# Prove that every continuous map $f: [0,1] \rightarrow [0,1]$ has a fixed point.

Prove that every continuous map $$f: [0,1] \rightarrow [0,1]$$ has a fixed point.

Suppose $$f$$ does not have a fixed point, then $$\forall x \in [0,1], f(x) \neq x$$.

Thus we have a well defined function $$g(x) = \frac{1}{f(x)-x}$$. Note that as $$g(x)$$ is the composition or continuous functions, it must be continuous.

However, $$g(0) > 0$$ and $$g(1) < 0$$, so by the intermediate value theorem $$\exists x \in [0,1]$$ such that $$g(x) = 0$$. This is clearly impossible.

Thus $$f$$ has a fixed point.

• This is correct but it is simpler to use $g(x)=f(x)-x$. – Kavi Rama Murthy Jun 23 at 0:22
• I guess the proof is okay, but in my opinion it's not natural. The natural g to define is $g(x)=f(x)-x;$ can you see the geometrical motivation behind defining this function ? – yousef magableh Jun 23 at 0:35

Your proof is correct. Anyway note that the function $$f$$ must meet the line $$y=x$$ in atleast one point, and that point is the fixed point. To see this geometrical idea into a formal proof, set $$g(x)=f(x)-x$$ and apply IVT to $$g$$.