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Math Help - L-1 space of integrable functions

  1. #1
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    L-1 space of integrable functions

    If f is in L1 and delta > 0, then the L1 norm of f(delta*x)-f(x) goes to zero as delta goes to 1. I have been studying my text for two hours and still really don't understand it. Any thoughts?

    By the way, I have some ideas--

    If I can show that f is a continuous function of compact support, then I will know that f is dense in L-1. That will guarantee that there exists a function g in L-1 such that the L1 norm of f minus g is less than any epsilon (epsilon greater than zero).

    rogerpodger
    Last edited by rogerpodger; November 4th 2008 at 01:12 PM.
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  2. #2
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    Quote Originally Posted by rogerpodger View Post
    If f is in L1 and delta > 0, then the L1 norm of f(delta*x)-f(x) goes to zero as delta goes to 1. I have been studying my text for two hours and still really don't understand it. Any thoughts?

    By the way, I have some ideas--

    If I can show that f is a continuous function of compact support, then I will know that f is dense in L-1. That will guarantee that there exists a function g in L-1 such that the L1 norm of f minus g is less than any epsilon (epsilon greater than zero).
    For f\in L^1(\mathbb{R}) and \delta>0, write M_\delta f(x) = f(\delta x). You want to prove that \|M_\delta f - f\|\to0 as \delta\to1. I think that the idea of proving this by doing it first for the case when f is continuous with compact support is the best way to go about it.

    The first step is to show that if f,\,g\in L^1(\mathbb{R}) then \|M_\delta f - M_\delta g\| = \delta^{-1}\|f-g\|. This is easily seen by making the substitution y=\delta x in the integral \int_{\mathbb{R}}|f(\delta x) - g(\delta x)|\,dx\ (= \|M_\delta f - M_\delta g\|). That shows that if \delta is close to 1 and f is close to g, then M_\delta f is close to M_\delta g.

    The next step is to show that if g is continuous with compact support then \|M_\delta g - g\| is small if \delta is close to 1. This is an easy consequence of the fact that g is uniformly continuous.

    Now you just have to put those two steps together and use the triangle inequality: \|f-M_\delta f\|\leqslant \|f-g\| + \|g-M_\delta g\| + \|M_\delta g - M_\delta f\|. If g is close enough to f and \delta is sufficiently close to 1, then all three terms on the right side of the inequality can be made arbitrarily small.
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  3. #3
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    Thanks!

    I was able to solve the problem! Thanks for you help! You gave me somewhere to go without doing the problem for me! I really appreciate it!

    rogerpodger
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