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

‖x‖1=n∑i=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.,

c‖x‖≤‖x‖1≤C‖x‖.

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}i∈I which we may assume to be normalized, i.e., ‖ei‖=1 for all i. Every vector x∈X has a unique representation x=∑i∈Ixiei with only finitely many nonzero entries (by definition of a basis).

Now choose a countable subset i1,i2,… of I. Then ϕ(x)=∑∞k=1k⋅xik defines a linear functional on x. Note that ϕ is not continuous, as 1√keik→0 while ϕ(1√keik)=√k→∞.

There can’t be a C>0 such that the norm ‖x‖ϕ=‖x‖+|ϕ(x)| satisfies ‖x‖ϕ≤C‖x‖ since otherwise ‖1√kek‖→0 would imply |ϕ(1√kek)|→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.*