Let $f:\mathbb{R}\to\mathbb{R}$ be differential function such that $f'(x)\leq r<1$ for all $x\in \mathbb{R}$. Then $f$ has at least one fixed point.

What I am considering is $h(x)=f(x)-x$ then $h(x)$ is strictly decreasing, but I did not get why it will hit the x axis???

  • $\begingroup$ Can you show that $\lim_{x \to -\infty}h(x)>0$ while $\lim_{x \to +\infty}h(x)<0$? $\endgroup$ – Jimmy R. Dec 12 '15 at 15:05
  • $\begingroup$ Try to draw the graph of h(x) (which is monotonically decreasing) without lifting up pencil. Can you avoid intersection with X-axis? $\endgroup$ – Nitin Uniyal Dec 12 '15 at 15:06

The idea is good: you have $h'(x)=f'(x)-1<0$, so the function is strictly decreasing.

Now you have to show that $h$ assumes positive and negative values. Suppose not; either $f(x)>x$ for every $x$ or $f(x)<x$ for every $x$.

Let's do the first case; by Lagrange's theorem you have, for $x>0$, $$ \frac{f(x)-f(0)}{x-0}=f'(c) $$ for some $c\in(0,x)$. Thus $$ f(x)=f'(c)x+f(0)>x $$ Rearranging it, we can write $$ f(0)>x(1-f'(c)) $$ so that, using $f'(c)\le r$, $$ f(0)>x(1-r) $$ for every $x>0$. This is of course a contradiction.

Suppose instead $f(x)<x$, for every $x$. Then, for $x<0$, $$ \frac{f(x)-f(0)}{x-0}=f'(c) $$ so $$ f(x)=xf'(c)+f(0)<x $$ that implies $f(0)<x(1-f'(c))$. Since $f'(c)\le r$, we have $1-f'(c)\ge1-r$ and so $x(1-f'(c))\le x(1-r)$ and therefore $$ f(0)<x(1-r) $$ for every $x<0$, a contradiction.

  • $\begingroup$ Your proof looks as if only $f'(x) < 1$ were needed, but in the last step you need actually $f'(c) \le r < 1$ because $c$ might depend on $y$. Or am I overlooking something? $\endgroup$ – Martin R Dec 12 '15 at 16:28
  • $\begingroup$ @MartinR Yes, you're right, it was implicitly used; I'll add the note. $\endgroup$ – egreg Dec 12 '15 at 16:34

(Elaborating on the comments to the question:)

The function $h(x) = f(x) - x$ satisfies $h'(x) \le r - 1 < 0$. So $h$ is not only decreasing, it decreases with a certain "minimum rate".

This vague formulation can be made into a precise statement using the mean value theorem, which states that for $x > 0$ $$ \begin{aligned} h(x) &= h(0) + (x-0) \, h'(t) \quad \text{for some } t \in (0, x) \\ &= f(0) + x \, h'(t) \\ &\le f(0) + x \, (r-1) \, . \end{aligned}$$ In particular, for any $x \ge a := \dfrac{|f(0)|}{1-r}$, $$ h (x) \le f(0) - |f(0)| \le 0 \, . $$ In a similar manner you can show that for any $x \le -a = \dfrac{|f(0)|}{r-1}$, $$ h (x) \ge f(0) + |f(0)| \ge 0 \, . $$

Now it follows from the intermediate value theorem that $h(c) = 0$ for some $c$ in the interval $[-a, a]$. So $f$ has a fixed point in this interval.


Banach fixed point theorem: the mean value theorem shows that $|f(x) - f(y)| = |f'(c)||x-y|$ for some $c$, so $|f(x) - f(y)| \le r|x-y|$. So apply Banach's theorem.

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    $\begingroup$ This is wrong. The derivative is bounded by one, not its module. The function can fail to be a contraction. $\endgroup$ – Aloizio Macedo Dec 12 '15 at 16:34

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