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I'm trying to solve this question.

Let $0\leq \lambda \lt 1$ and $f:X\subset\mathbb R\to \mathbb R$ a $\lambda$-contraction, i.e., $|f(x)-f(y)|\leq \lambda |x-y|$ for any $x,y \in X$. Prove that if $X$ contains a closed interval $[a-r,a+r]$ and $|f(a)-a|\leq (1-\lambda)r$, then $f$ has a fixed point in $[a-r,a+r]$.

In order to begin to solve this question, I'm trying to prove that this function is derivable, I need help, particularly in this part.

thanks a lot

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Why do you want to prove that $f$ is differentiable? –  Chris Eagle Feb 11 '13 at 18:23
    
1) Prove that if there is a fixed point, then it is unique. 2) Prove that there exists a fixed point by taking an arbitrary $x_0$ in $[a-r,a+r]$ and defining by induction $x_{n+1}=f(x_n)$. You will prove easily that $x_n$ converges to a fixed point. –  1015 Feb 11 '13 at 18:26
    
@julien but in order to prove that I have to prove first the differenciability of the function, no? –  user42912 Feb 11 '13 at 18:31

1 Answer 1

If the contraction $f$ is defined all over $\mathbb{R}$, then by Banach fixed-point theorem $f$ will have a fixed point, which can be obtained as the limit point of $x_{0} = a$, $x_{i+1} = f(x_{i})$. (That is, $x_{n} = f^{n}(a)$.)

Now $f$ is only defined over $X$. But the second condition guarantees that the sequence $(x_i)$ is all contained in $[a-r,a+r]$, so you are done.

To see this, first prove by induction that $$ \lvert f^{n+1}(a) - f^{n}(a)\rvert \le \lambda^{n} (1-\lambda) r, $$ so that $$ \begin{align} \lvert f^{n+1}(a) - a\rvert &\le \lvert f^{n+1}(a) - f^{n}(a)\rvert + \lvert f^{n}(a) - f^{n-1}(a)\rvert + \dots + \lvert f(a) - a\rvert \\&\le (\lambda^{n} + \lambda^{n-1} + \dots + \lambda + 1) (1 - \lambda) r = (1 - \lambda^{n+1}) r < r. \end{align}$$

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but to use this we have to prove the differentiability of $f$, no? when I proved this fixed-point theorem, I used a fact that the differentiability of $f$ is bounded. –  user42912 Feb 11 '13 at 18:34
1  
@user42912, no, you just need $f$ to be a contraction. (See the Wikipedia reference.) You are probably thinking of the converse implication. –  Andreas Caranti Feb 11 '13 at 18:36

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