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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.

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  • $\begingroup$ Start by choosing $\lambda$ such that $f'(x) = 1 - \lambda F'(x)$ is always between $-1$ and $1$. $\endgroup$ – Hans Engler Oct 23 at 2:38
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You have

$$0 < \frac{K_1}{K_1+K_2} \leq \frac{F'(x)}{K_1+K_2} \leq \frac{K_2}{K_1+K_2} < 1$$

It follows

$$0 < 1- \frac{K_2}{K_1+K_2} \leq 1- \frac{1}{K_1+K_2}F'(x) \leq 1 - \frac{K_1}{K_1+K_2} < 1$$

So, $\boxed{\lambda = \frac{1}{K_1+K_2}}$ is a good choice.

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Thank you your comment to my first answer. Following your request I took a look at the revised version of your question. I find several problems with it, so I copied it here and will insert comments in it.

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 $x, y \in [a,b]$.

The above definition of contraction mapping is incomplete.
You should say "for some $K$ with $0\le K<1$, and for all $x, y \in [a,b]$. "
The condition $K<1$ is very important, the essence of the contraction mapping theorem.

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:

The above is likely a typo, you mean "set $\lambda=\frac{2}{K_2}$ ". But, apart from that, I don't think this value is a good choice for $\lambda$, more comments below.

\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*}

Well, if $0<K_1\le K_2$ then $\frac1{K_1}\ge\frac1{K_2}$, not $\frac1{K_1}\le\frac1{K_2}$. So you end up with $1-\lambda K_1\ge-1$. But, the latter wouldn't be of much value (I don't see how one could use it, to contribute to the solution). (Also, $-\frac2{K_2}\le\frac2{K_1}$ is obviously a typo, although formally correct since the left side is negative and the right positive, you meant $-\frac2{K_2}\le-\frac2{K_1}$. But again, it would end up $-\frac2{K_2}\ge-\frac2{K_1}$.)

Therefore, given this choice of $\lambda$, we have \begin{align*} -1 \leq 1 - \lambda F'(c) \leq 1. \end{align*}

I don't see in the above how you got $1$ at the right hand side, given all your previous inequalities involve $-1$, not $1$. If your previous (in)equalities were correct, then you get $-1=1-\lambda K_2\le1-\lambda F'(c)\le1-\lambda K_1\le-1$ which would give you $1-\lambda F'(c)=-1$, well which indeed formally implies $-1\le1-\lambda F'(c)\le1$. So anyway, I am a bit lost here, where you were heading, and where you came from.

Thus, \begin{align*} |1 - \lambda F'(c)| \leq 1. \end{align*}

Even if you correctly had derived that $|1-\lambda F'(c)|\le1$ then this gives you nothing, it is useless. The contraction mapping theorem does not work with $|1-\lambda F'(c)|\le1$, it works with $|1-\lambda F'(c)|\le K<1$. To clarify this, consider the following example. Let your complete metric space be the unit circle in the plane (a sphere would also work the same way) and let $g(x)=-x$ (where $x=(x_1,x_2)$ with $|x|=\sqrt{x_1^2+x_2^2}=1$). Then $|g(x)-g(y)|\le1\cdot|x-y|$, indeed $|g(x)-g(y)|=|-x-(-y)|=|-1(x-y)|=|x-y|$. But we cannot use the contraction mapping theorem, and we cannot conclude that $g$ has a fixed point, indeed it doesn't since $g(x)=-x\neq x$ for every $x$ on the unit circle.
(Alternatively, consider the zero-dimensional "sphere" $\{-1,1\}$ with $g(1)=-1$, and $g(-1)=1$. Again $|g(x)-g(y)|\le1\cdot|x-y|$ but there is no fixed point.)

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.

You could try again. The choice of $\lambda$ in @trancelocation's answer would work, as well as the choice of $\lambda$ in my first answer. Here are some more details (mouse over to see...though formatting doesn't work as I wish, and I can't make each line of this hint invisible).

If $K=1-\frac{K_1}{2K_2}$ then clearly $K<1$.

Let $\lambda=\frac1{2K_2}$ so $K=1-\lambda K_1$.

Then $0\le\frac12=1-\frac12=1-\lambda K_2\le1-\lambda F'(c)\le1-\lambda K_1=K<1$.

In particular you get $|1-\lambda F'(c)|\le K<1$.

Then $|f(x)-f(y)|=|x-y|\,|1-\lambda F'(c)|\le |x-y| \cdot K$. Since $0\le K<1$, this is indeed the condition that says that $f$ is a contraction mapping and hence $f$ has a unique fixed point. Etc. (the rest of your answer is correct, implying that $F(z)=0$).

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  • $\begingroup$ Thank you again for the extremely helpful critique. I am working on an amended attempt, using your advice, and would greatly appreciate your feedback on it when it is finished. $\endgroup$ – Matt.P Oct 24 at 6:18
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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 $f'$).

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  • $\begingroup$ Thank you for the very helpful answer. I've modified my question somewhat with a revised attempt after reading these comments. Would you mind taking a look? $\endgroup$ – Matt.P Oct 23 at 17:20

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