# Does anybody know an estimation of L4 norm of fejer kernel ?

Hi, I need an estimation or an exact closed form expression for the following integral

$\int_{0}^{2\pi} K_N^4(s) ds$

where $K_N(s)= \frac{1}{N2\pi} (\frac{sin(Ns/2)}{sin(s/2)})^2$, the Fejer kernel.

I don't know how to obtain an estimation better than

$\int_{0}^{2\pi} K_N^4(s) ds < N^4$

Does anyone know a better estimation or some trigonometric tricks that can help me to improve my estimation?

Imma

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I agree with Davide Giraudo, I think the Fejer kernel has a square, so you are either missing a square, or using the wrong name. Have you tried just expressing with complex exponentials and evaluating directly? Since you're raising a trig. polynomial to a positive integer power, it has an exact closed form expression, which may be easier to estimate. –  Zen Harper Jul 29 '12 at 15:05
I agree with Davide and Zen that the question needs to be edited or clarified. Incidentally, what you claim to have is a bound on the 4th power of the L^4-norm, not the L^4-norm itself, so I think your title needs some minor corrections –  Yemon Choi Jul 29 '12 at 18:03
For any given integer power of $K_N$ you can get an exact formula by expanding the exponential sum, though this gets tiresome past the first few cases. For any power, you get an easy upper bound from $K_N(s) < \min(N, 1/\sin(s/2)^2)$ [I assume that the factor of $1/2\pi$ should be applied to the integral, not to $K_N(s)$], and this should be within a small factor of the truth. –  Noam D. Elkies Jul 29 '12 at 23:43

$\displaystyle \int_0^{2\pi} K_N^4(s)\ ds = \frac{c_{N-1}}{8 \pi^3 N^4}$ where $c_n$ is the coefficient of $z^{4n}$ in $(1 + z + \ldots + z^n)^8$.
$c_n$ appears to have the closed form $$c_n = \frac{\left( 315+1284 n + 2734 n^2+3300{n}^{3}+2335{n}^{4}+906{n}^{5} +151{n}^{6} \right) \left( n+1 \right)}{315}$$
$${\frac {45+49 {N}^{2}+ 70 N^4 + 151{N}^{6}}{2520 {N }^{3}{\pi }^{3}}}$$