Analysis Problem of the Day 4

The following nice problem comes from a problem set for 245C, the third quarter of graduate real analysis at UCLA:

Problem: If $f \in L^1(\mathbb{R}^d),$ $f$ is continuous at $0,$ and $\widehat{f} \geq 0,$ show that $\widehat{f} \in L^1(\mathbb{R}^d).$

Solution: This problem is actually quite a bit trickier than it looks. A classical fact about the Fourier transform that at first seems relevant here is that $\mathcal{F}: L^1 \to C_0,$ where $C_0$ is the space of continuous functions decaying to zero at infinity. However, this implies nothing about the integrability of $\widehat{f}.$ It is also not super clear how the positivity of the Fourier transform or the continuity of $f$ at $0$ are relevant, so it appears that this approach will not work.

The approach that works turns out to be a limiting argument, of which here is one version (inspired by the answer in this post):

Let $\phi$ be a Gaussian normalized so that $\|\phi\|_1 = 1,$ and consider the rescaling $\phi_\epsilon(x) = \frac{1}{\epsilon^d} \phi(\frac{x}{\epsilon}).$ This turns out to be an approximation to the identity, i.e. as $\epsilon \to 0,$ $\phi_\epsilon \to \delta,$ the Dirac delta distribution, in the sense of distributions. In particular, by properties of convolutions, $f_\epsilon := f * \phi_\epsilon$ is a collection of smooth $L^1$ functions such that $f_\epsilon \to f$ in $L^1$ as $\epsilon \to 0.$ On the Fourier side, convolution transforms to multiplication, so $\widehat{f_\epsilon}(\xi) = \widehat{f}(\xi)\widehat{\phi}(\epsilon \xi).$ In particular, $\widehat{f_\epsilon} \in L^1$ as a product of a bounded function and a Gaussian, so the Fourier inversion formula applies, i.e.

$(f*\phi_\epsilon)(x) = \frac{1}{(2\pi)^d}\int_{\mathbb{R}^d} \widehat{f}(\xi)\widehat{\phi}(\epsilon \xi) e^{2\pi i\langle x, \xi \rangle}d\xi.$

This holds for all $x$ since the functions on both sides are continuous.

Plugging in $x=0$ yields

$(f*\phi_\epsilon)(0) = \frac{1}{(2\pi)^d} \int_{\mathbb{R}^d} \widehat{f}(\xi)\widehat{\phi}(\epsilon \xi),$

and as $\epsilon \to 0,$ one applies the continuity of $f$ at 0 on the left and the monotone convergence theorem on the right (this is where we use the fact that the Fourier transform is positive) to obtain

$f(0) = \frac{1}{(2\pi)^d} \int_{\mathbb{R}^d} \widehat{f}(\xi) d\xi,$

i.e. $\|\widehat{f}\|_1 = f(0)(2\pi)^d.$ Thus, $\widehat{f} \in L^1(\mathbb{R}^d).$

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