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Given complete, locally convex Hausdorff vector spaces $E$ and $F$, let $$ E \otimes_i F, \qquad E \otimes_\pi F$$ denote the (completed) inductive and projective tensor products respectively. The inductive (resp. projective) tensor product has the universal property for separately (resp. jointly) continuous bilinear maps out of $E \times F$. By the universal property, there is a continuous linear map $$E \otimes_i F \to E \otimes_\pi F.$$ If $E$ and $F$ are Frechet, then this map is an isomorphism. This is because separate and joint continuity coincide in this case. I am interested in knowing if this map is an isomorphism in situations where the hypotheses on $E$ are strengthened and the hypotheses on $F$ are relaxed.

More specifically, I am interested in the case $E = C^\infty(M)$, the nuclear Frechet space of smooth functions on a compact manifold $M$. In this case, $$C^\infty(M) \otimes_\pi F \cong C^\infty(M,F),$$ the space of smooth functions on $M$ with values in $F$. So I would like to know what hypotheses, if any, are needed on $F$ to ensure that $$C^\infty(M) \otimes_i F \to C^\infty(M,F)$$ is an isomorphism. For example, is it an isomorphism if $F$ is an $LF$-space, that is, an inductive limit of Frechet spaces?

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If you mean by LF-space a strict inductive limit of Frechet spaces (as it was done by Dieudonne and Schwartz) I think the answer is yes.

Here is what I believe could be made a proof: Since the inductive tensor product respects inductive limits you have $C^\infty(M) \otimes_i F = \lim C^\infty(M) \otimes_i F_n$ if $F= \lim F_n$ is a strict LF-space. Since $F_n$ is a closed topological subspace of $F$ you get that $C^\infty(M) \otimes_i F_n = C^\infty(M) \otimes_\pi F_n$ and this implies that $C^\infty(M) \otimes_i F = \lim C^\infty(M) \otimes_\pi F_n$ is a dense topological subspace of $C^\infty(M) \otimes_\pi F$. On the other hand, it is complete because it remains strict (as for as I remember this is due either to Dieudnne-Schwartz or to Köthe).

EDIT: Sorry, I confused the Frechet case with the DF case -- only in the latter case $\lim E \otimes_\pi F_n$ is a topological subspace of $E\otimes_\pi \lim F_n$.

For strict LF-spaces the answer to your question was already given in Grothendieck's thesis (part I, page 47): If E is a proper (non-normable) Frechet space and $F=\lim F_n$ is a strict inductive limit with a strictly increasing sequence $F_n$ then $\lim E\otimes_\pi F_n$ is NEVER a topological subspace of $E \otimes_\pi \lim F_n$.

On the other hand, if the nuclear Frechet space $E$ has a continuous norm (as it is the case for $C^\infty(M)$) and $F=\lim F_n$ is a strict LF-space then $\lim E\otimes_\pi F_n = E\otimes_\pi F$ holds algebraically:

The latter space is the space of all continuous linear operators $T$ from $E'$ to $F$ and the former of those operators with values in some $F_n$. But $E'$ contains a bounded set (the polar of the $0$-neighborhood corresponding to the continuous norm) which is total, i.e., its linear span is dense in $E'$. For any continuous $T$ the bounded set $T(B)$ is contained in some $F_n$ (by the so called regularity of strict LF-spaces) and since $F_n$ is closed in $F$ one gets from the continuity $T(E') \subseteq \overline{[T(B)]} \subseteq F_n$.

There might be non-strict inductive limits where $\lim E \otimes_\pi F_n = E\otimes_\pi\lim F_n$ holds topologically but I do not know an example since both, the algebraic and the topological, conditions are very restrictive. If $F$ is the dual of a nuclear Frechet space $X$ then the arguments above should show that the algebraic coincidence is equivalent to $L(X,E) = LB(X,E)$, the space of all operators from mapping some $0$-neighborhood of $X$ into a bounded subset of $E$. This situation has been investigated by Vogt, J. Reine Angew. Math. 345 (1983), 182–200, and it implies Ext$^1(E,X)=0$. On the other hand, I believe results of Grothendieck and Vogt yield that the topological coincidence implies Ext$^1(X,E)=0$. If $X$ and $E$ are both power series spaces these conditions contradict each other.

In conclusion: The topological equality $C^\infty(M) \otimes_i F = C^\infty(M) \otimes_\pi F$ seems very unlikely whenever $F$ is not a Frechet space.

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  • $\begingroup$ Thanks for the reply. I think I am content with dealing with the LF case, and yes, I meant strict inductive limit. So if I understand correctly, the key is to see that the obvious map $\lim C^\infty(M) \otimes_\pi F_n \to C^\infty(M) \otimes_\pi F$ is a homeomorphism onto its image. This is certainly the case for the maps $C^\infty(M) \otimes_\pi F_n \to C^\infty(M) \otimes_\pi F$ by the properties of the injective tensor product and the nuclearity of $C^\infty(M)$. Is it automatic that the induced map from the inductive limit is a homeomorphism onto its image? $\endgroup$
    – Allan Yashinski
    Mar 7, 2013 at 20:46
  • $\begingroup$ After thinking about this some more, it seems to me that a proof of this would have to use compactness of $M$, and not just the fact that $C^\infty(M)$ is Frechet and nuclear. If we consider the case $F = \oplus F_n$, a topological direct sum of Frechet spaces, then $C^\infty(\mathbb{R}) \otimes_i (\oplus F_n) \cong \oplus C^\infty(\mathbb{R}, F_n)$ is strictly smaller than $C^\infty(\mathbb{R}) \otimes_\pi (\oplus F_n) \cong C^\infty(\mathbb{R}, \oplus F_n)$. $\endgroup$
    – Allan Yashinski
    Mar 7, 2013 at 21:17
  • $\begingroup$ @Allan: You are right, my answer was not correct. I hope that the new answer is better. $\endgroup$
    – Jochen Wengenroth
    Mar 11, 2013 at 16:00

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