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Math Help - Question involving a continuous function on a closed interval

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    Question involving a continuous function on a closed interval



    Intuitively this is obvious by graphing g(x) = x on [0,1] and seeing since f is continuous it has to intersect with g at some point. But I spent a long time and cannot figure out how to prove this.
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    Senior Member Sambit's Avatar
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    It is the well-known fixed point theorem. Suppose f is continuous on [0,1] and f(x)\epsilon[0,1] for every x\epsilon[0,1].

    If f(0)=0 or f(1)=1, the theorem is proved. So we try to prove the theorem assuming

    f(0)>0 and f(1)<1.

    Let g(x)=f(x)-x for all x\epsilon[0,1]. Hence g(0)>0 and g(1)<0 and g is continuous on [0,1], that is, 0 is an intermediate value of g on [0,1]. Hence by intermediate value theorem, there exists a point

    c\epsilon(0,1) such that g(c)=0 --which means f(c)=c. Hence the prrof.

    EDIT: c is equivalent to x_0
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by paulrb View Post


    Intuitively this is obvious by graphing g(x) = x on [0,1] and seeing since f is continuous it has to intersect with g at some point. But I spent a long time and cannot figure out how to prove this.
    Alternatively, suppose that f(x)\ne x then the mapping \displaystyle f:[0,1]\to\{-1,1}:\frac{|f(x)-x|}{f(x)-x} is a continuous surjection which is impossible.
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