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Today we were introduced to locally convex spaces, defined thusly:

A vector space is locally convex iff it has a family of semi-norms $(p_i)$ such that $x=0$ if and only if $p_i(x)=0$ for all $i$.

The professor then said we would limit ourselves to countable families $(p_i)$ and introduced the following function:

$$d(x, y) = \sum_{n=0}^\infty \frac 1 {2^n} \frac {p_n(x-y)}{1+ p_n(x - y)}$$

He explained that this is always a metric, and called it the "natural" metric. Naturally, it didn't seem that natural to any of us. This same formula is also found on Wikipedia.

Is there some property of this metric which characterizes it? Is it for instance the only metric having some form of compatibility with the family of semi-norms $(p_i)$?

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    $\begingroup$ There are lots of other metrics compatible with the family of seminorms. In some respects, $$d_s(x,y) = \max \biggl\{\frac{1}{2^n} \frac{p_n(x-y)}{1+p_n(x-y)}\biggr\}$$ is related closer to the family of seminorms, since a $d_s$-ball is the intersection of finitely many $p_n$-balls, which is not the case for $d$. $\endgroup$ – Daniel Fischer Sep 8 '16 at 15:02
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    $\begingroup$ @DanielFischer Do you have a precise definition in mind for "compatible"? $\endgroup$ – Jack M Sep 8 '16 at 15:04
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    $\begingroup$ The topology induced by the metric is the same as the topology induced by the family of seminorms. Since we're dealing with topological vector spaces, it would also be sensible to additionally require the metric to be translation-invariant. And since we're dealing with locally convex spaces, it would further be sensible to require that the balls of the metric be convex. (And while it's easy to see that that is the case for $d_s$, it's not easy to see for $d$ - I don't even know for sure whether it's always the case for $d$, need to think about that.) $\endgroup$ – Daniel Fischer Sep 8 '16 at 15:08
  • $\begingroup$ Anyway @JackM with the condition "$x=0$ if and only if $p_i(x)=0$" you are considering a family of norms $\endgroup$ – user288972 Sep 8 '16 at 17:31
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    $\begingroup$ @JohnMartin Yes, if we start with a countable separating family of seminorms, we get a metric. Whether the family is separating or not, if it's countable, the (pseudo)metric induces the same topology as the family of seminorms. $\endgroup$ – Daniel Fischer Sep 8 '16 at 17:45
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Local convexity is not really the property given in the question, but, rather, that $0$ (hence, every point in the vector space) has a local basis consisting of convex open sets. It is a not-completely-trivial theorem that there exists a set of semi-norms giving the topology. (That a "separating" family of seminorms, as in the question, gives a locally convex topology, is easy.)

If the collection of seminorms is countable (which is far from universally so!), then the topology is metrizable... but there is no canonical metric, as other comments and answers have mentioned. This illustrates the point that the map from "metrics" to "topologies" is many-to-one.

That is, that metric is not "natural", but the topology is natural.

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  • $\begingroup$ You say a local basis given by convex open sets, but shouldn't it be convex, open, balanced, and absorbent sets? $\endgroup$ – Jack M Sep 8 '16 at 15:18
  • $\begingroup$ @JackM, maybe by some conventions those further adjectives would be imposed, but I myself do not like to impose un-obvious implicit baggage on adjectives. That is, I read "locally convex" as literally having convex local basis (of opens, yes). The "balanced" and "absorbing" are separate issues, that don't really help or hinder the convexity, etc., to my perception. But, sure, other people may have other conventions. No doubt. $\endgroup$ – paul garrett Sep 8 '16 at 15:30
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The compatibility lies in the (much wanted) property that a sequence $x_n$ converges to $x$ for the metric iff it converges to $x$ for every fixed seminorm. As already mentioned there are many choices with that property. The construction is (only?) used in spaces where the topology does not allow for the construction of a norm compatible with the topology.

As mentioned by Daniel Fischer in a comment, there are more natural metrics like: $$ d(x,y)= \max_n \frac{c_n p_n(x-y)}{1+p_n(x-y)}$$ where $c_n$ is a strictly positive sequence converging to zero. For that metric the balls $B_r =\{ x: d(0,x)<r \}$ form a convex and balanced local base for the topology (also absorbing). With the sum instead of max the balls need not be convex.

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The Frechet norm or pseudo-norm it is defined starting with a separable family of seminorm $\lbrace p_n \rbrace_{n \in \mathbb{N}}$ (i.e. $p_n(x)=0$ implies $x=0$), and it is \begin{align*} \displaystyle \left \| x \right \|:= \sum_{n=1}^{\infty} 2^{-n} \frac{p_n(x)}{1+p_n(x)} \end{align*} this function does not define a norm but satisfies the following properties

(1). $x=0$ if $\left \| x \right \|=0$

(2). $\left \| - x \right \|=\left \| x \right \|$

(3). $\left \| x+y \right \| \leq \left \| x\right \| + \left \| y \right \|$.

In particular, the function $d(x,y):=\left \| x-y \right \|$ defines a pseudo-metric, in the sense that $d(x,y)=0$ not necessarily imply that $x-y = 0$. This is the only thing missing to $d$ is a metric.

Now on a locally convex space $E$, the family $\lbrace p_n \rbrace$ defines a vector topology of Hausdorff, where the family of finite intersections \begin{align*} \displaystyle U_{p_1}(\epsilon) \cap \cdot \cdot \cdot \cap U_{p_N}(\epsilon) = \lbrace x\in E: \max \lbrace p_1(x),...,p_N(x)\rbrace < \epsilon \rbrace \end{align*} is a local basis $\mathcal{U}$ for this topology $\mathcal{T}_P$. You can be shown that the topology defined from the pseudo-metric $d(x,y)$ is exactly the topology $\mathcal{T}_P$: formally it proves $B_d(0,\delta) \subset U$ and $U \subset B_d(0,\delta)$ with $U \in \mathcal{U}$. In other words we have a condition of "pseudo-metrizability" on the LCS $E$. Then this leads to the definition of Fréchet space, so if the locally convex space $E$ is complete with respect to pseudo-metric or equivalently if $p_n(x_k - x) \rightarrow 0$ for each seminorm ($\forall n \in \mathbb{N}$) as $k \rightarrow \infty$, then $E$ is said to Fréchet space.

Note that in general for LCS $E$ you can define them also considering $\lbrace p_j \rbrace_{j \in J}$ a family seminorm not necessarily countable. While this condition of countability of $\lbrace p_n \rbrace$ assumes when you consider the pseudo-norm, this allows also to take the increasing sequence $\lbrace p_n \rbrace$.

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  • $\begingroup$ Doesn't $x=0$ imply $\vert \vert x\vert \vert=0$? Note that a seminorm sends the origin to zero by the homogeneity. And hence if $x=0$, then $\vert \vert x\vert \vert=0$. $\endgroup$ – Sam Wong May 24 '18 at 18:49
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The seminorms define a topology on the vector space $X$, with subbasis $$V=V(p_i,n)=\{x\in X:p_i(x)<1/n\}.$$

The metric also defines a topology, with basis $$U=U(y,n)=\{x\in X:d(y,x)<1/n\},$$ where $y\in X$.

Compatibility means these two topologies coincide, but $d$ is not unique nor natural in any sense.

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  • $\begingroup$ Correct. In fact, the Wikipedia page cited by the OP mentions the term "natural topology" but not "natural metric". $\endgroup$ – GEdgar Sep 8 '16 at 16:34
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My favorite metric is $d(x,y)= \sup\lbrace p_n(x-y) \wedge \frac 1n: n\in\mathbb N\rbrace$ where $a\wedge b =\min\lbrace a,b \rbrace$.

The name locally convex space has traditional reasons because in Bourbaki's times one started with topological vector spaces of which locally convex ones are a particular case as explained in Paul Garret's answer. I only know of Helemskii who uses the term poly-normed space (he has a remark that poly-seminormed would be logically correct).

I have never seen an example of a space which is naturally a topological vector space but local convexity came out as a surprise (perhaps with the exception of finite-dimensional subspaces of Hausdorff topological vector spaces).

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