Trace class for operators

Let $\mathcal{H}$ be a Hilbert space and $T: \mathcal{H} \to \mathcal{H}$ a bounded linear operator. The $n$-th singular number ${\mu_{n}}(T)$ of $T$ is defined as the distance from $T$ to the space of operators of rank at most $n$. We say that $T$ is in the trace class if $$\sum_{n} {\mu_{n}}(T) < \infty.$$ Show that in this case, $T$ is compact and if $\{ \lambda_{n} \}$ are its eigenvalues, then $$\sum_{n} |\lambda_{n}| < \infty.$$ Also, show that, in general, the converse is not true.

I have not seen this definition of ‘trace class’ before. Can anyone give me some hints? Can I approximate $T$ with finite-rank operators?

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My "guess" is: because your sum is finite, you must have $\mu_n(T)\rightarrow 0$, hence I think you can use this fact to produce the desired aproximation. –  Tomás Jan 3 '13 at 22:33
To disprove the converse, take a diagonal operator on $\ell_2$ with diagonal entries $\lambda_n$ that converge to zero, but slowly enough so that $\sum \lambda_n=\infty$. –  user53153 Jan 4 '13 at 1:31
@PavelM I dont see this contradiction... isnt it $\sum|\lambda_n| < \infty$ in the converse statement? –  Johan Jan 4 '13 at 21:41
Indeed, I misunderstood what was meant by the converse here. Maybe look at some form of Volterra operator which is compact and has no eigenvalues at all. –  user53153 Jan 4 '13 at 22:36
The operator $Te_i=\frac{e_{i+1}}{i}$, where $\{e_i\}$ is a orthonormal sequence seems to work fine as a counter example? –  Jonas Wallin Jan 5 '13 at 0:41

In order to distinguish the new definition of ${\mu_{n}}(T)$ from the old one, let us call it ${\mu^{\text{New}}_{n}}(T)$.

We shall assume throughout this discussion that $\displaystyle \sum_{n=1}^{\infty} {\mu_{n}}(T) < \infty$.

By the Divergence Test from calculus (it’s hard to believe that something so simple can crop up here!), we have $\displaystyle \lim_{n \to \infty} {\mu_{n}}(T) = 0$. Hence, for any $\epsilon > 0$, there exists an $n \in \mathbb{N}$ sufficiently large so that ${\mu_{n}}(T) < \epsilon$, which means that we can find an $F \in B(\mathcal{H})$ of rank $\leq n$ such that $\| T - F \|_{B(\mathcal{H})} < \epsilon$. Therefore, $T$ can be approximated in the operator norm by bounded operators of finite rank, making it a compact operator.

Recall that ${\mu_{n}}(T)$ is defined as the $n$-th term of the null sequence that is formed by listing the eigenvalues of the positive compact operator $|T|$ in decreasing order, taking multiplicity into account. The Minimax Principle then says that ${\mu^{\text{New}}_{n}}(T) = {\mu_{n}}(T)$ (please click here to access a set of notes on trace-class operators that contains a proof of this result; see Lemma 12). Therefore, ‘trace-class’ in the new sense is the same as ‘trace-class’ in the old sense, and so $$\text{Tr}(|T|) \stackrel{\text{def}}{=} \sum_{n=1}^{\infty} {\mu_{n}}(T) = \sum_{n=1}^{\infty} {\mu^{\text{New}}_{n}}(T).$$

For any $T \in K(\mathcal{H})$, one can find orthonormal sequences $(\mathbf{v}_{n})_{n \in \mathbb{N}}$ and $(\mathbf{w}_{n})_{n \in \mathbb{N}}$, not necessarily complete, such that $$T = \sum_{n=1}^{\infty} {\mu_{n}}(T) \langle \mathbf{v}_{n},\bullet \rangle_{\mathcal{H}} \cdot \mathbf{w}_{n}.$$ We thus obtain a more explicit approximation of $T$ by bounded operators of finite rank. This is a standard result in the theory of compact operators; please refer to the Wikipedia article on Compact Operator or to Corollary 4 of the notes mentioned above.

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Thanks! can you please expand a little on how this implies the second statement? –  Johan Jan 4 '13 at 21:42
@Johan: I’ve done a revision of my original post. The links will explain some of the details that I haven’t managed to provide. –  Haskell Curry Jan 29 '13 at 4:54
Great! thank you very much! –  Johan Jan 29 '13 at 14:40
@Johan: One more thing. We already have $\forall n \in \mathbb{N}: ~ |{\lambda_{n}}(T)| \leq {\mu_{n}}(T)$, so by the Comparison Test, the convergence of $\displaystyle \sum_{n=1}^{\infty} {\mu_{n}}(T)$ yields the convergence of $\displaystyle \sum_{n=1}^{\infty} |{\lambda_{n}}(T)|$. –  Haskell Curry Jan 29 '13 at 22:11