Let $F$ a Fréchet space. This means that $F$ is a complete Hausdorff topological space whose topology can be generated by an increasing family of seminorms $\{ p_{n} \}_{n \in \mathbb{N}}$. Let's denote by $F_{n}$ the completion of of $F$ with respect to $p_{n}$.

Now, $F$ is nuclear if the family $\{ p_{n} \}_{n \in \mathbb{N}}$ can be chosen to consist of Hilbert seminorms (i.e. seminorms coming from pre-inner products) with the property that the natural maps $F_{i} \leftarrow F_{j}$ for $i < j$ are trace-class.

On the other hand, $F$ is countably normed if the family $\{ p_{n} \}_{n \in \mathbb{N}}$ can be chosen to consist of norms with the property that the natural maps $F_{i} \leftarrow F_{j}$ for $i < j$ are dense embeddings.

Suppose now that $F$ is both countably normed, and nuclear.

Can the family $\{p_{n}\}_{n \in \mathbb{N}}$ be chosen to consist of Hilbert norms, with the property that the natural maps $F_{i} \leftarrow F_{j}$ for $i < j$ are dense, trace-class embeddings?

In Holmströms paper "A Note on Countably Normed Nuclear Spaces" it is claimed that the family $\{p_{n}\}_{n \in \mathbb{N}}$ can be chosen to consist of Hilbert norms with the property that the natural maps are dense embeddings, however, the trace-class condition is not mentioned.

  • $\begingroup$ This can be proved in a couple of lines if one uses the (trivial) fact that any continuous linear mapping from a suitable lcs projective of a family (not necessarily countable) of Banach spaces into a Banach space factors over one of the components. This implies the well-known fact that any two such representations intertwine in the natural sense, a result which implies your claim and much more. $\endgroup$ – bathalf15320 Oct 10 at 6:46

Prolog. As the arguments below are somewhat technical and probably not too interesting for many readers, I would like to point out that such problems can be quite subtle. The crucial problem (which could also be interesting for Banach spacers) is that the unique extension of a linear injection between normed spaces to the completions need not be injective! (Take a discontinuous linear functional $f$ on a Banach space $(X,\|\cdot\|)$ and consider identity from $X$ endowed with the stronger norm $\|x\|+|f(x)|$ to $(X,\|\cdot\|)$.)

A consequence of this is that even if a Fréchet space has continuous norms it is in general not possible to write it as an intersection of Banach spaces (i.e., it need not be a countably normed space -- which is a rather unfortunate terminology). It was an observation of Pelczynski that an example of a nuclear Fréchet space with a continuous norm which is not countably normed would solve a problem of Grothendieck about the bounded approximation property for Fréchet spaces. Such examples where found in the eighties by Bellenot, Dubinsky, and Vogt.

The answer. Yes, this is possible. The main point in the argument is that any two reduced representations factorize through each other where a reduced representaion $F=\lim\limits_\leftarrow (F_i,f_j^i)$ is given by an increasing sequence of seminorms $p_i$ defining the topology such that $F_i$ is the Hausdorff completion of $(F,p_i)$ (first factor out the kernel of $p_i$ then take the completion) and $f_j^i:F_{j}\to F_i$ is the canonical map (associated to the identity $(F,p_j)\to (F,p_i)$) for $j\ge i$. Given two reduced representaions $F=\lim\limits_\leftarrow (F_i,f_j^i)$ and $F=\lim\limits_\leftarrow (H_i,h_j^i)$ there are $j(i)>i$ such that $f_{j(i)}^i$ factorizes through some $H_k$ and $h_{j(i)}^i$ factorizes through some $F_k$. Passing to a subsequence we can thus assume that we have factorizations $$ \cdots \leftarrow F_{i-1} \stackrel{\phi_{i-1}}{\leftarrow} H_i \stackrel{\eta_{i}}{\leftarrow} F_i \leftarrow \cdots $$ with $f_i^{i-1}=\phi_{i-1}\circ \eta_i$ and $h_{i+1}^i=\eta_i\circ \phi_i$.

Assume now that the $f_j^i$ are injective (because $F$ is countably normed) which implies that $\eta_i$ are also injective and that $h_j^i$ are trace class between Hilbert spaces (because $F$ is nuclear). Let $L_{i+1}$ be the kernel of $h_{i+1}^{i}$. The injectivity of $\eta_i$ implies that $L_{i+1}$ is the kernel of $\phi_i$ and hence the canonical map $\overline{h}_{i+1}^i:H_{i+1}/L_{i+1} \to H_i$ factorizes over $F_i$. Denoting the quotient map $q_{i+1} : H_{i+1}\to H_{i+1}/L_{i+1}$ we get the sequence $$ \cdots\leftarrow F_{i-1} \leftarrow H_i/L_i \stackrel{q_i}{\leftarrow} H_i \stackrel{\eta_i}{\leftarrow} F_i \stackrel{\overline{h}_{i+1}^i}{\leftarrow} H_{i+1}/L_{i+1} \leftarrow \cdots $$ One can then check that the map $H_{i+1}/L_{i+1}\to H_i/L_i$ is injective. Finally this implies that $H_{i+2}/L_{i+2}\to H_i/L_i$ is injective and trace class because it factorizes over a trace class operator.

| cite | improve this answer | |
  • $\begingroup$ Hi Jochen, thanks for your answer, it's been very enlightening! Especially the statement that any two reduced representations factorize through each other seems very useful for some things I've been thinking about, do you have a reference for this result? $\endgroup$ – Peter Oct 8 at 13:50
  • $\begingroup$ This is a kind of folklore and I do not a good reference (Proposition 3.3.8 in my Derived Functors in Functional Analysis contains it as a quite particular case). It is simple anyway: $F=\lim\limits_\leftarrow H_i$ implies that $q_i(x)=\| h^i(x)\|_{H_i}$ are a fundumaental system of semi-norms for $F$ where $h^i:F\to H_i$ is the canonical map into the local Banach space $H_i$. Given $i$ the canonical map $f^i:F\to F_i$ is clearly continuous and therefore $\|f^i(x)\|_{F_i} \le c\|h^j(x)\|_{H_j}$ for some $j$ and some constant $c$. In particular, $h^j(x)=0$ implies $f^i(x)=0$ and on ... $\endgroup$ – Jochen Wengenroth Oct 8 at 14:52
  • $\begingroup$ ... the dense subspace $M_j=\{h^j(x):x\in F\}$ you can define $\phi_j: M_j\to F_i$ by $h^j(x)\mapsto f^i(x)$. This is well-defined, linear, and continuous and has a unique extension $H_j\to F_i$. $\endgroup$ – Jochen Wengenroth Oct 8 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.