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An absolute value $|\cdot|$ of field $F$ is a function $F \to \Bbb R$ such that

  1. $|x|≥0$ for any $x∈V$ and $|x|=0$ iff $x=0$
  2. $|xy|=|x||y|$
  3. $|x+y|≤|x|+|y|$ for any $x,y∈F$

An absolute value can be viewed as a norm on $F$. The absolute value of $\Bbb R$ and the modulus of $\Bbb C$ are both not differentiable at 0. I am wondering if all absolute values defined on subfields of $\Bbb C$ are not differentiable at zero.

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    $\begingroup$ There are plenty of finite fields on which differentiability makes no sense... $\endgroup$ – Brian Borchers Feb 26 '17 at 4:58
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    $\begingroup$ Just because you can set a topology on $F$ using the absolute value doesn't mean you still have differentiability as a well-defined notion. The definition is of the form $\lim_{x\to y}{\frac{f(x)-f(y)}{x-y}}$; in your case, $f$ is the absolute value, so you're taking a quotient of something in $\mathbb{R}$ and something in $F$, which in general does not need to make sense. So at the very least you need to restrict to subfields of $\mathbb{R}$ or field extensions of $\mathbb{R}$ (say, $\mathbb{C}$ or $\mathbb{R}(x)$) $\endgroup$ – Hayden Feb 26 '17 at 5:08
  • $\begingroup$ Thanks for your comments! I have edited the question. $\endgroup$ – River Feb 26 '17 at 5:10
  • $\begingroup$ @Hayden : ​ No, that operation is fine on any field-extension of $\mathbb{R}$. ​ ​ ​ ​ $\endgroup$ – user57159 Feb 26 '17 at 5:11
  • $\begingroup$ @River : ​ See my above comment. ​ ​ ​ ​ $\endgroup$ – user57159 Feb 26 '17 at 5:11
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The answer to the question in the title is yes. More generally --

I am wondering if all absolute values defined on subfields of $\mathbb{C}$ are not differentiable at zero.

Yes, this is the case. Any subfield of $\mathbb{C}$ contains $\mathbb{Q}$. We show that as long as $| \cdot |$ is defined on $\mathbb{Q}$, it is not differentiable at $0$ in the following sense: $$ \lim_{\substack{h \to 0 \\ h \in \mathbb{Q}}} \frac{|h| - |0|}{h - 0} = \lim_{\substack{h \to 0 \\ h \in \mathbb{Q}}} \frac{|h|}{h} \tag{*} $$ does not exist as a complex number. (That is to say, the limit above does not exist with respect to the usual topology on complex numbers.)

First, note that property (2) implies $|1|^2 = |1|$, so $|1| \in \{0,1\}$, and by property (1) $|1| = 1$. Then by property (2) $|-1|^2 = |1| = 1$ so $|-1| = 1$. By property (2) we conclude that $|-x| = |x|$ for all $x$.

Consider two sequences converging to $0$: $a_n = \frac{1}{2^n}$, and $b_n = \frac{-1}{2^n}$. On the one hand, $$ \frac{|a_n|}{a_n} = \frac{|1/2|^n}{(1/2)^n} = 2^n \left|\frac12\right|^n = \left( \left|\frac12\right| + \left|\frac12\right| \right)^n \ge (|1|)^n = 1, $$ by property (3). On the other hand, $$ \frac{|b_n|}{b_n} = \frac{|1/2|^n}{-(1/2)^n} = -2^n \left|\frac12\right|^n = -\left( \left|\frac12\right| + \left|\frac12\right| \right)^n \le -(|1|)^n = -1. $$ We have two sequences which cannot possibly converge to the same number, so the limit in (*) does not exist.

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  • $\begingroup$ Note that this answer is with respect to the usual topology, not the one defined by |$\cdot$|. ​ The function that sends zero to itself and everything else to 1 induces the discrete topology (which makes the notion of limit undefined), and there might be other topologies (beyond usual and discrete) induced by other absolute values. ​ ​ ​ ​ $\endgroup$ – user57159 Feb 26 '17 at 5:55
  • $\begingroup$ @RickyDemer Thanks for the sidenote, yes. I clarified your point in my post. I haven't thought about using another topology, such as the one induced by the custom absolute value function we are considering. $\endgroup$ – 6005 Feb 26 '17 at 6:02
  • $\begingroup$ The argument in my post shows the limit does not exist as long as the real number intervals $(-\infty,1]$ and $[1,\infty)$ have disjoint closures. And I think these are already closed in any reasonable topology -- one that includes the order topology on $\mathbb{R}$, for instance. $\endgroup$ – 6005 Feb 26 '17 at 6:15
  • $\begingroup$ You also need your two sequences to actually be converging to 0. ​ ​ $\endgroup$ – user57159 Feb 26 '17 at 6:29
  • $\begingroup$ @RickyDemer Correct on both accounts. I fixed the limit indices. $\endgroup$ – 6005 Feb 26 '17 at 6:36

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