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Convolutions of Lebesgue integrable functions

by redbowlover
Tags: convolutions, functions, integrable, lebesgue
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redbowlover
#1
May4-11, 10:21 PM
P: 16
Hello,

Looking through Torchinksy's Real Variables text and I'm infinitely confused about convolutions.

For two integrable functions [tex] f, g \in \ L^1(R) [/tex] we define the convolution [tex] f*g=\int_R f(x-y)g(y) dy , \forall x\in R [/tex].

Then, apparently, [tex]f*g[/tex] is also integrable. But I'm not sure how to prove this. The text says, for now suppose f, g and the integrand are nonnegative. Then [tex]||f*g||_1=\int_R\int_R f(x-y)g(y) dydx[/tex].

But why is this true? Clearly [tex]||f*g||_1=\int_R\int_R f(x-y)g(y) dydm[/tex], where m is the Lebesgue measure. But what allows you to switch to dx? Maybe this speaks to my greater misunderstanding of integrals. Sigh...If anyone could explain this to me I'd be very grateful.

I thought in general you could only switch from Lebesgue integration to Riemann integration when you knew the thing was Riemann integrable.
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g_edgar
#2
May5-11, 02:02 PM
P: 607
He writes $\int_R f(x) dx$ to mean the Lebesgue integral.


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