What is the proper notion of an isomorphism between Fréchet spaces? Obviously it should be a linear map. I'm just worried about the analytic structure. Should one be able to order the seminorms on each space so that the isomorphism preserves each seminorm (i.e., $q_n(\phi (v))=p_n(v)$)? Should it preserve the translation invariant metric? Or should it just be a homeomorphism?

I'm leaning towards the last one, as the other two notions seem too strong, especially the first, but I figured I'd ask here to double check before I go ahead with what I'm doing.

Thanks much!

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    $\begingroup$ It is not clear to me that there ought to be a unique best notion. There are multiple reasonable notions of isomorphism between Banach spaces or metric spaces and I wouldn't expect Frechet spaces to be any different. It all depends on your application. $\endgroup$ Jul 22, 2011 at 2:32
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    $\begingroup$ I disagree with @Qiaochu. Contrary to Banach spaces a Fréchet space is not equipped with a metric, it is complete with respect to its locally convex topology which is assumed metrizable. It is clear that an isomorphism should be a linear homeomorphism. Neither the seminorms nor the metric are intrinsic in the usual definitions. Note that a continuous linear map is automatically uniformly continuous, hence completeness is preserved. $\endgroup$
    – t.b.
    Jul 22, 2011 at 2:37
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    $\begingroup$ @Theo: ah, I was a little confused about the definition. I guess it is more of a convenient collection of TVSes than anything else, so the isomorphisms should be TVS isomorphisms. $\endgroup$ Jul 22, 2011 at 2:43
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    $\begingroup$ @Qiaochu: Yes you can put it this way. Also, I don't know a single example of a non-normable Fréchet space which is equipped with a preferred metric. All the standard examples don't have a canonical candidate and the specific metric one chooses depends on the specific application one has in mind, while most Banach spaces can be defined with via some completion process which makes their norm very natural in some sense. $\endgroup$
    – t.b.
    Jul 22, 2011 at 2:49
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    $\begingroup$ Thanks, great and nice to hear! I was just wondering since you seem to have moved on to unbounded operators :) By the way, I think for unbounded operators you could do worse than work through chapter 5 of Pedersen's Analysis Now for the bare-bones basic theory (although he makes a point that there isn't such a thing). For applications, I strongly recommend Reed-Simon, Methods of modern mathematical physics, Vol. 1: Analysis of operators, Ch. VIII. $\endgroup$
    – t.b.
    Jul 28, 2011 at 2:24

1 Answer 1


To expand on my comment:

Fréchet spaces are a special class of topological vector spaces. Note that a topological vector space has a uniform structure coming from the underlying abelian topological group, so it makes sense to speak of completeness. A Fréchet space is a complete and metrizable locally convex topological vector space.

It is not hard to show that a continuous linear map between topological vector spaces is uniformly continuous. Thus, completeness and local convexity are both preserved under isomorphisms inside the category of topological vector spaces and continuous linear maps.

Metrizability of a locally convex topological vector space is equivalent to admitting a countable neighborhood base of $0$ consisting of convex, balanced and absorbing sets (it is often convenient to replace this base by a decreasing sequence of such sets). This property is also preserved under isomorphisms as topological vector spaces, hence in my opinion there is only one reasonable notion of isomorphism that can be considered: simply the linear homeomorphisms. By the open mapping theorem (as stated e.g. in Rudin's Functional Analysis, Theorem 2.11, page 47) a continuous linear bijection between Fréchet spaces is a homeomorphism.

As a further point, the (sequence of) semi-norms and metrics with which you can define a Fréchet space are usually not natural. That is, there are many convenient choices and it depends on the context which one is the most appropriate. This is in stark contrast to Banach spaces which come equipped with a preferred norm. As a corollary, neither preserving some (increasing) sequence seminorms nor preserving some translation invariant metric are natural in my opinion.

Added: in view of paul garret's second comment below:

One can quite easily show (as soon as the necessary language is developed) that every complete locally convex topological vector space $X$ is the (filtered projective) limit of Banach spaces (in the category of locally convex spaces).

To see this, choose a base $\{U_{\alpha}\}_{\alpha \in A}$ of the neighborhood filter of $0$, consisting of convex, balanced and absorbing sets and let $p_{\alpha}$ be Minkowski functional associated to $U_{\alpha}$. The Hausdorffification $X_{\alpha}$ of $(X, p_{\alpha})$ is easily seen to be a Banach space and because $A$ is directed by reverse inclusion so is $X_{\alpha}$. It is straightforward to check that $X = \varprojlim X_{\alpha}$ in the category of locally convex spaces. For details, I refer to Schaefer, Topological Vector Spaces, Chapter II.§5, page 51ff.

Now given that a Fréchet space admits a decreasing sequence of convex balanced and absorbing neighborhoods, it follows immediately that every Fréchet space is the projective limit of a sequence of Banach spaces. For further details see Schaefer, Chapter II.§4, page 48f (as well as Theorem I.6.1, page 28). Conversely, any limit of a sequence of Banach spaces is a Fréchet space.

In view of this, it is even less natural to expect that a (general) Fréchet space comes equipped with a preferred metric.

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    $\begingroup$ I'd second Theo B's emphasis that there's not necessarily a natural metric on a Frechet space, it's merely metriIZABLE, with its canonical topology. After all, already on a countable product of metric spaces, there's no canoncial metric, but the topology is uniquely determined. $\endgroup$ Jul 22, 2011 at 14:03
  • $\begingroup$ @paul: Related to this: do you know a good example of a non-normable Fréchet space that is naturally equipped with a metric, that is one that comes along with a metric that is preferred in some sense? I couldn't think of any. $\endgroup$
    – t.b.
    Jul 22, 2011 at 14:09
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    $\begingroup$ @Theo... no non-normable Frechet spaces with preferred metrics come to mind. If anything, the pattern that Frechet spaces naturally occur as countable (proj) limits of Banach spaces seems to militate against expecting preferred metrics. (It is unfortunate that so many texts on functional analysis shy away from admitting this and developing a suitable vocabulary.) $\endgroup$ Jul 22, 2011 at 15:11
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    $\begingroup$ @paul: Yes, that was probably me... Mnemonic (that's how I remember it): $\ell^{\infty} = \prod_{\mathbb{N}} \mathbb{R}$ and $\ell^1 = \coprod_{\mathbb{N}} \mathbb{R}$ in the category of Banach spaces and contractions (or $c_0 = \ell^1$ for the coproduct if you prefer to work in the ultrametric situation). But it's a good point: if you want to be in the additive setting, you have to consider things up to NONisometric isomorphism both in metric vector spaces and in Banach spaces (in the archimedean setting). $\endgroup$
    – t.b.
    Jul 22, 2011 at 20:32
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    $\begingroup$ @paul: In fact, $\ell^1(S)$ is the free Banach space over the set $S$ (if you have a map from $S$ to the unit ball of another Banach space then it extends to a contraction). Similarly, $\ell^{\infty}$ is the co-free Banach space (Hahn-Banach!). In other words, there are adjuctions $\ell^{1} \dashv \operatorname{Hom}_{\leq 1}(\mathbb{R},{-})$ (unit ball functor) and similarly for $\ell^{\infty}$ and the "unit ball in the dual space". If you want to insist: the ground field is a projective generator and an injective co-generator (Hahn-Banach again!). Sorry, I'll stop now... $\endgroup$
    – t.b.
    Jul 22, 2011 at 20:47

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