in H$, hence it follows that $L_n$ is linear and bounded for any n (is this correct?). The solution to point one follows by applying the uniform boudedness principle (or Banach-Steinhaus theorem) using the fact that $lim_n  (L_nx)$ is well defined for all $x \in H$.
Concerning point (ii): I would figure out a $L$ bounded using the argumentation of the point (i), and on the other hand I would construct a $L$ compact by using the fact that we can find a compact operator by summing $L_n$ (using the approximation propriety of compact operators). Probably this answer is a total mess.
Thanks!
 A: (i) $L_n$ are finite rank, so by definition are bounded. Since $L_nx$ converges, let $L(x):=\lim_{n\to\infty}L_nx$. That $L$ is linear is easy to verify.
Also, $\|L_nx\|\le c_x$ since $L_nx$ is a convergent sequence. Thus by the uniform boundedness theorem, $\|L_n\|\le c$ independent of $n$. Taking the limit of $L_nx\to L(x)$ then shows $L$ is bounded $$\|Lx\|=\lim_{n\to\infty}\|L_nx\|\le c\|x\|$$
(ii) For an example where $L$ is compact, take $H=\ell^2$ with $L_N(a_n):=(a_n/n)$ for $n\le N$ and $0$ otherwise. It converges to a compact operator.
For an example where $L$ is not compact take $L_N(a_n):=(a_n)$ for $n\le N$ and $0$ otherwise. It converges pointwise to the identity operator $L=I$ which is not compact.
A: Yes, Uniform Boundedness Principle give the first part.
Take $L_n$ independent of $n$ to get an example where the limit is compact.
Let $(e_n)$ be an orthonormal basis for $H$. Let $ P_n$ be the projection on the span of $\{e_1,e_2,...,e_n\}$. Then $P_nx \to Ix$  for every $x$ and the identity operator $I$ is never compact on an infinite dimensional space.
