Minimal displacement for isometries composition

Question

i was reading this book of Athanase Papadopoulos, Metric Spaces, Convexity and Nonpositive Curvature and in the isometries chapter it's defined the following concept. Given $f:X\rightarrow X$ an isometry define the minimal displacement as \begin{align*} \lambda(f)=\inf_{x\in X}|x-f(x)| \end{align*} It's a very intuitive definition, and there are obvious examples like in the euclidean plane, $\lambda(T_{(x,y)})=||(x,y)||$ where $T$ is the traslation in the $(x,y)$ direction or $\lambda(R_{p,\alpha})=0$ for any rotation centered in any point $p$ and for any angle $\alpha$. My problem start with the Proposition 11.1.3 that say given two isometries $f,g$ of the same metric space $X$ then $\lambda(f\circ g)\leq \lambda(f)+\lambda(g)$. Now, in the euclidean plane, consider two distinct points, $p$ and $p'$ and two rotations $R_{p,\pi}$ and $R_{p',\pi}$ in $\pi$ centered on those points. For me, it's clear $S=R_{p,\pi}\circ R_{p',\pi}$ is a traslation because has no fixed points. But if this was true then $\lambda(S)>0$. Although $\lambda(R_{p,\pi})=\lambda(R_{p',\pi})=0$ which is a contradiction to the proposition. For sure i'm misunderstanding something but i can't figure it out what it is. If someone can help me i would be thankful.

Answer 1

1. As stated, the claim is simply false. For instance, let $X$ be the real line with the usual metric and $f(x)=-x, g(x)=1-x$. Then $\lambda(f)=\lambda(g)=0$ since both have fixed points in $X$. On the other hand, $g\circ f(x)= x+1$, hence, $\lambda(g\circ f)=1$.

2. On the other hand, suppose that $X$ has nonpositive curvature (actually, convex distance function will suffice).

Lemma. Suppose that $f, g$ are commuting isometries of $X$. Then $$ \lambda(g\circ f)\le \lambda(f) + \lambda(g). $$ Proof. For of all, in view of convexity of the distance function, if $h\in Isom(X)$ and $Y\subset X$ is a closed convex $h$-invariant subset then $\lambda(h)=\lambda(h|Y)$. Now, for $\epsilon>0$ define
$$ M_{f,\epsilon}= \{x\in X: d(x, f(x))\le \lambda(f) + \epsilon\}$$ This subset of $X$ is closed convex and $f$-invariant. Moreover, since $f$ and $g$ commute, this subset is also $g$-invariant and $h$-invariant for $h=g\circ f$. Applying the same reasoning to $M_{g,\epsilon}$, we see that $$ \lambda(h)= \lambda(h|M_{f,\epsilon}\cap M_{g,\epsilon}). $$ Therefore, take $$ x\in M_{f,\epsilon}\cap M_{g,\epsilon}. $$ For such $x$ we have: $$ d(gf(x), x)\le d(gf(x), f(x)) + d(f(x), x) \le \lambda(g) +\epsilon + \lambda(f) +\epsilon = \lambda(g) + \lambda(f) + 2\epsilon. $$ Since $\epsilon>0$ was arbitrary, we conclude that $$ d(gf(x), x)\le \lambda(g) + \lambda(f). $$ Taking the infimum over all $x\in M_{f,\epsilon}\cap M_{g,\epsilon}$, we conclude that $$ \lambda(h)\le \lambda(f) + \lambda(g). $$