Using the Banach Fixed Point Theorem instead of Intermediate Value Theorem to find a root

Question

I'm given the following problem in the "Fixed Point Theorem" chapter of my textbook:

Suppose $F:[a,b] \rightarrow \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and satisfies $F(a) < 0 < F(b)$, and $0<K_1\leq F'(x)\leq K_2$ for some real numbers $K_1,K_2$. Show there is a unique solution to $F(x) = 0$. [Hint: Consider the equation $f(x) = x$ where $f(x) = x - \lambda F(x)$ for some suitably chosen $\lambda$.

I'm pretty sure I'm able to solve this without using any fixed point stuff. The Intermediate Value Theorem tells us that there has to be a root of $F$ in $[a,b]$, and the condition on the derivative gives uniqueness.

My question is, can anyone help me figure out how I would go about getting the solution using the "hint"? I know that to apply the fixed point theorem I need my function to be a contraction on a set. I know that the function $f$'s preimage has to be $[a,b]$, but I'm having trouble finding a way to get it to map into $[a,b]$ as well...

Answer 1

Let's simplify life by ensuring that $f'(x) \geq 0$ for $x \in (a,b)$. For this

$$ f'(x) = 1 - \lambda F'(x) \geq 1 - \lambda K_2(b-a). $$ So, if we choose any $\lambda \leq \frac{1}{K_2(b-a)}$ we'll be set on that end.

Now, with $f(x)$ monotonically increasing, we need to make sure that $a \leq f(a) \leq b$ and $a \leq f(b) \leq b$ to ensure that $f([a,b]) \subset [a,b]$. To that end, you will need (remembering that $F(a) < 0$ and $F(b) > 0$):

• $a \leq a - \lambda F(a) \leq b \implies \lambda \leq \frac{b-a}{|F(a)|}$
• $a \leq b - \lambda F(b) \leq b \implies \lambda \leq \frac{b-a}{F(b)}$

Finally, we know that $K_2 \geq K_1$. So if we set $\lambda \leq \frac{1}{K_2}$. Then, for any $x, y \in (a,b)$ with $x > y$, you have $$ 0 \leq \left[ 1 - \lambda K_2\right](x-y) \leq x - y - \lambda \big[ F(x) - F(y) \big] \leq \left[ 1 -\lambda K_1\right](x-y) $$

So, in the end, choose $\lambda$ as some value $$ \lambda < \min\left(\frac{1}{K_2(b-a)}, \frac{b-a}{|F(a)|}, \frac{b-a}{F(b)}, \frac{1}{K_2}\right) $$

Answer 2

$|f(x)-f(y)|=|(x-y)-\lambda(F(x)-F(y))|$

$=|(x-y)-\lambda F'(c)(x-y)|=|1-\lambda F'(c)||x-y|$ where $c\in (x,y)$.

Let $n,m$ and integer suc that ${1\over{nK_1}}<{1\over mK_2}$, We take $\lambda$ such that $\lambda\in [{1\over {nK_1}},{1\over{mK_2}}]$, this implies that ${1\over{nK_1}}F'(c)<\lambda F'(c)<{1\over{mK_2}}F'(c)$ and ${1\over n}<\lambda F'(c)<{1\over m}$ We deduce that $1-{1\over m}<1-\lambda F'(c)<1-{1\over n}$ and $|f(x)-f(y)|\leq (1-{1\over n})|x-y|$. The fixed point theorem implies that $f(x)=x$ has a unique solution.