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Let $f:\mathbb R\rightarrow \mathbb R$ be a continuous function such that $$|f(x)-f(y)|\geq \frac12|x-y|$$ for all$x,y\in \mathbb R$. Then is $f$ one-one and onto?

Let $f(x)=f(y)$ i.e. $0=|f(x)-f(y)|\geq (1/2)|x-y|$ i.e $x=y$

Hence $f$ is injective.

But I am unable to conclude whether $f$ is onto.Any help

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3 Answers 3

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Hint. Prove that $f$ is monotonic and unbounded above and below. Then apply the intermediate value theorem (again).

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$f$ is one to one and continue, therefore, $f$ is stricly increasing or strictly decreasing. Let us show that $f$ is not bounded, Let $y=0$, then $-|f(x)-f(0)|\leq \frac{x}{2}\leq |f(x)+f(0)|$

Therefore $|f(x)|\geq \frac{x}{2}-|f(0)|$ and so there is no upper bound and $|f(x)|\leq \frac{x}{2}+|f(0)|$ therefore $f$ is not low bounded, therefore $f$ is onto.

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Set $y=0.$ Choose any real value $c > 0$ you like.

Can you choose $x$ so that you can guarantee that $f(x) > c$?

Can you choose $x$ so that you can guarantee that $f(x) < -c$?

What does that say about the existence of $x$ such that $f(x) = r$ for $-c \leq r \leq c$?

Given a number $r \in \mathbb R,$ how might you go about showing that $\exists x. f(x) = r$?

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