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I am trying to prove the following statement: Let $A \subseteq \mathbb{R}^n$ be a closed set, $f$ $: A \rightarrow A$ a contraction. Then, the sequence $\{x_n\}$, where $x_1 = a \in A$, $x_n = f(x_{n-1})$ $\forall $n $\geq 2$, is convergent.

My attempt at proof:

Let $L\in(0, 1) $ be Lipschitz constant of $f$.

$f$ is Lipschitz $\Rightarrow$ $d(x_m, x_{m+1}) \leq L \cdot d(x_{m-1}, x_m)$

$\Rightarrow$ $d(x_m, x_{m+1}) \leq L \cdot d(x_{m-1}, x_m) \leq$ ... $\leq L^{m-1}\cdot d(x_{1}, x_2)$

Notice that $\lim_{{m \to \infty}}L^{m-1}=0$.

Let $\epsilon > 0, \exists m\in\mathbb{N}$ such that $d(x_m, x_{m+1}) \leq L \cdot d(x_{m-1}, x_m) \leq$ ... $ \leq L^{m-1}\cdot d(x_{1}, x_2) < \epsilon$

Let $k,l\geq m$, $k<l$

$\Rightarrow$ $d(x_k, x_{l}) \leq d(x_k, x_{k+1}) + d(x_{k+1}, x_{k+2})+$...$+ d(x_{l-1}, x_{l}) < (l-k)\epsilon$

$\Rightarrow \{x_n\}$ is Cauchy sequence

$\Rightarrow \{x_n\}$ is convergent

$\tag*{$\square$}$

I am skeptical with implication that $\{x_n\}$ is Cauchy sequence because $(l-k)\epsilon$ is not good for all $k,l\geq m$.

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  • $\begingroup$ The defect of your proof is the looseness of your inequality $d(x_{l-1},x_{l}) \leq L^{m-1} d(x_1, x_2)$. $\endgroup$
    – Functor
    Commented Jan 16 at 3:12

1 Answer 1

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This question is known as Banach’s Fixed Point Theorem. To see this, you need to know that $A$ is a complete metric space.

Your proof is close to the correct proof. You have deduced the key inequality $$ d(x_m, x_{m+1}) \leq L^{m-1} d(x_1,x_2), \forall m>1 $$ The next step is for $k<l$, $$ d(x_k,x_l) \leq (L^{k-1}+ L^k +\dots+ L^{l-2})d(x_1,x_2) = \frac{L^{k-1}-L^{l-1}}{1-L} d(x_1,x_2) \leq \frac{L^{k-1}}{1-L} d(x_1,x_2) $$ So just take $N$ to be big enough such that $\frac{L^{N-1}}{1-L} < \epsilon/2$. Then for $N < k < l$, $d(x_k,x_l) \leq d(x_N,x_k) + d(x_N, x_l) < \epsilon$. The rest thing is just using the completeness of the space $A$.

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