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Recently I wondered, if convergence in some given metric $d$ on $\mathbb R^n$ induces convergence in norm. Of course, if $d(x,y) = \|f(x)-f(y)\|$, where $f$ is a bijection on $\mathbb R^n$, then this does not hold in general provided $f^{-1}$ is discontinuous in at least one point.

Now, I stepped over topological vector spaces (TVS), which have a topology that interacts well with addition and scalar multiplication. In this case we have that the topology is induced by a norm if and only if there is a convex and bounded neighborhood of $0$.

That's a useful result but a bit to much for my intentions, as I want only to know if the metric induces a stronger convergence than the norm.

So what am I looking for? I'm curious if there is a metric on $\mathbb R^n$ (possible even on $\mathbb R$) which makes $\mathbb R^n$ to a topological vector space and which has one of the following properties:

  1. It produces a convex and bounded neighborhood of $0$ and induces therefore a stronger convergence than convergence in norm.
  2. It does not provide (1) but nevertheless induces a stronger convergence than convergence in norm.
  3. It doesn't induce a stronger convergence than convergence in norm.

To make the case more exiting, the discrete metric is not allowed and also metrics that are "too normy" are excluded. By the rather vague "too normy" I mean that it should not be trivial to deduce the mentioned properties, as it would if

  • $d(x,y) = \|f(x)-f(y)\|$ for some linear $f$ on the reals,
  • $d(x,y) = \|x-y\|/(1+ \|x-y\|)$,
  • $d$ is the British Rail metric or some other metric induced by special paths (which are "normy" in my sense, but also they often don't induce a TVS).

So, I'm looking for examples that show that TVS-metrics are not as boring as discrete or normy ones. (But maybe, there are indeed just normy and discrete ones, but I don't think so.)

Thank you in advance!

Edit: As pointed out in the comments, the discrete metric doesn't induce a TVS over $\mathbb R$, where the underlying scalar field $\mathbb R$ is endowed with the standard topology. If the underlying scalar field is induced with the standard topology, the case is closed.

So one should rather ask, if there are metrics for which the addition on $\mathbb R^n$ is continuous and for which the scalar multiplication is continuous with respect to some other topology on the underlying scalar field (e.g. the discrete topology).

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    $\begingroup$ A finite-dimensional vector space over $\mathbb{R}$ allows only one Hausdorff vector space topology. So if you have a metric on $\mathbb{R}^n$ that makes it a topological vector space, it induces the standard topology. If you move to infinite-dimensional spaces, things get more interesting. $\endgroup$ – Daniel Fischer May 6 '16 at 13:56
  • $\begingroup$ The discrete metric doesn't induce the standard topology, as every set is open. Then the discret metric induces no vector space topology? $\endgroup$ – Dreipunkt May 6 '16 at 14:07
  • $\begingroup$ Right. Assuming we still use the standard topology on the scalar field. If we allow endowing the scalar field with different topologies, things change. $\endgroup$ – Daniel Fischer May 6 '16 at 14:10
  • $\begingroup$ Is the topology on the underlying scalar field involved? Maybe I confused some definitions ... I had in mind the following: Let $V$ be a vector space endowed with a topology $\tau$. The pair $(V,\tau)$ is called a TVS if all singletons are closed and if the vector space operations are cts. with respect to $\tau$. $\endgroup$ – Dreipunkt May 6 '16 at 14:25
  • $\begingroup$ Oh. I had the wrong definition for cts. scalar multiplication in my mind. >_> Thank you very much! $\endgroup$ – Dreipunkt May 6 '16 at 14:52

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