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.


3 Answers 3


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.

  • $\begingroup$ 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. $\endgroup$
    – Marc Palm
    Commented Jul 22, 2011 at 8:07
  • $\begingroup$ This only for an interval in the real line, though=( $\endgroup$
    – Marc Palm
    Commented Jul 22, 2011 at 9:11
  • 2
    $\begingroup$ @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... $\endgroup$
    – BS.
    Commented Jul 22, 2011 at 16:19
  • $\begingroup$ When is a group with Haar measure a standard measure space? $\endgroup$
    – Marc Palm
    Commented Jul 25, 2011 at 8:10
  • $\begingroup$ 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. $\endgroup$
    – Marc Palm
    Commented Sep 9, 2011 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.

  • 1
    $\begingroup$ Some guidance as to the purported unhelpfulness of this answer would be appreciated, if not inconvenient. $\endgroup$ Commented Jul 22, 2011 at 17:45
  • 1
    $\begingroup$ I'd also prefer an explanation for downvotes here. $\endgroup$
    – Marc Palm
    Commented Jul 25, 2011 at 8:09
  • $\begingroup$ @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. $\endgroup$
    – GH from MO
    Commented Jan 29, 2012 at 21:38
  • $\begingroup$ Why not define trace class as those in the range of the map $H\widehat{\otimes} H^* \to B(H)$, given cokernel norm? $\endgroup$
    – Yemon Choi
    Commented Feb 4, 2012 at 2:09
  • $\begingroup$ 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 $\endgroup$
    – Yemon Choi
    Commented Feb 4, 2012 at 2:52

A remark on (Q3): There is this famous example of T.Carleman (1916 Acta Math link) where he constructs a (normal ) operator with a continuous kernel such that it belongs to all Schatten p-classes if and only if $p\geq 2.$

More precisely it's possible to construct $k(x)=\sum_n c_ne^{2\pi i n x}$ continuous and periodic with $\sum_n|c_n|^p=\infty$ for $p<2$. Then $Tf=f\ast k$ acting on $L^2(\mathbb T)$ yields the desired result.

Provided some extra regularity on the kernel, the trace formula works fine (there are a lot of results in the literature)

Regarding (Q4) I personally find C. Brislawn's result very interesting but rather difficult to implement in practice.


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