Traces of operators in nuclear spaces

Question

I am currently reading up on nuclear spaces in Jarchow, "Locally Convex Spaces", but I got confused and don't seem to find my mistake. In said book, theorem 21.5.9 states:

Let $F$ be a nuclear Frechet space. Then, $F'_\beta \otimes_{\pi} F = L_\beta (F,F)$. i.e. the projective tensor product of $F$ with its strong dual is homeomorphic to the operators on $F$ with the strong topology.

Now, that means every operator on $F$ has a trace, right?

So take the the space $X=\Pi_\mathbb{N} \mathbb{R}$. This is a nuclear Frechet space as it is a countable infinite product of those. Let $id_k : \mathbb{R} \rightarrow \mathbb{R}$ denote the identity on the k'th component of $X$. We have $id_X = \Sigma_k id_k$. Now, by above theorem $id_X$ and all $id_k$ must be trace class, but clearly, the sum can not converge to $id_X$ as otherwise the trace must be infinite.

If I read correctly, the strong topology has a 0-basis of $L(B,U)$, maps from some bounded set into an open set. The partial sums $f_n=\Sigma_{1≤k≤n} id_k$ however satisfy that $f_n-id_X$ is zero on the first $n$ components and thus maps any set into a chosen open set for n big enough. Thus, the partial sum converges and there is a contradiction.

I simply can't spot my mistake, perhaps you can help me?

Answer 1

It is indeed true that for a nuclear Frechet space $F$, the complete projective tensor product $F'_\beta \tilde{\otimes}_\pi F$ is isomorphic to $L_\beta(F,F)$. The mistake is your claim that the complete projective tensor product $E\tilde{\otimes}_\pi F$ consists of convergent series $\sum\limits_{n=1}^\infty \lambda_n e_n \otimes f_n$. This is true if $E$ and $F$ are both Frechet spaces or both strong duals of Frechet spaces. In the mixed case as in your situation the complete inductive tensor product can be described by such series but it is much smaller than the projective tensor product.

EDIT: I meant the inductive tensor product (the injective coincides indeed with the projective one because of nuclearity).

Answer 2

Your answer is wrong (unfortunately, I can't comment, so here is another answer instead). Now, I'm quoting from "Notes on locally convex vector spaces" by J.L. Taylor.

Since $F$ is nuclear, the (completed, you are right, I meant that) injective and projective tensor products agree, so "is smaller than" doesn't apply here.

Furthermore, theorem 21.5.9 in Yarchow explicitely states that the maps are homeomorphisms, and not just an isomorphism. We have:

$F'_\beta \hat{\otimes}_\pi F=F'_\beta \hat{\otimes}_\epsilon F=L_\beta(F,F)$

(corrolary 5.19 in the aforemention lecture notes). Thus the injective tensor product is just as big, "is smaller than" does not make much sense here.

Also, we have (proposition 3.5):

Any bilinear continuous map $\phi : F'_\beta \times F \rightarrow \mathbb{R} $ extends to a map $\phi' : F'_\beta \hat{\otimes}_\pi F \rightarrow \mathbb{R}$.

Which brings me to the real answer to the problem, which I just found myself:

Now, this would be fine if the trace was indeed continuous. But: Let $U$ be an open set in $\mathbb{R}^\mathbb{N}$, $V$ be an open set in $\mathbb{R}_\mathbb{N}$. Then, we always have: $U(V)=\mathbb{R}$, as can always be seen by choosing a sufficiently small open subset. Hence, the trace is not a bilinear continous map, hence can not be extended! Please correct me if you see any mistake, but I think that is the answer, no?