Continuous function with no fixed point

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 $f(x_0)\not= x_0 \forall x_0$.

• Dont you mean ''$=$'' – Marios Gretsas Sep 13 '17 at 2:01
• No, the function that have not fixed point – Kalawa Sep 15 '17 at 15:43

Hint: consider polynomials of degree $1$ such that $f(1)=1$.

• @Kalawa You asked for $f: [0,1) \to [0,1)$. That doesn't allow $1$ as a value. – Robert Israel Sep 15 '17 at 20:29
• Still doesn't take $[0,1)$ into $[0,1)$. – Robert Israel Sep 17 '17 at 7:20
• f I tried the function $f(x)=\dfrac{x^2+1}{2}$, would it work? – Kalawa Sep 17 '17 at 18:20
• Yes, that would also work. – Robert Israel Sep 17 '17 at 20:27
• 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, $y=0.99x+0.01$ – nature1729 Oct 25 '18 at 7:57

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.

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 $$f$$.