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Math Help - Existence of Convolution

  1. #1
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    Existence of Convolution

    Show that if f is locally integrable and that g\in C^{\infty}_{c}(\mathbb{R}), then f*g(x)<\infty.

    OK, so first thing I'm thinking to do is to use the commutativity of the convolution operator and then say since g has compact support (which in this setting means it vanishes at positive and negative infinity, or that it has support on a bounded interval), we can write the convolution in terms of a finite integral in this way:

    f*g = \int_{-\infty}^{\infty}f(y)g(x-y)dy

    =\int_{-\infty}^{\infty}f(x-y)g(y)dy

    =\int_a^bf(x-y)g(y)dy

    For some interval [a, b].

    And now, trying to figure out the bounds has me confused, since we know nothing about f other than local integrability (so we can't say that we have less than or equal to f\cdot\max\{g(x): x\in [a,b]\} or anything of that sort), so I'm having issues. :-/
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  2. #2
    Super Member
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    Notice \int_{\mathbb{R}}\left( \int_{\mathbb{R} } |f(y)g(x-y)|dx\right) dy <\infty so by Tonelli's theorem (x,y)\mapsto f(y)g(x-y) \in L^1(\mathbb{R} ^2) then a standard theorem tells you that the convolution is almost everywhere finite, but since (it's easy to prove) it is continous it cannot blow up at any number (only at plus/minus infinity).
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