Show the subspace of finite rank operators (defined on a Hilbert space) is dense
in the space of all compact operators.
I just started reading about compact operators and I saw this questions.
So far I learnt any finite rank operator on a Hilbert space is compact, to show it's dense in space of all compact operators, there must exist a sequence of finite rank operator $(T_n)$ such that it converges to compact operator $T$, but I'm not sure how I can construct this sequence, I may need to think about complete orthogonal sequence $\{e_1\:,e_2\dots\} $ of the Hilbert spaces, so I get
$$T(x)=\sum_{i=1}^{\infty}\langle T(x),e_i\rangle e_i$$
If I define $T_n(x)=\sum_{i=1}^{n}\langle T(x),e_i\rangle e_i$, this is off course a finite dimensional and compact operator. Can I say it's the sequence that converges to $T$? Or maybe I'm totally wrong, but I appreciate, if you give me some hints about this.
Best Answer
Here's a proof I learned from some notes of Richard Melrose. I just noticed another answer was posted while I was typing. This uses a different characterization of compactness so hopefully it is interesting for that reason.
First, I claim that a set $K\subset H$ of a Hilbert space is compact if and only if it is closed, bounded, and satisfies the equi-small tail condition with respect to any orthonormal basis. This means that given a basis $\{e_k\}_{k=1}^\infty$, for any $\varepsilon>0$ we can choose $N$ large enough such that for any element $u\in H$, $$\sum_{k>N} |\langle u , e_k \rangle |^2 < \varepsilon.$$
The main point here is that this condition ensures sequential compactness. The proof is a standard "diagonalization" argument where you choose a bunch of subsequences and take the diagonal. This is done in detail on on page 77 of the notes I linked.
With this lemma in hand, the proof is straightforward. I repeat it from page 80 of those notes. Fix a compact operator $T$. By definition [this is where we use a certain characterization of compactness] the image of the unit ball $T(B(0,1))$ is compact. Then we have the tail condition that, for given $\varepsilon$, there exists $N$ such that
$$\sum_{k>N} |\langle Tu , e_k \rangle |^2 < \varepsilon.$$ for any $u$ such that $\| u \| < 1$.
We consider the finite rank operators
$$T_nu = \sum_{k\le n} \langle Tu , e_k \rangle e_k.$$
Now note the tail condition is exactly what we need to show $T_n \rightarrow T$ in norm. So we're done.