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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

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

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  • 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.

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  • $\begingroup$ Good points. Applicable ones. +1 $\endgroup$ – mrs Jan 22 '14 at 14:29
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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.

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  • $\begingroup$ All the counterexamples are from geometrical visualisations. But to write concrete proof (probably method of contradiction will be fruitful) can someone give any idea? $\endgroup$ – user121418 Jan 22 '14 at 13:36
  • $\begingroup$ $f(x) = x/4$ 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! $\endgroup$ – 5xum Jan 22 '14 at 13:38
  • $\begingroup$ 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. $\endgroup$ – user121418 Jan 22 '14 at 13:47
  • $\begingroup$ 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. $\endgroup$ – 5xum Jan 22 '14 at 13:53
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Or use the trivial function $f(x) = 0$.

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