# If $d(f(x), f(y)) \leq kd(x, y)$ for all $x, y \in X$. Then $f$ is continuous and has exactly one fixed point

Let $d$ be a complete metric for $X$. Let $f: X \to X$ be a function. Suppose there is a number $k$, with $0 < k < 1$, such that $d(f(x), f(y)) \leq kd(x, y)$ for all $x, y \in X$. Then $f$ is continuous and has exactly one fixed point.

To show that $f$ is continuous, it felt like a good idea to use sequential continuity. So we take a convergent sequence $x_n$ with limit $x$. Since $X$ iss complete, we know that $x_n$ is a Cauchy sequence. So for every $\varepsilon/k > 0$ we have some $N \in \mathbb N$ such that for every $n, m \geq N$ we have $d(x_n, x_m) < \varepsilon/k$. But, if $d(x_n, x_m) < \varepsilon/k$, then $d(f(x_n), f(x_m)) < \varepsilon$. So $f(x_n)$ is a Cauchy sequence which converges to $f(x)$. Hence $f$ is continuous.

My problem is how to show that $f$ has exactly one fixed point. What I want to do is assume that $f$ has no fixed points, and then assume that $f$ has at least two, and obtain contradictions. However, I'm not sure how to proceed after assuming $f$ has no fixed points.

As for continuity, there is a simpler way: Show that for each $\varepsilon>0$ and $x\in X$, there exists some $\delta>0$ such that if $y\in X$ and $d(y,x)<\delta$, then $d(f(y),f(x))<\varepsilon$. For fixed $\varepsilon>0$ and $x\in X$, take $\delta\equiv \varepsilon/k$. Therefore, if $y\in X$ and $d(y,x)<\delta$, then $$d(f(y),f(x))\leq kd(y,x)<k\delta=\varepsilon,$$ as claimed.
The other claim is Banach’s famous fixed-point theorem, also known as the contraction mapping theorem. See a proof, for example, here. The main idea is to construct a special Cauchy sequence, which, in turn, will be convergent, given that the metric space is complete. The limit of this special Cauchy sequence will be a fixed point of the function $f$.
Uniqueness is actually easier than existence, let me show it here. Suppose that the existence of a fixed point $x^{\star}\in X$ of $f$ has already been proven. Suppose also that $y^{\star}\in X$ is another fixed point of $f$. Then, $x^{\star}=f(x^{\star})$ and $y^{\star}=f(y^{\star})$. Therefore, $$d(x^{\star},y^{\star})=d(f(x^{\star}),f(y^{\star}))\leq kd(x^{\star},y^{\star}).$$ Rearranging the two extreme sides of this inequality, $$(k-1)d(x^{\star},y^{\star})\geq0.$$ Since $k-1<0$, this is possible only if $d(x^{\star},y^{\star})\leq 0$. Since $d$ is a metric, one now has $d(x^{\star},y^{\star})=0$, so $x^{\star}=y^{\star}$. It follows that the fixed point—if any exists—is unique.