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Your answer confuses me (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 to me here?

Also, we have (proposition 3.5):

Any bilinear continuous map $\phi : F'_\beta \times F \rightarrow \mathbb{R} $ extends to a map $\phi' : F'_\beta \hat{\otimes}_\pi F \rightarrow \mathbb{R}$. We take $\phi$ as the trace and can thus extend it, or no?

I just don't see how your arguments can be applied, especially since they disagree with these things here? Perhaps my mistake is of topological nature, that the sum doesn't converge in the topology to the identity?

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?

Your answer confuses me (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?

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 to me here?

Also, we have (proposition 3.5):

Any bilinear continuous map $\phi : F'_\beta \times F \rightarrow \mathbb{R} $ extends to a map $\phi' : F'_\beta \hat{\otimes}_\pi F \rightarrow \mathbb{R}$. We take $\phi$ as the trace and can thus extend it, or no?

I just don't see how your arguments can be applied, especially since they disagree with these things here? Perhaps my mistake is of topological nature, that the sum doesn't converge in the topology to the identity?

Your answer confuses me (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 to me here?

Also, we have (proposition 3.5):

Any bilinear continuous map $\phi : F'_\beta \times F \rightarrow \mathbb{R} $ extends to a map $\phi' : F'_\beta \hat{\otimes}_\pi F \rightarrow \mathbb{R}$. We take $\phi$ as the trace and can thus extend it, or no?

I just don't see how your arguments can be applied, especially since they disagree with these things here? Perhaps my mistake is of topological nature, that the sum doesn't converge in the topology to the identity?

Your answer confuses me (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?

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 to me here?

Also, we have (proposition 3.11):

$(F'_\beta\hat{\otimes}_\pi F)'=B(F'_\beta,F)$, the space of continuous bilinear forms, of which the trace is an element of, thus we can take the trace of any operator, whether injective or projective tensor product, since they agree.

I just don't see how your arguments can be applied, especially since they disagree with these things here? Perhaps my mistake is of topological nature, that the sum doesn't converge in the topology to the identity?

Your answer confuses me (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?

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 to me here?

Also, we have (proposition 3.5):

Any bilinear continuous map $\phi : F'_\beta \times F \rightarrow \mathbb{R} $ extends to a map $\phi' : F'_\beta \hat{\otimes}_\pi F \rightarrow \mathbb{R}$. We take $\phi$ as the trace and can thus extend it, or no?

I just don't see how your arguments can be applied, especially since they disagree with these things here? Perhaps my mistake is of topological nature, that the sum doesn't converge in the topology to the identity?