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(Edit: Replaced $||\cdot||_0$ with $||\cdot||_1$ to clarify.)

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 $$\|x\|_1=\sum_{i=1}^n|\alpha_i|$$ 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 $\|\cdot\|_1$, i.e., $$c\|x\|\leq\|x\|_1\leq C\|x\|.$$ 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?


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
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
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
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
up vote 16 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
@GEdgar Is it true in ZF that all infinite dimensional vector space can be given a norm.(Then only we can talk about your question otherwise an infinite dimensional vector space which doesn't have norm will trivially satisfy the condition)? I don't know whether we can define norm on each vector space in ZF, Please tell – Sushil Jul 8 '15 at 10:33
@t.b. can we prove in ZF only that countable no. of linearly independent vectors in a infinite dimensional vector space,(Because any set of finite vectors can't be maximal linearly independent set, right?) – Sushil Jul 8 '15 at 13:18

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

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
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

Here's a proof of the equivalence of norms in finite dimension that does not use compactness arguments. It breaks down into three steps.

We set some notation first: given a normed finite dimensional real vector space $E$, write $E'$ for its algebraic dual space (set of all linear forms) and $E^{*}$ for its topological dual space (set of all continuous linear forms).

  • All linear forms on $E$ are continuous, i.e $E'=E^{*}$, whatever the norm $\|.\|$ on $E$.

The Hahn-Banach theorem states that every continuous form on a subspace of $E$ can be extended to all of $E$. In particular, given $x\in E$, there is a $\varphi:E\rightarrow \mathbb{R}$ for which $\varphi(x)\neq 0$, since there are non trivial forms $\mathbb{R}x\rightarrow \mathbb{R}$ and these can be extended. As a consequence, the map \begin{align*} E &\rightarrow (E^{*})' \\ x &\mapsto (\phi\rightarrow \phi(x)) \end{align*} is injective. We also have isomorphisms $E'\simeq E$ and $E^{*}\simeq (E^{*})'$. Composing the three maps we have an injection $E' \rightarrow E^{*}$ so that $\dim E' \leqslant \dim E^{*}$. On the other hand since $E^{*}\subset E'$ we must have $\dim E^{*}=\dim E'$ and $E^{*}=E'$ as desired.

  • All linear maps between finite dimensional normed spaces are continuous.

Let $A:F\rightarrow E$ be such a map and $e_{i},i\in I$ a basis for $E$. Let $(e_{i}^{*})$ be the dual basis of $(e_{i})$ and $\tilde{e_{i}}$ the linear map $\mathbb{R}\rightarrow E$ that maps $1$ to $e_{i}$. Then $Id_{E}=\sum_{i\in I}\tilde{e_{i}}e_{i}^{*}$. In particular, $A=\sum \tilde{e_{i}}e_{i}^{*}A$. The the $\tilde{e_{i}}$ are continuous and so are the $e_{i}^{*}A$, since they are linear forms. Hence $A$ is continuous as a sum of composites of continuous maps.

  • All norms are equivalent.

Given two norms $\|.\|_{1}, \|.\|_{2}$ on $E$, the identity, being linear, defines an homeomorphism between $E,\|.\|_{1}$ and $E,\|.\|_{2}$. It is easy to check that this means that the norms are equivalent.

This proof is really a way of saying that the topology induced by a norm on a finite-dimensional vector space is the same as the topology defined by open half-spaces; in particular, all norms define the same topology and all norms are equivalent. There are other ways to prove that using the Hahn-Banach theorem.

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