Proof about isometries

Question

Suppose $f\colon\mathbb R\to\mathbb R$ is an isometry of the reals. Prove $f$ is a non-trivial translation iff $f$ has no fixed points.

Assumption: $f$ is a non-trivial translation (trivial translation: $f$ s.t. $f(x)=x$ for all $x \in\mathbb R$). Conclusion: $f$ has no fixed points (definition of fixed point: $f(x)=x$ for all $x\in \mathbb R$).

I'm not sure how to use this information to get the proof started.

Answer 1

$f$ is an isometry so for all $x,y\in\Bbb R$ we have

$$|f(x)-f(y)|=|x-y|$$ so for $y=0$ we have

$$|f(x)-f(0)|=|x|$$ hence $$f(x)=\pm x+a,\quad\forall x\in\Bbb R$$ and since $f$ hasn't a fixed point then the only possibility is $$f(x)=x+a,\quad a\ne0$$

Answer 2

If $f$ is a non-trivial translation that means $\exists x_0 \in \mathbb{R}$ such that $f(x_0) \neq x_0$. Then $c=f(x_0)-x_0$ is the amount of translation.

Now you can show that no $x$ is fixed.

Answer 3

Let $f$ be an isometry of $\mathbb R$ that is not the identity. Then there exists $a\in\mathbb R$ with $b:=f(a)\ne a$. Let $c=f(b)$. Then for any $x\in \mathbb R$ we have both $$\tag1 |f(x)-b|=|f(x)-f(a)|=|x-a|$$ and $$\tag2|f(x)-c|=|f(x)-f(b)|=|x-b|.$$ For all $x\in \mathbb R$, we have $$\tag3\begin{align}f(x)&\in\underbrace{\{b+|x-a|,b-|x-a|\}}_{\text{from (1)}}\cap \underbrace{\{c+|x-b|,c-|x-b|\}}_{\text{from (2)}}\\&=\{b-a+x,b+a-x\}\cap\{c-b+x,c+b-x\}.\end{align}$$ Especially for $x=b$ this yields $$ c=f(b)\in\{2b-a,a\}\cap\{c\}, $$ that is $c\in \{a,2b-a\}$.

• If $c=a$ then $f(x)=a+b-x$ for all $x\in\mathbb R$, because $a\ne b$ implies that there cannot be an $x$ with $f(x)=b-a+x=a-b+x$.

• If $c=2b-a$ then $f(x)=b-a+x$ for all $x\in\mathbb R$, because $a\ne b$ implies that there cannot be an $x$ with $f(x)=b+a-x=3b-a-x$

In other words, $f$ is either the reflection about $\frac{a+b}2$ (which is a fixpoint, or $f$ is the translation by $b-a\ne 0$.