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.

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    $\begingroup$ This is correct but it is simpler to use $g(x)=f(x)-x$. $\endgroup$ – Kavi Rama Murthy Jun 23 '19 at 0:22
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    $\begingroup$ 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 ? $\endgroup$ – I0_0I Jun 23 '19 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 $g$.


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