Example of a function, so that $g(x)\neq x$ I'm trying to find an example of a function $g:\mathbb{R}\to \mathbb{R}$ (or $g:[1,\infty) \to \mathbb{R}$), so that 
$$|g(x_1)-g(x_2)|<|x_1-x_2|$$
for all $x_1, x_2\in \mathbb{R}$ ( or $x_1,x_2\in [1,\infty)$) and $x_1\neq x_2$ but
$$g(x)=x$$
doesn't have a solution
Any ideas?
 A: I think $g(x)=\frac{x}{2}$ works.
A: $g(x)=\sqrt{x^2+1}$ works on $\mathbb{R}.$ It has no fixed point, since squaring would give $x^2+1=x^2.$ And the derivative of $g$ is strictly increasing. More explicitly if $a>b$ so that also $g(a)>g(b)$ then the inequality desired becomes equivalent to $\sqrt{a^2+1}-a>\sqrt{b^2+1}-b,$ and then since $\sqrt{x^2+x}-x$ has an everywhere negative derivative, the inequality follows.
Added: Above it is OK to assume WLOG that $a>b,$ however it then doesn't follow that the absolute value sign may be just dropped in $\sqrt{a^2+1}-\sqrt{b^2+1}.$ If this quantity is $\ge 0$ the absolute sign may be dropped and treated as above. In the opposite case the absolute value becomes $\sqrt{b^2+1}-\sqrt{a^2+1},$ which we still wish to show is less than $a-b$ since we're assuming $a>b.$ So here we want to wind up with $\sqrt{b^2+1}+b<\sqrt{a^2+1}+a,$ which is clear from $b<a$ and that $x+\sqrt{x^2+1}$ is increasing.
Simpler proof of inequality: If $a \neq b$ then to show $|g(a)-g(b)|<|a-b|$ one can divide by $[a-b|$ and use that, via the mean value theorem, there is a $c$ between $a$ and $b$ for which
$$\frac{g(a)-g(b)}{a-b}=g'(c).$$ Here $g'(c)=c/\sqrt{c^2+1},$ which has magnitude strictly less than $1.$
