I want to solve the following problem from Arthur Mattuck's Introduction to Analysis:
If $a_1$ is the first positive zero of $J_p(x)$ (assume $p>0$), prove $a_1>p$. (Hint: let $c$ be the maximum between $0$ and $a_1$. Show $c>p$ by substituting $x=c$ in Bessel's equation.)
My attempt:
I tried following the hint. First since $a_1$ is positive by definition, I don't see the point of defining $c=\max (0,a_1) =a_1$. Second, with Bessel's equation being $$x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + \left(x^2 - p^2 \right)y = 0, $$ I got $$a_1 J_p''(a_1)+J_p'(a_1)=0. $$ The thing is, substituting $x=p$ gives the same constraint: $$p J_p''(p)+J_p'(p)=0. $$
I tried plotting some graphs, and indeed both $a_1$ and $p$ are fixed points of $$x \mapsto -\frac{J_p'(x)}{J_p''(x)}. $$
I can't see how to get the requested bound. Any help would be appreciated. Thanks.
The hint was supposed to guide us towards defining $c := \textrm{argmax}\{f(x), x \in [0,a_1]\}$, i.e. (one of) the argument(s) where $f$ reaches it's maximum between $0$ and $a_1$.
We know that $f(c)>0$ because, as a comment noticed, $f'(0) > 0$ implies that $f$ takes at least some positive values. Therefore we have $f'(c) = 0$ and $f''(c) \leq 0$. But this implies that $0 = c^2 f''(c) + cf'(c) + (c^2-p^2)f(c) \leq (c^2 - p^2)f(c)$, and thus, since $f(c)>0$, that $c^2 - p^2 \geq 0$, yielding $a_1 \geq c \geq p$ as desired.