A proof of $|J_{\nu}(x)|\leq x/(2\nu-1)$

Question

I am looking for a proof of the following inequality for Bessel functions : $$|J_{\nu}(x)|\leq \frac{x}{2\nu-1} \quad \left(\text{for}~\nu>1,~0\leq x \leq \frac{\pi}{2}\right).$$

Many thanks !

Answer 1

Since: $$ J_\nu(x)=\sum_{m\geq 0}\frac{(-1)^m}{m!\Gamma(m+\nu+1)}\left(\frac{x}{2}\right)^{2m+\nu}$$ over the interval $\left[0,\frac{\pi}{2}\right]$ we obviously have: $$ \left|J_\nu(x)\right|\leq \sum_{m\geq 0}\frac{1}{m!\Gamma(m+\nu+1)}\left(\frac{x}{2}\right)^{2m+\nu}=I_\nu(x)$$ but $\frac{I_\nu(x)}{x}$ is an increasing (and convex) function on $\left[0,\frac{\pi}{2}\right]$, so: $$ \frac{\left|J_\nu(x)\right|}{x} \leq \frac{2}{\pi}\cdot I_\nu\left(\frac{\pi}{2}\right) $$ where $I_\nu\left(\frac{\pi}{2}\right)$ drops very fast to zero, since such a coefficient is a coefficient of the Fourier cosine series of $e^{\frac{\pi}{2}\cos\theta}$, that is an analytic function. So the original factor $\frac{1}{2\nu-1}$ can be replaced by $\frac{1}{\nu!}$.