Continuity of one Norm w.resp. to Another. Meaning?

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SUMMARY

The discussion centers on proving the equivalence of norms in finite-dimensional normed spaces, specifically addressing the continuity of norms as functions of one another. The key assertion is that every norm in a finite-dimensional space V can be expressed as a continuous function of another norm. The proof involves demonstrating that if two norms are equivalent on the unit sphere, they are also equivalent throughout the space. The participants explore the implications of this continuity and the conditions under which it holds.

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  • Understanding of finite-dimensional normed spaces
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Mathematicians, students of functional analysis, and anyone interested in the properties of norms in finite-dimensional spaces will benefit from this discussion.

Bacle2
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Hi, All:

I am working on a proof of the fact that any two norms on a f.dim. normed space V are equivalent. The idea seems clear, except for a statement that (paraphrase) any norm in V is a continuous function of any other norm. For the sake of context, the whole proof goes like this:

1)Show that , if any two norms f,g on the unit sphere on V are equivalent, then f,g are equivalent in the whole of V. Easy; just rescaling.

2)**Every norm is a function of some other norm **

3) In particular, from 2, f,g are both cont. functions of another norm,say, h.

4)We define the function j on the unit sphere S^1 of V by : j=f/g . Then, by compactness of


S^1 , there are constants m, M with 0<m<=M with m<j<M , i.e., m<f/g < M

Not too hard.

Still, I'm having trouble pinning-down the meaning of the statement in 2. Any Ideas?

Thanks.
 
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It just means that for every norm f on V, there exists a continuous function F:ℝ→ℝ and a norm h on V, such that f=F\circ h.
 
But this seems like a very stringent condition. After some reflection with my friend

double-espresso , I think it is more reasonable that , given a fixed norm N and other

norms f,g as above, that we can say that f,g are continuous functions in the

topology that N gives rise to , when we use the metric d(x,y)=N(x-y) , don't

you think this seems more reasonable ?
 

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