Equivalent Norms in an infinite vector space?

[andyb177]

Can some one explain why not all norms are equivalent in and infinite vector space?
Examples/Counter examples?
How would you go about proving/disproving this?
Any or all of the above or just any help anyone has to offer would be great
Thanks

 

[micromass]

Take the space \mathcal{C}([0,1]). The norms

\|f\|_1=\int_0^1{|f(x)|dx}~\text{and}~\|f\|_\infty=\sup_{x\in [0,1]}{|f(x)|}

are not equivalent. Indeed, the space equipped with \|f\|_\infty is complete and the space equipped with \|f\|_1 is not.

 

[andyb177]

Is it possible for you to describe why they are equivalent in more detail?
I understand the definitions for the norms etc
If one is complete then do all equivalent others have to be complete as well?

 

[disregardthat]

What do you mean by equivalent norms? Obviously, if two norms are the same function they would share the same properties.

 

[andyb177]

I mean there exists a c st 1/cmod1(v)<= mod2(v) <= cmod1(v)
I think i understand it now, please correct me if I'm wrong if the vector space is of infinite dimensions there isn't a c big enough or there can't exsit a c that can multiply an infinite amount of vector elements?

 

[micromass]

If you work with equivalent norms, then most of the metric properties are preserved. That is: convergent sequences, Cauchy sequences, boundedness, completeness, compactness,... are all preserved under equivalent norms. That is not very hard to check...

 

[andyb177]

Yeap your right, just got my head round the definition, which shows it for a standard norm, but you can replace that for a norm which you want to check for.
Thanks alot, apologies for the blatent question.

 

[Landau]

@Jarle: Two norms n_1 and n_2 are called equivalent iff they induce the same topology. This happens iff the identity map (between the same space considered with one and the other norm) is a homeomorphism. And it happens iff there are constants c,C>0 such that for all x

cn_1(x)\leq n_2(x)\leq Cn_1(x).
It is obviously an equivalence relation.

 

[micromass]

@Landau: so you say that the identity map is a homeomorphism iff there exists c,C such that

cn_1(x)\leq n_2(x)\leq Cn1(x)

Can you give me a reference of the proof of this? I know it isn't true for metrics, but it would be interesting to see that it's true for norms...

 

[Landau]

If I am not mistaking, the proof is trivial. Since for linear operators, continuity is equivalent to boundedness, the identity map being a homeo just means

n_1(x)\leq Cn_2(x)
for some C (id continuous)
n_2(x)\leq cn_1(x)
for some c (id^{-1} continuous).

 

[lugita15]

Take the space \mathcal{C}([0,1]). The norms

\|f\|_1=\int_0^1{|f(x)|dx}~\text{and}~\|f\|_\infty=\sup_{x\in [0,1]}{|f(x)|}

are not equivalent. Indeed, the space equipped with \|f\|_\infty is complete and the space equipped with \|f\|_1 is not.

Completeness is not in general preserved by homeomorphism. For instance, R is homeomorphic to (0,1), and the former space complete while the rather is not. So the mere fact that completeness is not preserved does not imply that the metrics are not equivalent.

 

[lugita15]

If you work with equivalent norms, then most of the metric properties are preserved. That is: convergent sequences, Cauchy sequences, boundedness, completeness, compactness,... are all preserved under equivalent norms. That is not very hard to check...

I agree that topological properties like compactness and convergence are preserved, but why are metric properties like boundedness and completeness preserved?

 

[Landau]

@lugita15: this is immediate from the criterion that both norms can be estimated in terms of each other:

cn_1(x)\leq n_2(x)\leq Cn1(x)

For example, if a set A is bounded w.r.t. n_1, then there exists a K>0 such that for all x in A: n_1(x)\leq K. But then

n_2(x)\leq Cn_1(x)\leq CK.

Convergence of a sequence x_k to x w.r.t. n_1 means that n_1(x-x_k) is small when k is large. But then n_2(x-x_k)\leq Cn_1(x-x_k) is also small. Etc.