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In Wikipedia's definition of Fréchet space, it is stated that a Fréchet space is a topological vector space that satisfies the following:

  • It is locally convex
  • Its topology can be induced by a translation-invariant metric
  • Any (hence every) translation-invariant metric inducing the topology is complete

1) My first question lies with the "hence every" clause. It is known that completeness is a property of the metric, not topology, the standard example being $(0,1)$ under the absolute value and arctan metrics. As such how is it that we can conclude completeness here for every translation-invariant metric inducing the topology?

2) Wikipedia gives yet another equivalent condition for the completeness property - namely, let $\{p_k\}_{k\in\mathbb{N}}$ denote a countable family of seminorms defining the topology of $X$; then the space is complete with respect to the family of seminorms (i.e. if $(x_n)$ is a sequence in $X$ which is Cauchy with respect to each seminorm $p_k$, then there exists $x\in X$ such that $(x_n)$ converges to $x$ with respect to each seminorm $p_k$). How does the completeness with respect to any translation-invariant metric imply the completeness with respect to any countable family of seminorms and vice versa?

Remark: I suppose an answer to Q2 will answer Q1.

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1 Answer 1

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Let $F$ be your Fréchet space.

Suppose that $d$ is a translation invariant metric that induced the topology of $F$ and that $F$ is complete under $d$. Let $\mathcal P=\{p_k\}_{k\in\mathbb N}$ be any family of seminorms inducing the topology of $F$. Let $$d'(x,y)=\sum_{k=1}^\infty 2^{-k}\,\frac {p_k(x-y)}{1+p_k (x-y)}$$ be the metric induced by $\mathcal P$. It is easy to check that this metric induces the same topology that $\mathcal P$, that is the topology of $F$.

Looking at the respective unit balls, we can find (since both metrics induce the same topology) $\alpha,\beta>0$ such that $$\alpha\,d(x,0)\leq d'(x,0)\leq \beta\,d(x,0).$$ Because the metrics are translation invariant, for any $x,y\in F$ we have $d(x,y)=d(x-y,0)$ and the same for $d'$ so $$\tag1\alpha\,d(x,y)\leq d'(x,y)\leq \beta\,d(x,y)$$for all $x,y\in F$. Thus a sequence will be complete for $d$ if and only if it is complete for $d'$.

As $(1)$ is an equivalence relation, we have that completeness is the same for any translation invariant metric that induces the topology for $F$, and also agrees with completeness with respect to any sequence of seminorms that induces the topology.

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  • $\begingroup$ So in essence the argument is: every pair of equivalent translation-invariant metrics on a linear space is strongly equivalent (in the sense (1)) and strong equivalence of metrics preserves completeness (as opposed to simple topological equivalence). $\endgroup$ Mar 3, 2019 at 7:03
  • $\begingroup$ Yes, if by "equivalent" you mean that the two metrics determine the same open sets around $0$ (or, what is the same, they generate the same topology). $\endgroup$ Mar 3, 2019 at 12:14
  • $\begingroup$ yes, that’s what I call topological equivalence. $\endgroup$ Mar 3, 2019 at 12:15
  • $\begingroup$ $d'$ isn't a metric (the series may diverge to $\infty$). The usual modification is $d'(x,y)= \sum\limits_{k=1}^\infty 2^{-k} \frac{p_k(x-y)}{1+p_k(x-y)}$. $\endgroup$
    – Jochen
    Mar 4, 2019 at 12:36
  • $\begingroup$ You are totally right. Editing right now. $\endgroup$ Mar 4, 2019 at 13:04

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