# Prove that the first Bessel function is continuous

• Feb 19th 2009, 12:08 PM
Amanda1990
Prove that the first Bessel function is continuous
Let $\displaystyle J_0(x) =2/\pi \int_{0}^{\pi/2}{\cos(x \cos{t})}dt$

I need to show that $\displaystyle J_0$ is a continuous bounded function. The fact that it is bounded seems completely obvious since cos is bounded. The continuity is causing some problems however. Clearly, cos is continuous and the process of "integrating" is continuous, but this doesn't give a proof.
• Feb 19th 2009, 02:59 PM
ThePerfectHacker
Quote:

Originally Posted by Amanda1990
Let $\displaystyle J_0(x) =2/\pi \int_{0}^{\pi/2}{\cos(x \cos{t})}dt$

I need to show that $\displaystyle J_0$ is a continuous bounded function. The fact that it is bounded seems completely obvious since cos is bounded. The continuity is causing some problems however. Clearly, cos is continuous and the process of "integrating" is continuous, but this doesn't give a proof.

Define $\displaystyle f(x,t) = \cos (x\cos t)$ for $\displaystyle 0\leq x < \infty, 0\leq t\leq \tfrac{\pi}{2}$. Let$\displaystyle x_0$ be any point. If we can show that for any $\displaystyle \epsilon > 0$ there is $\displaystyle \delta > 0$ so that if $\displaystyle |x-x_0| < \delta \text{ and }x\in [0,\infty)$ we have that $\displaystyle |f(x,t) - f(x_0,t)| < \epsilon$ for all $\displaystyle t\in [0,\tfrac{\pi}{2}]$ then this will complete the proof. This is because $\displaystyle |J_0(x) - J_0(x_0)| = \left| \tfrac{2}{\pi}\int_0^{\pi/2} f(x,t) - f(x_0,t) dt \right| \leq \tfrac{2}{\pi}\int_0^{\pi/2} |f(x,t) - f(x_0,t)| dt \leq \epsilon$ if $\displaystyle |x-x_0| < \delta \text{ and }x\in [0,\infty)$.

Thus, all that remains is to (hopefully) establish this continuity condition on $\displaystyle f$.
• Feb 19th 2009, 11:32 PM
Amanda1990
Can we not let t be arbitrary (in [0,pi/2]) so that cos t is fixed, then appeal to continuity of cos to get continuity of f? This seems a little dodgy; am I missing something? Or could we perhaps use a clever substitution?
• Feb 20th 2009, 09:09 AM
ThePerfectHacker
Quote:

Originally Posted by Amanda1990
Can we not let t be arbitrary (in [0,pi/2]) so that cos t is fixed, then appeal to continuity of cos to get continuity of f? This seems a little dodgy; am I missing something? Or could we perhaps use a clever substitution?

Let $\displaystyle |x-x_0|<\delta \text{ and }x\in [0,\infty)$. We want to bound $\displaystyle |\cos (x\cos t) - \cos (x_0\cos t)|$. By trigonometric identities we get, $\displaystyle \cos (x\cos t) - \cos (x_0\cos t) = - 2 \sin \left( \tfrac{\cos t (x-x_0)}{2}\right) \sin \left( \tfrac{\cos t (x+x_0)}{2} \right)$. Use the identity that $\displaystyle |\sin \alpha| \leq |\alpha$ and therefore $\displaystyle |\cos (x\cos t) - \cos (x_0 \cos t)| \leq \left| \sin \left( \tfrac{\cos t (x-x_0)}{2} \right) \right| \leq |\cos t| |x-x_0| \leq \delta$. Therefore, if $\displaystyle \delta$ is sufficiently small then we can make the functions sufficiently close together for all $\displaystyle t\in [0,\tfrac{\pi}{2}]$.