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Math Help - Real Analysis, continuity on [0,1]

  1. #1
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    Real Analysis, continuity on [0,1]

    Q: let f be continuous from [0,1] to [0,1]. Show their is an x on [0,1] for which f(x) = x

    Our hint was to use g(x) = f(x) - x and use the IVT

    We figure that since our Range is from [0,1] with our max = 1 and min = 0 that if we could show that the function attains a value above y=x and one below y=x, that continuity implies that it mus cross y=x attaining our value.

    Problem is how do we do this, my intial thinking was using the defintition of continuity using our

    |f(x) - f(a)| < epsilon where we take 2 cases where f(a) is our min and f(a) is our max, where working it down would take us to

    |f(x) - f(max)| < epsilon
    |f(x) - f(min)| < epsilon

    |f(x) - 1| < epsilon
    |f(x) - 0| < epsilon

    putting our x

    |f(x)| < x < |f(x) - 1|


    hence where I stopped and got stuck. Thanks for any help
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  2. #2
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    Quote Originally Posted by pila0688 View Post
    Q: let f be continuous from [0,1] to [0,1]. Show their is an x on [0,1] for which f(x) = x

    Our hint was to use g(x) = f(x) - x and use the IVT
    Note g(0) = f(0) - 0 = f(0)\geq 0 and g(1) = f(1) - 1 \leq 1 -1 = 0. Therefore there is 0\leq x_0\leq 1 so that g(x_0) = 0 \implies f(x_0) = x_0.

    This is Mine 18th Post!!!
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  3. #3
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    (1) Suppose f(a) - a < 0 but f(b) - b > 0 for some 0 < a < 1, 0 < b < 1.

    Then by the IVT, there is c such that a < c < b, and hence 0 < c < 1, with f(c) - c = 0.

    (2) Suppose f(x) - x > 0 for all 0 <= x <= 1. Then f(1) - 1 > 0, or f(1) > 1, which is impossible.

    (3) Suppose f(x) - x < 0 for all 0 <= x <= 1. Then f(0) - 0 < 0, or f(0) < 0, which is impossible.

    Then if situation (1) does not exist, either f(x) - x <= 0 for all 0 <= x <= 1 or
    f(x) - x >= 0 for all 0 <= x <= 1, with equality occurring at least once.
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  4. #4
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    Quote Originally Posted by pila0688 View Post
    Q: let f be continuous from [0,1] to [0,1]. Show their is an x on [0,1] for which f(x) = x

    Our hint was to use g(x) = f(x) - x and use the IVT

    We figure that since our Range is from [0,1] with our max = 1 and min = 0 that if we could show that the function attains a value above y=x and one below y=x, that continuity implies that it mus cross y=x attaining our value.

    Problem is how do we do this, my intial thinking was using the defintition of continuity using our

    |f(x) - f(a)| < epsilon where we take 2 cases where f(a) is our min and f(a) is our max, where working it down would take us to

    |f(x) - f(max)| < epsilon
    |f(x) - f(min)| < epsilon

    |f(x) - 1| < epsilon
    |f(x) - 0| < epsilon

    putting our x

    |f(x)| < x < |f(x) - 1|


    hence where I stopped and got stuck. Thanks for any help
    This is called as Fixed Point Theorem.
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