Let $|\cdot|_1$ and $|\cdot|_2$ be two norms on a field $\mathbb F$. We call the two norms equivalent if every Cauchy-sequence with respect to $|\cdot|_1$ is also a Cauchy-sequence with respect to $|\cdot|_2$. Prove the following statement:

$$|\cdot|_1\sim|\cdot|_2\quad\Leftrightarrow\quad\exists \alpha\in\mathbb{R}_{>0}: \forall x\in\mathbb F: |x|_1=|x|_1^\alpha.$$

The direction "$\Leftarrow$" is straightforward: Let there an $\alpha$ with the property above, $(a_i)_{i\in\mathbb N}$ be a Cauchy sequence with respect to $|\cdot|_1$ and $\varepsilon_2\in\mathbb{R}_{>0}$ be arbitrary. Set $\varepsilon_1:=\varepsilon_2^\alpha$ and follow from the fact that $(a_i)_{i\in\mathbb N}$ is a Cauchy sequence, that there exists an $N\in\mathbb N$ such that $$\forall n,m>N: |a_m-a_n|_1<\varepsilon_1.$$ Since $|\cdot|_1\sim|\cdot|_2$ and the definition of $\varepsilon_1$ we get that $$\forall n,m>N: |a_m-a_n|_2^\alpha<\varepsilon_1^\alpha$$ which is equivalent to $$\forall n,m>N: |a_m-a_n|_2<\varepsilon_1$$ which means that $(a_i)_{i\in\mathbb N}$ is a Cauchy sequence with respect to $|\cdot|_2$.

For the other direction (which is probably harder) we may assume that the being a Cauchy sequence is the same property for both norms but I don't see how I can construct such an $\alpha$ from this fact.

Remark 1: This is an exercise number 5 on page 7 in the book "p-adic Numbers, p-adic Analysis and Zeta-Functions" (Second Edition) by Neal Koblitz.

Remark 2: The right side is the notion of norm-equivalence that I am familiar with, but in this book it is explicitly defined in the way from this post.


To prove the direction in question, assume otherwise. That is, there exists $x, y \in \Bbb{F}^{\times}$ such that

$$ \frac{\log|x|_1}{\log|x|_2} = \alpha(x) \neq \alpha(y) = \frac{\log|y|_1}{\log|y|_2}. \tag{1} $$

The naive idea of the proof is to exaggerate this difference in a deliberate way.

The condition $(1)$ means that two vectors

$$ v_x = (\log|x|_1, \log|x|_2) \quad \text{and} \quad v_y = (\log|y|_1, \log|y|_2) $$

are linearly independent, forming a basis of $\Bbb{R}^2$. In particular,

$$L = v_x \Bbb{Z} \oplus v_y \Bbb{Z} = \{ m v_x + n v_y : m, n \in \Bbb{Z}\}$$

is a lattice on $\Bbb{R}^2$. Then for any vector $v = pv_x + qv_y \in \Bbb{R}^2$, the lattice point $w = [p]v_x + [q]v_y \in L$ satisfies

$$ \|v - w\| \leq \|v_x\| + \|v_y\| =: R.$$

In particular, any closed ball of radius $R$ contains at least one lattice point of $L$. Thus for each $k = 1, 2, 3, \cdots$, we can find some a sequence of pairs of integers $(m_k, n_k)$ such that

$$ m_k v_x + n_k v_y \in [2kR, 2(k+1)R] \times (-\infty, -kR]. $$

Then for

$$z_k = x^{m_k}y^{n_k} \in \Bbb{F}$$

we have

$$ e^{2kR} \leq |z_k|_1 \leq e^{2(k+1)R} \quad \text{and} \quad |z_k|_2 \leq e^{-kR}. $$

Now, for any $l+2 \leq k$ we have

$$|z_k - z_l|_1 \geq |z_k|_1 - |z_l|_1 \geq e^{2Rk} - e^{2R(l+1)} \geq e^{2R(l+2)} - e^{2R(l+1)} \geq e^{2R} - 1 > 0$$

and $(z_k)$ is not Cauchy in $|\cdot|_1$. But clearly $(z_k)$ is Cauchy in $|\cdot|_2$. Thus two norms are not equivalent and therefore the proof is completed.

  • $\begingroup$ there is something wrong in the line starting with "is a lattice on $\mathbb{R}^2$: Do you mean for all $p,q\in\mathbb R$ define $v$ and $w$ in the way you did and then use the triangle-inequality? $\endgroup$ – born Aug 30 '12 at 15:15
  • $\begingroup$ And then I don't get the line with $k\rightarrow\infty$: Do you mean that the relation in the line befor ($m_kv_x+n_kv_y\in[2kR,2(k+1)R)\times(-\infty,-kR)$ holds for all $k$? And then: Why does it hold for all $k$? :-) $\endgroup$ – born Aug 30 '12 at 15:16
  • $\begingroup$ In the line after you chose $l$ with $l+2\leq k$ there is a $\leq$ in the middle where (imho) should be an $\geq$. $\endgroup$ – born Aug 30 '12 at 15:39
  • $\begingroup$ the $=$-sign after that should also be a $\geq$. $\endgroup$ – born Aug 30 '12 at 15:42
  • $\begingroup$ @born, Sorry for some typos I made. I wrote it when I was drinking 2 cans of beers. I also fixed some expressions so that the proof becomes clearer. I hope that this will be an appropriate answer for your questions. $\endgroup$ – Sangchul Lee Aug 30 '12 at 17:09

Hint: \begin{equation} |a_m-a_n| < \varepsilon \Leftrightarrow |a_m-a_n|_{2}^{\alpha}<\varepsilon \Leftrightarrow |a_m-a_n|_{2}< \varepsilon^{1/\alpha}. \end{equation}


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