Finite rank nilpotent operators Let $H$ be a separable complex Hilbert space
Let $\mathcal{F(H)} =\{T \in \mathcal{B(H)}: \text{rank}(T) < \infty \}$ ,
and $\mathcal{N(H)} = \{ T \in \mathcal{F(H)}: T^k=0 \text{ for some }  k \in \mathbb{N} \}$
Find the norm closure $\overline{\mathcal{N(H)}}$.
Ideas:
I know that $\overline{\mathcal{F(H)}} = \mathcal{K(H)}$, where $\mathcal{K(H)}$ is the set of compact operators.
So the closure will be some compact operators.
And $\mathcal{N(H)} \subset \mathcal{K(H)} $.
Also, through some research, I've found that the quasinilpotnets are limits of nilpotent.
So I'm suspecting the that the norm close of finite rank nilpotents will be compact and quasinilpotents.
Any help will be appreciated!
Thank you in advance!
 A: Let $T$ be quasi-nilpotent and compact. Then $T\subset \overline {\mathcal N(H)}$. (This is the proof of Douglas mentioned in the Hilbert space problems paper by Halmos).
First, the spectrum is upper semicontinuous: For every $\epsilon>0$ there is $\delta>0$ such that $\sigma(S) \subset B_\epsilon(0)$ for all $S$ with $\|S-T\|<\delta$. (This can be seen as follows: If $|\lambda|\ge\epsilon$ then $\lambda I-T$ is continuously invertible by $\sigma(T)=\{0\}$, and so is $\lambda I-S$ for $S$ close to $T$.)
Take $\epsilon>0$ and let $\delta>0$ be as above.
Since $T$ is compact and $H$ is Hilbert, there is a finite-rank operator $S$ with $\|T-S\|<\delta$.
Since $S$ has finite-rank, we can triangularize $S=D+N$, where $D$ is diagonal and $N$ is nilpotent:
Write $S(x) = \sum_{i=1}^n \langle x,e_i\rangle f_i$, then build the complex Schur form of the matrix representation of $S$ on $span(e_1\dots e_n,f_1\dots f_n)$. Note that the resulting coordinate transform is unitary.
Since the diagonal $D$ contains the non-zero eigenvalus of $S$, it follows $\|D\|<\epsilon$, where we used again that the underlying basis is orthonormal. And $\|T-N\|< \|T-S+D\|<\epsilon+\delta$.
This proves one inclusion. I do not know how to prove the reverse.
