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Thread: 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; Nov 4th 2008 at 12:12 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 $\displaystyle f\in L^1(\mathbb{R})$ and $\displaystyle \delta>0$, write $\displaystyle M_\delta f(x) = f(\delta x)$. You want to prove that $\displaystyle \|M_\delta f - f\|\to0$ as $\displaystyle \delta\to1$. 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 $\displaystyle f,\,g\in L^1(\mathbb{R})$ then $\displaystyle \|M_\delta f - M_\delta g\| = \delta^{-1}\|f-g\|$. This is easily seen by making the substitution $\displaystyle y=\delta x$ in the integral $\displaystyle \int_{\mathbb{R}}|f(\delta x) - g(\delta x)|\,dx\ (= \|M_\delta f - M_\delta g\|)$. That shows that if $\displaystyle \delta$ is close to 1 and $\displaystyle f$ is close to $\displaystyle g$, then $\displaystyle M_\delta f$ is close to $\displaystyle M_\delta g$.

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

    Now you just have to put those two steps together and use the triangle inequality: $\displaystyle \|f-M_\delta f\|\leqslant \|f-g\| + \|g-M_\delta g\| + \|M_\delta g - M_\delta f\|$. If $\displaystyle g$ is close enough to $\displaystyle f$ and $\displaystyle \delta$ 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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