condition for equivalence of norms on vector spaces

Let us call two norms $|x|_1$ and $|x|_2$on a finite-dimensional vector space equivalent if they set the same topology on that space. I need to show that this definition is equivalent to the existence of two constants $C_1$ and $C_2$ such that $C_1|x|_2\le|x|_1\le C_2|x|_2$. Attention: the space can be over any field, not only $\mathbb R$.

• What does being a norm mean «over any field»? – Mariano Suárez-Álvarez Jan 14 '15 at 20:26
• I meant that I want to prove this for spaces over all fields not only for spaces over $\mathbb R$ – anna abasheva Jan 14 '15 at 20:31
• You must be able to make sense of $\|\alpha x\|=|\alpha|\|x\|$ for all $\alpha$ in the field. And how do you define the absolute value of an element in a general field? A norm is defined only over $\mathbb{R}$ or $\mathbb{C}$. – Disintegrating By Parts Apr 1 '16 at 4:39
• @TrialAndError: Why? You can have the concept of a norm on an arbitrary field also, the simplest being the trivial norm.. For a non-trivial example look at $p$-adic norm on $\mathbb Q$. – Sameer Kulkarni Apr 1 '16 at 18:58
• @SameerKulkarni : You don't have a norm for an arbitrary field. You need an absolute value function, and that absolute value function must play nice with the norm. – Disintegrating By Parts Apr 1 '16 at 19:02

I think that if you want to work with norms on vector spaces over fields in general, then you have to use the concept of valuation.

Valued field: Let $K$ be a field with valuation $|\cdot|:K\to\mathbb{R}$. This is, for all $x,y\in K$, $|\cdot|$ satisfies:

1. $|x|\geq0$,
2. $|x|=0$ iff $x=0$,
3. $|x+y|\leq|x|+|y|$,
4. $|xy|=|x||y|$.

The set $|K|:=\{|x|:x\in K-\{0\}\}$ is a multiplicative subgroup of $(0,+\infty)$ called the value group of $|\cdot|$. The valuation is called trivial, discrete or dense accordingly as its value group is $\{1\}$, a discrete subset of $(0,+\infty)$ or a dense subset of $(0,+\infty)$. For example, the usual valuations in $\mathbb{R}$ and $\mathbb{C}$ are dense valuations.

Norm: Let $(K,|\cdot|)$ be a valued field and $X$ be a vector space over $(K,|\cdot|)$. A function $p:X\to \mathbb{R}$ is a norm iff for each $a,b\in X$ and each $k\in K$, it satisfies:

1. $p(a)\geq0$ and $p(a)=0$ iff $a=0_X$,
2. $p(ka)=|k|p(a)$,
3. $p(a+b)\leq p(a)+p(b)$

Proposition (Non-trivial valuation case): Let $(K,|\cdot|)$ be a valued field with non-trivial valuation and let $X$ be a vector space over $(K,|\cdot|)$. Two norms $p_1,p_2$ on $X$ are equivalent iff there are constants $c_1$ and $c_2$ such that $c_1p_1\leq p_2\leq c_2p_1$.

Proof: If there are constants $c_1$ and $c_2$ such that $c_1p_1\leq p_2\leq c_2p_1$, then it is clear that $p_1$ and $p_2$ are equivalent.

Now suppose that $p_1$ and $p_2$ are equivalent.

Then there exists $\delta>0$, such that, for all $a\in X, p_2(a)<\delta\implies p_1(a)<1.$

Let $a\in X-\{0_X\}$, then $p_2(a)\neq0$.

Claim: There exists $x\in K-\{0\}$ such that $|x|<1$.

Since the valuation on $K$ is non-trivial, there exist $y\in X-\{0\}$ such that $|y|\neq1$. Because of the equality $|y||y^{-1}|=1$, we can choose $|x|=\min\{|y|,|y^{-1}|\}$.

Claim: There exist $n\in\mathbb{Z}$ such that $\frac{\delta|x|}{2}< |x|^np_2(a)<\delta$.

The claim is direct application of the following:

Lemma: $(\forall 0<s<1)(\forall \delta>0)(\forall w>0)(\exists n\in\mathbb{Z})(\frac{\delta s}{2}<s^nw<\delta).$

Proof: Let $0<s<1, 0<\delta<1$ and $w>0$ be given. Consider $n$ such that $s^n w$ is as close to $\delta/2$ as possible while still being smaller or equal to $\delta/2$. Since $0<s<1$, this is $n=\lceil \log_s(\delta/2w) \rceil$. Then $n<\log_s(\delta/2w)+1$ so $\frac{\delta s}{2}<s^nw\leq\frac{\delta }{2}<\delta$.

(Thanks to Ian for helping me with the lemma).

Now from the last claim we have that $\frac{1}{|x|^n}<\frac{2}{\delta|x|}p_2(a)$ and $p_2(x^n a)=|x|^np_2(a)<\delta$. The last inequality implies that $p_1(x^n a)<1.$

Finally, $p_1(a)<\frac{1}{|x|^n}<\frac{2}{\delta|x|}p_2(a)$. Hence, $p_1(a)\leq\frac{2}{\delta|x|}p_2(a)$, $\forall a\in X$.

By symmetry, we can find a constant $c>0$ such that $p_2(a)\leq cp_1(a)$, $\forall a\in X$. $\blacksquare$

Notice that we didn't need any information about the dimension of $X$ in the proof.

Cases with trivial valuation: If $X$ is a finite-dimensional space, then the proposition is still true when the valuation is trivial. In fact, in that case, for any norm $p$ there are constants $c_1$ and $c_2$ such that $c_1q\leq p\leq c_2q$, where $q$ is the trivial norm (i.e. $q(a)=1$ if $a\neq0$). See the proof here.

If $X$ is an infinite-dimensional space and the valuation is trivial, then the proposition is false in general. See a counterexample here.

Since all the possible cases are considered, and any field can be considered as a valued field (for example, we always may consider the trivial valuation), then the proof is complete.

This link might help. Have a think about what it means to 'set the same topology'...

http://math.mit.edu/~stevenj/18.335/norm-equivalence.pdf

• They prove this theorem only for spaces over $\mathbb C$ with use of the compactness of the unit sphere. The unit sphere is not always compact. For example, that's not true for incomplete spaces – anna abasheva Jan 14 '15 at 22:05
• the unit sphere is compact for finite dimensional spaces, which is one of the assumptions in the origins question. books.google.com.au/… – beedge89 Jan 14 '15 at 22:58
• The proof is given for $\mathbb R$ and $\mathbb C$. Think of the unit sphere in $\mathbb Q^2$ - it is not sequentially compact as there is sequence there that converges to an irrational point, so you can't choose a convergent subsequence. – anna abasheva Jan 14 '15 at 23:18

If the two norms satisfy $\mid x\mid_1=C_2\mid x\mid_2$, then $\mid x-a \mid_1 <d$ if and only if $\mid x-a \mid_2 <d/C_2$. Hence the open balls are related by $B_1(a,d)=B_2(a,d/C_2)$, with $B_i(a,d)=\{ x\in K:\mid x-a\mid_i<d \}$ for $i=1,2$. Thus the basis of open neighborhoods of $a$ for $\mid \cdot \mid_1$ and $\mid \cdot \mid_2$ are identical. By definition then the induced topologies coincide.

• Equivalence of norms does not imply one is a scalar multiple of the other. – Arun Kumar Mar 30 '16 at 5:50