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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.

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  • $\begingroup$ Just to be clear, you want to find $$\| D_n \|_p = \left(\int_{-\pi}^{\pi} \left| \frac{\sin(nx + x/2)}{\sin(x/2)} \right|^p dx \right)^{1/p},$$ right? $\endgroup$ – JavaMan Mar 17 '11 at 6:04
  • $\begingroup$ yes thats right $\endgroup$ – jack Mar 17 '11 at 14:52
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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 $$ \int_{- n \pi}^{n \pi} \left| \frac{\sin(y + \frac{y}{2 n}) - \sin y}{\frac{y}{2}} \right|^p dy \leq \int_{- n \pi}^{n \pi} \left| \frac{\frac{y}{2 n}}{\frac{y}{2}} \right|^p dy = \frac{2 \pi}{n^{p - 1}} = o(n) . $$

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