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For each $0<\alpha<1$ the series $$\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^{\alpha}}$$ converges for every $x$ but is not the Fourier series of a Riemann integrable function.

a) If the conjugate Dirichlet kernel is defined by

$$\bar {D}_{N}(x)=\sum_{|n|\le N}\operatorname{sign}(x)e^{inx} \quad \text{where } \operatorname{sign}(x)=\begin{cases} 1 & \text{if } n>0 \\ 0 & \text{if } n=0 \\ -1\ & \text{if } n<0.\end{cases}$$ then show that $$\bar {D}_{N}(x)=\frac{\cos(x/2)-\cos((N+1/2)x)}{\sin(x/2)},$$ and $$\int_{-\pi}^{\pi}|\bar{D}_{N}(x)|dx\le c\log(N).$$ b) As as result, if $f$ is a Riemann integrable, then $$(f*\bar{D}_{N})(0)=O(\log N).$$ c) In the present case, this leads to $$\sum_{n=1}^{N}\frac{1}{n^\alpha}=O(\log N),$$

which is a contradiction.

I proved that $$\bar {D}_{N}(x)=\frac{\cos(x/2)-\cos((N+1/2)x)}{\sin(x/2)},$$ by using: $$D_{N}(x)=\sum_{n=-N}^{N}e^{inx}=\frac{\sin((N+1)x)}{\sin(x/2)}$$ But I don't know how to prove $$\int_{-\pi}^{\pi}|\bar{D}_{N}(x)|dx\le c\log(N).$$

Theo Bendit
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John He
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1 Answers1

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Since $\bar{D}_N(x) = \sum_{n=1}^N \sin nx $ we have the estimate

$$\tag{1}|\bar{D}_N(x)| \leqslant N$$

Another estimate (using $\sin (x/2) \geqslant x/\pi$ for $0 \leqslant x \leqslant \pi$) is

$$\tag{2}|\bar{D}_N(x)| = \left|\frac{\cos(x/2) - \cos((N+1/2)x)}{2\sin(x/2)} \right| = \left|\cos(x/2) - \cos((N+1/2)x\right|\left|\frac{x/2}{\sin(x/2)}\right|\frac{1}{|x|}\\ \leqslant \frac{\pi}{|x|} $$

Split the integral into contributions from intervals $[-\pi, - \pi/N]$, $[-\pi/N,0]$, $[0,\pi/N]$ and $[\pi/N,\pi]$ and obtain the estimates

$$\int_0^{\pi/N}|\bar{D}_N(x)| \, dx \leqslant \int_0^{\pi/N}N \, dx = \pi, \\ \int_{\pi/N}^{\pi}|\bar{D}_N(x)| \, dx \leqslant \pi\int_{\pi/N}^{\pi} \frac{dx}{x} = \pi\log N, \\ \text{etc.}$$

Collecting together we have for an appropriate choice of $c$,

$$\int_{-\pi}^{\pi}|\bar{D}_N(x)| \, dx \leqslant 2\pi + 2\pi\log N \leqslant c \log N$$

RRL
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