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Suppose that g is continuous on an interval [a,b] and that $g(x) ∈ [a,b]$ for all $x ∈ [a,b]$.

(a) Use the intermediate value theorem to prove that is at least one number $c ∈ [a,b]$ with $g(c) = c$.

Here is my attempt:

Define $G: [a,b] \to $$\mathbb R$ by $G(x) = x - g(x).$ Then $G$ is continuous on $[a,b].$ Since $a \leq g(a)\leq b$ and $a \leq g(b)\leq b$ we find $G(a) = a -g(a) \leq 0 $ and $G(b) = b -g(b) \geq 0 $. By the intermediate value theorem, there is a $c ∈ [a,b]$ such that $G(c) = 0$ or $c - g(c) = 0.$ Thus, $g(c) = c.$

(b) Suppose further that $g$ is differentiable and that there exists a $\lambda <1$ with $|g'(x)|\leq \lambda$ for all $x ∈ [a,b]$. Prove that there is exactly one number $c ∈ [a,b]$ with $g(c) = c$.

Here is my attempt:

Since there exists two different fixed points $\xi<\xi'$. Then as stated we can use Lagrange's theorem on $[\xi,\xi']$:

We get $$1=\frac{g(\xi)-g(\xi')}{\xi-\xi'}=f'(\nu)$$ for some $\nu$, which contradicts $|f'(x)|<1$.

(c) For any initial value $x_{0} ∈ [a,b]$, define a sequence ${x_{n}} = x_{0}, x_{2}, x_{3} ...$ by $x_n = g(x_{n-1})$ for $n \geq 0$, define $E_{n} = |x_{n}-c|$ and $D_{n} = |x_{n+1} - x_{n}|$.

Here is my attempt:

I suppose that we can say set $g(c)=c$ and $g'(c)=0$. Then let the sequence $x_{n} = g(x_{n-1}) = g(c) + g'(c)(x_{n-1} - c) +$ $ \dfrac{g''(Thi)}{2} (x_{n-1}-c)^{2}$ for Thi between c and x.

For E_{n}

$E_{n} = |x_{n}-c| = |g(x_{n-1} - g(c)| = $ $ \dfrac{g''(Thi)}{2} |x_{n-1}-c|^{2} =$ $\dfrac{g''(Thi)}{2} {E_{n-1}}^{2} $

For D_{n}

$D_{n} = |x_{n+1}-X_{n}| = |g(x_{n} - g(X_{n})| = $ $ \dfrac{g''(Thi)}{2} |x_{n}-X_{n}|^{2} =$ $\dfrac{g''(Thi)}{2} {E_{n}}^{2} $

Is this how we would define this? If not, can you please show me?

Out of curiosity from a question I just thought of. From question (c):

Is it possible that we can prove that $E_{n} < \lambda^{n} E_{0}$ which would make $(x_{n})$ converge to $c$ and $D_{n} < \lambda^{n} D_{0}$. If so can you please show me how to? It seems good to know.

(ii) Prove that for $n < k$, $|x_k - x_n|\leq $ $$ \dfrac{\lambda^{n} - \lambda^{k}}{1-\lambda}(D_{0}) $$

(iii) Prove that for $E_{n} \leq $ $$ \dfrac{\lambda^{n}}{1-\lambda}(D_{0}) $$

As you can probably tell, I learned most of this by myself. I am a self taught person who is trying to show some effort. I am sorry if this is not enough, but still can you show me the proofs to each section. I will be able to learn and understand them. Thanks to all the people that do help!

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I just editted my question. I am still stuck on c. Is there anyone there who can help me please. This is frustrating since I did try but can not get through the parts of c. –  9599 Oct 7 '13 at 19:13

1 Answer 1

The claim is false as stated. You want $\lambda <1$, that is, strictly smaller than $1$, since $g(x)=x$ is a counterexample. What you might want to do is suppose to the contrary, that there existed two different fixed points $\xi<\xi'$. Then use Lagrange's theorem on $[\xi,\xi']$:

We would have that $$1=\frac{g(\xi)-g(\xi')}{\xi-\xi'}=f'(\nu)$$ for some $\nu$, which contradicts $|f'(x)|<1$.

In more generality, you may read about Banach's fixed point theorem.

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Sorry I edited it to $\lambda\ <1$. The question was wrong before. This is what I want to prove now. –  9599 Oct 5 '13 at 23:50
    
For (b), I suppose we can use that [0,1] is continuous and that (0,1) is differentiable. –  9599 Oct 6 '13 at 0:00
    
@9599 You mean that "g is continuous on..." and "g is diff. on...", and yes, that is what I am suggesting: Lagrange's theorem. –  Pedro Tamaroff Oct 6 '13 at 0:03
    
I edited my proof now. Can you take a look at it? If possible, can you please write your own solution? –  9599 Oct 6 '13 at 0:36
    
Hey! I think we got the samething using what you told me! I was working on it a while ago be posting it. –  9599 Oct 6 '13 at 0:54

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