A function $$f: \mathbb{R} \to \mathbb{R}$$ be increasing and yet be bounded?
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Of course. 1 - 1/x is an obvious one. Okay that is only defined increasing on x > 0 but it can be tweak. $1- (1/2)^x$ if x $\ge$ 0; $-(1 - (1/2)^{|x|}$ if $x < 0$. Is another . There are gazillions of them. Any function strictly below a horizontal asymptote. Any bounded increasing sequence can be extended to such a function. – fleablood Mar 25 '16 at 06:44
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Yes. $f(x)=\arctan x$ has derivative $1/(1+x^2)$ which is positive for all $x$, hence $f$ is strictly increasing (proof: mean value theorem). But $f$ is bounded above and below, by $\pi/2$ and $-\pi/2$ respectively.
grand_chat
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$f(x) = 1 - a^x$, $0 < a < 1; x \ge 0$; $f(x) = -f(|x|); x < 0$.
There's millions of them.
fleablood
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