(Cross posted to MO https://mathoverflow.net/questions/496508/)
Let $J_\nu$ be the Bessel function of the first kind of order $\nu$. Does the inequality \begin{equation} \label{eq:1} \tag{1} J_{\nu}(x)^2 \leq J_{\nu-1/2}(x)^2 + J_{\nu+1/2}(x)^2 \end{equation} hold for every $x \in (0, \infty)$ and $\nu \in [0, \infty)$?
- I am particularly interested in the cases $\nu = n$ and $\nu = n+1/2$ for $n \in \mathbb{N}$.
- It is not difficult to see that for each $\nu \in [0, \infty)$, there exists $C_\nu \in (0, \infty)$ such that \begin{equation} J_{\nu}(x)^2 \leq C_\nu ( J_{\nu-1/2}(x)^2 + J_{\nu+1/2}(x)^2 ) \end{equation} for every $x \in (0, \infty)$. I want to know if the constant can be taken independently of $\nu$.
- As Timothy Budd commented (see MO), more generally \begin{equation} J_{\mu}(x)^2 \leq J_{\nu-1/2}(x)^2 + J_{\nu+1/2}(x)^2 \end{equation} seems true for every $x \in (0, \infty)$, $\nu \in [0, \infty)$, and $\mu \in [\nu-1/2, \nu+1/2]$.
Graphs
The graphs of \begin{equation} \label{eq:2} \tag{2} [0, 3\nu+10] \ni x \mapsto \pi x ( J_{\nu-1/2}(x)^2 + J_{\nu+1/2}(x)^2 - J_{\nu}(x)^2 ) \end{equation} and \begin{equation} \label{eq:3} \tag{3} [0, 10] \ni x \mapsto ( \Gamma(\nu+1/2) )^2 (x/2)^{-2\nu+1} ( J_{\nu-1/2}(x)^2 + J_{\nu+1/2}(x)^2 - J_{\nu}(x)^2 ) \end{equation} are here: the graph of \eqref{eq:2}, the graph of \eqref{eq:3}.
The case $\nu = 0$
In the case $\nu = 0$, the inequality \eqref{eq:1} holds for every $x \in (0, \infty)$. This follows from \begin{gather} \pi x( J_{-1/2}(x)^2 + J_{1/2}(x)^2 ) = 2 , \\ \sup_{x \in (0, \infty)} \pi x J_{0}(x)^2 = 2 . \end{gather}
Limits as $x \to \infty$
It is known that the following hold:
\begin{equation}
\lim_{x \to \infty} \lvert \pi x( J_{\nu-1/2}(x)^2 + J_{\nu+1/2}(x)^2 ) - 2 \rvert = 0 , \\
\lim_{x \to \infty} \lvert \pi x J_{\nu}(x)^2 - \sin{ (2x - \nu \pi) } - 1 \rvert = 0 .
\end{equation}
Therefore, for each $\nu \in [0, \infty)$ and $\varepsilon > 0$, there exists $R_{\nu, \varepsilon} \in (0, \infty)$ such that
\begin{equation}
\pi x J_{\nu}(x)^2 - \varepsilon \leq \sin{ (2x - \nu \pi) } + 1 \leq 2 \leq \pi x( J_{\nu-1/2}(x)^2 + J_{\nu+1/2}(x)^2 ) + \varepsilon
\end{equation}
holds for every $x \in (R_{\nu, \varepsilon}, \infty)$.
Limits as $x \downarrow 0$
It is known that the following hold:
\begin{equation}
\lim_{x \downarrow 0} (x/2)^{-(2\nu-1)} ( J_{\nu-1/2}(x)^2 + J_{\nu+1/2}(x)^2 ) = \frac{1}{ ( \Gamma(\nu+1/2) )^2 } > 0 , \\
\lim_{x \downarrow 0} (x/2)^{-(2\nu-1)} J_{\nu}(x)^2 = 0 .
\end{equation}
Therefore, for each $\nu \in [0, \infty)$, there exists $\delta_{\nu} \in (0, \infty)$ such that
\begin{equation}
(x/2)^{-(2\nu-1)} J_{\nu}(x)^2 \leq (x/2)^{-(2\nu-1)} ( J_{\nu-1/2}(x)^2 + J_{\nu+1/2}(x)^2 )
\end{equation}
holds for every $x \in (0, \delta_{\nu})$.
Integral representations
We have the following integral representation for each $\nu, x \in (0, \infty)$: \begin{equation} \label{eq:4} \tag{4} J_{\nu-1/2}(x)^2 + J_{\nu+1/2}(x)^2 - J_{\nu}(x)^2 = \frac{2}{\pi} \int_0^{\pi/2} \left( \frac{2\nu}{x \cos{\theta}} - 1 \right) J_{2\nu} (2x \cos{\theta}) \, d\theta . \end{equation} This follows from \begin{gather} J_{\mu}(x)^2 = \frac{2}{\pi} \int_0^{\pi/2} J_{2\mu} (2x \cos{\theta}) \, d\theta , \\ J_{\mu+1}(z) + J_{\mu-1}(z) = \frac{2\mu}{z} J_{\mu}(z) . \label{eq:5} \tag{5} \end{gather} See http://dlmf.nist.gov/10.22.E13 and http://dlmf.nist.gov/10.6.E1.
Induction for $\nu \in \mathbb{N}$ (failed)
I tried induction as suggested by K.defaoite's comment (see MSE), but could not find a way. Fix $n \in \mathbb{N}$, and assume that the inequality \eqref{eq:1} holds for every $x \in (0, \infty)$ when $\nu = n, n+1$. Now we need to prove it for $\nu = n+2$. Using the formula \eqref{eq:5} and the induction hypothesis, we get \begin{align} &\quad J_{n+3/2}(x)^2 + J_{n+5/2}(x)^2 - J_{n+2}(x)^2 \\ &= J_{n-1/2}(x)^2 + J_{n+1/2}(x)^2 - J_{n}(x)^2 \\ &\quad + \frac{1}{x^2} ( (2n+1)^2 J_{n+1/2}(x)^2 + (2n+3)^2 J_{n+3/2}(x)^2 - (2n+2)^2 J_{n+1}(x)^2 ) \\ &\quad - \frac{2}{x} \left( (2n+1) J_{n-1/2}(x) J_{n+1/2}(x) - (2n+2) J_{n}(x) J_{n+1}(x) + (2n+3) J_{n+1/2}(x) J_{n+3/2}(x) \right) \\ &\geq \frac{1}{x^2} ( (2n+1)^2 J_{n+1/2}(x)^2 + (2n+3)^2 J_{n+3/2}(x)^2 - (2n+2)^2 J_{n+1}(x)^2 ) \\ &\quad - \frac{2}{x} \left( (2n+1) J_{n-1/2}(x) J_{n+1/2}(x) - (2n+2) J_{n}(x) J_{n+1}(x) + (2n+3) J_{n+1/2}(x) J_{n+3/2}(x) \right) \\ &=: f_n(x) . \end{align} However, $f_n(x) \geq 0$ fails unfortunately. Indeed, we have \begin{gather} \lim_{x \downarrow 0} (x/2)^{-(2n-1)} f_n(x) = - \frac{1}{\Gamma(n+1/2)^2} < 0 , \\ \lim_{x \to \infty} \lvert x^2 f_n(x) - 4 (-1)^{n-1} ( (n+1) \cos{2x} - \sin{2x} ) \rvert = 0 . \end{gather}
