Let $F(x)$ be a continuously differentiable function defined on the interval $[a,b]$ such that $F(a)<0$ and $F(b)>0$ and $$ 0<K_1\leq F'(x)\leq K_2\quad (a\leq x\leq b) $$ Find the unique root of equation $F(x)=0$.

The given hint is to use the contraction mapping theorem i.e., if $f(.)$ is a contraction mapping it has a fixed point. Define $f(x)=x-\lambda F(x)$ $$ |f(x)-f(y)|=\Bigg|(x-y)\Big(1-\lambda\frac{F(x)-F(y)}{x-y}\Big)\Bigg|=|x-y|\Big|1-\lambda\frac{F(x)-F(y)}{x-y}\Big| $$ So if I can show $ \Big|1-\lambda\frac{F(x)-F(y)}{x-y}\Big|$ is smaller than $1$, I can say $f(x)=x$. However, I can't proceed from thereon. Many thanks for any help!

  • $\begingroup$ What does the mean value theorem say about $$\frac{F(x)-F(y)}{x-y}$$ $\endgroup$ – Daniel Fischer Nov 5 '13 at 11:09
  • $\begingroup$ Well, $\exists z\in (x,y)$ so that $F'(z)=\frac{F(x)-F(y)}{x-y}$, but this only shows as long as $\lambda>0$ $f$ is contraction mapping. I still can't see unique root of $F(x)=0$ $\endgroup$ – user64066 Nov 5 '13 at 11:15

There are two things you have to impose on $\lambda$ in order to apply the Banach fixed point theorem (and the unique fixed point of $f$ that the theorem guarantees is the unique zero of $F$),

  1. you must choose $\lambda$ so that there is a $c < 1$ with $$\left\lvert1 - \lambda\frac{F(x)-F(y)}{x-y} \right\rvert \leqslant c$$ for all $x\neq y\in[a,b]$, and
  2. you must choose $\lambda$ so that $f([a,b]) \subset [a,b]$.

By the mean value theorem, the difference quotient is a value of the derivative, so the given bounds of $F'$ yield

$$1 - \lambda K_2 \leqslant 1 - \lambda \frac{F(x)-F(y)}{x-y} \leqslant 1 - \lambda K_1$$

for $\lambda \geqslant 0$. Thus choosing $0 < \lambda < \frac{2}{K_2}$ will satisfy the first requirement. It remains to show that for all sufficiently small $\lambda > 0$ you have

$$a \leqslant x - \lambda F(x) \leqslant b$$

for all $x\in [a,b]$.


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