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how to prove, $f$ is onto if $f$ is continuous and satisfying $|f(x) - f(y)| ≥ |x - y|$ for all $x,y$ in $\mathbb{R}$

we are given that $f$ is continous on reals, and $|f(x)-f(y)|\ge |x-y|\forall x,y$, we need to find out: is $f$ surjective from $\mathbb{R}\rightarrow\mathbb{R}$?

I think the statement is false, is it not violating definition of continuity itself?I have picked it from a question paper. thank you.

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marked as duplicate by Jonas Meyer, Brandon Carter, ncmathsadist, Douglas S. Stones, Alexander Gruber Jan 15 '13 at 2:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

It's surely not violating the definition of continuity since $f(x)=x$ satisfies exactly that condition. – JSchlather Jan 14 '13 at 22:55
If it violated the definition of continuity, then the statement would be neither false nor true: it would be ill-defined. – Hurkyl Jan 15 '13 at 1:35
up vote 5 down vote accepted

$f$ is 1-1 as $x \neq y$ implies $ | f(x) - f(y) | \geq | x - y | > 0.$

Any 1-1 continuous function from R to R is strictly monotonic (*).

If $f$ is strictly increasing then for any $ x, y $ with $ x > y $, $f(x) \geq f(y) + (x - y) $, keeping $y$ fixed and letting $x$ tend into $\infty$, we have $ \lim_{x \to \infty} f(x) = \infty $, and keeping $x$ fixed and letting $y$ tend to $-\infty$ we get $ \lim_{y \to -\infty} f(y) = -\infty.$ Hence $f$ is surjective. For strictly decreasing $f$ redo the argument with $-f$ instead of $f$.

(*) For completeness: Let $f$ mapping reals to reals be continuous and 1-1, then $ g(x,y) = ( x - y ) ( f(x) - f(y) ) $ maps the connected set $ \{ (x,y) : x > y \}$ to a connected subset of $ \mathbb{R} - \{0\}$, hence the image of g lies in $(0,\infty) $ in which case f is strictly increasing, or it lies $(-\infty,0)$ in which case $f$ is strictly decreasing.

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I love your proof that 1-1 and continuous implies monotonic! :) – Bruno Joyal Feb 19 '13 at 7:19

Assume $f$ is bounded from above, $M:=\sup f<\infty$. Select $x_0$ with $f(x_0)>M-1$. Then $|f(x_0\pm1)-f(x_0)|\ge 1$ implies $f(x_0\pm1)\le f(x_0)-1$ as we cannot have values $\ge f(x_0)+1$. By the intermediate value theorem, there is $x_1\in(x_0-1,x_0)$ with $f(x_1)=f(x_0)-\frac12$. Similarly, there is $x_2\in(x_0,x_0+1)$ with $f(x_2)=f(x_0)-\frac12$. But then $0=|f(x_1)-f(x_2)|\ge |x_1-x_2|>0$, contradiction. We conclude that $f$ is not bounded from above.

Similarly, $f$ is not bounded from below. Now given $y\in\mathbb R$, we therefore find $x_3$ with $f(x_3)<y$ and $x_4$ with $f(x_4)>y$ and by the intermediate value theorem again, some $x_5$ with $f(x_5)=y$.

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I claim that either $f(y) - f(x) \leq -(y - x)$ for all $y > x$ or $f(y) - f(x) \geq (y - x)$ for all $y > x$. For suppose not. Then you could find $x_1 < y_1$ with $f(y_1) - f(x_1) \leq -(y_1 - x_1) < 0$ and $x_2 < y_2$ with $f(y_2) - f(x_2) \geq (y_2- x_2) > 0$. So by the intermediate value theorem on $t \in [0,1]$ there would be some $t \in (0,1)$ such that $f(ty_1 + (1-t)y_2) - f(tx_1 + (1-t)x_2) = 0$. Since $ty_1 + (1-t)y_2 > tx_1 + (1-t)x_2$ this would contradict the condition you're given.

So replacing $f$ by $-f$ if necessary assume that $f(y) - f(x) \geq (y - x)$ for all $x < y$. Fixing $x = 0$ this means $f(y) \geq y + f(0)$ for all $y > 0$, so $\lim_{y \rightarrow \infty} f(y) = \infty$. Similarly $f(y) \leq -y + f(0)$ for all $y < 0$, so $\lim_{y \rightarrow -\infty} f(y) = -\infty$. Since $f$ is continuous the intermediate value theorem tells you the range of $f$ is all of ${\mathbf R}$

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