In a proof of Ostrowski's theorem (The only nontrivial norms on $\mathbb{Q}$ are $\|\cdot\|_{p}$ and $\|\cdot\|_{\infty}$), we come to the point where we have shown a certain norm, $\|\cdot\|$, has $\|n\|=n^{\alpha}$ for all natural numbers $n$ and a posititve constant $\alpha$. How is it that we can immediately conclude that $\|\cdot\|$ is equivalent to $\|\cdot\|_{\infty}$? (Two norms are equivalent if they induce equivalent metrics. Two metrics are equivalent if they have the same Cauchy sequences.) Thank you for any help or insight you can give.
Equivalence of Euclidean norm used in Ostrowski's Theorem
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0$||\cdot||_{\infty}$ is the Euclidean norm on $\mathbb{Q}$ – 2012-01-07
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0...isn't it usually the max norm? – 2012-01-07
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0http://en.wikipedia.org/wiki/Ostrowski%27s_theorem The definition in the link says that two absolute values on a field are equivalent if one is a power of the other. Maybe the definition used in your proof is the same. Then there is nothing to prove. – 2012-01-07
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0Thanks Beni. Now that I see we have proved $||n||=||n||_{\infty}^{\alpha}$ for all natural numbers, we have it for all rationals, and I am just left with changing my defintion of equivalence – 2012-01-07
1 Answers
Given $\| n \| = n^{\alpha}$ and the property $\| x y \| = \| x \| \cdot \|y \|$, we can derive $\| x \| = \| x \|_{\infty}^{\alpha}$ for all $x \in \mathbb Q$. To prove this, take $x = \frac{m}{n}$, or $m = x \cdot n$; then $\| m \| = \| x \| \cdot \| n \|$.
It remains to show that if two norms $\| \cdot \|$ and $\| \cdot \|_{\infty}$ on a field are (positive) powers of each other, then they are topologically equivalent.
Suppose the sequence $(x_n)$ is Cauchy w.r.t. $\| \cdot \|$. Therefore, given $\varepsilon > 0$, there exists $N$ such that for all $n, m \geqslant N$, we have $\| x_n - x_m \| \leqslant \varepsilon$. Therefore, $\| x_n - x_m \|_{\infty}^{\alpha} \leqslant \varepsilon$, or $\| x_n - x_m \|_{\infty} \leqslant \varepsilon^{1/\alpha}$ for all $n, m \geqslant N$. Since the final statement is true for all $\varepsilon > 0$, it follows that $(x_n)$ is Cauchy w.r.t. $\| \cdot \|_{\infty}$.
Reversing the above argument, we conclude that if $(x_n)$ is Cauchy w.r.t. $\| \cdot \|_{\infty}$, then it is Cauchy w.r.t. $\| \cdot \|$ as well. Thus the norms $\| \cdot \|$ and $\| \cdot \|_{\infty}$ have the same set of Cauchy sequences; i.e., they are topologically equivalent.
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0Very nice. Thanks for the help! – 2012-01-07