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Let $f,g$ and $f\cdot g$ be a functions with $\int_{-\infty}^\infty|h(x)|dx<\infty$ for $h=f,g,fg$.

I have now given in a proof $$\int_{-\infty}^\infty\left(\int_{-\infty}^\infty f(u)g(w)\exp(-iwt)du\right)dw=\int_{-\infty}^\infty\left(\int_{-\infty}^\infty f(u)g(w)\exp(-iwt)dw\right)du$$

without saying why this equation is right.

So when is this right? In other words: When can you change the order of integration? What are the conditions? And why?

Add: I know for Lebesgue integrable functions the theorem of Fubini but above I don't have given functions in $\mathscr L^1$.

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This is Fubini's Theorem. Both $f$ and $g$ are integrable functions and $e^{-iwt}$ has absolute value one (assuming $t$ is real). So

$$ \int_{-\infty}^\infty\left(\int_{-\infty}^\infty |f(u)g(w)\exp(-iwt)|du\right)dw = ||f||_1 ||g||_1 $$ where $||\cdot||_1$ is the $L^1$-norm.

First we can take out $g(w)$ since it is independent of $u$ and then take out $||f||_1$ since it is a constant:

$$ \begin{split} &\int_{-\infty}^\infty\left(\int_{-\infty}^\infty |f(u)g(w)|du\right)dw = \int_{-\infty}^\infty |g(w)| \left(\int_{-\infty}^\infty |f(u)|du\right)dw \\ &= \int_{-\infty}^\infty |g(w)| ||f||_1 dw = ||f||_1 \int_{-\infty}^\infty |g(w)| dw = ||f||_1 ||g||_1 \end{split} $$ So $f(u)g(w)$ is thus integrable in $\mathbb{R} \times \mathbb{R}$ (but remember that this is different from $fg$ being integrable in $\mathbb{R}$ which was one your assumptions). I have by the way used the version of Fubini where one checks the iterated integrals (it is in the Wikipedia page).

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  • $\begingroup$ but I need measure spaces, and $\mathbb R$ isn't. $\endgroup$
    – user81109
    Jun 6 '13 at 21:07
  • $\begingroup$ @jul8 $\mathbb{R}$ certainly has a measure, the Lebesgue measure: en.wikipedia.org/wiki/Lebesgue_measure $\endgroup$
    – N.U.
    Jun 6 '13 at 21:09
  • $\begingroup$ okay. then Fubini holds since $|fg|$ is absolutely integrable. last question: I know $$||f||_1||g||_1=\left(\int_{-\infty}^\infty |f(u)| du\right)\cdot\left(\int_{-\infty}^\infty |g(w)| dw\right)$$ but why can you multiplicate these integrals like you did? and in general: can you follow of $f,g$ absolut integrable that $f\cdot g$ is also absolut integrabel? $\endgroup$
    – user81109
    Jun 6 '13 at 21:18
  • $\begingroup$ @jul8 I answered you in the post since I had to write all these integrals. $\endgroup$
    – N.U.
    Jun 7 '13 at 5:43

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