Does $f(0) = +\infty$ when $\hat f \geq 0$ and $\int \hat f (s) \ ds = +\infty$?

Question

Throughout, $f \in L^1(\mathbb{R})$ and $\hat f \in C_0(\mathbb{R})$ is its Fourier transform $s \mapsto \int e^{its} f(t) \ dt$.

Motivation: If $\hat f \in L^1(\mathbb{R})$ too, then, by Fourier inversion, $f$ is continuous (possibly after making changes on a null set) and we have the formula \begin{align*} f(0) = \frac{1}{2 \pi} \int \hat f(s) \ ds. && && (*) \end{align*} Now suppose instead that $\hat f \notin L^1(\mathbb{R})$, but $\hat f \geq 0$. In this situation, measure theory traditionally assigns the value $+ \infty$ to the integral $\int \hat f (s) \ ds$. So, at least one side of $(*)$ makes sense. Two natural questions arise:

• Does $f(0)$ always make sense in this situation?
• In instances where $f(0)$ does make sense, does it equal $+ \infty$?

Here is a precise question, though I'd be interested in answers to other questions in the same spirit.

Question 1: Suppose $\hat f\geq 0$ and $\int \hat f(s) \ ds = + \infty$. Does it follow that $f$ is "essentially infinite" at zero in the sense that, for all $M > 0$ $$ \lim_{\epsilon \to 0^+} \frac{ \mu \big( \{ t \in [-\epsilon, \epsilon] : f(t) < M \} \big) }{2 \epsilon} =0. $$

If the answer to the above question is "no" then we can worry about the 2nd bullet point failing badly. For instance, I think we would all agree "$f$ continuous" $\Longrightarrow$ "$f(0)$ makes sense". A positive answer to the following question would be a sort of "worst case scenario".

Question 2: Is it possible to have $f$ continuous (in particular $f(0) <\infty$), and yet also have $\hat f$ nonnegative with divergent integral?

Answer 1

Question 1. Let's use the language of tempered distributions. Then we can unambiguously talk about the Fourier transform without worrying if the function is in $L^1$. (If you don't know about tempered distributions, just let it mean that most anything is O.K.)

Now actually the definition of the Fourier transform of a tempered distribution is $$ \int_{\mathbb R} \hat f(x) \phi(x) \, dx = \int_{\mathbb R} f(x) \hat \phi(x) \, dx \qquad(1)$$ where $\phi$ is a test function (a smooth function of which all derivatives converge rapidly to zero at infinity).

Consider $f$ such that $\hat f \ge 0$, and $\int_{\mathbb R} \hat f(x) dx = \infty$. Let $\phi = \frac1{\sqrt{2\pi}} e^{-\sigma^2 x^2/2}$. Then $\hat \phi(x) = \frac1{\sqrt{2 \pi} \sigma} e^{-x^2/2\sigma^2}$. (The $2\pi$ may be out of place, depending upon which definition of Fourier transform we use.) Now, as $\sigma \to 0$, it is clear that the left hand side of $(1)$ converges to infinity. Therefore the right hand side also converges to infinity.

I don't think this quite gets you to answering question 1. But it does show that the infinity at $x=0$ is, in some sense, not isolated.

Answer 2

The answer to Question 2 is no. $ f\in L^1(\mathbb{R}) $, $ f $ is continuous at $ 0 $, $ \widehat{f}\geq 0 $, show that $ \widehat{f}\in L^1(\mathbb{R}) $. tells us that for $f\in L^1(\mathbb R)$, continuity at $0$ and spectral non-negativity is enough to force $\hat f \in L^1$.