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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2$\begingroup$ Dont you mean ''$=$'' $\endgroup$– Marios GretsasSep 13, 2017 at 2:01
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$\begingroup$ No, the function that have not fixed point $\endgroup$– KalawaSep 15, 2017 at 15:43
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3 Answers
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Hint: consider polynomials of degree $1$ such that $f(1)=1$.
answered Sep 13, 2017 at 2:00
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$\begingroup$ @Kalawa You asked for $f: [0,1) \to [0,1)$. That doesn't allow $1$ as a value. $\endgroup$ Sep 15, 2017 at 20:29
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$\begingroup$ Still doesn't take $[0,1)$ into $[0,1)$. $\endgroup$ Sep 17, 2017 at 7:20
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$\begingroup$ f I tried the function $f(x)=\dfrac{x^2+1}{2}$, would it work? $\endgroup$– KalawaSep 17, 2017 at 18:20
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$\begingroup$ 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, $y=0.99x+0.01$ $\endgroup$ Oct 25, 2018 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 $f$.
