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Suppose $f:\mathbb{R}\to\mathbb{R}$ be a continuous function satisfying $|f(x)-f(y)|\ge \frac{1}{2}\cdot|x-y|$ for all $x,y\in \mathbb{R}$, is $f$ onto?

I know this is true if $|f(x)-f(y)|\ge |x-y|$

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Mathronaut
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    How do you prove the case you know the answer of? Once you have explained this, you might want to add what is stopping you in applying the same strategy in the case at hand? You might also want to check whether a very simple transformation of $f$ would not show that the $\frac12$ case is equivalent to the $1$ case... – Did Dec 10 '13 at 17:45
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    @Did IMO this is enough hints for a full fledged answer... – gt6989b Dec 10 '13 at 17:47
  • @gt6989b Not everybody agrees, it seems (but I do). – Did Dec 10 '13 at 18:54
  • Got something from my answer? You have been remarkably silent about it... – Did Dec 21 '13 at 10:29
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    do you mean the transformation $g(x)=2f(x)$? – Mathronaut Dec 22 '13 at 07:16

1 Answers1

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How does one prove the case you know the answer of? Once you have explained this, you might want to add what is stopping you in applying the same strategy in the case at hand.

You might also want to check whether a very simple transformation of $f$ would not show that the $\frac12$ case is equivalent to the $1$ case.

Did
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