Equivilant norms of a Vector Space

[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

 

[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.

 

[M98Ranger]

First of all, don't worry, it's completely normal to get stuck on a math problem and have it "kill" you for a while. It's all part of the learning process and it's important to keep pushing through and seeking help when needed.

To understand why the two norms on a finite dimensional Vector space X are equivalent if and only if there exist positive real constants c_1 and c_2 satisfying the given inequality, let's break it down step by step.

First, let's define what it means for two norms to be equivalent. Two norms are equivalent if there exists a constant c such that for all x in X, \|x\|_1 \le c\|x\|_2 and \|x\|_2 \le c\|x\|_1. Essentially, this means that the two norms are "comparable" in the sense that one is always bounded above and below by a multiple of the other.

Now, let's consider the Cauchy criterion for a sequence in a vector space. A sequence (x_n) in a vector space X is Cauchy if for any \epsilon > 0, there exists an N such that for all m,n > N, \|x_m - x_n\| < \epsilon. In other words, the terms of the sequence are getting closer and closer together as n increases.

Now, let's assume that the two norms on X are equivalent, i.e. there exists a constant c such that for all x in X, \|x\|_1 \le c\|x\|_2 and \|x\|_2 \le c\|x\|_1. This means that the two norms are "comparable" in the sense that one is always bounded above and below by a multiple of the other.

Now, let's consider a sequence (x_n) in X that is Cauchy under the norm \| \cdot \|_1. By the Cauchy criterion, for any \epsilon > 0, there exists an N such that for all m,n > N, \|x_m - x_n\|_1 < \epsilon. But since the norms are equivalent, we also have \|x_m - x_n\|_2 \le c\|x_m - x_n\|_1 < c\epsilon. This means that the sequence (x_n) is also Cauchy under the