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I am trying to show that for $f,g\in L_1(\mathbb{R}^d)$, $f*g\in L_1(\mathbb{R}^d)$.

Somewhere along the way I need to switch the order of integration in the following integral (I know this for sure because it is literally a step out of my professor's notes).

$$\int_{\mathbb{R}^{d}}f(x-y)g(y)e^{-i\xi\cdot x}\;dy\;dx.$$

In the notes it says "By Fubini's Theorem". But I can't verify the hypothesis of the theorem which says the integrals may be switched if the following is true:

$f(\cdot-y)g(y)e^{-i\xi\cdot \cdot}$ and $f(x-\cdot)g(\cdot)e^{-i\xi\cdot x}$ are both in $L_1(\mathbb{R}^d)$.

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Hmm.. since $f$ and $g$ are bounded right? I guess that would make this trivial... –  Kyle Schlitt Nov 18 '11 at 2:03
    
If my previous comment is not mistaken, then I don't need an answer but just a confirmation would be correct. I never trust my logic in Measure spaces. :P –  Kyle Schlitt Nov 18 '11 at 2:04
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Why would they be bounded? Not all functions in $L_1$ are bounded. –  Michael Hardy Nov 18 '11 at 2:06
    
Of course not. $1/x^2$ is not bounded near $0$, and it is an $L_{1}$ function. I don't know what I was thinking. –  Kyle Schlitt Nov 18 '11 at 3:47
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1 Answer

up vote 4 down vote accepted

We want to think about $$\int_{\mathbb{R}^d} \left(\int_{\mathbb{R}^d} f(x-y)g(y)\;dy\right) e^{-i\xi\cdot x} \;dx.$$ The expression inside the parentheses is the convolution. Suppose for now that we already know that $y \mapsto f(x-y)g(y)$ is in $L_1$. The expression above is equal to $$ \int_{\mathbb{R}^d} \left(\int_{\mathbb{R}^d} f(x-y)g(y)e^{-i\xi\cdot x}\;dy\right) \;dx $$ (the factor $e^{-i\xi\cdot x}$ does not depend on $y$, so it is "constant"). Here we have an iterated integral. Fubini's theorem says this is equal to the double integral $$ \int_{\mathbb{R}^d\times\mathbb{R}^d} f(x-y)g(y)e^{-i\xi\cdot x} (dy\;dx) $$ if the latter exists. Since $x$ and $\xi$ are real, the factor $e^{-i\xi\cdot x}$ has absolute value $1$. Hence we have $$ \int_{\mathbb{R}^d\times\mathbb{R}^d} |f(x-y)g(y)e^{-i\xi\cdot x}| (dy\;dx) = \int_{\mathbb{R}^d\times\mathbb{R}^d} |f(x-y)g(y)| (dy\;dx). $$ All we need now is that that last integral is finite. If you've seen a theorem saying $L_1$ is closed under convolution, you've got it.

Later note: If $L_1$ is closed under convolution, that says $$ \int_{\mathbb{R}^d} |(f*g)(x)|\;dx < \infty, $$ which is the same as saying that the iterated integral $$ \int_{\mathbb{R}^d} \left(\int_{\mathbb{R}^d} |f(x-y)g(y)| \;dy\right)\;dx $$ is finite. That means where I wrote "...you've got it" above, you'd probably need to cite one more fact about integrals.

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I'm so sorry for the major typo in my question! the result you mention is in fact what I was trying to prove: that $f*g\in L_{1}(\mathbb{R^{d}})$. But I still think I can use your explanation to answer my question! –  Kyle Schlitt Nov 18 '11 at 3:52
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This answer is evidence of the overall trend of under-voting on MSE... –  The Chaz 2.0 Apr 17 '12 at 4:54
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