Be a continuous function $f:[0,1] \rightarrow [0,1]$.
Show that there is a $\zeta \in [0,1]$ with $f(\zeta)=\zeta$ ($\zeta$ is called fixed point).
Consider the function $g:[0,1] \rightarrow [-1,1]$, $g(x):= f(x)-x$.
$g$ is continuous.
Because of $f(0),f(1) \in [0,1]$ is $g(0)\geq0$ and $g(1)\leq 0$.
Because $f(0)$ has a value between $0$ and $1$, $f(0)\geq 0$.
$g(0) = f(0)-0= f(0) \geq 0 – 0 = 0$.
Because $f(1)$ has a value between $0$ and $1$, $ f(1) \leq 1 $.
$g(1)=f(1)-1 \leq 1 – 1 = 0$ $\Leftrightarrow g(0)\geq 0$ and $g(1)\leq 0 $
After the IVT: $\exists \zeta \in [0,1]:g(\zeta)=0 \Leftrightarrow f(\zeta) = \zeta $
$\zeta$ is a fixed point of $f$. $\Box$
My questions are:
Is this proof done in the correct way or have I missed something?
Is there something I can improve?
Best Answer
Yes, your proof is pretty much correct. I might change it slightly so that it's presented better.
Consider the function $g:[0,1] \rightarrow [-1,1]$ defined by $g(x):= f(x)-x$. Note that since $f$ is continuous and polynomials are continuous, $g$ is also continuous.
Now observe that since $f(0)\in [0,1] \implies f(0) \ge 0$, we have: $$ g(0) = f(0)-0= f(0) \geq 0 $$ Likewise, since $f(1) \in [0,1] \implies f(1) \leq 1$, we have: $$ g(1)=f(1)-1 \leq 1 - 1 = 0 $$ Thus, since $g(1) \leq 0 \leq g(0)$, we know by IVT that $\exists \zeta \in [0,1]$ such that $g(\zeta)=0 \iff f(\zeta) = \zeta $. Hence, $\zeta$ is a fixed point of $f$, as desired. $\Box$