## Is every norm in R^n a continuous function?

Is every norm in R^n a continuous function?

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 I would've forgiven you if you'd entered in different body text, but as it stands this looks like a homework question. – Qiaochu Yuan Oct 19 2009 at 21:17 Well, if it was then the solutions given below probably wouldn't count anyway for invoking too many other results ;). – Akhil Mathew Oct 19 2009 at 21:35

The exaplanation depends on what is meant by "continuous". Let's agree that "the norm || || is continuous" means "if (x_n) is a sequence of vectors such that the coordinates x_n converge to the coordinates of some vector x, then ||x_n|| converges to ||x||".

Let e_1, ..., e_n be the standard basis of R^n. First one shows that for every i, the function t -> ||t*e_i|| is continuous (t real). Then, after writing every vector as a_1 * e_1 + ... + a_n * e_n, the result follows easily from the triangle inequality.

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I agree that this looks like a homework question, but since some people have already bitten, I'd just like to point out what might be said for infinite-dimensional spaces. So suppose you have an infinite-dimensional real or complex vector space, equipped with a norm || . ||

What does it mean for a function on V to be continuous? Well, you have to specify a topology on V, and it's natural to use the one defined by the norm. But then it's an immediate corollary of the triangle inequality that the norm function is continuous with respect to the topology it defines. (In some sense, if this weren't true, then we wouldn't bother studying normed vector spaces!)

However, V might also carry some weaker topology (such as a w*-topology induced by some predual) and then the norm will not in general be continuous with respect to that topology.

(Silly remark: equip R^n with the indiscrete topology, i.e. the one with only two members. Then the usual norm is not continuous. Of course, that's a ridiculous topology to put on the space. I have a feeling that every Hausdorff topology on R^n for which translations and dilations are continuous, is equivalent to the usual one, but I'd need to check in something like Rudin's book to be sure.)

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 "I have a feeling that every Hausdorff topology on R^n for which translations and dilations are continuous, is equivalent to the usual one, but I'd need to check in something like Rudin's book to be sure.)" I don't think this is true- the discrete topology is an example. – Akhil Mathew Oct 19 2009 at 21:33 The statement is that any Hausdorff topology for which translations and dilations are continuous is the usual one, where you require that dilations R \times R^n \to R^n be continuous if you put the usual topology on R. – Eric Wofsey Oct 19 2009 at 22:26

Yes, and this is true more generally (same reason as John Cook) for norms over a finite-dimensional vector space over a field complete with respect to an absolute value. It doesn't work for infinite-dimensional spaces.

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It is important that the field is required to be complete. Exercise: find two norms on the vector space Q^2 (over the rationals Q with the usual absolute value) that are not topologically equivalent. – Gerald Edgar Oct 27 2009 at 12:05

Yes, because in finite dimensional spaces all norms are topologically equivalent.

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