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 $$\sum \lambda\_n<\infty.$$
> 
> 
> 


Can someone please help me prove it?  

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.



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


