# Random stuff 2026 week 23
## A complement to contraction, expansion?
Suppose $f:X \to X$ is continuous on *compact* metric space be an **expansion**, namely for any $a,b \in X$, we have $d(f(a),f(b)) \geqslant d(a,b)$. Then $f$ is surjective.
In fact, this $f$ is also injective. In fact, by folklore, $f$ is an isometry (where $d(f(a),f(b)) = d(a,b)$ for all $a,b \in X$) and a homeomorphism! Let us show $f$ is an isometry.
Proof of $f$ is an isometry. Note this does not require $f$ to be continuous in the hypothesis.
Suppose to the contrary that there exists distinct $x, y$ such that $d(f(x) , f(y)) > d(x,y)$.
Consider sequences $(x_{n})$ and $(y_{n})$ such that $x_{n} = f^{(n)}(x)$ and $y_{n} = f^{(n)}(y)$ for $n \geqslant 0$.
Claim: We can extract subsequences from $(x_{n})$ and $(y_{n})$ that converge to $x$ and $y$.
Since $X$ is compact, there exists a convergent subsequence $(x_{n(k)})$ of $(x_{n})$. Now using the increasing indexing $n(k)$, we look at the subsequence $(y_{n(k)})$ and take a convergent sub-subsequence $(y_{n(k(j))})$. This gives two subsequences of common indexing that converges. Since the sub-sub-indexing $n(k(j))$ strictly increases, we can pick a further sub-indexing $m(i)$ such that the gap $m(i+1) - m(i)$ is strictly increasing as well.
The result is now we have an indexing $m(i)$ where
$$
x_{m(i)} \to x^{\ast}, \quad y_{m(i)} \to y^{\ast}
$$
as $i \to \infty$, with $m(i+1) - m(i) \uparrow \infty$.
Let us now write $\ell(i) = m(i+1) -m(i)$, which is strictly increasing. Note, $(x_{\ell(i)})$ is still some subsequence of $(x_{n})$.
Note by expansion property
$$
d(x_{m(i+1)} , x_{m(i)}) \geqslant d(x_{m(i+1) -m(i)},x) = d(x_{\ell(i)},x)
$$
Since $d(x_{m(i+1)},x_{m(i)}) \to d(x^{\ast}, x^{\ast}) = 0$ as $i \to \infty$, we see that $d(x_{\ell(i)},x) \to 0$. So we have $x_{\ell(i)} \to x$. Similarly we also have $y_{\ell(i)} \to y$. This gives our claim.
And so now by expansion property,
$$
d(x, y) \leqslant d(f(x),f(y)) \leqslant d(f^{(\ell(i))}(x), f^{(\ell(i))}(y)) = d(x_{\ell(i)}, y_{\ell(i)}) \to d(x,y)
$$
Hence $d(x,y) = d(f(x),f(y))$! $\blacksquare$
And since $f$ is an isometry of $X$, it is also a homeomorphism onto itself.
This is interesting to me. As it is not entirely obvious to me. A similar sounding theorem is *Edelstein*, but for *weak contractions*:
**Edelstein fixed point theorem.** Suppose $f: X \to X$ is continuous on a _compact_ metric space, and for all distinct $a,b$ we have $d(f(a),f(b)) < d(a,b)$, then there exists a unique fixed point $p \in X$ such that $f(p) = p$.
Proof of Edelstein.
Define a new function $h: X \to \mathbb{R}$ where $h(x) = d(x, f(x))$, which is a continuous function (exercise!) that measures the distance between $x$ and its image $f(x)$. Since $h$ is real-valued continuous function on a compact set $X$, there exists some $p \in X$ such that $h(p) = \min_{X} h \geqslant 0$ by extreme value theorem.
If $h(p) = 0$, then $d(p,f(p)) = 0$ and hence $p=f(p)$ is a fixed point of $f$.
Suppose to the contrary that $h(p) = d(p,f(p)) > 0$, where $p$ and $f(p)$ are distinct. Then note that $h(f(p)) = d(f(p),f(f(p))) < d(p, f(p)) = h(p)$. This contradicts the minimality of $h(p)$. Hence $h(p) = 0$.
Now suppose there are two such distinct fixed points $p,q$, then
$$
d(p,q) \leqslant d(p,f(p)) + d(f(p), f(q)) + d(q,f(q)) < d(p,q)
$$
where we get $d(p,q) < d(p,q)$, a contradiction. Hence the fixed point is unique. $\blacksquare$
A similar looking theorem is the contraction mapping theorem, or Banach fixed point theorem.
**Banach fixed point theorem.** Suppose $f: X \to X$ is continuous on *complete* metric space, and there exists some $\lambda < 1$ such that for any $a,b \in X$, we have $d(f(a),f(b)) \leqslant \lambda d(a,b)$. Then there exists a fixed point, some $p \in X$ such that $f(p) = p$.
And finally we have also Brouwer fixed point theorem:
**Brouwer fixed point theorem.** Suppose $f:X \to X$ is continuous on a compact convex set $X \subset \mathbb{R}^{n}$ (finite dimensional), then there exists a fixed point $p \in X$ such that $f(p) = p$.