# Weakly shrinking functions on compact spaces

I have been dealing with this problem for some days without any progress:

"Let $$f:K \rightarrow K$$ be a weakly shrinking function defined on a compact metric space $$K$$ (i.e.: $$\forall x,y \in K, x \neq y \implies d(f(x), f(y)) \le d(x,y)$$); then, prove that $$f$$ is surjective if and only if $$\forall x,y \in K, x \neq y \implies d(f(x), f(y)) = d(x,y)$$ (that is, the equality always holds in the definition of weakly shrinking function)"

My Attempt

So far I have only been able to prove that $$f$$ is uniformly continuous. Also, in trying to show the direction (surjectivity $$\implies$$ equality) I had the idea to separate the points on which the equality holds from those on which the strict inequality holds into two sets A and B respectively; then, I thought of reaching a contradiction in the two possible cases: B finite (in which case I should use finiteness somehow) and B infinite (in which case I would probably use the existence of a limit point by the characterization of compact sets). However I haven't made any progress in this direction yet.

Any hint on how to proceed or any comment on my idea are highly appreciated as always!

• Mar 15, 2022 at 9:23
• @MartinR Thank you for the link! However, that question deals with only one direction of the equivalence and it does so with a stronger condition. Mar 15, 2022 at 9:52

Claim: Given a function $$f:K \to K$$ which does not increase distances on a compact metric space $$(K,d)$$, surjectivity is equivalent to preserving distances.

Proof: (a) First suppose $$f$$ preserves distances but is not surjective, so there exists $$z_0 \in K \setminus f(K)$$. Let $$K_0=K$$. Since $$K_1:=f(K)$$ is compact, we have $$\delta:=d(z_0,K_1)>0 \,.$$ For $$n \ge 1$$, define inductively $$K_n=f(K_{n-1})$$ and $$z_n=f(z_{n-1}) \in K_n$$. Then $$d(z_{n-1},K_n)=\delta$$ for all $$n \ge 1$$, and for positive integers $$\ell we have $$z_n \in K_{\ell+1}$$, so $$d(z_\ell,z_n) \ge \delta \,.$$ Thus the sequence $$\{z_n\}$$ does not have a convergent subsequence, contradicting compactness.

(b) Suppose that $$f$$ is surjective, and $$x_0,y_0 \in K$$ are distinct. Given $$\epsilon>0$$, there is a finite set $$S \subset K$$ such that $$\cup_{s \in S} B(s,\epsilon) =K$$, where $$B(s,\epsilon)$$ is the open ball of radius $$\epsilon$$ centered at $$s$$. For $$n \ge 1$$, define inductively preimages $$x_n,y_n$$ such that $$f(x_n)=x_{n-1}$$ and $$f(y_n)=y_{n-1}$$. For each $$n$$, there exist $$\tilde{x}_n, \tilde{y}_n \in S$$ such that $$d(\tilde{x}_n,x_n)<\epsilon$$ and $$d(\tilde{y}_n,y_n)<\epsilon$$. In the infinite sequence $$\{(\tilde{x}_n, \tilde{y}_n) \}_{n \ge 1}$$, some element of the finite set $$S \times S$$ must repeat. Thus there exist $$\ell>k>0$$ such that $$(\tilde{x}_k, \tilde{y}_k)=(\tilde{x}_\ell, \tilde{y}_\ell) \,,$$ so $$d(x_k,x_\ell)<2\epsilon \quad \mbox{ and} \quad d(y_k,y_\ell)<2\epsilon \,.$$ Applying $$f^{k+1}$$, we infer that $$d(f(x_0),x_{\ell-k-1})<2\epsilon \quad \mbox{ and} \quad d(f(y_0),y_{\ell-k-1})<2\epsilon \,.$$ Therefore, by the triangle inequality, $$d(x_{\ell-k-1}, y_{\ell-k-1}) < d(f(x_0),f(y_0))+4\epsilon \,.$$ Applying $$f^{\ell-k-1}$$ yields that $$d(x_{0}, y_{0}) < d(f(x_0),f(y_0))+4\epsilon \,.$$ Since $$\epsilon>0$$ was arbitrary, this proves that $$d(x_{0}, y_{0}) = d(f(x_0),f(y_0)) \,,$$ so $$f$$ is distance preserving.

Edit: According to [1], The equivalence in the claim is due to [2].

[1] Hu, Thakyin, and W. A. Kirk. "On local isometries and isometries in metric spaces." In Colloquium Mathematicum, vol. 44, no. 1, pp. 53-57. Institute of Mathematics Polish Academy of Sciences, 1981.

[2] Freudenthal, Hans, and Witold Hurewicz. "Dehnungen, verkürzungen, isometrien." Fundamenta Mathematicae 26, no. 1 (1936): 120-122.

• Thank you very much for the precise and clear answer! Mar 20, 2022 at 8:35
• @Matteo Menghini: How did this problem arise? I really enjoyed thinking about it. Do you know any reference where this question was discussed? Mar 20, 2022 at 22:57
• I searched some more, one direction is also proved in math.stackexchange.com/questions/12285/… Mar 20, 2022 at 23:15
• @Yuvale Peres actually, it was taken from a supplement to Rudin’s “Principle of Mathematical Analysis” which contains many problems as thought-provoking as this one, I’ll leave the link here: math.berkeley.edu/~gbergman/ug.hndts/m104_Rudin_exs.pdf Mar 20, 2022 at 23:17
• Apparently this equivalence was proved in 1936. I added a reference to my answer. Mar 20, 2022 at 23:21