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Math Help - Prove that the first Bessel function is continuous

  1. #1
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    Prove that the first Bessel function is continuous

    Let J_0(x) =2/\pi \int_{0}^{\pi/2}{\cos(x \cos{t})}dt

    I need to show that 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.
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    Quote Originally Posted by Amanda1990 View Post
    Let J_0(x) =2/\pi \int_{0}^{\pi/2}{\cos(x \cos{t})}dt

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

    Thus, all that remains is to (hopefully) establish this continuity condition on f.
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    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?
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    Quote Originally Posted by Amanda1990 View Post
    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 |x-x_0|<\delta \text{ and }x\in [0,\infty). We want to bound |\cos (x\cos t) - \cos (x_0\cos t)|. By trigonometric identities we get, \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 |\sin \alpha| \leq |\alpha and therefore |\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 \delta is sufficiently small then we can make the functions sufficiently close together for all t\in [0,\tfrac{\pi}{2}].
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