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I am searching an example of a continuous function $f: [0,1) \to [0,1)$ but f has no fixed point, that is, there is no point $x_0 \in [0,1)$ such that $f(x_0)\not= x_0 \forall x_0$.

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    $\begingroup$ Dont you mean ''$=$'' $\endgroup$ – Marios Gretsas Sep 13 '17 at 2:01
  • $\begingroup$ No, the function that have not fixed point $\endgroup$ – Kalawa Sep 15 '17 at 15:43
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Hint: consider polynomials of degree $1$ such that $f(1)=1$.

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  • $\begingroup$ @Kalawa You asked for $f: [0,1) \to [0,1)$. That doesn't allow $1$ as a value. $\endgroup$ – Robert Israel Sep 15 '17 at 20:29
  • $\begingroup$ Still doesn't take $[0,1)$ into $[0,1)$. $\endgroup$ – Robert Israel Sep 17 '17 at 7:20
  • $\begingroup$ f I tried the function $f(x)=\dfrac{x^2+1}{2}$, would it work? $\endgroup$ – Kalawa Sep 17 '17 at 18:20
  • $\begingroup$ Yes, that would also work. $\endgroup$ – Robert Israel Sep 17 '17 at 20:27
  • $\begingroup$ Simply take, a linear polynomial, as @RobertIsrael has suggested. Any two none-parallel line, $y=mx+c$, $y=x$ will intersect in a single point. Make that intersection point as $(1,(f1))$. As an example, take, $y=0.99x+0.01$ $\endgroup$ – nature1729 Oct 25 '18 at 7:57
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Always try to take motivation from picture.Draw straight line above line y=x which starts at (0,y) and ends at (1,1).y will be any value in (0,1) and you know equation of line passing through two points.

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Such function does not exist.

Let $\phi(x)= f(x)-x$, $\phi(0)=f(0)\ge 0$ and $\phi(1)=f(1)-1<0$, therefore by ITV theorem there must be a $x_0\in [0;1)$ so that $\phi(x_0)=0$ which simply mean $f(x_0)=x_0$.

If you want a such function, you should avoid continuity, or the stability of $[0,1)$ by $f$.

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