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I am currently reading up on nuclear spaces in Yarchow, "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?

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?

I am currently reading up on nuclear spaces in Yarchow, "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 mapa 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?

I am currently reading up on nuclear spaces in Yarchow, "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?

I am currently reading up on nuclear spaces in Yarchow, Locally Convex Spaces, but I got confused an don't seem to find my mistake :( So, 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 mapa 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?

I am currently reading up on nuclear spaces in Yarchow, "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 mapa 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?