# Maximal minimum of Bessel functions

This comes from a scattering problem. Consider the usual non singular Bessel functions of the first kind, $J_n(x)$. It is known that their zeros are countable, and all zeros are distinct. My question is the following: is there a lower bound to $$f(N)=\sup_{x>0}\inf_{n\in\mathbb{N},n\leq N}|J_n(x)|$$ It is clear $f(N)\to0$ as $N\to\infty$, so it must depend on $N$.

• Is it a correct expression? It means we first take any x, then for this x find inf in n as function of x, and then take sup over x? Or it is better to change sup and inf? – Sergei Oct 30 '14 at 12:59
• @Sergei It is the correct expression. If you invert the sup and inf, the answer is known: $n\to n^{1/3}\sup_{x>0} |J_n(x)|$ is a monotone function, growing from $0$ to $0.675..$ as $n\to \infty$ from a paper of Landau, J. London Math. Soc. (2) 61 (2000) 197-215. – username Oct 30 '14 at 15:07
• Thank you. What I missed. Take such $x=a$ that $J_1(a)=0$. Then inf mod over all n will be zero, not so? – Sergei Oct 30 '14 at 15:52

f := proc (N) options operator, arrow; DirectSearch:-Search(proc (x) options operator, arrow; min(seq(BesselJ(n, x), n = 1 .. N)) end proc, {0 < x}, maximize) end proc:

$$.5564145611, .3883486929, .4924980215, .5720147732, .6487478567, .7617480189, .8208958156, .8474981352$$ suggest the dependence $$\log(f(N))\asymp -N\log(N),\,N\to\infty.$$