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Math Help - [SOLVED] Metric Space and Inequality

  1. #1
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    [SOLVED] Metric Space and Inequality

    Problem:
    For any two functions f and g in C([0,1],R), define

    a. Prove that this defines a metric on C([0,1],R)
    b. Prove that the following inequality relating this metric and the uniform metric:

    c. Compare the concepts of convergence of a sequence of functions in this metric and in the uniform metric.
    ================================================== ================================
    Attempt:
    a. First, C([0,1],R) denotes that the function is continuous.
    We need to prove that

    i. if and only if f = g. If f = g, then



    ii. d*(f,g) = d*(g,f) (symmetry): This is true since , thus

    iii. Triangle Inequality:



    Thus, d*(f,g) is a metric on C([0,1],R)

    b.We need to prove that





    I'm unsure where to start. I try to divide both sides by (b-a) which yields





    Here, I use the Mean Value Theorem for integrals, since the function on [a,b] is continuous, then there exists a c such that





    c. I'm not sure.


    Thank you for your time.
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  2. #2
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    Quote Originally Posted by Paperwings View Post
    a. Prove that this defines a metric on C([0,1],R)
    You almost proved this. You did not prove the "only if" part in the second part of being a metric space.
    If, \int_a^b |f(x)-g(x)| dx = 0 then it means f(x)-g(x)=0 i.e. f=g.
    Because if f,g are continous then |f-g| is continous.

    To show d^*(f,g) \leq (b-a)d(f,g). I think here f,g\in \mathcal{C}([a,b],\mathbb{R}). I am also assuming that you define d(f,g) = \sup\{ |f(x)-g(x)| : x\in [a,b]\}. Clearly, |f(x)-g(x)|leq d(f,g) for all x\in [a,b]. And thus, \smallint_a^b |f-g| \leq \smallint_a^b d(f,g) \implies d^*(f,g)\leq (b-a)d(f,g).
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  3. #3
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    Thank you very much TPH. I also understand part (c) more clearly as well. Cheers.
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