prove $f(x)=x$ has a unique solution Question:  Let $f$ be a continuous function from $\mathbb{R^2} \rightarrow \mathbb{R^2}$ such that $| f (x)− f (y)| ≤  \frac {1}{3} |x−y|$.
Prove $f(x)=x$ has a unique solution. 
My sketch: There exists $\epsilon, \delta$ such that the following holds because $f$ is continuous
 $|f(x)-f(y)| \leq \epsilon \leq \frac {1}{3} |x-y| \leq |x-y|< \delta$
Thus by the squeeze lemma, if $\delta$ goes to $0$, then $f(x)=f(y)$

Assume there are two such points $x_1,x_2$ such that $f(x_1)=x_1$ and $f(x_2)=x_2$. Then $|f(x_1)-f(x_2)|=|x_1-x_2|\leq \frac{1}{3}|x_1-x_2|$. This is true if and only if $|x_1-x_2|=0$. Therefore $x_1=x_2$ and the fixed point, if it exists, is unique.
To show that a solution does exist consider the sequence $\{x_n\}$ where $x_1=x$ for some point $x$, and $x_{n+1}=f(x_n)$. The limit of this sequence will end up being your fixed point!
