# A positive “Fourier transform” is integrable

Let $f\in L^1_\mathbb C(\mathbb R^n)$. I once read, in one of my old exam, that if $\hat{f}(\mathbb R^n)\subset\mathbb R_+$, then $\hat{f}\in L^1(\mathbb R^n)$. As far as I remember, the professor gave an identity using mollifiers but I'm not 100% sure. Any hints on how to prove that ?

Thanks!

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What does the complex plane in the subscript mean? – Glen Wheeler Oct 10 '12 at 17:09
Hello, It means that $f$ is a complex-valued function. – Philippe Malot Oct 10 '12 at 17:13

If $f(x)$ is continuous at $x = 0$ it is true: Let $\phi(x)$ be a Schwartz function with positive Fourier transform, such as a Gaussian, and define $\phi_{\epsilon}(x) = {1 \over \epsilon^n} \phi({x \over \epsilon})$. Then $f \ast \phi_{\epsilon}(x)$ converges to $f(x)$ for almost all $x$ (this is a property of approximations to the identity), while ${\scr{F}}(f \ast \phi_{\epsilon}) = \hat{f}(\xi)\hat{\phi}(\epsilon x)$.

Notice that $\hat{f}(\xi)\hat{\phi}(\epsilon \xi)$ is in $L^1$ since $\hat{f}$ is bounded and $\hat{\phi}(x)$ is a Schwartz function. So the Fourier inversion formula applies to $\hat{f}(\xi)\hat{\phi}(\epsilon \xi)$ and you have that $${\scr F}^{-1}(\hat{f}(\xi)\hat{\phi}(\epsilon \xi)) = f \ast \phi_{\epsilon}(x)$$ This holds at all $x$ since both sides are continuous functions.
Plugging in $x = 0$ gives $$\int_{{\mathbb R}^n} \hat{f}(\xi)\hat{\phi}(\epsilon \xi)\,d\xi = (2\pi)^{n \over 2}f \ast \phi_{\epsilon}(0)$$ Now let $\epsilon$ go to zero and you get $$\int_{{\mathbb R}^n} \hat{f}(\xi)\,d\xi = (2\pi)^{n \over 2}f (0)$$ I don't believe it's true necessarily if $f(x)$ is not continuous at $x = 0$... you can take a function $g(x)$ with $g(x) = c|x|^{-\epsilon}$ near the origin and then let $f(x) = g(x) \ast \bar{g}(-x)$; its Fourier transform will be $|\hat{g}(\xi)^2|$ which will be nonnegative.

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That looks good! So the fact that $\widehat{f}(\mathbb R^n)\subset \mathbb R_+$ is used when you let $\epsilon$ go to zero in the integral of the left hand side. Thank you! – Philippe Malot Oct 10 '12 at 18:27

Correct me if I'm wrong, but I don't think the identity you're seeking is true in the first place:

$\hat{f}\in L^1(\mathbb R^n) \implies {\int_S |\hat{f}|\;\mathrm{d}\mu} <\infty$

${\int_S |\hat{f}|\;\mathrm{d}\mu} = {\int_S \hat{f}\;\mathrm{d}\mu}$, since $\hat{f}(\mathbb R^n)\subset\mathbb R_+$

When we consider that:

${\int_S \hat{f}\;\mathrm{d}\mu} = f(0)$

None of the properties given above stop $f$ from being equal to the Dirac delta function and when this is the case $\hat{f} \notin L^1(\mathbb R^n)$.

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In this rigorous mathematical setting, it is considered (very) bad practice to speak of the "Dirac delta function", since this is in fact a measure, or rather a so-called distribution. For any proper function $f \in L^1_{\Bbb C}$, $f(0) \in \Bbb C$. – Lord_Farin Oct 10 '12 at 18:03
@Lord_Farin I agree that this approach is not rigorous enough for a proof, but I was just trying to indicate that the search for a proof might be in vain, since it is possible to find counter examples. Zarrax's answer addresses this issue by adding the restriction that $f$ needs to be continuous at $0$. – enobayram Oct 10 '12 at 22:17