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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.

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  • $\begingroup$ You can choose the orthonormal basis of eigenvectors of $T$. $\endgroup$ – C.Ding Jul 22 '17 at 2:49
  • $\begingroup$ @C.Ding Can I find a sequence of eigenvectors which forms a basis for the Hilbert space? $\endgroup$ – Parisina Jul 22 '17 at 2:52
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    $\begingroup$ Yes, you can, if $T$ is a normal compact operator. $\endgroup$ – C.Ding Jul 22 '17 at 3:01
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    $\begingroup$ @C.Ding but it's not talking about only normal compact operators! $\endgroup$ – Parisina Jul 22 '17 at 3:08
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    $\begingroup$ But you only have to show normal compact operator is in the space of finite rank operators, since $T$ can be represented as $T=R+iS$ where $R$ and $S$ are hermitian. $\endgroup$ – C.Ding Jul 22 '17 at 3:13
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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.

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This answer comes from Murphy's book $C^*$ algebra and operator theory(Theorem 2.4.5).

It suffices to show that if $T$ is a hermitian compact operator, then $T\in F(H)^-$(see comments above),where $F(H)$ is the set of all finite rank operators.

Let $E$ be an orthonormal basis of $H$ consisting of eigenvectors of $T$, and let $\varepsilon>0$. We know taht the set $S$ of eigenvalues $\lambda$ of $T$ such that $|\lambda|>\varepsilon$ is finite (Since $\ker(T-\lambda)$ is finite-dimensional for any $\lambda\neq 0$). It is therefore that the set $S'$ of elements of $E$ corresponding to elements of $S$ is finite.

Now define a finite-rank diagonal operator $u$ on H by setting $u(x) = \lambda x$ if $x\in S'$ and $\lambda$ is the eigenvalue corresponding to $x$, and setting $u(x) = 0 $ if $x \in E \backslash S'$. It is easily checked that $\lVert u — T\rVert \leq \sup_{\lambda\in \sigma(T)\backslash S }|\lambda|\leq\varepsilon$. This shows that $T\in F(H)^-$.

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  • $\begingroup$ if we don't talk about complex field, just consider when the compact operator $T=S+i0$, where $S$ is Hermitian. By Hermitian do you mean $\bar{T^*}=T$?Doesn't it mean $T$ is selfadjoint? $\endgroup$ – Parisina Jul 22 '17 at 19:09
  • $\begingroup$ @Parisina here, hermitian = selfadjoint, replace 'hermitian' with 'selfadjoint' as you want. $\endgroup$ – C.Ding Jul 22 '17 at 19:43
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This is indeed the idea. My understanding is you need to assume the Hilbert space in question is separable, though I'm not familiar with any counterexamples.

Suppose $\{e_{i}\}_{i \in \mathbb{N}}$ is a complete orthonormal system. For each $N \in \mathbb{N}$, consider the operator $T_{N}$ defined by $$T_{N}(x) = \sum_{j = 1}^{N} \langle T(x), e_{j} \rangle e_{j}.$$
$T_{N}$ has range contained in $\text{span}\{e_{1},\dots,e_{N}\}$ so it's a finite rank operator.

Recall that $e_{i} \rightharpoonup 0$. Since $T$ is compact, it follows that $T(e_{i}) \to 0$ in norm.

We claim $\lim_{n \to \infty} \|T_{n} - T\| = 0$. Pick $\epsilon > 0$. By the preceding remark, we can find $\mathcal{I} \in \mathbb{N}$ such that $$\|T(e_{i})\|^{2} < \epsilon^{2}$$ if $i \geq \mathcal{I}$. Moreover, since $\{1,2,\dots,\mathcal{I} - 1\}$ is finite, we can find $N \in \mathbb{N}$ such that $$\sum_{j = n + 1}^{\infty} |\langle T(e_{i}), e_{j} \rangle|^{2} < \epsilon^{2}$$ if $i \in \{1,2,\dots,\mathcal{I}\}$ and $n \geq N$.

Now suppose $x$ is any vector in the Hilbert space. If $n \geq N$, then \begin{align*} \|(T - T_{n})(x)\|^{2} &= \sum_{j = n + 1}^{\infty} |\langle T(x), e_{j} \rangle|^{2} \\ &= \sum_{i = 1}^{\infty} |\langle x, e_{i} \rangle|^{2} \left(\sum_{j = n + 1}^{\infty} |\langle T(e_{i}),e_{j}\rangle|^{2}\right) \\ &\leq \sum_{i = 1}^{\mathcal{I} - 1} |\langle x,e_{i} \rangle|^{2} \left(\sum_{j = n + 1}^{\infty} |\langle T(e_{i}),e_{j} \rangle|^{2}\right) + \sum_{i = \mathcal{I}}^{\infty} |\langle x,e_{i} \rangle|^{2} \|T(e_{i})\|^{2} \\ &\leq \epsilon^{2} \sum_{i = 1}^{\infty} |\langle x,e_{i} \rangle|^{2} \end{align*} by the choice of $\mathcal{I}$ and $N$. From this, we deduce that $\|T - T_{n}\| \leq \epsilon$. Since $\epsilon$ was chosen arbitrarily, we conclude $\lim_{n \to \infty} \|T - T_{n}\| = 0$.

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    $\begingroup$ No, there is no need to assume $H$ is separable. $\endgroup$ – C.Ding Jul 22 '17 at 3:28
  • $\begingroup$ your second inequality is far away from being correct. $\endgroup$ – Lin Xuelei Oct 19 '18 at 15:56
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We can reduce to the separable case since $T\in \mathcal{L}_c(X)$ compact implies that $X = Ker(T)\oplus \overline{Ran(T)}$ and $\overline{Ran(T)}$ is separable since $T$ is compact.

Fix $\{e_k\}_{k\in \mathbb{N}}$ Hilbert basis of $X$, let $P_n $ be the metric projector onto $Span(e_1,\dots, e_n)$, i.e. $P_n(x) = \sum_{k=1}^n \langle x, e_k\rangle e_k$. I say that $P_n\circ T\to T$ in operator norm. Indeed $$||P_n\circ T - T||_{\infty, B_X} = ||P_n - I||_{\infty, \overline{T(B_X)}} \to 0$$ where the last convergence is to be intended as strong operator convergence (we have this convergence since $\{e_k\}_{k\in \mathbb{N}}$ is an Hilbert basis). But since $\{P_n - I\}_{n\in \mathbb{N}}$ is an equicontinuos family and $\overline{T(B_X)}$ is compact, strong operator convergence implies norm convengence. Obviously the $P_n\circ T$ are finite rank operators so we are done.

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