g continuous on [a,b] using intermediate value theorem

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}) $$As you can probably tell, I learned most of this by myself. I am a self taught person who is trying to show some effort. I am sorry if this is not enough, but still can you show me the proofs to each section. I will be able to learn and understand them. Thanks to all the people that do help! - I just editted my question. I am still stuck on c. Is there anyone there who can help me please. This is frustrating since I did try but can not get through the parts of c. – 9599 Oct 7 '13 at 19:13 1 Answer The claim is false as stated. You want \lambda <1, that is, strictly smaller than 1, since g(x)=x is a counterexample. What you might want to do is suppose to the contrary, that there existed two different fixed points \xi<\xi'. Then use Lagrange's theorem on [\xi,\xi']: We would have that$$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. - 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