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

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

    Hello

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

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

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

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

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

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

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

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

    Quote Originally Posted by issacnewton View Post
    function $\displaystyle f :[0,1]\rightarrow [0,1]$ is continuous. Also range of $\displaystyle f$ is $\displaystyle [0,1]$. I have to use intermediate value theorem to prove that $\displaystyle f$ must have a fixed point in it domain. Now let's define $\displaystyle g(x) = f(x) - x$. So $\displaystyle g(0) = f(0)$ and $\displaystyle g(1) = f(1)-1$. Now since $\displaystyle f$ is continuous on $\displaystyle [0,1]$, $\displaystyle g$ is also continuous on $\displaystyle [0,1]$.
    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
    $\displaystyle \begin{align*}0&\le f(1)\le 1\\-1&\le f(1)-1\le 0 \\-1&\le g(1)\le 0\\g(1)&\le 0\le g(0) \end{align*}$
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  3. #3
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    Re: Proving that continuous function f:[0,1]-> [0,1] has a fixed point from IVT

    Thanks Plato, that was short
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