Fixed point and period of continuous function

Prove/ Disprove:

1. Let $f:(0,1)\to(0,1)$ be such that $|f(x)-f(y)|\leq 0.5|x-y|$ for all $x ,y.$ Then f has a fixed point.

2.Let $f:\mathbb R\to\mathbb R$ be continuous and periodic with period $T>0.$Then there exists a point $x_0\in\mathbb R$ such that

$f(x_0)=f(x_0+T/2).$

• Let $f(x)=\frac{x}{2}$ so $f$ hasn't a fixed point in $(0,1)$.

• Let $g(x)=f(x+T/2)-f(x)$ then $g$ is continuous and $g(0)g(T/2)\le0$ so use the intermediate value theorem to conclude.

• Good points. Applicable ones. +1 Jan 22, 2014 at 14:29

The counterexample for $1$ is the function $f(x) = x/4$ which would only have a fixed point if $0$ was included in the interval.

• All the counterexamples are from geometrical visualisations. But to write concrete proof (probably method of contradiction will be fruitful) can someone give any idea? Jan 22, 2014 at 13:36
• $f(x) = x/4$ suffices the condition $|f(x) - f(y)|\leq 0.5 |x - y|$ and does not have a fixed point on $(0,1)$. What more do you need, there's no geometry here!
– 5xum
Jan 22, 2014 at 13:38
• 5xum See the counterexamples $x/2,x/4,0$ are all the lines in $\mathbb R^2$ with slopes less than the slope of $y=x$ satisfying the given criteria and it is also obvious that all the lines cut $y=x$ line only at (0,0). This is the picture in background but not a formal proof. Jan 22, 2014 at 13:47
• No, but you can easily MAKE a formal proof. Like this: Take any pair of $x,y\in\mathbb R$. For this pair, $f(x) - f(y) = (x-y)/4$, meaning that $|f(x) - f(x)| = 0.25 |x - y| < 0.5 |x-y|$. If you cannot prove that $x/4$ does not have a fixed point on $(0,1)$, you should do some more studying before trying to solve the task in your question.
– 5xum
Jan 22, 2014 at 13:53

Or use the trivial function $f(x) = 0$.