(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 in 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{align} \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 ) ,\\ \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{align} 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. Now note that $J_{\nu}(x) \geq 0$ holds for every $x \in (0, \nu]$ (http://dlmf.nist.gov/10.21.i). Combining this fact and the identity \eqref{eq:4}, we conclude that the desired inequality \eqref{eq:1} holds in the case $x \in (0, \nu]$, since the integrand of the right-hand side of the identity \eqref{eq:4} is non-negative in that case.
Induction for $\nu \in \mathbb{N}$ (failed)
I tried induction as suggested by K.defaoite's comment in 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}
Conjecture on stationary points
Let $g_{\nu} \colon (0, \infty) \to \mathbb{R}$ be \begin{equation} g_\nu(x) := \frac{J_{\nu}(x)}{ ( J_{\nu-1/2}(x)^2 + J_{\nu+1/2}(x)^2 )^{1/2}} . \end{equation} As we see in the graph of the $x \mapsto \lvert g_\nu(x) \rvert$, the following conjecture seems to be true for every $\nu \in [0, \infty)$.
Conjecture. Let $a, b \in (0, \infty)$ be stationary points of $g_\nu$ satisfying $a < b$. Then we have $\lvert g_\nu(a) \rvert \leq \lvert g_\nu(b) \rvert$.
I know it is true when $\nu = 0$. Furthermore, for each fixed $\nu \in [0, \infty)$, the desired inequality \eqref{eq:1} holds for every $x \in (0, \infty)$ if the conjecture is true. The following arguments are based on Theorems 7.31.1 and 7.31.2 on G. Szego's book "Orthogonal polynomials", respectively.
Proof of the conjecture when $\nu = 0$. Let \begin{gather} h_0(x) := \frac{1}{1 + 1/(4x^2)} , \\ G_0(x) := g_0(x)^2 + h_0(x) \left( \frac{d g_0}{dx}(x) \right)^2 . \end{gather} Since we have \begin{equation} G_0(a) = g_0(a)^2 \end{equation} whenever $a \in (0, \infty)$ is a stationary point of $g_0$, it is enough to show that $G_0$ is non-decreasing on $(0, \infty)$. To see this, notice that \begin{equation} g_0(x) = \sqrt{\frac{\pi x}{2}} J_{0}(x) \end{equation} satisfies \begin{equation} h_0 \frac{d^2 g_0}{d x^2} + g_0 = 0 , \end{equation} which easily follows from Bessel's differential equation. Moreover, it is clear that $h_0$ is increasing on $(0, \infty)$, so that \begin{equation} \frac{d h_0}{ dx } \geq 0 \end{equation} holds. Hence, we have \begin{align} \frac{d G_0}{dx} &= 2 g_0 \frac{d g_0}{d x} + \frac{d h_0}{d x} \left( \frac{d g_0}{dx} \right)^2 + 2 h_0 \frac{d g_0}{dx} \frac{d^2 g_0}{d x^2}\\ &= \left( 2 g_0 + \frac{d h_0}{d x} \frac{d g_0}{dx} + 2 h_0 \frac{d^2 g_0}{d x^2} \right) \frac{d g_0}{dx} \\ &= \frac{d h_0}{d x} \left(\frac{d g_0}{dx} \right)^2 \\ & \geq 0 , \end{align} this shows that $G_0$ is non-decreasing.
The conjecture implies the inequality \eqref{eq:1}. Fix $\nu \in [0, \infty)$ and assume that the conjecture is true. Since we know that \begin{equation} \lim_{x \downarrow 0} g_\nu(x) = 0 , \quad \limsup_{x \to \infty} \lvert g_\nu(x) \rvert = 1 , \end{equation} it is enough to show that \begin{equation} \lvert g_\nu(a) \rvert \leq 1 \end{equation} holds for every stationary points $a \in (0, \infty)$. To see this, recall that we have \begin{equation} \lim_{x \to \infty} \lvert g_\nu(x)^2 - ( \sin(2x - \nu \pi) + 1 )/2 \rvert = 0 . \end{equation} Thus, there exists a sequence of stationary points $\{ a_n \}_{n \in \mathbb{N}} \subset (0, \infty)$ satisfying \begin{equation} \lim_{n \to \infty} a_n = \infty , \quad \lim_{n \to \infty} g_\nu(a_n) = 1 . \end{equation} Now it is easy to see that the conjecture implies $\lvert g_\nu(a) \rvert \leq 1$ for every stationary points. In fact, for each fixed stationary point $a$, there exists $N \in \mathbb{N}$ such that $a < a_n$ holds for every $n \geq N$, so that $\lvert g_\nu(a) \rvert \leq \lvert g_\nu(a_n) \rvert$ also holds for every $n \geq N$ by the conjecture. Therefore, we conclude that \begin{equation} \lvert g_\nu(a) \rvert \leq \liminf_{n \to \infty} {\lvert g_\nu(a_n) \rvert} = 1 \end{equation} holds.