Let be $F_N$ the sequence of the arithmetic means of Dirichlet kernels $D_N (x)$ defined by $$ F_N := \frac{1}{N+1} (D_0 (x) +D_1 (x)+..+D_N(x)) $$

Where the Dirichlet kernel is defined by $$D_N (x)= \sum_{n=-N}^N e^{inx} $$

I have no ideas of ways to prove that

$$ F_N \geq 0 $$

and that even for $x \not\in 2\pi \mathbb{Z} $ it holds that: $$ \lim_{N \rightarrow \infty} F_N (x)=0 $$

I appreciate any help of you guys.

  • $\begingroup$ $F_N(x)=\frac 1 {N+1}[ \sin (\frac {(N+1)x} 2)]^{2}]/ [\sin (\frac x 2)]^{2}$. You can find this in any book on Fourier series, in particular on p. 79 of Edwrads book. $\endgroup$ Commented Oct 29, 2019 at 10:23
  • $\begingroup$ It is Fejér's kernel (en.wikipedia.org/wiki/Fej%C3%A9r_kernel) ; I have added the corresponding tag. $\endgroup$
    – Jean Marie
    Commented Oct 29, 2019 at 11:43

1 Answer 1


Available in many places, the idea is to sum the $D_N$ using a geometric series fomula \begin{align} D_N(x) &= e^{-iNx}\sum_{k=0}^{2N} e^{ikx} \\ & =e^{-iNx}\frac{e^{(2N+1)x}-1}{e^{ix}-1} \\ &= \frac{e^{(N+1)x}-e^{-iNx}}{e^{ix}-1}\\ & = \frac{e^{(N+1/2)x}-e^{-i(N+1/2)x}}{e^{ix/2}-e^{-ix/2}}\\ & = \frac{\sin((N+1/2)x)}{\sin(x/2)}\end{align} Alternatively, there's a nice telescoping sum argument here. Then $ \sum_{n=0}^N \sin((n+1/2)x) = \Im \sum_{k=0}^Ne^{i(n+1/2)x}$, and $$ \sum_{k=0}^Ne^{i(n+1/2)x} = e^{ix/2}\sum_{k=0}^Ne^{in x} = \frac{e^{i(N+1)x}-1}{e^{ix/2}-e^{-ix/2}} =\frac{e^{i(N+1)x}-1}{2i \sin(x/2)} $$ and therefore

$$F_N(x) = \frac1{(N+1)\sin(x/2)^2}\times (-1)\times \Re (\frac{e^{i(N+1)x}-1}2) =\frac{1-\cos((N+1)x)}{2(N+1)\sin(x/2)^2}$$ this already proves $F_N(x)\ge 0$, to put it in the "standard form" just use highschool trig $ \cos(2\theta) = 1 - 2\sin(\theta)^2$ to get $$ F_N(x) = \frac1{N+1} \left(\frac{\sin((N+1)x/2)}{\sin(x/2)}\right)^2\ge0.$$

For the convergence to zero, at every $x$ that is not a zero of $\sin(x/2)$, we have

$$ F_N(x) = \underbrace{\frac{1}{\sin(x/2)^2}}_{\text{constant in $N$}}\times \underbrace{\sin((N+1)x/2)^2}_{\le 1} \times \frac1{N+1}\to 0$$


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