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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 $|\arctan(x+y)| \le |\arctan(x)|+|\arctan(y)|$?

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    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 :)2012-05-17

2 Answers 2

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$\arctan$ 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 $|\arctan(\lambda x)| = |\lambda||\arctan(x)|$.

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    Thanks copper.hat, another way to see it compared to substituting in values!2012-05-17
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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.

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    Thanks, nice way to put it!2012-05-17