Understanding a proof idea for fourier inversion

Question

In the book of Real Analysis by G. B. Folland, there's a short proof of the validity of $\mathcal{F^{-1}}(\hat{f})=f,$ when both $f$ and $\hat{f}$ lie in $L^1(\mathbb{R^n}).$

A first lemma used, since both are in $L^1:$ $$ \int \hat{f}g = \int f\hat{g} \tag{1} $$

Writing down $\mathcal{F^{-1}}(\hat{f})$ direclty does not lead to much since the integrand is not in $L^1(\mathbb{R^n}\times \mathbb{R^n}).$

Instead the author introduces a convergence factor first, only then invokes Fubini's theorem in order to pass to the limit. A rough sketch is given below:

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Given $t>0,$ Using the lemma in (1):

$$ \int e^{-\pi t^2 |\xi|^2}e^{2 \pi i \xi \cdot x} \hat{f}(\xi) \, d\xi = \int \hat{f}\Phi= \int f\hat{\Phi}=f*g_t(x) \tag{2} $$

Where $*$ stands for convolution, and $g_t(x):=t^{-n}\exp{(-\pi |x|^2/t^2)}.$

Since $\int e^{-\pi |x|^2} dx=1$, we have $f*g_t \to f$ in the $L^1$ norm as $t\to 0.$ Finally as $\hat{f}\in L^1$ the dominated convergence theorem yields:

$$ \lim_{t\to 0} \int e^{-\pi t^2 |\xi|^2}e^{2 \pi i \xi \cdot x} \hat{f}(\xi) \, d\xi = \int e^{2\pi i \xi \cdot x} \hat{f}(\xi) \, d\xi = \mathcal{F^{-1}}[\hat{f}](x) \tag{3} $$

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• I fail to grasp the whole idea of introducing a convergence factor as was done in (2), and how at the end it allows to prove the validity of fourier inversion in $L^1$ by taking the limit of $t\to 0$ and using the dominated convergence theorem. It would be definitely interesting if someone could shine a light on the intuition behind this approach.

Answer 1

This is what you would like to do, but can't: $$\begin{align} \int e^{2\pi i x\xi}\hat f(\xi)\,d\xi&=\int e^{2\pi i x\xi}\Bigl(\int e^{-2\pi i z\xi}f(z)\,dz\Bigr)\,d\xi\\ &=\int f(z)\Bigl(\int e^{-2\pi i (x-z)\xi}\,d\xi\Bigr)\,dz\\ &=f\ast\delta(x)\\ &=f(x), \end{align}$$ where $\delta$ is Dirac's delta. To make this argument rigorous you multiply by the convergence factor, which is an approximation of the identity (that is, $g_t\to\delta$.) This gives you equation (3). Letting $t\to0$, the right hand side converges to $f$, while the left hand side, by the dominated convergence theorem, to the inverse Fourier transform of $\hat f$.