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Suppose that $\chi$ is the characteristic function of the unit ball in $\mathbb{R}^d$. It is known that $\chi$ has Fourier transform (up to a constant) given by

$$\displaystyle \mathcal{F}(\chi)(\xi) = \frac{J_{d/2}(|\xi|)}{|\xi|^{d/2}},$$

where $J_{\nu}$ denotes the Bessel function of the first kind, and $|\cdot|$ denotes the Euclidean norm on $\mathbb{R}^d$. Then $\mathcal{F}(\chi) \in L^{2}(\mathbb{R}^d)$ and clearly $\chi \in L^{p}(\mathbb{R}^d)$ for any $p \in \mathbb{N}$. Note also that $\mathcal{F}(\chi) \not \in L^{1}(\mathbb{R}^d)$. I'd like to know whether the following norm is finite:

$$\displaystyle \|\mathcal{F}(\chi) \ast \mathcal{F}(\chi)\|^2_2.$$

By Plancherel's theorem the above expression is equal to $\|\mathcal{F}(\mathcal{F}(\chi) \ast \mathcal{F}(\chi))\|_2^2,$ since $\mathcal{F}(\chi) \in L^2(\mathbb{R}^d).$ Is this sufficient to say that the norm is finite?

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    Note that $\chi\cdot\chi=\chi$. So $\mathcal F(\chi)*\mathcal F(\chi)=\mathcal F(\chi\cdot\chi)=\mathcal F(\chi)$ is in $L^2$. It follows the norm must be finite. – 2017-02-19
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    I don't understand the first equality: I assume you used the convolution theorem here. Don't we need $\mathcal{F}(\chi) \in L^1$ for that to be valid? – 2017-02-19
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    You are right, I mixed up the order of whats what in the convolution theorem. – 2017-02-19
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    @s.harp By Young's inequality for convolutions, since $\mathcal{F}(\chi) \in L^{2}(\mathbb{R}^d)$, can we say that $\|\mathcal{F}(\chi) \ast \mathcal{F}(\chi)\|_{\infty} \leqslant \|\mathcal{F}(\chi)\|_{2} \cdot \|\mathcal{F}(\chi)\|_{2} < \infty,$ which implies that $\mathcal{F}(\chi) \ast \mathcal{F}(\chi) \in L^{\infty} \subset L^{2}$, so $\|\mathcal{F}(\chi) \ast \mathcal{F}(\chi)\|_{2}$ must be finite? Or is this wrong? – 2017-02-19
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    $L^\infty(\mathbb R^d)$ is not a subset of $L^2(\mathbb R^d)$, and $\mathcal F(\chi)$ is not in $L^2$, but in $L^p$ where $p>\frac{-4+d}{1+d}$ (if I didn't make a miscalculation in $d-1-p(d+1)/2<1$) – 2017-02-19
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    But isn't the Fourier transform of an $L^2$ function always in $L^2$, by Plancherel's theorem? – 2017-02-19
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    Uhhh yes, guess today isn't a day where I can think very clearly. $p>\frac{2(d-2)}{1+d}$ was the condition, and $2$ always satisfies it. But $L^\infty$ is not a subset of $L^2$, I didn't mess that part up at least. The convolution of two $L^2$ functions need not be $L^2$, cf http://math.stackexchange.com/questions/118230/on-the-closedness-of-l2-under-convolution – 2017-02-19
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    Hmm, I see (I forgot that that inclusion only holds on finite measure spaces). Thanks for the link -- it doesn't seem like there is going to be an easy way to show that the norm is finite, in that case. – 2017-02-19
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    Also, could you tell me where you got your calculation from? It seems that your inequality is saying that if $d = 2$ then $\mathcal{F}(\chi) \in L^{1}(\mathbb{R}^d)$ -- but my calculations suggest that this is not the case. – 2017-02-19
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    $J_{d/2}(x)$ has an enveloping asymptote $\frac C{\sqrt x}$. Then $\left(\frac{J_{d/2}(x)}{x^{d/2}}\right)$ has asymptotic behaviour $x^{-(d+1)/2}$. If $x^{d-1}x^{-p(d+1)/2}$ is integrable then the function was in $L^p(\Bbb R^d)$. Where I notice that I forgot about the $x^{d-1}x^{-p d/2}$ singularity at zero... – 2017-02-19

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