3
$\begingroup$

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 $L_1(\mathbb{R}^d)$.

  • 0
    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.2011-11-18

1 Answers 1

7

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.

  • 1
    This answer is evidence of the overall trend of *under*-voting on MSE...2012-04-17