3
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.