# If $F$ is a countably normed, nuclear Fréchet space, can I then find a fundamental system which exhibits both of these properties at once?

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.

• 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. – 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 $$(X,\|\cdot\|)$$.)
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.
• 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 ... – Jochen Wengenroth Oct 8 at 14:52
• ... 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 $H_j\to F_i$. – Jochen Wengenroth Oct 8 at 14:56