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](https://www.jstor.org/stable/2045495?seq=1)" 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.



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


