# 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.

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

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.

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 $$0 then $$\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 $$F(z)=0$$).

• 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. – Matt.P Oct 24 at 6:18

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 $$f'$$).

• 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? – Matt.P Oct 23 at 17:20