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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 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
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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 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*}
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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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