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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 the textbook. First, pick a norm for $X$, say $$\|x\|_0=\sum_{i=1}^n|\alpha_i|$$ where $x=\sum_{i=1}^n\alpha_ix_i$, and $(x_i)_{i=1}^n$ is a basis for $X$. Then show that every norm for $X$ is equivalent to $\|\cdot\|_0$, i.e., $$c\|x\|\leq\|x\|_0\leq C\|x\|.$$ 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 strategy 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, is the following also true: If all norms for a vector space $X$ are equivalent, then $X$ is of finite dimension?

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It's not really different from Bolzano-Weierstrass, but we can use the compacity of $\left\{x\in X, \lVert x\rVert =1\right\}$ and the fact that the map $x\mapsto \lVert x\rVert_0$ is continuous. –  Davide Giraudo Aug 15 '11 at 22:03
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Well, Bolzano-Weierstrass is essentially equivalent to compactness of the unit ball with respect to the norm which you call $\|\cdot\|_0$ and everyone else I know calls $\|\cdot\|_1$. See also Fabian's proof here where the maximum norm is used. –  t.b. Aug 15 '11 at 22:10
    
@Theo: As I understand, Bolzano-Weierstrass is essentially equivalent because the unit sphere is a special closed bounded set(hence compact in the finite dimensional case) and $\|\cdot\|$ is homogeneous. Correct? Or how should one prove the equivalence? –  Jack Aug 16 '11 at 1:23
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I'm not really talking about the unit ball directly :) Starting at "Now" I argue why a bounded set $S$ contains a convergent subsequence: If $S$ is bounded and $v_n \in S$ then using Bolzano-Weierstrass I can extract a subsequence that converges coordinate-wise, hence with respect to $\|\cdot\|_1$. If $S$ is in addition closed, then the limit point will belong to $S$. Does that help? –  t.b. Aug 16 '11 at 1:43
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Side note: if $F$ is a field with absolute value, then it makes sense to define a norm on an $F$-vector space. If $F$ is complete, then the same theorem is true: for a finite-dimensional vector space all norms are equivalent. Note: compactness is not available for the proof! 2nd note: this theorem may fail if $F$ is not complete. –  GEdgar Aug 16 '11 at 14:51
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4 Answers

up vote 8 down vote accepted

To answer the question in the update:

If $(X,\|\cdot\|)$ is a normed space of infinite dimension, we can produce a non-continuous linear functional: Choose an algebraic basis $\{e_{i}\}_{i \in I}$ which we may assume to be normalized, i.e., $\|e_{i}\| = 1$ for all $i$. Every vector $x \in X$ has a unique representation $x = \sum_{i \in I} x_i \, e_i$ with only finitely many nonzero entries (by definition of a basis).

Now choose a countable subset $i_1,i_2, \ldots$ of $I$. Then $\phi(x) = \sum_{k=1}^{\infty} k \cdot x_{i_k}$ defines a linear functional on $x$. Note that $\phi$ is not continuous, as $\frac{1}{\sqrt{k}} e_{i_k} \to 0$ while $\phi(\frac{1}{\sqrt{k}}e_{i_k}) = \sqrt{k} \to \infty$.

There can't be a $C \gt 0$ such that the norm $\|x\|_{\phi} = \|x\| + |\phi(x)|$ satisfies $\|x\|_\phi \leq C \|x\|$ since otherwise $\|\frac{1}{\sqrt{k}}e_k\| \to 0$ would imply $|\phi(\frac{1}{\sqrt{k}}e_k)| \to 0$ 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.

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Aha! But now: is it consistent with ZF that there is some infinite-dimensional vector space where all norms are equivalent... –  GEdgar Aug 16 '11 at 12:14
    
@GEdgar: Apparently I was too obviously trying to avoid that question... Short answer: I don't know. –  t.b. Aug 16 '11 at 12:41
    
@GEdgar Apparently there isn't any contradiction. Note that $\|\cdot\|_\phi$ is defined not on $X$, but on a subspace of $X$ that has a countable basis. –  user1551 Dec 19 '12 at 3:35
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The converse of the theorem is certainly not true. Let $X$ be any infinite dimensional normed space. Set $\| x\|^*=2\|x\|$. The there two norms are equivalent.

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You are going to need something of this nature. A Banach Space is a complete normed linear space (over $\mathbb{R}$ or $\mathbb{C}$). The equivalence of norms on a finite dimensional space eventually comes down to the facts that the unit ball of a Banach Space is compact if the space is finite-dimensional, and that continuous real-valued functions on compact sets achieve their sup and inf. It is the Bolzano Weirstrass theorem that gives the first property.

In fact, a Banach Space is finite dimensional if and only if its unit ball is compact. Things like this do go wrong for infinite-dimensional spaces. For example, let $\ell_1$ be the space of real sequences such that $\sum_{n=0}^{\infty} |a_n| < \infty $. Then $\ell_1$ is an infinite dimensional Banach Space with norm $\|(a_n) \| = \sum_{n=0}^{\infty} |a_n|.$ It also admits another norm $\|(a_n)\|' = \sqrt{ \sum_{n=0}^{\infty} |a_{n}|^2}$ , and this norm is not equivalent to the first one.

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Apparently there are some generalizations -- see this MO-thread. Admittedly I was unable to follow the discussion (but I didn't really bother to try). I'm happy with what paul garrett explained there. –  t.b. Aug 15 '11 at 22:18
    
In that MO thread @Theo B. mentions, the questioner was apparently as much interested in very weak hypotheses on the topological field... didn't want it to have a topology defined by a norm, etc. All those complications (which I barely followed) are irrelevant to the present question. –  paul garrett Aug 15 '11 at 22:35
    
@both: yes, the other thread seems to deal with issues which I don't normally worry about. –  Geoff Robinson Aug 15 '11 at 23:05
    
How do you know that every finite dimensional normed space is Banach? –  André Caldas Feb 3 '13 at 21:51
    
@Andre: You know that because the equivalence of norms on a f.d. space gives isomorphism to $\mathbb{R}^{n}$, with the usual normal $\|(x_{1},\ldots ,x_{n}) \| = \sum_{i=1}^{n}|x_{i}|.$ The proof of this only uses completeness of $\mathbb{R}^{n}$ with respect to this norm- it does not assume completeness of the other space. –  Geoff Robinson Feb 4 '13 at 2:21
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One doesn't really need a different argument for each side of the inequality. If $\vert\vert \cdot \vert\vert_1,\vert\vert \cdot \vert\vert_2$ are two norms on a finite-dimensional vector space (over $\mathbb{R}$ or $\mathbb{C}$), then the restriction of $\vert\vert \cdot \vert\vert_1$ to the closed unit ball of $\vert\vert \cdot \vert\vert_2$ is a continuous function on a compact set (here the finite-dimensionality is used) and is therefore bounded from above by some $M > 0$. By positive homogeneity, it follows that $\vert\vert \cdot \vert\vert_1 \le M \vert\vert \cdot \vert\vert_2$. Switching the roles of $\vert\vert \cdot \vert\vert_1$ and $\vert\vert \cdot \vert\vert_2$, you get $\vert\vert \cdot \vert\vert_2 \le m \vert\vert \cdot \vert\vert_1$, hence $\frac{1}{m} \vert\vert \cdot \vert\vert_2 \le \vert\vert \cdot \vert\vert_1$, for some $m>0$.

Don't take this theorem too seriously though. This kind of equivalence relation between norms is pretty weak and two normed spaces with $\mathbb{R}^n$ as the underlying vector space can be completely different as far as their geometry is concerned (for instance, some norms come from an inner product [hence satisfy the nice geometric property which we call the "Parallelogram law"] and some don't).

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Hm. I'm not convinced. Isn't it much more cumbersome to prove that closed and bounded implies compact with respect to both norms than using the triangle inequality once? –  t.b. Aug 16 '11 at 0:23
    
To add to that, you seem to be assuming that the norms are already continuous with respect to each other. How do you know that? –  t.b. Aug 16 '11 at 0:35
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I guess I was tacitly assuming that one knows that all normed topologies on $\mathbb{R}^n$ are the product topology. –  Mark Aug 16 '11 at 3:05
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