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Let $\phi\colon\mathbb{R}\to\mathbb{R}$ be a function that satisfies $\lvert \phi(x)\rvert \geq 1$ for all real $x$ and $\lvert\phi(x) - \phi(y)\rvert \leq \lvert x - y\rvert $ for all real $x,y$ (for example $\phi(x) = \cos(x)$). Let

$f(x) = \left\{ \begin{array}{ll} 0 & \mbox{if } x = 0 \\ x\phi(\frac{1}{x}) & \mbox{if } x \neq 0 \end{array} \right.$

Show that $f$ is a uniformly continuous mapping from $\mathbb{R} \to \mathbb{R}$.

I got that $\phi(x)$ is Lipschitz with a constant of 1. It seems like one of the Carathéodory's theorems could be involved as well.

Thanks!

tomasz
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Anon
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  • If $\Phi$ was Lipschitz with constant $1$, you would have $|\Phi(x) - \Phi(y)| \le 1*|x - y|$, $\forall x,y\in\Bbb R$. Even the example that you give is the other way around, $|cos(x)|\le 1$, $\forall x,y\in\Bbb R$ – Ana Galois Nov 18 '13 at 23:47
  • I think you have a typo. I think you meant $|\Phi(x) - \Phi(y)| \le |x-y|$. – Stephen Montgomery-Smith Nov 18 '13 at 23:47

1 Answers1

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The assumption on $|\phi|$ should be $|\phi|\le 1$, not the other way around.

The continuity of $f$ at $0$ follows from $\phi$ being bounded. Hence, $f$ is uniformly continuous on every bounded interval.

Concerning the behaviour of $f$ ar $\infty:$ begin with $$x\phi(1/x)-y\phi(1/y)\le x(\phi(1/x)-\phi(1/y)) + (x-y)\phi(1/y)$$ where we may assume $|x|\le |y|$ and $|y|\ge 1 $.

The second term, $(x-y)\phi(1/y)$, is bounded by $|x-y|$.

The first one is bounded by $$|x|\frac{|x-y|}{|xy|} = \frac{|x-y|}{| y|} \le |x-y|$$

Thus, $f$ is $2$--Lipschitz outside of $(-1,1)$.

Post No Bulls
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