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I had some trouble to approach the question above. Especially (2) and (3). I appreciate if you can help!

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Where is the problem? –  Amzoti Dec 16 '12 at 22:38

2 Answers 2

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This is a weaker version of the Banach fixed point Theorem.

For 1 use the Intermediate Value Theorem on $g(x)=f(x)-x$

For 2. Let suppose that $g$ has two zeroes (two fixed points for $f$) at $x_1$,$x_2$. By Rolle Theorem, $\exists \xi\in (x_1,x_2)$ so that $$g^{\prime}(\xi)=0\Leftrightarrow f^{\prime}(\xi)=1$$ which is a contradiction.

For 3 Since $f^{\prime}$ is continuous on $[0,1]$ we have that $f([0,1])=[a,b]$. Because $\left|f^{\prime}(x)\right|<1$ we must have $\left|f^{\prime}(x)\right|<\theta<1$. The functions $g(x)=f(x)-x$ and $h(x)=f(x)+x$ are decreasing and increasing respectively and so $$\left|f(x)-f(y)\right|\le \theta\left|x-y\right|$$ Therefore, $$\left|x_{n+1}-x_n\right|=\left|f(x_n)-f(x_{n-1})\right|\le \theta\left|x_n-x_{n-1}\right|\le ...\le \theta^n\left|x_1-x_0\right|$$ Thus, because $\theta^n\to 0$, $(x_n)$ is a Cauchy sequence and as such it converges. Therefore, $x_n\to l$. Then $f(x_{n+1})=x_n\Rightarrow f(l)=l$ by continuity and so $l=p$

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To prove $1$, use the intermediate value theorem: $f(1)\leq 1$ so $f(1)-1\leq0$ and $f(0)\geq0$, so $f(0)-0\geq 0$. By The IVT, there is some $x^*$ where $f(x^*)-x^*=0$, i.e., $f(x^*)=x^*$.

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