I am trying to show that the function $f(x) = 2\pi+x-\tan^{-1}x$ is contractive but has no fixed points. Finally I wish to conclude that it does not contradict the
contraction mapping theorem.
$f$ is contractive:
by the mean value theorem we have:
\begin{align}
|f(x)-f(y)| =& |f'(x)(x-y)|
\end{align}
It is sufficient to show that $f'(x) < 1$:
\begin{align}
f'(x) =& 1 – \frac{1}{1+x^2} \\
=& \frac{x^2}{1+x^2}
\end{align}
$f$ has no fixed points:
Suppose $f$ does have a fixed point. Then $\lim_{x_n\to\infty} = x:$
\begin{align}
x_{n+1} =& 2\pi + x_n – \arctan(x_n) \\
\lim x_{n+1} =& \lim 2\pi + x_n – \arctan(x_n) \\
x =& 2\pi +x – \arctan(x) \\
0 =& 2\pi – \arctan(x)
\end{align}
This is a contradiction as $\frac{-\pi}{2} < \arctan(x) < \frac{\pi}{2}$ for all $x$. Thus $f$ does not have a fixed point.
Now I am stuck on how to show that this does not contradict the contraction mapping theorem which says:
Let $C$ be a closed subset of the real line. If $F$ is a contractive mapping of $C$ into $C$ then $F$ has a unique fixed point. Morever, this fixed point is the limit of
every sequence obtained from $x_n+1 = F(x)$ starting with $x_0\in C$.
For $f(x) = 2\pi+x-\arctan(x)$ the domain is all reals and so is the codomain. I don't see another premise that is violated by $f(x)$.
Thanks for all the help.
Best Answer
A contraction in a metric space is a mapping $f:X\to X$ such that there exists $0<c<1$ such that for all $x,y\in X$, $x\neq y$, it holds that $d(f(x),f(y))\leq cd(x,y)$. Indeed, the Banach fixed-point theorem shows that such mappings on complete metric spaces always have a unique fixed point.
However, if one relaxes the conditions to require only $d(f(x),f(y))<d(x,y)$ (for all $x\neq y$), the function in question no longer necessarily has a fixed point. Indeed, the example you provide demonstrates that principle well: note that where you show that $|f(x)-f(y)|=|f^\prime(x)(x-y)|$, based (I presume) on the mean-value theorem, in fact all that can be known is that there exists some $z\in(y,x)$ (or $(x,y)$, naturally) such that $f(x)-f(y)=f^\prime(z)(x-y)$. Hence, the fact that--for the provided function--$\lim_{z\to\infty}f^\prime(z)=1$ implies that no $c<1$ exists to be used in the conditions to the Banach fixed-point theorem.
It remains true, however, that such a mapping cannot have more than a single fixed point, as that part of the proof relies solely on the strict inequality. As an aside, if one accepts the assumption that $X$ is compact (hence, as @MattE reminds us, necessarily complete), then even such mappings still have a (unique) fixed-point, as the only possible minimum to the continuous function $F(x)=d(x,f(x))$.