an example of a complete space $X$ and a such mapping $T$ without fixed points & Show that if $X$ is compact then such $T$ has a unique fixed point.

Question

Suppose a mapping $T:X \to X$ has the property: $d(T_x,T_y)< d(x,y)\ \forall x,y \in X $ with $ x\neq y$

a) Give an example of a complete space $X$ and a such mapping $T$ without fixed points

b) Show that if $X$ is compact then such $T$ has a unique fixed point.

For the solution, $T$ is a contraction operator by the description.

For a, the hint is using $f(x) = \ln (e^x+1)$ For b, the hint is to consider $\inf d(x,Tx)$

Answer 1

Hint for (a): Apply the Mean Value Theorem to that function you were given.

Hint for (b): Since $d(T(x),T(T(x))) < d(x, T(x))$ if $x\ne T(x)$, you must have $\inf d(x,T(x))=0$. (You should probably explain that a bit.) Now compactness shows (why?) that there exists $x$ with $d(x,T(x))=0$.

Answer 2

Another, perhaps easier example for $(a)$ would be $f : \mathbb{R} \to \mathbb{R}$ given as $$f(x) = \sqrt{1+x^2}$$ for all $x \in \mathbb{R}$.

$f$ has no fixed points:

$$x = \sqrt{1 + x^2} \implies x^2 = 1 + x^2 \implies 0 = 1$$

However, for any $x, y \in \mathbb{R}$, $x \ne y$ holds $\left|f(y) - f(x)\right| < |y - x|$.

Indeed, assume $y > x$:

\begin{align}&\quad\quad\,\,\sqrt{1+y^2} - \sqrt{1+x^2} < y - x\\ &\iff \left(\sqrt{1+y^2} - \sqrt{1+x^2}\right)^2 < \left(y - x\right)^2\\ &\iff 2 + x^2 + y^2 - 2\sqrt{\left(1+x^2\right)\left(1+y^2\right)} < x^2 - 2xy + y^2\\ &\iff 1+xy < \sqrt{\left(1+x^2\right)\left(1+y^2\right)} \end{align}

This is true by CSB. The strict inequality is because $(1,x)$ and $(1,y)$ are not proportional.