Given $f: X \to \mathbb{R}$ ($X \subseteq \mathbb{R})$ such that $f$ is differentiable on the domain. Similarly, $x_0$ is a fixed point that lies in the interior of $X$. We are given that $f'$ is continuous at $x_0$ with $|f'(x_0)|< 1.$ Show that there exists a closed interval $K$ centered at $x_0$ such that $f$ is contraction on $K$.

I used the definition of continuity to start with: Fix $\epsilon > 0.$ Then, there exists $\delta > 0$ such that whenever $x \in (x_0 - \delta , x + \delta), |f'(x) - f'(x)| < \epsilon.$ I think now I should try to show that for all $x \in (x_0 - \delta , x + \delta), |f'(x)| \leq r$ such that $|f'(x_0)| \leq r < 1.$ Then, I could use mean value theorem to find the interval needed. But I am not able to proceed ahead.

Does this look like a right direction?

  • $\begingroup$ it appears that $X$ should be, say, an interval, in any case a subset of $\mathbb R$ $\endgroup$
    – Will Jagy
    Feb 26, 2021 at 1:29
  • $\begingroup$ I just edited! Thanks! $\endgroup$
    – lyatit
    Feb 26, 2021 at 1:42
  • $\begingroup$ Can you prove that if $g$ is continuous at $x_0$ and $g(x_0)>0$ then there exists an open interval $I$ containing $x_0$ such that $g(x)>0$ for all $x\in I$? If so, apply this fact to $g(x)=1-|f'(x)|$ and see where that leads. $\endgroup$
    – Matthew H.
    Feb 26, 2021 at 1:48
  • $\begingroup$ I dont understand what would $g$ be here $\endgroup$
    – lyatit
    Feb 26, 2021 at 2:07
  • $\begingroup$ You could also use triangle inequality to get $|f'(x)| = |f'(x)-f'(x_0)+f'(x_0)| \leq |f'(x_0)| + |f'(x) - f'(x_0)|$. You have bounds on RHS values. $\endgroup$
    – Dylan
    Feb 26, 2021 at 2:27

1 Answer 1


Notice that there exists $ \epsilon >0$ such that $|f'(x_0)|<1-\epsilon$, for that $\epsilon > 0$ there exists $\delta_1 >0$ such that $|f'(x)-f'(x_0)|< \epsilon$ for all $x \in (x_0-\delta_1, x_0+\delta_1)$, then $|f'(x)|\leq |f'(x_0)|+|f'(x)-f'(x_0)|<|f'(x_0)|+\epsilon~~~(<1-\epsilon+\epsilon=1)$ for all $x \in (x_0-\delta_1, x_0+\delta_1)$. Let $\delta=\frac{1}{2}\delta_1$, for any $x_1<x_2 \in [x_0-\delta, x_0+\delta]$ and by MVT in $[x_1,x_2]$, $|f(x_1)-f(x_2)|=|f'(\xi)|\cdot|x_1-x_2|\leq(|f'(x_0)|+\epsilon)|x_1-x_2|$ since $\xi \in (x_0-\delta_1, x_0+\delta_1)$. Thus $f$ is a contraction on $[x_0-\delta, x_0+\delta]$.


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