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Math Help - Metric Spaces

  1. #1
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    Metric Spaces

    Problem:

    Let X = {the set of all continuous functions f: [a,b] to R}. Let d(f,g) = the integral from a to b of |f(x) - g(x)|dx. Show that d is a metric on X, and therefore, X,d is a metric space.

    I also have a hint: Recall that if h(x) is greater than or equal to 0 on [a,b] then the integral from a to b of h(x)dx is greater than or equal to 0; also if the integral from a to b of h(x)dx = 0 for h(x) greater than or equal to 0 then h(x) = 0 for all x in [a,b]

    Here are the three conditions for a metric space:

    1. d(x,y) = 0 if and only if x=y
    Meaning that the distance from function to itself is 0

    2. d(x,y) = d(y,x)
    Reflexive property

    3. d(x,y) is less than or equal to d(x,z) + d(z,y)
    Triangle Inequality

    I have no clue where to even start on this problem so any help would be greatly appreciated!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by hannahs View Post
    Problem:

    Let X = {the set of all continuous functions f: [a,b] to R}. Let d(f,g) = the integral from a to b of |f(x) - g(x)|dx. Show that d is a metric on X, and therefore, X,d is a metric space.

    I also have a hint: Recall that if h(x) is greater than or equal to 0 on [a,b] then the integral from a to b of h(x)dx is greater than or equal to 0; also if the integral from a to b of h(x)dx = 0 for h(x) greater than or equal to 0 then h(x) = 0 for all x in [a,b]

    Here are the three conditions for a metric space:

    1. d(x,y) = 0 if and only if x=y
    Meaning that the distance from function to itself is 0
    we need to show that \int_a^b |f(x) - g(x)|~dx = 0 \implies f(x) = g(x)

    and that f(x) = g(x) \implies \int_a^b |f(x) - g(x)|~dx = 0

    these are trivial, since |a| = 0 iff a = 0. here we can take a = f(x) - g(x)


    2. d(x,y) = d(y,x)
    Reflexive property
    Again, by the definition of the absolute value function, it is obvious that |f(x) - g(x)| = |g(x) - f(x)| hence \int_a^b |f(x) - g(x)|~dx = \int_a^b |g(x) - f(x)|~dx


    3. d(x,y) is less than or equal to d(x,z) + d(z,y)
    Triangle Inequality
    Let h(x): [a,b] \to \mathbb{R} be continuous, then h \in X and we have that

    \int_a^b |f(x) - g(x)|~dx = \int_a^b |f(x) - h(x) + h(x) - g(x)|~dx \le \int_a^b|f(x) - h(x)|~dx + \int_a^b |h(x) - g(x)|~dx by the Triangle Inequality

    thus we have shown d is a metric
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    Thank you so much!
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