(New page: Back to The Pirate's Booty Given that <math>f\in L^1(\mathbb{R})</math> and <math>\int_{\mathbb{R}}\int_{\mathbb{R}}f(4x)f(x+y)dxdy = 1</math>. Calculate <math>\int_{\mathbb{R}}f(x)d...) |
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| + | == Problem #7.9, MA598R, Summer 2009, Weigel == | ||
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<math>\Rightarrow \int_{\mathbb{R}}f(x)dx =\pm 2</math> | <math>\Rightarrow \int_{\mathbb{R}}f(x)dx =\pm 2</math> | ||
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Revision as of 04:38, 11 June 2013
Problem #7.9, MA598R, Summer 2009, Weigel
Back to The Pirate's Booty
Given that $ f\in L^1(\mathbb{R}) $ and $ \int_{\mathbb{R}}\int_{\mathbb{R}}f(4x)f(x+y)dxdy = 1 $. Calculate $ \int_{\mathbb{R}}f(x)dx $
Proof: Since $ \mathbb{R} $ is $ \sigma $-finite we can apply Fubini's Theorem. Hence,
$ \int_{\mathbb{R}}\int_{\mathbb{R}}f(4x)f(x+y)dxdy =\int_{\mathbb{R}}f(4x)\bigg(\int_{\mathbb{R}}f(x+y)dy\bigg)dx= $ $ \int_{\mathbb{R}}f(4x)dx\cdot \int_{\mathbb{R}}f(y')dy'=\frac{1}{4}\int_{\mathbb{R}}f(x')dx'\cdot \int_{\mathbb{R}}f(y')dy'= $ $ \frac{1}{4}\bigg(\int_{\mathbb{R}}f(x)dx\bigg)^2 = 1 $
$ \Rightarrow \int_{\mathbb{R}}f(x)dx =\pm 2 $
