# Let $a_{n}\downarrow 0$ and if series $\sum a_{n}\sin nx$ is a Fourier series of function $f\in L^{1}$ then $\sum \frac{a_{n}}{n}<+\infty.$

I want to show that if $\displaystyle\sum a_{n}\sin nx (a_{n}\downarrow 0)$ is a Fourier series of $f\in L^{1}$ then $\displaystyle \sum \frac{a_{n}}{n}<+\infty.$ I know i have to use some property of Dirichlet's kernel but i am stuck how to use them to derive my result.

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See the comments to this question. – Julián Aguirre Apr 18 '12 at 15:35
@JuliánAguirre The comments in that question assume $f\in L^1([-\pi,\pi]) \subset L^2([-\pi,\pi]),$ a stronger condition than what I assumed. – Ragib Zaman Apr 18 '12 at 15:58
@RagibZaman You assumed $f\in L^2$, which is stronger than $f\in L^1$. – Julián Aguirre Apr 18 '12 at 16:06
@JuliánAguirre My bad, I see now. – Ragib Zaman Apr 18 '12 at 16:13

We will use the function on $(0,2\pi)$ $$\varphi(x)=\sum_{n=1}^\infty\frac{\sin(nx)}{n}=\frac{\pi-x}{2}\in L^\infty$$ Let $$f(x)=\sum_{n=1}^\infty a_n\sin(nx)$$ Then we get $$\left|\pi\sum_{n=1}^\infty\frac{a_n}{n}\right|=\left|\int_0^{2\pi}f(x)\varphi(x)\,\mathrm{d}x\right|\le\|f\|_{L^1}\|\varphi\|_{L^\infty}$$ Since $\|\varphi(x)\|_{L^\infty}=\frac\pi2$, we get $$\left|\sum_{n=1}^\infty\frac{a_n}{n}\right|\le\frac12\|f\|_{L^1}$$

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By Parseval's identity, we can deduce $$\sum_{n=1}^{\infty} |a_n|^2 < \infty.$$ Now by the Cauchy-Schwarz inequality we have $$\biggr| \sum_{n=1}^{\infty} \frac{a_n}{n} \biggr| \leq \sum_{n=1}^{\infty} \frac{|a_n|}{n^2} \leq \left( \sum_{n=1}^{\infty} |a_n|^2 \right)\left( \sum_{n=1}^{\infty} \frac{1}{n^2} \right)$$ and since both quantities on the right are finite, so is the sum on the left.

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If u don't mind can you please write some step? So i can understand this properly. Thanks! – Kns Apr 18 '12 at 15:01
@Kunjanshah See my edit. – Ragib Zaman Apr 18 '12 at 15:19
You do not know that it is the Fourier series of an $L^2$ function, so you do not know that $\sum|a_n|^2<\infty$. – Julián Aguirre Apr 18 '12 at 15:32
But how you know that $\displaystyle \sum a_{n}^{2}<+\infty$ ? Does there any result? @ Ragib Zaman – Kns Apr 18 '12 at 15:40
@Kunjanshah In Parseval's identity, the integral on the right is finite for $L^2$ functions, which I assumed was the class of functions you were working in. – Ragib Zaman Apr 18 '12 at 15:57

Let $f(x)=\sum a_k \sin kx.$ The integral of $f(x),$ say $F(x)=-\sum \frac{a_k}{k}\cos (kx),$ is absolutely continuous. A standard theorem says that the Fourier series of an absolutely continuous function converges to it uniformly so taking $x=0$ you get the result. By the way you have proved that the series with $a_k=1/(\ln k)$ is NOT a Fourier series.

After getting some good hints from my friends i am also trying to solve this question on my own way. But i don't know that am i approach this question correctly? Let $\displaystyle S_{N}(x)=\sum_{n=1}^{N}a_{n}\sin nx$ denotes the $n$th partial sum of series $\displaystyle \sum a_{n}\sin nx$. Now applying Abel's summation formula to right hand side of $S_{N}(x)$, we get, $$\sum_{n=1}^{N}a_{n}\sin nx=a_{N}\widetilde{D_{N}(x)}+\sum_{n=1}^{N-1}\Delta a_{n}\widetilde{D_{n}(x)}$$, where $\Delta a_{n}=a_{n}-a_{n+1}$. Now \begin{align*} f(x)&=\lim_{N\to\infty}\S_{N}(x)\\ &=\sum_{n=1}^{\infty}\Delta a_{n} \widetilde{D_{n}(x)}\ \text{e.w}(\because a_{n}\downarrow 0) \end{align*} Define $$f^{*}(x)=\sum_{n=1}^{\infty}\Delta a_{n}\widetilde{D_{n}(x)}^{*}$$ $\displaystyle \therefore f(x)-f^{*}(x)=\sum_{n=1}^{\infty}\Delta a_{n}\sin nx$. Since $\displaystyle \sum_{n=1}^{\infty}|\Delta a_{n}|<+\infty (\because a_{n}\downarrow 0)$ It follows that by Wierstrass M-test $f-f^{*}$ is uniform limit of continuous function and hence $f-f^{*}$ is a continuous function. Hence $f-f^{*}\in L^{1}.$ Thus $f\in L^{1}\Leftrightarrow f^{*}\in L^{1}$. Now, \begin{align*} \frac{1}{2\pi}\int_{-\pi}^{\pi}|f^{*}(x)| dx &= \frac{1}{\pi}\int_{0}^{\pi}|f^{*}(x)|dx\\ &=\frac{1}{\pi}\int_{0}^{\pi}|\sum_{n=0}^{\infty}\Delta a_{n}\widetilde{D_{n}(x)}^{*}|dx\\ &= \frac{1}{\pi}\int_{0}^{\pi}\sum_{n=0}^{\infty}\Delta a_{n}\widetilde{D_{n}(x)}^{*} dx\\ &=\frac{1}{\pi}\sum_{n=1}^{\infty}\Delta a_{n}\int_{0}^{\pi}\widetilde{D_{n}(x)}^{*}dx\\ &=\sum \Delta a_{n}\|\widetilde{D_{n}}^{*}\|_{1} \end{align*} Thus, $\displaystyle \|f^{*}\|_{1}=\sum_{n=1}^{\infty}\Delta a_{n}\|\widetilde{D_{n}}^{*}\|_{1}$......(1) Since $\displaystyle \|\widetilde{D_{n}}^{*}\|_{1}=c\log n+O(1)$ $\therefore$ The right hand side of (1) is finite if and only if $\displaystyle \Delta a_{n}\log n<+\infty$. Which is equivalent to $\displaystyle \sum_{n=1}^{\infty}\frac{a_{n}}{n}<+\infty$.