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I am stuck on the following problem that says:

Let $\,f \colon \Bbb R \to \Bbb R$ be a continuous function such that $\,|f(x)-f(y)|\ge \frac12 |x-y|, \forall x,y \in \Bbb R$ . Then which of the following options is correct?

  1. $f$ is both one-to-one and onto

  2. $f$ is one-to-one but may not be onto

  3. $f$ is onto but may not be one-to-one

  4. $f$ is neither one-to-one nor onto

I do not know how to approach this particular problem even though I know about one-to-one and onto functions. Can someone explain? Thanks and regards to all.

Community
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learner
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  • the condition rules out a function which is not 1-1, since if $x \ne y$ then $\mid x-y \mid \gt 0$, but if $f(x)=f(y)$ then $f(x)-f(y)=0$ – David Holden Dec 11 '13 at 18:00
  • The given inequality says that for $x \neq y$, we have $f(x) \neq f(y)$, so the function must be 1-to-1. – yoknapatawpha Dec 11 '13 at 18:01
  • you can then see the function must be strictly monotone. so if it is not onto, it must be bounded above or below. how does the condition relate to this? – David Holden Dec 11 '13 at 18:04
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    http://math.stackexchange.com/q/601587/6179 – Did Dec 11 '13 at 18:16

3 Answers3

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Another idea for the onto part. You can see immediately that $f$ is one to one (injective). You know that a continuous function which is injective is monotone. Suppose that the function is not onto. Then one of the limits $\lim_{x \to \pm\infty} f(x)$ exists and is finite.

Suppose that $L=\lim_{x \to \infty}f(x)$ is finite (the same argument works if the other limit is finite). Then take $y$ arbitrary and $x \to \infty$ in the initial inequality and you'll get $|L-f(y)| \geq \infty$, contradiction.

Or you could work directly in the initial inequality. Fix $x_0 \in \Bbb{R}$ and let $y$ go to $\pm\infty$. Then $|f(y)-f(x_0)|$ goes to $+\infty$. This implies that both the limits in $\pm \infty$ are infinite, and the monotonicity implies that their sign is different. By the intermediate value theorem we conclude that the image of $f$ is $\Bbb{R}$ so $f$ is onto.

Beni Bogosel
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Suppose on the contrary, that $f$ is not one-to-one. Then there are two distinct $x$ and $y$ for which $f(x) = f(y)$, which is impossible given that $f(x) - f(y) \neq 0$.

Suppose $f$ is not onto. By continuity, it is either bounded above or below, and by the previous result it's either strictly increasing or strictly decreasing. Suppose it's increasing and bounded above, take a $x_0$ such that $f(x_0) > \sup_x f(x) - \delta$, then $f(x_0 + 2\delta) > \sup_x f(x)$ which is absurd.

rewritten
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  • I am having trouble to understand the last line "then $f(x_0 + 2\delta) > \sup_x f(x)$ which is absurd. " Otherwise,the rest of the explanation is fine .Can you explain a bit more? – learner Dec 11 '13 at 18:13
  • $f(x_0)>\sup_xf(x)$ is what the last line says. In other words, there is a point in the domain of $f$, $x_0$, for which $f(x_0)$ is strictly greater than the supremum of $f$. This is impossible. – Laars Helenius Dec 11 '13 at 19:36
  • We assume that $f$ is increasing (otherwise reverse signs, it doesn't really matter), so $f(x_0+2\delta)-f(x_0) = |f(x_0+2\delta)-f(x_0)| ≥ \frac12|(x_0+2\delta)-x_0| = \delta$, so $f(x_0+2\delta) ≥ f(x_0)+\delta$, but we assumed that $f(x_0) + \delta > \sup_x f(x)$. – rewritten Dec 11 '13 at 21:07
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$f$ is $\DeclareMathOperator{\Ima}{Im}$ injective. Suppose $f(x) = f(y).$

$$0 = |f(x) - f(y)| \ge \frac12 |x - y| \implies x = y$$

We shall show that $f$ is also surjective.

Since $f$ is a continuous injection, it follows that $f$ is strictly monotonic. WLOG suppose that $f$ is strictly increasing.

Let $c = f(0)$ and let $y \in \mathbb{R}$. Let's show that $y \in \Ima f$.

If $y > c$, then $x_0 = 2(y - c) > 0$ so $f(0) \le f(x_0)$. We have:

$$f(x_0) - f(0) = |f(x_0) - f(0)| \ge \frac12 |x_0 - 0| = y - c$$

Hence $f(x_0) \ge y$ so $y \in \langle f(0), f(x_0)]$. Since $f$ is continuous, there exists a value $x_1 \in \langle 0, x_0]$ such that $f(x_1) = y$.

If $y < c$, then $x_0 = 2(y-c) < 0$ so $f(0) \ge f(x_0)$. We have:

$$f(0) - f(x_0) = |f(0) - f(x_0)| \ge \frac12 |0 -x_0| = c - y$$

Hence $f(x_0) \le y$ so $y \in [f(x_0), f(0)\rangle$. Since $f$ is continuous, there exists a value $x_2 \in \langle x_0, 0]$ such that $f(x_2) = y$.

If $y = c$, then $y = f(0)$.

In either case, $y \in \Ima f$.

We conclude that $f$ is surjective.

mechanodroid
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