# Contraction Mapping and fixed point

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?

• it appears that $X$ should be, say, an interval, in any case a subset of $\mathbb R$ Feb 26, 2021 at 1:29
• I just edited! Thanks! Feb 26, 2021 at 1:42
• 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. Feb 26, 2021 at 1:48
• I dont understand what would $g$ be here Feb 26, 2021 at 2:07
• 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. Feb 26, 2021 at 2:27

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 $$x_1 and by MVT in $$[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]$$.