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Math Help - Metric of C[0,1]

  1. #1
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    Metric of C[0,1]

    This is from my classnote:

    Define C[0,1] to be all continuous functions on [0,1], and define its metric be p(f,g)= \int ^1 _0 \mid f-g \mid dx

    Show that if f \neq g , then p(f,g) \neq 0

    Proof.

    if  f \neq g , then  f(x_0) \neq g(x_0) for some  x_0 \in [0,1]

    Consider the function h(x) = f(x)-g(x) , we see that [tex]h(x) [tex] is continuous on x_0

    Now, let  \epsilon = \mid f(x_0)-g(x_0) \mid , then  \exists \delta > 0 \ s.t. \ \mid x - x_0 \mid < \delta

    with  \mid [ f(x) - g(x)]-[f(x_0)-g(x_0)] \mid < \frac { \epsilon }{2}

    Then we have  \mid  f(x) - g(x) \mid \geq \frac {\epsilon }{2} on  [x_0 - \delta , x_0 + \delta ] (Why is that?)

    Then we have p(f,g)= \int ^1_0 \mid f-g \mid dx \geq \int ^{x_0+ \delta} _{x_0 - \delta } \mid f-g \mid dx \geq 2 \delta \frac { \mid f(x_0) - g(x_0) \mid }{2} > 0
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  2. #2
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    Quote Originally Posted by tttcomrader View Post

    Then we have  \mid  f(x) - g(x) \mid \geq \frac {\epsilon }{2} on  [x_0 - \delta , x_0 + \delta ] (Why is that?)

    [/tex]
    By the triangle inequality, and properties of the absolute value we have:

    \vert f(x_0)-g(x_0) \vert - \vert f(x)-g(x) \vert \leq \vert f(x)-g(x) - [f(x_0)-g(x_0)] \vert < \frac{ \epsilon }{2} now, using the first and last parts of the inequality we have: \vert f(x_0)-g(x_0) \vert -\frac{ \epsilon }{2} < \vert f(x)-g(x) \vert and using the fact that \vert f(x_0)-g(x_0) \vert = \epsilon you're finished
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