Banach fixed point theorem with lip constant possibly attaining $1$ An important condition of Banach fixed point theorem is that the map is strict contraction with Lipschitz constant strictly less than 1. Then the convergence is exponentially fast. But what if we relax the condition to less and equal to 1. Are there still a convergence (albeit not at exponential speed)?
 A: I think you're looking for the following generalization:

If $(S,d)$ is a compact metric space and $f:S\rightarrow S$ is such that $d(f(x),f(y)) < d(x,y)$ for every pair of distinct $x$ and $y$, then there exists some unique fixed point $z$ of $f$ and, additionally, for every $x\in S$ we have $$\lim_{n\rightarrow\infty}f^n(x)=z.$$

The proof of this is harder than the proof of the contraction mapping theorem. I would prove this in two parts.
Part 1: Any two fixed points of $f$ must be equal.
Much as in the original theorem, this just follows by noting that if $f(x)=x$ and $f(y)=y$, then, if distinct, they would have $d(x,y) = d(f(x),f(y)) < d(x,y)$, which is a contradiction.
Part 2: The sequence $f^n(x)$ converges to a fixed point.
This bit is trickier - how I would go about it is to consider the set of limit points $L$ of the sequence. Note that $L$ must satisfy $L=f(L)$ since $f$ is continuous and $S$ is compact and the image of the given sequence under $f$ is just a shift of the original sequence. By compactness, $L$ is non-empty.
Note that $L$ itself is compact, therefore there must be some pair $a,b\in L$ maximizing $d(a,b)$. However, we can then choose $\tilde a$ and $\tilde b$ so that $f(\tilde a)=a$ and $f(\tilde b)=b$ due to the relation $L=f(L)$. Note that, if $a\neq b$, then $d(\tilde a,\tilde b) > d(a,b)$, contradicting maximality - so, to the contrary, we must have $a=b$ - and, by maximality, this means that $d(a',b') \leq d(a,b) = 0$ for every $a',b'\in L$. Otherwise said, $L$ has only one element.
The existence of a unique element of $L$ implies that this element is the limit $\lim_{n\rightarrow\infty}f^n(x)$ and is also a fixed point.

Side note: Another way to prove this theorem is to consider the set
$$I=\bigcap_{n=0}^{\infty}f^n(S).$$
This set also satisfies $I=f(I)$, so can be shown to be a single point similarly. If you do it this way, you can additionally prove that the functions $f^n$ converge uniformly to a constant function.
A: An alternative proof to Milo's theorem using variational ideas (original argument for Edelstein theorem):
Check that any two fixed points of $\mathrm{f}$ must be equal. Consider $Q(x):=d(x, f(x)) .$ This is continuous function on a compact space so it must admit a minimizer z. We can check that minimizer must be a fixed point (and thus unique by first sentence). For any sequence $\left\{f^{n}(x)\right\},$ there exists a subsequence $\left\{f^{n_{i}}(x)\right\}$ that converges to some limit $a$. By continuity of $f, f^{n_{i}+1}(x) \rightarrow f(a) .$ Meanwhile, $d\left(f^{n}(x), z\right)$ is a decreasing sequence lower bounded by 0 so it must converge to some limit $l$. This implies that $d\left(f^{n_{i}}(x), z\right) \rightarrow l, d\left(f^{n_{i}+1}, z\right) \rightarrow l$. By continuity of metric, $d(a, z)=l=$ $d(f(a), z)=d(f(a), f(z)) .$ This implies $a=z .$ So the subsequence converges to the unique fixed point. For any $n_{i}<n<n_{i+1}, d\left(f^{n}(x), z\right) \leq d\left(f^{n_{i}}, z\right) .$ This implies that the entire sequence converges to z.
