Understanding of the theorem that all norms are equivalent in finite dimensional vector spaces

The following is a well-known result in functional analysis:

If the vector space X is finite dimensional, all norms are equivalent.

Here is the standard proof in one textbook. First, pick a norm for X, say
x1=ni=1|αi|
where x=ni=1αixi, and (xi)ni=1 is a basis for X. Then show that every norm for X is equivalent to 1, i.e.,
cxx1Cx.
For the first inequality, one can easily get c by triangle inequality for the norm. For the second inequality, instead of constructing C, the Bolzano-Weierstrass theorem is applied to construct a contradiction.

The strategies for proving these two inequalities are so different. Here is my question,

Can one prove this theorem without Bolzano-Weierstrass theorem?

UPDATE:

Is the converse of the theorem true? In other words, if all norms for a vector space X are equivalent, then can one conclude that X is of finite dimension?

Answer

To answer the question in the update:

If (X,) is a normed space of infinite dimension, we can produce a non-continuous linear functional: Choose an algebraic basis {ei}iI which we may assume to be normalized, i.e., ei=1 for all i. Every vector xX has a unique representation x=iIxiei with only finitely many nonzero entries (by definition of a basis).

Now choose a countable subset i1,i2, of I. Then ϕ(x)=k=1kxik defines a linear functional on x. Note that ϕ is not continuous, as 1keik0 while ϕ(1keik)=k.

There can’t be a C>0 such that the norm xϕ=x+|ϕ(x)| satisfies xϕCx since otherwise 1kek0 would imply |ϕ(1kek)|0 contrary to the previous paragraph.

This shows that on an infinite-dimensional normed space there are always inequivalent norms. In other words, the converse you ask about is true.

Attribution
Source : Link , Question Author : Community , Answer Author : t.b.

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