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$\|f(x)-f(y)\|\geqslant \frac 1{2} \|x-y\|$ for any $x, y \in W$


$f$ is injective in $W$

How to prove this? If that inequality is right is it mean that the images are equal or not?

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If the distance between two points is more than zero, then they're not the same point. – Michael Hardy Jun 6 '12 at 16:24
Just to make a comment, I think I saw this in a proof of the Inverse Function Theorem. – Aden Dong Jun 6 '12 at 16:52
@AdenDong, you are right. Thanks – Kamil Hismatullin Jun 6 '12 at 17:24
up vote 8 down vote accepted

Assume $f(x) = f(y)$. What is $\|f(x)-f(y)\|$? What can you conclude about $\|x-y\|$?

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Thanks. It's mean that the proof of this expression is in definition of injective function. $f(x) = f(y) => x=y $ – Kamil Hismatullin Jun 6 '12 at 17:35

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