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Thread: Modulus of continuity is continuous?

  1. #1
    Junior Member RaisinBread's Avatar
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    Modulus of continuity is continuous?

    Hey,

    I'm currently reading a book on convergence of probability measures, and there is a property that they assert without too many details that I can't manage to work out for myself.

    To put you in context, we're in the space $\displaystyle C[0,1]$ of functions $\displaystyle f:[0,1]\to\mathbb{R}$ that are continuous with respect to the standard euclidean metric $\displaystyle d(x,y)=|x-y|$, and we define the metric on $\displaystyle C[0,1]$ to be $\displaystyle \rho(f,g)=\sup_{t\in[0,1]}|f(t)-g(t)|$.

    For every $\displaystyle \delta>0$ and $\displaystyle f\in C[0,1]$, define the modulus of continuity $\displaystyle w(f,\delta)$ as

    $\displaystyle w(f,\delta)=\sup_{|x-y|<\delta}|f(x)-f(y)|$


    In the book, they say that for any fixed $\displaystyle \delta>0$, the function $\displaystyle w(\cdot,\delta):C[0,1]\to\mathbb{R}$ is continuous. Their only argument is that
    for any $\displaystyle f,g\in C[0,1]$, we have $\displaystyle |w(f,\delta)-w(g,\delta)|\leq 2\rho(f,g)$. It's easy to see how this implies continuity, but I can't manage to show this inequality myself. Any help or hints would be greatly appreciated.
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  2. #2
    Super Member girdav's Avatar
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    Re: Modulus of continuity is continuous?

    We have $\displaystyle |f(x)-f(y)|\leq 2\rho(f,g)+|g(x)-g(y)|$ then take supremum. Switch the roles of $\displaystyle f$ and $\displaystyle g$ to get what you want.

    P.S. Which book are you studying?
    Thanks from RaisinBread
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  3. #3
    Junior Member RaisinBread's Avatar
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    Re: Modulus of continuity is continuous?

    Thanks for the answer, I'll try this out.

    I'm studying Convergence of Probability Measures 2nd edition by Patrick Billingsley, I'm trying to gain a better understanding of Donsker's Theorem, which is basically a version of the Central Limit Theorem for random variables taking values in C[0,1].
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