I recently took a mastercourse on functional analysis and I was wondering why we 'skip' the metric structure on vector spaces. I have seen that $$\{\text{vector spaces}\}\supsetneq\{\text{topological vector spaces}\}\supsetneq\{\text{locally convex vector spaces}\}\supsetneq\{\text{normed vector spaces}\}\supsetneq\{\text{innerproduct spaces}\}.$$ Isn't it natural to include vector spaces endowed with a metric (that may not be induced by a norm) in this sequence? I can't seem to find any literature about it.

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    $\begingroup$ Perhaps there is no literature on that because no one found it useful or interesting. How might your metric play with the vector space structure if it does not come from a norm? $\endgroup$ Dec 16, 2019 at 23:36
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    $\begingroup$ And, indeed, there are many important Frechet spaces of natural functions that have metrics not given by a single norm. So your premise is not quite accurate... $\endgroup$ Dec 16, 2019 at 23:40
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    $\begingroup$ "The reader is warned that universal agreement has not been reached on terminology in the theory of metric linear spaces, so that care must be taken in consulting the literature to see exactly what is assumed. Some authors, for example, do not include translation invariance in the definition of a metric linear space, since they use a theorem of Kakutani to show that as non-translation invariant metric may be replaced by a translation invariant metric which yields the same metric topology." (Maddox, Elements of Functional Analysis, p90.) The theorem was also proved independently by G. Birkhoff. $\endgroup$ Dec 17, 2019 at 0:09
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    $\begingroup$ The inclusions are turned the wrong way. There are F-spaces (metric linear) between your second and third entry, and Frechet spaces (metric locally convex) between third and forth. $\endgroup$
    – Conifold
    Dec 17, 2019 at 1:13
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    $\begingroup$ A typical example is $\mathbb{R}^\omega$ in the metric $d(x,y)=\sum_n \frac{1}{2^n}\min(|x_n-y_n|, 1)$ which induces a TVS structure, is loc. convex and complete in the metric, but is not normable. $\endgroup$ Dec 17, 2019 at 8:08

2 Answers 2


If the metric is not related with the vector space structure, there is not much to talk about.

As you say, we could require that the metric is translation invariant. And there is another operation on a vector space, which is multiplication by scalars: does it make any sense to say that $2x$ is not at twice the distance from the origin than $x$ is? So you want to assume that the metric scales with scalar multiplication. That is, the metric satisfies

  • $d(x,y)=d(x+z,y+z)$

  • $d(\lambda x,\lambda y)=|\lambda|\,d(x,y)$

With those two assumptions, $\|x\|=d(x,0)$ is a norm that induces the metric $d$.

So, there is very little room for endowing a vector space with a meaningful metric that does not come from a norm but still somehow interacts naturally with the vector space structure. And if the metric does not match with the vector space structure, then you have no reason to pay attention to the topology and the vector space structure at the same time.

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    $\begingroup$ BTW, if a TVS is metrisable, it has a translation invariant metric. The scalars are the "issue". $\endgroup$ Dec 17, 2019 at 8:03
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    $\begingroup$ I'm not convinced by this. An example that would come to mind: modular-arithmetic vector spaces. These have basically a torus topology, which gives rise to a metric – but that metric does not correspond to a norm. $\endgroup$ Dec 17, 2019 at 10:33
  • $\begingroup$ The second equation presupposes that $\lvert\lambda\rvert$ is defined. But vector spaces are defined over general fields of scalars, which need not be equipped with an absolute value function. However one can express a scaling law also without an absolute value function: Demand that for any vectors $x$ and $y$ and any scalar $\lambda$, we have $d(\lambda x,0)\,d(y,0)=d(x,0)d(\lambda y,0)$. This in turn allows to define an “absolute value” $\lvert\lambda\rvert = d(\lambda x,0)/d(x,0)$ where $x\ne 0$. Whether that function then necessarily has the properties of an absolute value, I don't know. $\endgroup$
    – celtschk
    Dec 17, 2019 at 17:12
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    $\begingroup$ @celtschk: I'm not claiming any generality. But note that the context of the question is functional analysis, where the scalars are either $\mathbb C$ or $\mathbb R$. $\endgroup$ Dec 17, 2019 at 17:25
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    $\begingroup$ I'm out of my depth here. (If it turns out I'm just being thick-headed, I'll delete this comment to avoid clutter, as well as embarrassment!) In the last sentence: if the metric does not match with the vector space structure, then where does the topology of the plain topological vector space come from? Also, I can only read your answer as implying that there is no reason to study the F-spaces and Frechet spaces mentioned in Conifold's comment on the question. Can you add something (in a comment or the answer) to explain why it doesn't imply that? I might not be the only reader who is confused. $\endgroup$ Dec 17, 2019 at 21:19

In general topology, a metric space has much more structure than a topological space, and metrization theorems form a rich area of research. In topological vector spaces, things are rather different: a tvs is metrizable if and only if it is (Hausdorff and) first countable. If this is the case, then there also is a translation-invariant metric. (Both of these results can be found, for instance, in [Köt83, §15.11.(1)]. Further properties of metrizable tvs can be found in some of the more advanced functional analysis textbooks; for instance, see [Köt83, §15.11 and §18.2] or [Jar81, §2.8].) Topological vector spaces are already uniform spaces, which perhaps explains why being metrizable is not as important as it is for general topological spaces.

While it is true that the metrizable theory is not nearly as rich as the locally convex theory, metrization properties do occasionally come up in functional analysis. A few examples:

  • As pointed out by others, the open mapping theorem holds for arbitrary (not necessarily locally convex) F-spaces. More generally, the Baire category theorem is a very powerful topological tool. It requires a complete metric (a complete uniform structure is not enough), but local convexity is not needed.

  • If an ordered tvs $E$ is completely metrizable and has a closed and generating positive cone, then every positive linear functional on $E$ is automatically continuous. (See [AT07, Corollary 2.34].)

  • If $E$ is a separable metrizable lcs (in particular, if $E$ is a separable normed space), then $E'$ is weak-$*$ separable (cf. [Köt83, §21.3.(5)]). In the proof, it is crucial that equicontinuous subsets of $E'$ are weak-$*$ metrizable (under the present assumptions), so here the metrizability of a subset of a topological vector space matters too.

  • The spaces $L^p[0,1]$ for $0 < p < 1$ form an important example of a topological vector space $E$ whose dual $E'$ does not separate points on $E$. (In fact, these spaces have no continuous linear functionals, apart from $0$.) Nevertheless, these spaces are completely metrizable.

In a master's course, you can only do so much, so it is reasonable to focus on the most important concepts and theorems. Metrizable spaces do not play a crucial role in the theory — when metrizable tvs have special properties, it is usually because they belong to a larger class (every metrizable lcs is bornological, every F-space is a Baire space, etc.). On the other hand, locally convex spaces are central to the theory, and this is why your masters course or textbooks treat locally convex spaces in great detail but do not say much about metric vector spaces.


[AT07]: Charalambos D. Aliprantis, Rabee Tourky, Cones and Duality (2007), Graduate Studies in Mathematics 84, American Mathematical Society.

[Jar81]: Hans Jarchow, Locally Convex Spaces (1981), Mathematische Leitfäden, Teubner.

[Köt83]: Gottfried Köthe, Topological Vector Spaces I, Second revised printing (1983), Grundlehren der mathematischen Wissenschaften 159, Springer.


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