# Equivilant norms of a Vector Space

1. May 1, 2005

### snoble

This is killing me that I can't see this. Why is it two norms on a finite dimensional Vector space X are equivilant if and only if there exist positive real constants $$c_1,\, c_2$$ such that $$\forall x\in X$$, $$\|x\|_2 \le c_1 \|x\|_1$$ and $$\|x\|_1 \le c_2 \|x\|_2$$.

Here equivilant means that a sequence in the VS is cauchy under one norm iff it is cauchy under the other norm.

Alright, if you assume the inequality and assume a sequence is Cauchy under one norm then it is easy to show it is Cauchy under the other norm. Just scale the $$\epsilon$$ appropriately.

Now the other direction is what has me stumped. So assuming that the v.s. norms are equivilant you can get that the norms on the underlying fields are equivilant; just take a sequence in the field that is cauchy under the first norm, multiply the sequence by any non-zero vector - that'll be cauchy in the V.S. under the first norm so cauchy under the second. Then just divide by the value of the vector under the second norm. Then it is clear that the original sequence is Cauchy in the field under the second norm.

I realize that the above is kind of fuzzy but the point is the norms on the underlying field are equivilant. And I know that if two field norms are equivilant then there exists a positive real $$\alpha$$ such that $$\forall k\in F$$, $$\|k\|_1^\alpha = \|k\|_2$$.

But now I can't seem to get the inequality. I assume it's easiest by trying to show the converse ie if the inequality hold for no constant then you can get a sequence that is Cauchy under one norm but not under the other. But I still can't seem to see it.

Any hope would be appreciated

Thanks,
Steven

2. May 2, 2005

### matt grime

So to clarify, you're asking to show

two norms have the same cauchy sequences if and only if there are constants k and m such that |x|<k|x|' and |y|'<m|y| for all x and y. i'll use | | and | |' for the two norms.

I'm only asking because to me the second of these two (the one with the constants) is how i've usually seen the norms defined to equivalent.

as you say, the <= is trivial.

Now let's try the other way.

Let's just consider the unit ball in | |, ie the x such that |x| <=1.

If we can show this is bounded in | |' we're done, since it sufices to consider only vectors of length 1 by linearity.

If it's not bounded, then there are points x(n) such that |x(n)|' > n for each n, but the x(n) are a sequence in a closed totally bounded subset of a metric space, ie it is compact. (nb, i'm actually just assuming this, as it's the only point at which we need finite dimensionality, so it should be checked: the unit sphere in a finite dimensional normed vector space is compact)

Compact implies sequentially compact in a metric space so it must have a convergent subsequence that is cauchy, relabelling as necessary we may assume x(n) is cauchy in | |, but it must be cauchy in | |', which is a contradiction given the way they were constructed.

Note, all norms on a finite dimensional metric space are equivalent.

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