Real continuous function with attractive fixed point has range subset of domain

Question

Let $f:\mathbb{R}\to \mathbb{R}$ be $C^1$ with fixed point $p$, such that $|f'(p)|< 1$. Then there exists an interval $I$ containing $p$ such that $f:I\to I$.

I'm trying to prove this rigorously, but there's a roadblock. My best attempt is as follows:

Let $\varepsilon > 0$, such that $\varepsilon$ is small, and consider $f(p+\varepsilon)=f(p)+f'(p)\varepsilon+\mathcal{O}(\varepsilon^2)$ and $f(p-\varepsilon)=f(p)-f'(p)\varepsilon+\mathcal{O}(\varepsilon^2)$.

Then $$|f(p+\varepsilon)-f(p-\varepsilon)|=|\varepsilon(2f'(p))+O(\varepsilon^2)|$$$$\le 2\varepsilon|f'(p)|+|\mathcal{O}(\varepsilon^2)|<2\varepsilon+|\mathcal{O}(\varepsilon^2)|$$

Everything would be as desired if not for the $|\mathcal{O}(\varepsilon^2)|$ term. How does one get rid of it? I could argue that since $\varepsilon$ is small, we can neglect the term, but that is not very rigorous. Would appreciate your insight.

Answer 1

If $f$ is differentiable and $f'$ is continuous at $p$ then there is some interval $I$ containing $p$ in the interior such that $|f'(x)| < 1$ for $x \in I$. Without loss of generality we can presume that $I$ is symmetric about $p$, and so has the form $I=(p-\delta, p+\delta)$ for some $\delta >0$.

If $x \in I$ then the mean value theorem gives $f(x) = f(p) + f'(\xi) (x-p)$ for some $\xi \in I$ and so $|f(x)-p| = |f(x)-f(p) | = |f'(\xi)| |x-p| < |x-p|< \delta$, and so $f(x) \in I$.

Answer 2

Your argument hasn't exploited the fact that the derivative is continuous; so far you're only using differentiability.

You can instead try something like this which directly exploits that continuity:

By continuity of $f'$, there is a closed interval $I=[p-t,p+t]$ such that $0 \le |f'| < 1$ on $I$. We claim that if $x\in I$, then $f(x) \in I$. Equivalently, we'll show if $x\in I$, then $|f(x)-f(p)| \le t$. To see this, we compute $$ |f(x)-f(p)| = |f(x) - p| \le |f'(c)|\cdot |x-p| \le |x-p| \le t $$ by our choice of $x$. This shows that locally $f:I\to I$.

Answer 3

For $C^1$ function, estimating or otherwise controlling the magnitude of the $O(\varepsilon^2)$ error term can be quite difficult, so I would recommend using integration as opposed to a Taylor expansion, thusly:

Since

$f(x) \in C^1(\Bbb R, \Bbb R), \tag 1$

we have that

$f'(x) \in C^0(\Bbb R, \Bbb R), \tag 2$

i.e., $f'(x)$ is continuous on $\Bbb R$. Now if $p \in \Bbb R$ with

$\vert f'(p) \vert < 1, \tag 3$

then by the continuity of $f'(x)$ there is a $\delta > 0$ such that

$\vert f'(s) \vert < 1 \tag 4$

when

$s \in I = [p - \delta, p + \delta]. \tag 5$

Since $I$ is compact, $\vert f'(s) \vert$, itself a continuous function on $I$, takes a maximum value $\mu$ on $I$, and this maximum $\mu < 1$. For $x \in I$ we have

$f(x) - f(p) = \displaystyle \int_p^x f'(s) ds, \tag 6$

whence for $x \ge p$,

$\vert f(x) - f(p) \vert = \vert \displaystyle \int_p^x f'(s) ds \vert \le \int_p^x \vert f'(s) \vert ds \le \int_p^x \mu ds = \mu (x - p) = \mu \vert x - p \vert, \tag 7$

whilst if $x \le p$,

$\vert f(p) - f(x) \vert = \vert \displaystyle \int_x^p f'(s) ds \vert \le \int_x^p \vert f'(s) \vert ds \le \int_x^p \mu ds = \mu (p - x) = \mu \vert x - p\vert; \tag 8$

thus, for any $x \in I$,

$\vert f(x) - f(p) \vert \le \mu \vert x - p\vert < \vert x - p\vert, \tag 9$

and since $f(p) = p$ we have

$\vert f(x) - p \vert < \vert x - p\vert \le \delta. \tag {10}$

(10) implies

$f(x) \in [p - \delta, p + \delta] = I; \tag{11}$

in fact, (10) shows that

$f(x) \in (p - \delta, p + \delta) = I^\circ, \tag{12}$

the interior of $I$. So we have that $f$ maps the open interval $I^\circ$ into itself as well.

Answer 4

You don't need $C^1$, just differentiable at the fixed point $p$ with $|f'(p)| < 1$. Choose $\epsilon > 0$ with $|f'(p)| < 1 - \epsilon$. By definition of derivative, there is $\delta > 0$ so that for all $x$ with $|x - p| < \delta$, $$ \left| \frac{f(x)-f(p)}{x - p} - f'(p) \right| < \epsilon$$ Thus since $f(p)=p$, if $|x-p|<\delta$ we have $$f(x) \le p + |f'(p) (x-p)| + \epsilon |x-p| < p + |x-p| < p + \delta $$ and similarly $$f(x) \ge p - |f'(p)(x-p)| - \epsilon |x-p| > p - |x-p| > p - \delta$$ Thus $f$ maps the interval $(p-\delta, p+\delta)$ into itself.