Convolution $f \ast f$, where $f \in L^{2}(\mathbb{R}^d)$ but $f \not \in L^{1}(\mathbb{R}^d)$

Question

Suppose that I have a real-valued function $f$ that is in $L^{2}(\mathbb{R}^d)$ but not in $L^{1}(\mathbb{R}^d)$. It is known that $\mathcal{F}^{-1}(f) \in L^{p}(\mathbb{R}^d)$ for all $p \in \mathbb{N}$. I want to know if this is sufficient enough to deduce that the expression

$$\displaystyle \| f \ast f \|_{2}$$

is finite. Since $f$ is in $L^2$, then by Plancherel's theorem we have

$$\|f \ast f\|_{2} = \| \mathcal{F}(f \ast f) \|_2,$$

where $\mathcal{F}$ denotes the Fourier transform. I'd like to use the convolution theorem at this point, but as far as I understand, we would need $f$ to be in $L^1$. This suggests that $f$ does not have a Fourier transform in the traditional sense. Where can we go from here?

It is also known that $f$ is the Fourier transform of a characteristic function. In general, it is not true that $f \ast f \in L^{2}(\mathbb{R}^d)$. But for which functions does it hold?

Answer 1

One of the ideas would be (napkin math incoming!) to take $f\in L^{4/3}$, then $\hat f \in L^4$, then $|\hat f|^2 \in L^2$, then by the inverse FT we can hope that $f\ast f \in L^2$.