# Application of Fubini's Theorem

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... – roo 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 – roo Nov 18 '11 at 2:04
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. – roo Nov 18 '11 at 3:47

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.
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! – roo Nov 18 '11 at 3:52