# Thread: Proving that continuous function f:[0,1]-> [0,1] has a fixed point from IVT

1. ## Proving that continuous function f:[0,1]-> [0,1] has a fixed point from IVT

Hello

function $f :[0,1]\rightarrow [0,1]$ is continuous. Also range of $f$ is $[0,1]$. I have to use intermediate value theorem to prove that $f$ must have a fixed point in it domain. Now let's define $g(x) = f(x) - x$. So $g(0) = f(0)$ and $g(1) = f(1)-1$. Now since $f$ is continuous on $[0,1]$, $g$ is also continuous on $[0,1]$. Consider the product $g(0)g(1)$. We have three possibilities here.
Case 1) $g(0)g(1) > 0$

So, we have two sub cases here. Sub case 1 is $g(0) > 0$ and $g(1) >0$. $g(0) > 0\Rightarrow f(0) >0$ and $g(1) > 0 \Rightarrow f(1) > 1$. But since range of $f$ is $[0,1]$, $f(1)\ngtr 1$. So this sub case is not possible. Consider the second sub case here. $g(0) <0$ and $g(1) <0$. $g(0) <0 \Rightarrow f(0) < 0$. Since range of $f$ is $[0,1]$, $f(0) \nless 0$, hence the second sub case is also not possible. Which means, case 1 is impossible for function $f$. Let's go to the case two

Case 2) $g(0)g(1) < 0$

First sub case is that $g(0) < 0$ and $g(1) > 0$. Since $g(0) < 0 < g(1)$, from intermediate value theorem, $\exists~ c \in (0,1)$ such that $g(c) = 0$, which means we have $f(c) = c$ and $c \in \text{Dom}(f)$. So $c$ is the fixed point of function $f$. Second sub case is $g(0) >0$ and $g(1) < 0$. The proof will be on similar lines. Now let's go to the last case

Case 3) $g(0)g(1) = 0$

Here either $g(0) = 0$ or $f(1)=1$. So either $f(0) = 0$ or $f(1) = 1$. Since $0$ and $1$ are in domain of $f$, we proved that there is a fixed point of $f$.

Since all cases are exhausted, we proved that $f$ must have a fixed point.
Is my proof correct ?

Thanks

2. ## Re: Proving that continuous function f:[0,1]-> [0,1] has a fixed point from IVT

Originally Posted by issacnewton
function $f :[0,1]\rightarrow [0,1]$ is continuous. Also range of $f$ is $[0,1]$. I have to use intermediate value theorem to prove that $f$ must have a fixed point in it domain. Now let's define $g(x) = f(x) - x$. So $g(0) = f(0)$ and $g(1) = f(1)-1$. Now since $f$ is continuous on $[0,1]$, $g$ is also continuous on $[0,1]$.
It is true that $g(0)=f(0)$ and $f(0)\in[0,1]$ so that $0\le g(0)\le 1$.
Now because $g(1)=f(1)-1$ AND
\begin{align*}0&\le f(1)\le 1\\-1&\le f(1)-1\le 0 \\-1&\le g(1)\le 0\\g(1)&\le 0\le g(0) \end{align*}

3. ## Re: Proving that continuous function f:[0,1]-> [0,1] has a fixed point from IVT

Thanks Plato, that was short