# Proof that such sequence is convergent.

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

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

$$\Rightarrow \{x_n\}$$ 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 $$k,l\geq m$$.

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

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 $$k, $$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$$.