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Given a measure space $(X, \mu)$ and a measurable integral kernel $k : X \times X \rightarrow \mathbb{C}$, the operator $$ K f(\xi) =\int_{X} f(x) k(x,\xi) d \mu(x),$$ the operator $K$ is Hilbert Schmidt iff $k \in L^2(X \times X, \mu \otimes\mu)$!

Q1:The main point of this questions, what are necessary and sufficient conditions for it to be trace class?

I know various instances, where $$ \mathrm{tr} K = \int_X k(x,x) d \mu(x).$$

Q2:What are counterexamples, where $x \mapsto k(x,x)$ is integrable, but the operator is not trace class?

Q3:What are counterexamples for a $\sigma$ finite measure space, where $k$ is compactly supported and continuous, but the kernel transformation is not trace class and the above formula fails?

Q4: Is there a good survey/reference for these questions.

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2 Answers 2

up vote 5 down vote accepted

There are many results of the kind you ask about in the book

I. C. Gohberg and M. G. Krein, Introduction to the theory of linear nonselfadjoint operators. Providence, RI: American Mathematical Society, 1969.

It contains both necessary and sufficient conditions, and counter-examples.

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Okay, there is a chapter "Tests for nuclearity of integral transforms and computation of the trace" pg.112ff. Thanks a lot. I will have to check, wether this does the complete job here. –  Marc Palm Jul 22 '11 at 8:07
    
This only for an interval in the real line, though=( –  Marc Palm Jul 22 '11 at 9:11
1  
@pm : there are many $(X,\mu)$ measurably isomorphic to an interval of the real line. See "standard probability space" in wikipedia. Of course, this is not good when you consider continuity of the kernel... –  BS. Jul 22 '11 at 16:19
    
When is a group with Haar measure a standard measure space? –  Marc Palm Jul 25 '11 at 8:10
    
A group with Haarmeasure is a standard measure space, if and only if the group is a Polish group. In fact, Polish group with $G$ quasi invariant measures are locally compact. –  Marc Palm Sep 9 '11 at 9:51

It may be worth noting the phenomena that can appear in Hilbert spaces, where study of the things is more decisive, both positive and negative.

First, I like the "definition" of "trace class" $T:X\rightarrow Y$ with Hilbert spaces $X,Y$ to be that $T$ is a composition of two Hilbert-Schmidt operators (which are defined as being in the HS-norm completion of the algebraic tensor product $X^*\otimes_{\mathrm {alg}}Y$. This gives an intrinsic definition... which, if desired, is provably equivalent to the (ugly) requirement that $\sum |\langle Tx_i,y_i\rangle| <\infty$ for every pair of orthonormal bases.

The reason I recall this cliche is that, in many applications of interest (to me!), natural operators are visibly Hilbert-Schmidt (if compact at all), and the issue becomes to prove trace-class. In practice (for me) it often happens that we know that every one of these integral operators is a finite sum of compositions of two such, proving trace-class.

Sometimes proof of the latter is highly non-trivial, as in the Cartier/Dixmier-Malliavin proof that test functions on Lie groups are finite linear combinations of convolutions of pairs of such. The totally-disconnected group analogue is trivial.

That summing or integrating down the diagonal fails is easy to illustrate with not-normal operators: the shift operator on one-sided or two-sided $\ell^2$ might seem to have trace absolutely summing to $0$, but it is not trace class at all. Integral analogues of this are clear.

Edit: in response to question about reference, etc.: in Lang's "SL(2,R)" the equivalence of the coordinate-dependent definition of "trace class", and the definition as composition of two Hilbert-Schmidt, are carefully compared. Further, in that same source, various conditions on a kernel assuring that its trace is equal to its integral over the diagonal are carefully treated. (I must say "... in contrast to dangerously glib treatments elsewhere").

Further edit: in response to Yemon Choi's comments: yes, the space of trace-class operators is also the closure of finite-rank operators with respect to the "trace norm"... At the moment, verification of the equivalence seems straightforward.

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Some guidance as to the purported unhelpfulness of this answer would be appreciated, if not inconvenient. –  paul garrett Jul 22 '11 at 17:45
    
I'd also prefer an explanation for downvotes here. –  Marc Palm Jul 25 '11 at 8:09
    
@Paul: I just upvoted this response of yours. Some references to precise theorems (e.g. to your alternative definition of trace class) would be helpful. –  GH from MO Jan 29 '12 at 21:38
    
Why not define trace class as those in the range of the map $H\widehat{\otimes} H^* \to B(H)$, given cokernel norm? –  Yemon Choi Feb 4 '12 at 2:09
    
That said, I find the observation in your third paragraph, and the results mentioned in the fourth paragraph, quite interesting; so I am not trying to deny the merits of the definition you suggest –  Yemon Choi Feb 4 '12 at 2:52

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