5
$\begingroup$

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 $\mathbb R$.

$\endgroup$
  • 1
    $\begingroup$ What does being a norm mean «over any field»? $\endgroup$ – Mariano Suárez-Álvarez Jan 14 '15 at 20:26
  • $\begingroup$ I meant that I want to prove this for spaces over all fields not only for spaces over $\mathbb R$ $\endgroup$ – Anna Abasheva Jan 14 '15 at 20:31
  • 1
    $\begingroup$ 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 $\mathbb{C}$. $\endgroup$ – DisintegratingByParts Apr 1 '16 at 4:39
  • 1
    $\begingroup$ @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 $\mathbb Q$. $\endgroup$ – Sameer Kulkarni Apr 1 '16 at 18:58
  • 3
    $\begingroup$ @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. $\endgroup$ – DisintegratingByParts Apr 1 '16 at 19:02
6
+50
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Anna Abasheva Jan 14 '15 at 22:05
  • $\begingroup$ the unit sphere is compact for finite dimensional spaces, which is one of the assumptions in the origins question. books.google.com.au/… $\endgroup$ – beedge89 Jan 14 '15 at 22:58
  • $\begingroup$ 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. $\endgroup$ – Anna Abasheva Jan 14 '15 at 23:18
0
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ Equivalence of norms does not imply one is a scalar multiple of the other. $\endgroup$ – Arun Kumar Mar 30 '16 at 5:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.