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let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function such that $|f(x) - f(y)| ≥ |x - y|$ for all $x,y$ in $\mathbb{R}$. then $f$ is onto.

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What have you tried? What theorems about continuous functions do you think might be relevant? Do you know what possible sets can be the range of a continuous function from $\mathbb R$ to $\mathbb R$? Have you noticed any properties of functions satisfying this inequality? For example, is such an $f$ necessarily injective? –  Jonas Meyer Sep 27 '12 at 4:19
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4 Answers

up vote 4 down vote accepted

Let $A=f(\mathbb R)$. The inequality implies that $f$ is injective with continuous inverse defined on $A$, hence by assumed continuity of $f$, $f:\mathbb R\to A$ is a homeomorphism. The inequality along with continuity of $f$ and completeness of $\mathbb R$ also implies that $A$ is complete with the metric restricted from $\mathbb R$. The only subspace of $\mathbb R$ that is homeomorphic to $\mathbb R$ and complete (with the restricted metric) is $\mathbb R$, so $A=\mathbb R$.

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This seems likely beyond the scope of the course (which is not specified) to me. Maybe it will encourage some to go further. +1 in the hope that it will. –  Ross Millikan Sep 27 '12 at 4:44
Ross: Thank you. I do not intend to match the scope of a course with this answer. –  Jonas Meyer Sep 27 '12 at 4:48
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Assume $f$ is not onto, implying that there exist $y\in Y$, here i denote Y as the codomain of $f$, where no $x\in X$, X is the domain of $f$, satisfies $f(x)=y$. Note that $f$ is continuous in the whole real number line. In another words, for any $y\in Y$ as well as in real number line, due to continunity, there should exist and $x$ s.t.$\lim_{t\rightarrow x}f(t)=y$ implies $x$ satisfies $f(x)=y$ for any $y$ in real which is a contradiction

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I don't see an answer. Where is the inequality used? Not all continuous functions from $\mathbb R$ to $\mathbb R$ are onto. –  Jonas Meyer Sep 27 '12 at 4:17
That's no true. $f:\Bbb R\to \Bbb R$ given by $f(x)=e^x$ is continuous in the whole real line, but there is no $x$ such that $\lim_{t\to x} e^x=-1$ for example. –  leo Sep 27 '12 at 4:17
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The given inequality implies that $f$ must be one-to-one. Without loss of generality, suppose $f$ is increasing.

Suppose now that $f$ does not take the value $M>f(0)$ with $M$ positive. Then by continuity we must have $|f(x)-f(0)|\le M-f(0)$ for all $x\ge 0$. This, however, contradicts your given inequality (consider the inequality with $y=0$ and $x>M-f(0)$). Thus, $f$ is not bounded above.

Now argue in a manner similar to the above that $f$ is not bounded below.

So, you have a continuous function which is neither bounded above nor bounded below...

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If $M$ is connected and $f$ is continuous, then $f(M)$ is connected. $\mathbb{R}$ is connected. Hence $f(\mathbb{R})$ must be an interval. Now you only have to prove that $f(\mathbb{R})$ cannot be bounded from above or below. That's up to you.

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While this is true, I think it is clear from OP's question that he is doing introductory real analysis and that the (albeit cleaner) topological solution isn't appropriate. –  nullUser Oct 2 '12 at 15:33
I think that very little is clear from the OP's question. He neglected to share context (beyond tags) or thoughts about the problem, and we do not owe guesses to accommodate unstated needs more appropriately. Also, answers are for more than just the OP. Also, both of the other answers practically rely on connectedness without explicitly saying it. –  Jonas Meyer Jan 5 '13 at 20:12
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