I read the following statement in a book:
$$ \frac{J_{\frac{n-2}{2}}(2\pi r)}{r^{\frac{n-2}{2}}}\text{ is in }L^p\text{ if and only if }p>\frac{2n}{n-1}\text{ ;} $$where $n\geq3$, $r$ is non-negative real number.
Let's call the above fraction the function $K$.
I have proved that if $p>\frac{2n}{n-1}$ then $K\in L^p$. But I am stuck on the reverse side.
Particularly, if $p\leq\frac{2n}{n-1}$ then how one can rigorously show that $K\notin L^p$ ?
I know that one must use the following asymptotic expansion of the Bessel function : $$ J_{\nu}(t)=O(t^{-\frac{1}{2}})\text{ as }t\to\infty\text{ ; where }\nu,t\geq0. $$
But the above asymptotic expansion just gives an upper bound on our $K$. If one would have an appropriate lower bound on $K$ then, perhaps, one could do something to prove that $K\notin L^p$.
Thanks in advance.
The statement that you wrote is not exactly incorrect although one should write the more precise following statement :
Let $d$ be the dimension of the real Euclidean space $\mathbb{R}^d$ with $d\geq3$. Then $$\text{The function }K(x)=\frac{J_{\frac{d-2}{2}}(2\pi|x|)}{|x|^{\frac{d-2}{2}}}\text{ (where }x\in\mathbb{R}^d\text{) is in }L^p(\mathbb{R}^d)\text{ if and only if }p>\frac{2d}{d-1}. $$
Proof :
First of all, it is not so hard to see that the function $K$ on $\mathbb{R}^d$ is smooth and bounded.
Since $J_{\nu}(r)=O(r^{-\frac{1}{2}})$ as $r\to\infty$ (where $\nu,r\geq0$), it is easy to see that $K\in L^p(\mathbb{R}^d)$ for $p>\frac{2d}{d-1}$.
Now we recall the following $\textit{oscillatory}$ behaviour of the Bessel function :
$$ J_{\nu}(r)\sim\sqrt{\frac{2}{\pi r}}\cos\left(r-\frac{\pi\nu}{2}-\frac{\pi}{4}\right)\text{ as }r\to\infty\text{ ;} $$where $r,\nu\geq0$. So, in our case, we get $$ K(x)=\frac{J_{\frac{d-2}{2}}(2\pi|x|)}{|x|^{\frac{d-2}{2}}}\sim\frac{1}{\pi|x|^{\frac{d-1}{2}}}\cos\left(2\pi|x|-(d-1)\frac{\pi}{4}\right)\text{ as }|x|\to\infty.\text{ }\text{ }\text{ }\ldots(\ast) $$
Notice that, for large enough $N$,$$\int_{\mathbb{R}^d\setminus B\left(0;\frac{4N+d+1}{8}\right)}\left(\left\lvert\cos\left(2\pi|x|-(d-1)\frac{\pi}{4}\right)\right\rvert\cdot\frac{1}{|x|^{\frac{(d-1)}{2}}}\right)^pdx$$ \begin{align} & = C_d'\int_{\frac{4N+d+1}{8}}^{\infty}\left(\left\lvert\cos\left(2\pi r-(d-1)\frac{\pi}{4}\right)\right\rvert^p\cdot\frac{1}{r^{\frac{p(d-1)}{2}-(d-1)}}\right)dr \\ & = C_d'\sum_{k\geq N}\left[\int_{\frac{4k+d+1}{8}}^{\frac{4k+d+5}{8}}\left(\left\lvert\cos\left(2\pi r-(d-1)\frac{\pi}{4}\right)\right\rvert^p\cdot\frac{1}{r^{\frac{(p-2)(d-1)}{2}}}\right)dr\right] \\ & \geq C_d'\sum_{k\geq N}\left[\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{\left(\frac{4k+d+3}{8} \right)^{\frac{(p-2)(d-1)}{2}}} \right] \\ & = C_d\sum_{k\geq N}\left(4k+d+3 \right)^{\frac{-(p-2)(d-1)}{2}}<\infty\text{ }\text{ if and only if }p>\frac{2d}{d-1}.\text{ }\text{ }\text{ }\ldots(\ast\ast) \end{align}
Hence, by $(\ast)$ and $(\ast\ast)$, we assert that if $p\leq\frac{2d}{d-1}$ then $K\notin L^p(\mathbb{R}^d)$.