Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is $f(x) = |\arctan(x)|$ a norm on $\mathbb{R}$?

Im checking if the properties of a norm holds for $f(x) = |\arctan(x)|$.

$1. \ f(x) \ge 0 \Leftrightarrow |\arctan(x)| \ge 0 \\ 2. \ f(x)=0 \Leftrightarrow |\arctan(x)| =0 \Leftrightarrow x=0 \\$

But does $f(\lambda x)=|\arctan(\lambda x)|\Leftrightarrow |\lambda||\arctan(x)|?$ For some $\lambda \in \mathbb{K}$.

Also, how would I check if $|\arctan(x+y)| \le |\arctan(x)|+|\arctan(y)|$?

share|cite|improve this question
Have you tried checking whether $|\arctan(\lambda x)|=|\lambda||\arctan(x)|$ for some actual values of $\lambda$ and $x$? – Chris Eagle May 17 '12 at 15:30
Echoing Chris, take an example where $\lambda\neq 0$ and $x\neq 0$. – Jonas Meyer May 17 '12 at 15:31
You also need $f(x)=0 \implies x=0$, don't you? It is true, but you should state it. – Ross Millikan May 17 '12 at 15:33
Thanks, so by giving (1) counterexample with actual values of $\lambda$ and $x$ and showing the equality does not hold, is sufficient? Thanks Ross Millikan, I stated it above :) – Steven May 17 '12 at 15:48
up vote 9 down vote accepted

$\arctan$ can be used a distance, but not as a norm. (As in $d(x,y) = |\arctan(x)-\arctan(y)|$, which produces an incomplete metric space.)

$\arctan$ is bounded, so it cannot satisfy $|\arctan(\lambda x)| = |\lambda||\arctan(x)|$.

share|cite|improve this answer
Thanks copper.hat, another way to see it compared to substituting in values! – Steven May 17 '12 at 15:51

arctan is an odd function and its values at positive arguments are positive, so questions about the inequality $|\arctan(x+y)|\le|\arctan x| + |\arctan y|$ are reducible to questions where $x$ and $y$ are positive and no absolute values are considered.

So if $x,y>0$, how do we know $\arctan (x+y) \le \arctan x + \arctan y$?

Just notice that the growth rate of the arctan function gets smaller as $x$ gets bigger. If we fix $x>0$ and let $y$ grow from $0$ to some positive number, the right side of the inequality is always growing faster than the left side, since on the right side you're taking the arctan of something closer to $0$.

Therefore the inequality is true.

share|cite|improve this answer
Thanks, nice way to put it! – Steven May 17 '12 at 15:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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