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Young's inequality says that if $f \in L^{p}(\mathbb{R}^d)$ and $g \in L^{q}(\mathbb{R}^d)$, then the convolution $f \ast g \in L^{r}(\mathbb{R}^d)$, and

$$\displaystyle \|f \ast g\|_r \leqslant \|f\|_{p}\|g\|_{q},$$

where $\frac{1}{p} + \frac{1}{q} = \frac{1}{r} + 1$.

I'm currently trying to estimate the $L^{2}(\mathbb{R}^d)$ norm of the convolution $f \ast f$, for a particular function $f$. I have shown that $f \in L^{2}(\mathbb{R}^d)$. Unfortunately, $f \not \in L^{1}(\mathbb{R}^d)$, but the integral of $f$ (without modulus signs) converges. Is there an analogue of Young's inequality that can deal with this situation -- where the integral of $f$ converges conditionally but not absolutely?

Note: the function I have in mind is:

$$\displaystyle f(x) := \frac{J_{d/2}(|x|)}{|x|^{d/2}},$$

and it is known that $\displaystyle \mathcal{F}(\chi) = f$, where $\chi$ is the characteristic function of the unit ball on $\mathbb{R}^d$ (up to a constant).

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    You might be interested in the so called Calderon-Zygmund-kernels. – 2017-02-13
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    For $L^2$, you may be able to exploit Plancherel. $$\lVert f\ast f\rVert_{L^2}^2 = \lVert\mathscr{F}(f\ast f)\rVert_{L^2}^2 = \lVert \mathscr{F}(f)^2\rVert_{L^2}^2 = \lVert \mathscr{F}(f)\rVert_{L^4}^4.$$ (Possibly with some normalising constants, depends on the used convention for the Fourier transform.) – 2017-02-13
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    @DanielFischer When you use the convolution theorem in the second equality, what assumptions are needed on either $f$ or its Fourier transform for that to hold? – 2017-02-19
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    @DanielFischer If $f \not \in L^{1}(\mathbb{R}^d)$, doesn't that mean we can't use the convolution theorem? – 2017-02-19
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    The convolution theorem holds in wider situations. You can consider $f,g \in L^2$ as tempered distributions. The bilinear map $(f,g) \mapsto \mathscr{F}(f)\cdot \mathscr{F}(g)$ is continuous $L^2\times L^2 \to L^1$, the inclusion of $L^1$ into the space of tempered distributions is also continuous. The bilinear map $(f,g) \mapsto f\ast g$ is continuous $L^2\times L^2\to C_0$, the inclusion of $C_0$ into the space of tempered distributions is continuous, the Fourier transform of tempered distribution is continuous. – 2017-02-20
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    The two continuous bilinear maps $(f,g) \mapsto \mathscr{F}(g)\cdot \mathscr{F}(g)$ and $(f,g) \mapsto \mathscr{F}(f\ast g)$ from $L^2\times L^2$ to the space of tempered distributions agree on the dense subspace $(L^1\cap L^2)\times (L^1\cap L^2)$, hence everywhere. – 2017-02-20
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    This sounds very interesting, although I am not sure I completely understand why we can use it for this particular $f$. Can $f$ always be considered as a tempered distribution? Is there a restriction on the particular features $f$ must have in order for the convolution theorem to hold in this wider case? (For instance, in my particular case I know that $f \not \in S(\mathbb{R}^d)$.) – 2017-02-20
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    $L^p$ functions ($1 \leqslant p \leqslant \infty$) define tempered distributions. When $f\in L^r$ and $g\in L^s$ such that $f\ast g \in L^t$, you have $\mathscr{F}(f\ast g)$ as a tempered distribution. On the other end, for the pointwise product $\mathscr{F}(f)\cdot \mathscr{F}(g)$ to make sense, $\mathscr{F}(f)$ and $\mathscr{F}(g)$ must both be nice enough functions and their product must be locally integrable, and mustn't grow too fast. [That's not strictly true, additional regularity of one factor allows lesser regularity of the other, – 2017-02-20
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    e.g. the product of a function and a measure often makes sense.] The $L^2$ case is particularly nice because a function is in $L^2$ if and only if its Fourier transform is, and the product of two $L^2$ functions is in $L^1$, whereas for $p\neq 2$ there no such nice $f\in L^p \iff \mathscr{F}(f)\in L^q$. – 2017-02-20
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    I have edited the OP to show the function $f$ I am working with. Since $f = \mathcal{F}(\chi)$ then what I would like to say is $$\|f \ast f\|_{2}^{2} = \|\mathcal{F}(\chi) \ast \mathcal{F}(\chi)\|^2_2 = \|\mathcal{F}(\chi^2)\|_2^2 = \|\mathcal{F}(\chi)\|_2^2.$$ Do you think in this particular instance, that manipulation can be justified? The pointwise product $\mathcal{F}(\chi) \cdot \mathcal{F}(\chi)$ is locally integrable and decays with oscillations tending to zero at infinity. Is this a sufficiently nice function? – 2017-02-20

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