# Trace-class operator satisfies $\sum |\lambda_n|<\infty$?

Here's an "exercise" which I thought should be easy, but which I find myself unable to do.

Let $V$ be a Banach space.

Recall that an operator $f:V\to V$ is trace-class if it is in the image of the natural map $V\otimes_\pi V'\to \mathcal L(V,V)$, where $\otimes_\pi$ denotes the projective tensor product, and $V'=\mathcal L(V,\mathbb C)$ is the continuous dual of $V$.

Exercise:  [Edit: this is probably FALSE; thank you Mateusz Wasilewski]
Let $f:V\to V$ be a trace-class linear map which is "diagonalizable", in the sense that the linear span of its eigenspaces is dense in $V$. Write $\lambda_1,\lambda_2,\ldots$ for the eigenvalues of $f$ (enumerated with multiplicities). Then $$\sum |\lambda_n|<\infty.$$

Given that my previous "exercise" turned out to be probably false, here's a new one:

Exercise v2:
Let $f:V\to V$ be a trace-class linear map which is diagonalizable (in the same sense as above), and whose eigenvalues $\lambda_1,\lambda_2,\ldots$ are all positive real numbers. Then $$\sum \lambda_n<\infty.$$

And if this second exercise also too difficult, here's an even simpler one (which I also don't know how to prove):

Exercise v3:
Let $f:V\to V$ be a trace-class linear map which is diagonalizable (in the same sense as above), with eigenvalues $\lambda_1,\lambda_2,\ldots$. Then $$\lim_{n\to\infty} \lambda_n=0.$$

PS: I'm also interested in the same question when $V$ is Frechet, or when $V$ is a general complete locally convex topological vector space.

• This happens only for Hilbert spaces, see Eigenvalues of p-summing and I_p-type operators in Banach spaces by Johnson, König, Maurey and Retherford, Journal of Functional Analysis, vol. 32 (1979), 353-380. – Mateusz Wasilewski Feb 19 '18 at 0:13
• We proved that if every nuclear (what you call trace class are more usually called nuclear) operator on $X$ has absolutely summable eigenvalues, then $X$ is isomorphic to a Hilbert space. It is not clear to me that one can get a nuclear operator s.t. the eigenvectors have dense span. If you just want an example, it is natural to restrict the operator to the closed span $Y$ of the eigenvectors, but there is no reason that the resulting operator on $Y$ will be nuclear. – Bill Johnson Feb 19 '18 at 1:15
• You have an exotic use of the word "exercise"... – YCor Feb 19 '18 at 1:38
• Anyway, I was moaning about the use of "exercise" because I found it confusing: it can deter people from looking for a counterexample, thinking you know for some reason that the statement holds true. If you don't know whether a statement is true, "exercise" is not the right wording. If you're afraid that it might be "too easy", there are plenty of other ways to formulate it. – YCor Feb 19 '18 at 14:31
• Nuclear operators are $2$-absolutely summing and hence the eigenvalues are square summable. – Bill Johnson Feb 19 '18 at 14:40

As I mentioned in a comment, exercise 3 has a positive answer: Nuclear operators are absolutely $2$-summing, and $2$-summing operators have $2$-summable eigenvalues (see, e.g., Tomczak's book).
Exercise 2 has a negative answer, but I am not satisfied with the example and do not know what happens in various classical spaces. First, note that it is sufficient to find for each $n$ a space $X_n$ on which there is a rank $n$ projection $P_n$ that has nuclear norm of order $n^{1/2}$. One can on $(I-P_n)X_n$ construct a diagonalizable (in the sense of the OP) operator $U_n$ with positive eigenvalues and nuclear norm as small as desired, so $S_n :=P_n + U_n(I-P_n)$ will have nuclear norm of order $n^{1/2}$ but the eigenvalues sum to about $n$. Multiply the $S_n$ by suitable constants and take an $\ell_2$ direct sum to see that exercise 2 has a negative answer.
The problem is that on a space that has the approximation property, the nuclear norm dominates the trace for finite rank operators, so the desired $X_n$ must fail the approximation property. Fortunately (or unfortunately, depending on one's point of view) Pisier constructed a space on which the nuclear norm of every finite rank operator is of order the norm of the operator. Now use the fact that in every Banach space there is a projection of norm at most $n^{1/2}$ onto every $n$ dimensional subspace (also in Tomczak's book).