# Let $f: \Bbb R \to \Bbb R$ be a continuously differentiable function such that $|f(x)-f(y)| \ge |x-y| \forall x,y \in \Bbb R$ Then $f'(x)=\frac 12$

Let $f: \Bbb R \to \Bbb R$ be a continuously differentiable function such that $|f(x)-f(y)| \ge |x-y| \forall x,y \in \Bbb R$ Then $f'(x)=\frac 12$

A. has exactly one solution

B. has no solution

C. has a countably infinite number of solutions

D. has uncountably many solutions

I choose option B as the answer.

This is because $\frac {|f(x)-f(y)|}{|x-y|} \ge 1 \forall x,y \in \Bbb R \Rightarrow |f'(x)| \ge 1 \forall x \in \Bbb R \Rightarrow f'(x) \le -1$ or $f'(x) \ge 1 \forall x \in \Bbb R$.

Is this a correct way to approach?

• @lulu But $f'(x)=1$ in that case. – Error 404 Dec 14 '15 at 14:28
• @Stef Thanks a lot for pointing that out. I am editing the post. :D – Error 404 Dec 14 '15 at 14:29
• Oh. In that case, I agree with your reasoning. – lulu Dec 14 '15 at 14:30
• Your reasoning seems correct, only one step needs perhaps a limit $x\to y$ on both sides (the RHS is of course constant and so remains 1 or -1) for justification. – Jimmy R. Dec 14 '15 at 14:31
• @lulu Apologies for wrong information. But now I have edited. :) – Error 404 Dec 14 '15 at 14:32

## 1 Answer

Let $x > y$ be arbitrary numbers and suppose by contradiction that $f'(t) = \frac{1}{2}$.

Then $\frac{1}{2} = \frac{1}{2}\frac{x - y}{x-y} = \frac{1}{x - y} \int_{y}^{x}\frac{1}{2} dt = \frac{1}{x-y}\int_{y}^{x} f'(t) dt = \frac{f(x) - f(y)}{x - y}$.

It follows that $\frac{1}{2} = \left|\frac{1}{2} \right| = \left| \frac{f(x) - f(y)}{x - y} \right| \geq 1 > \frac{1}{2}$, which is absurd. Hence there is no solution.

• This was awesome!! – Error 404 Dec 14 '15 at 16:24
• there is a typo, the absurd assumption should be $f'(x)=1/2$ instead of $f'(t)=1/2$. – Surb Dec 14 '15 at 16:44
• If $x$ is arbitrary, doesn't that include the possibility that $x = t$? – akech Dec 14 '15 at 16:48