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How to show that $$ \sum_{n=2}^\infty \frac{\sin{(nx)}}{\log n} $$ not the Fourier series of any function?

I have shown that the series is convergent by Dirichlet test.

Let $a(n)=\frac{1}{\log n}$. What is $\sum (a(n))^2$, to apply Parseval's theorem?

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See Katznelson chapter I, 4.2. Briefly, a sine series $\sum a_n \sin{(nx)}$ is not a Fourier series of an integrable function if $a_n \gt 0$ and $\sum_n \frac{a_n}{n} = \infty$. –  t.b. Apr 9 '12 at 8:21
    
Do you mean $\sin \frac{nx}{ \log n}$ or $\frac{\sin nx}{\log n}$? –  Rudy the Reindeer Apr 9 '12 at 8:26
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@MattN: the former has no chance of being a Fourier series... –  t.b. Apr 9 '12 at 8:29
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@t.b.: google link to An Introduction to Harmonic Analysis 4.2 (perhaps... :-)) –  Raymond Manzoni Apr 9 '12 at 8:35

1 Answer 1

Suppose that it is the Fourier series of $f(x)$ in $(-\pi,\pi)$, namely, $$ f(x)=\sum_{n=2}^\infty \frac{\sin(nx)}{\log n},\quad x\in(-\pi,\pi).$$ Then by Pareraval's Identity, $$\int_{-\pi}^\pi f(x) \, dx=\sum_{n=2}^\infty\frac{1}{\log^2n}=\infty.$$

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This answer is incomplete. Parseval's identity applies when $\int_{-\pi}^\pi |f(x)|^2\,dx<\infty$. What about the case when that integral is infinite but $\int_{-\pi}^\pi |f(x)| \, dx<\infty$? Those functions also have Fourier series. –  Michael Hardy Mar 11 '13 at 18:31
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. . . also, did you intend $\int_{-\pi}^\pi |f(x)|^2\,dx$ where you wrote $\int_{-\pi}^\pi f(x)\,dx$? –  Michael Hardy Mar 11 '13 at 18:33

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