<s>[INST] Fixed Point Theorems and Contraction Mappings I am trying to solve this following exercise. 

Let $F(x)$ be a continuously differentiable function defined on the interval $[a,b]$ such that $F(a) < 0$, $F(b) > 0$, and 
  \begin{align*}
0 < K_1 \leq F'(x) \leq K_2 \; \; \; (a \leq x \leq b).
\end{align*}
  Use Theorem 1 (given below) to find the unique root of the equation $F(x) = 0$. 
Hint: Introduce the auxiliary function $f(x) = x - \lambda F(x)$, and choose $\lambda$ such that the theorem works for the equivalent equation $f(x) = x$
(Theorem 1: Every contraction mapping $A$ defined on a complete metric space $R$ has a unique fixed point.)

Here is what I have so far: 

Define the auxiliary function $f(x) = x - \lambda F(x)$. We first must show that $f$ is a contraction mapping, meaning that $|f(x) - f(y)| \leq K |x - y|$ where $x, y \in [a,b]$.
Thus, let $x, y \in [a,b]$. We have: 
  \begin{align*}
|f(x) - f(y)| & = |(x - \lambda F(x)) - (y - \lambda F(y))| & & \text{definition of $f(x)$} \\
& = |(x - y) - \lambda (F(x) - F(y))| & & \text{rearrange} 
\end{align*}
  Since $F(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, we invoke the mean value theorem. Thus, $\exists c \in (x,y)$ such that $F'(c) = \frac{F(x) - F(y)}{x-y}$. (This requires us to impose the restriction that $x \neq y$, if $x = y$, then $f(x) = f(y)$, and $|f(x) - f(y)| = 0$, and the result trivially holds for any choice of $K$.) This implies that $F' (c) (x - y) = F(x) - F(y)$. Using this, we get: 
  \begin{align*}
& = |(x - y) - \lambda F'(c) (x-y)| \\
& = |(x-y)(1 - \lambda F'(c))| & & \text{take out factor of $(x-y)$} \\
& = |x-y||1 - \lambda F'(c)| & & \text{properties of abs. value.}  
\end{align*}
  Using our assumption, we have:
  $K_1 \leq F'(c) \leq K_2$. For $\lambda \geq 0$, we have
  \begin{align*}
\lambda K_1 \leq \lambda F'(c) \leq \lambda K_2 & \iff - \lambda K_1 \geq - \lambda F'(c) \geq - \lambda K_2 \\
& \iff - \lambda K_2 \leq - \lambda F'(c) \leq - \lambda K_1 \\
& \iff 1 - \lambda K_2 \leq 1 - \lambda F'(c) \leq 1 - \lambda K_1 
\end{align*}
  Now, set $\lambda - \frac{2}{K_2}$. Then, we have: 
  \begin{align*}
1 - \lambda K_2 = 1 - \frac{2}{K_2} K_2 = 1 - 2 = -1 \\
\end{align*}
  Since $K_1 \leq K_2$, $\frac{1}{K_1} \leq \frac{1}{K_2}$, so $\frac{2}{K_2} \geq \frac{2}{K_1}$ and, hence, $- \frac{2}{K_2} \leq \frac{2}{K_1}$. Thus:
  \begin{align*}
1 - \lambda K_1 = 1 - \frac{2}{K_2} K_1  \leq 1 - \frac{2}{K_1} K_1 = 1 - 2 = -1.
\end{align*}
  Therefore, given this choice of $\lambda$, we have
  \begin{align*}
-1 \leq 1 - \lambda F'(c) \leq 1. 
\end{align*}
  Thus, 
  \begin{align*}
|1 - \lambda F'(c)| \leq 1.
\end{align*}
  Putting all of this together, we have: 
  \begin{align*}
|f(x) - f(y)| = |x-y||1 - \lambda F'(c)| \leq |x-y| \cdot 1 = |x-y|.
\end{align*}
  Therefore, $f$ is a contraction mapping, meaning that, by Theorem 1, it has a unique fixed point. This implies that $\exists z \in [a,b]$ such that $f(z) = z$. By the definition of $f$, this implies that
  \begin{align*}
z - \lambda F(z) = z,
\end{align*}
  and then that
  \begin{align*}
z - \frac{2}{K_2} F(z) = z. 
\end{align*}
  Subtracting $z$ from both sides gives
  \begin{align*}
- \frac{2}{K_2} F(z) = 0. 
\end{align*}
  Finally, multiplying both sides by $- \frac{K_2}{2}$ gives: 
  \begin{align*}
F(z) = 0.
\end{align*} 
  Thus, there exists a unique root, $z$, to the equation $F(z) = 0$. 
  Thanks. 

 [/INST] Pick $\lambda$ so that $0<\lambda F'<1$. E.g. take $\lambda=\frac1{2K_2}$. Then $0<\frac{K_1}{2K_2}\le\frac {F'}{2K_2}\le\frac{K_2}{2K_2}=\frac12<1$. Using that $f'=1-\lambda F'$ we obtain $0<1-\frac12=\frac12\le f'=1-\frac {F'}{2K_2}\le1-\frac{K_1}{2K_2}<1$. If $\mu=1-\frac{K_1}{2K_2}$ then $0<\frac12\le f'\le\mu<1$ (so now you could invoke the mean value theorem for $f'$). 
</s>