Weak Contraction of continuous function show $\lim_{n \to \infty} (x_n, y_n) \to (0,0)$

Question

I am currently studying for a calculus exam and stumbled across the following problem.

Let $B = \{(x, y) \in \mathbb{R}^2 : \|(x, y)\| < 1\}$, and let $f: B \to B$ be a continuous function. We define the sequence $\{(x_n, y_n)\}_{n \geq 1}$ by $(x_n, y_n) = f(x_{n-1}, y_{n-1})$ for $n \geq 1$ and $(x_0, y_0) \in B$. We define the norm as $ ||(x,y)|| = \sqrt{x^2+y^2} $.

We want to prove that if $\|f(x, y)\| < \|(x, y)\|$ for all $(x, y) \neq (0, 0)$ in $B$, then $\lim_{n \to \infty} (x_n, y_n) = (0, 0)$.

My first attemps tried to use the banach fixed point theorem, however they all failed. Later I tried looking at a compact subset of $ B $ since there exists an $ \epsilon > 0 $ such that $ \epsilon = ||(x_0, y_0)|| $ and from $|| f(x,y)|| < ||(x,y)||$ it follows that all $ (x_n, y_n) $ are on the closed $ \epsilon $ ball centered at $ (0,0) $. However, I wasn't able to show that every sequence $ (x_n, y_n) $ needs to converge to $ (0,0) $.

Any help would be much appreciated.

Answer 1

The sequence $\|(x_n,y_n)\|$ is decreasing and nonnegative, hence convergent. Let $$m=\lim_n\|(x_n,y_n)\|=\inf_n\|(x_n,y_n)\|$$ The sequences $x_n$ and $y_n$ are bounded. By the Bolzano-Weierstrass theorem there is a subsequence $n_k$ such that $x_{n_k}$ and $y_{n_k}$ are convergent. Let $x_{n_k}\to x$ and $y_{n_k}\to y.$
Then $$\|(x,y)\|=\lim_k\|(x_{n_k},y_{n_k})\|=m$$ $$ \|f(x,y)\|=\lim_k \|f(x_{n_k},y_{n_k})\|=\lim_k\|(x_{n_k+1},y_{n_k+1})\|=m$$ Hence $\|f(x,y)\|=\|(x,y)\|,$ which implies $(x,y)=(0,0),$ i.e. $m=0.$

Answer 2

I think the solution is to note that $r_0:=\Vert (x_0,y_0)\Vert<1$ and that $\{(x_n,y_n) \}\subseteq \overline{B_{r_0}(0)}$ which is compact. Denote $r_n:=\Vert( x_n,y_n )\Vert$, and note that $r_{n+1}<r_n$. Hence $r_n\to \rho$ for some $0\leq \rho <r_0$.

By compactness, there exists a subsequence converging to some $(x_\infty,y_\infty)\in \overline{B_{r_0}(0)}$. Then $\Vert(x_\infty,y-\infty) \Vert=\rho$, since $(x,y)\mapsto\Vert f(x,y)\Vert$ is continuous. By continuity of the last function, we get that some subsequence satisfies $$r_{n_k+1}\to \Vert f(x_\infty,y_\infty)\Vert \quad \text{while}\quad r_{n_k}\to \rho=\Vert (x_\infty,y_\infty)\Vert.$$ So $\rho<\rho$, if $(x_\infty,y_\infty)\neq(0,0)$. So $\rho=0$.

Answer 3

Firstly, note that $f\left((0,0)\right)=(0,0)$. For, choose a sequence $\left((s_{n},t_{n})\right)$ in $B$ such that $(s_{n},t_{n})\neq(0,0)$ and $(s_{n},t_{n})\rightarrow(0,0)$. Observe that \begin{eqnarray*} & & \left\Vert f\left((s_{n},t_{n})\right)\right\Vert \\ & \leq & \left\Vert (s_{n},t_{n})\right\Vert \\ & \rightarrow & 0 \end{eqnarray*} as $n\rightarrow\infty$. Therefore, by continuity of $f$, we have that $f\left((0,0)\right)=\lim_{n\rightarrow\infty}f\left((s_{n},t_{n})\right)=(0,0)$.

We denote $z_{n}=(x_{n},y_{n})$. Let $b=||z_{0}||<1$. Observe that $\left(||z_{n}||\right)_{n}$ is a non-negative, decreasing sequence, so $a:=\lim_{n\rightarrow\infty}||z_{n}||$ exists. We go to prove that $a=0$ by contradiction and it will follow that $z_{n}\rightarrow(0,0)$. Suppose the contrary that $a>0$. Let $K_{1}=\{z\in\mathbb{R}^{2}\mid a\leq||z||\leq b\}$. Clearly $z_{n}\in K_{1}$ for all $n$. Define $g:K_{1}\rightarrow\mathbb{R}$ by $g(z)=\frac{||f(z)||}{||z||}$. Note that $g(z)<1$ for each $z\in K_{1}$ and $g$ is continuous. Since $K_{1}$ is compact, $c=\max_{z\in_{K_{1}}}g(z)$ exists. Note that $c<1$. Now, \begin{eqnarray*} & & \left\Vert z_{n}\right\Vert \\ & = & \left\Vert g(z_{n-1})\right\Vert \left\Vert z_{n-1}\right\Vert \\ & \leq & c\left\Vert z_{n-1}\right\Vert \\ & \leq & \cdots\\ & \leq & c^{n}||z_{0}||\\ & \rightarrow & 0 \end{eqnarray*} as $n\rightarrow\infty$. This contradicts to the fact that $z_{n}\in K_{1}$ for all $n$.