8
$\begingroup$

Exercise: Find a smooth function $f:\mathbb{R}\to\mathbb{R}$ such that $|f'(x)| < 1$ and $f(x) \neq x$ for all $x\in\mathbb{R}$

I got this exercise from the book "Curso de Análise: volume 1", by Elon Lages Lima. (In Portuguese).

My attempts include

$1$) integrate $\frac{2\text{arctan}(x)}{\pi}$, but I get this. (Adding larger constants doesn't seem to help.)

$2$) $f(x) = \text{sin}(x/2) + x + 2$ but its derivative gets too large.

Any ideas?

$\endgroup$
1
  • $\begingroup$ Notice that by Banach's fixed point theorem, $|f'(x)|$ must go to $1$ for $x\rightarrow\infty$ or $x\rightarrow-\infty$. I know it's not really of any help towards a solution, but it is a nice fact. $\endgroup$ Feb 27 '13 at 22:58
6
$\begingroup$

Take any branch of the hyperbola $y^2-x^2=1$. It doesn't cross the line $y=x$ and satisfies the required condition.

$\endgroup$
1
  • $\begingroup$ Thanks! Here's a comparison between the derivatives in: my first attempt, N.S.'s answer and your answer. As you can see, mine takes too long to get close to 1 (as x goes to infinity), which kind of explains why it didn't work. $\endgroup$ Feb 27 '13 at 23:22
3
$\begingroup$

$$f(x)=x-\arctan(x)+\frac{\pi}{2}$$

$\endgroup$
2
  • $\begingroup$ Thanks, N.S.! May I ask you how you got this function? Did you integrate the function $\frac{x^2}{1 + x^2}$? I'm asking this because a friend of mine was trying to use "rotation" (I'm not sure what he means about that) and he mentioned the function arctan$(x)$, so I'm just wondering if you used this alternative method. $\endgroup$ Feb 28 '13 at 0:09
  • $\begingroup$ @TuringMachine If you are familiar with Banach Fixed point theorem, it implies that for any such example, $f(x)$ needs to get arbitrarily close to $1$. So I looked a function of the type $1-g(x)$, where $0 \leq g(x) \leq 1$ and $\lim_{x \to \infty}g(x)=0$ and which is easy to integrate...Note that to get no fixed point, all you need is that the antiderivative of $g(x)$ has no root. $\endgroup$
    – N. S.
    Feb 28 '13 at 2:31
0
$\begingroup$

One solution is a function that tends to 0 at $-\infty$, and tends to $x$ at $+\infty$. For instance, a hyperbola with asymptotes y=0 and y=x.

Do you know how to construct such a hyperbola?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.