Let $\mathcal{J}\_{n+1/2}$ be the Bessel function of order $n+1/2$. Let $j'\_{n+1/2,1}$ denote the first zero of its derivative, which is also the location of the first maximum of $\mathcal{J}\_{n+1/2}$.




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My question: Given a non-negative integer $n$, is it true that $|\mathcal{J}\_{m+1/2}(j'\_{n+1/2,1})|<|\mathcal{J}\_{n+1/2}(j'\_{n+1/2,1})|$ holds for all non-negative integers $m\neq n$?




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EDIT: The answer is yes and the proof can be found in the appendix of the paper <http://arxiv.org/abs/1504.07306> Thanks for the helpful comment @Jung Wen Chen!



Is a Bessel function larger than all other Bessel functions when evaluated at its first maximum?
We suppose $0<m<n$. It is known, from Landau's paper quoted above that
$
n\to n^{1/3}J\_n(x\_n)
$
increases (to some universal constant $b$). Thus
$$
J\_n(x\_n)>\left(\frac{m}{n}\right)^{1/3}J\_m(x\_m)>\left(\frac{m}{x\_n}\right)^{1/3}J\_m(x\_m).
$$
It is also known that for $x>n$,
$
x\to \sqrt{x^2-n^2}((J\_n(x))^2+(Y\_n(x))^2) 
$
increases to $2/\pi$. Therefore, as $x\_n>n>m$
$$
|J\_m(x\_n)|^2\leq \frac{2}{\pi}\frac{1}{\sqrt{x\_n^2 -m^2}}.
$$
Therefore, the inequality holds at least for all $n,m$ such that
$$
\frac{2}{\pi m^{2/3}(J\_m(x\_m))^2}<\sqrt{x\_n^{2} -m^{2}} \frac{1}{x\_n^{2/3}},
$$ 
and the left hand-side is between $1.39$ and $1.89$. 
If we write $$m=\frac{x\_n}{1+\lambda x\_n^{-2/3}}$$ we find that this inequality holds for all $x\_n\geq40$, and $\lambda>11/10$ for example. The asymptotic of $x\_n$ is quite well known, and it is precisely of that sort of magnitude, $x\_n=n+\beta n^{1/3}+O(1)$. So, refining the argument and using $n$ instead of $x\_n$, you would be very close. For $m=n-1$, you get 
$$
\frac{2}{\pi m^{2/3}(J\_m(x\_m))^2}<\sqrt{\beta^2 n^{2/3} +2\beta n^{4/3}+2n-1} \frac{1}{n^{2/3}},
$$ 
Asymptotically, this is not enough, because $\beta\approx 0.8$ and $\sqrt{1.6}=1.26<1.39$, but that does limit the range of $m$ to investigate, to $m\geq n- n^{1/3}/5$, so you are only have to consider the case when $J\_m(x\_n)>0$, $J\_m^\prime(x\_n)<0$, as the first root is further away ($m+ 1.8m^{1/3}$). 


