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I'm having trouble showing that the $L^p$ norm of the $n$-th order Dirichlet kernel is proportional to $n^{1-1/p}$.

I've tried brute force integration and it didn't work out. I would be grateful for any hints.

Thanks.

  • 0
    yes thats right2011-03-17

1 Answers 1

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Assuming the notation of DJC, apply the transformation $y = n x$ to obtain $ \frac{1}{n^{p-1}} \left\| D_n \right\|^p_p = \int_{- n \pi}^{n \pi} \left| \frac{\sin(y + \frac{y}{2 n})}{n \sin(\frac{y}{2 n})} \right|^p dy . $ Fatou's Lemma now shows that $ \frac{1}{n^{p-1}} \left\| D_n \right\|^p_p \leq \int_{\mathbb{R}} \left| \frac{2 \sin(y)}{y} \right|^p dy . $ For the other side of the inequality, note, that $ \frac{1}{n^{p-1}} \left\| D_n \right\|^p_p \geq \int_{- n \pi}^{n \pi} \left| \frac{\sin(y + \frac{y}{2 n})}{\frac{y}{2}} \right|^p dy, $ so using the triangular inequality of the $L^p$ norm we obtain $ \frac{1}{n^{p-1}} \left\| D_n \right\|^p_p \geq \left( \left( \int_{\mathbb{R}} \left| \frac{2 \sin(y)}{y} \right|^p dy \right)^{1/p} - \left( \int_{- n \pi}^{n \pi} \left| \frac{\sin(y + \frac{y}{2 n}) - \sin y}{\frac{y}{2}} \right|^p dy \right)^{1/p} \right)^p . $ The inequality now follows from $ \int_{- n \pi}^{n \pi} \left| \frac{\sin(y + \frac{y}{2 n}) - \sin y}{\frac{y}{2}} \right|^p dy \leq \int_{- n \pi}^{n \pi} \left| \frac{\frac{y}{2 n}}{\frac{y}{2}} \right|^p dy = \frac{2 \pi}{n^{p - 1}} = o(n) . $