A uniform bound by an integrable function for a Fourier series' partial sums.

Question

Consider \begin{equation} \sum\limits_{n=1}^\infty\frac{\cos(nx)}{n}=-\log|2\sin x/2|~~~ \big(x\in(0,2\pi)\big), \end{equation} and its $2\pi$-periodic extension $f$ (for a proof of the above identity see this MSE post.) Notice that $f\in L^1(0,\pi)$, since $f(x)\sim\log(x)~(x\rightarrow0)$. This Fourier series is not absolutely or uniformly convergent.

My problem is to show that every one of the partial sums \begin{equation} s_N(x)=\sum\limits_{n=1}^N\frac{\cos(nx)}{n} \end{equation} is bounded in absolute value by the same function $h\in L^1(0,\pi)$. I.e., $|s_N(x)|\leq h(x)$ for every $N\in\mathbb{N}$ and $x\in(0,\pi)$.

The various things I have tried so far mostly involve writing the partial sums using the Dirichlet kernel \begin{equation} D_N(x)=\frac{\sin(N+1/2)x}{2\sin x/2}=\frac{1}{2}+\sum_{n=1}^N\cos(nx). \end{equation} Then, using that $f$ is even and $2\pi$-periodic, \begin{align} \pi s_N(x) &=\int_{-\pi}^\pi f(t)D_N(t-x)\text dt \\ &=\int_0^\pi\big(f(y+x)+f(y-x)\big)D_N(y)\text dy \\ &=\int_0^\pi \underbrace{\log\left|\frac{\sin(y-x)/2}{\sin(y+x)/2}\right|}_{\displaystyle{:=g(x,y)}} D_N(y)\text dy. \end{align} We may differentiate $g$ to find \begin{equation} \partial_yg(x,y)=\frac{\sin x}{\cos x-\cos y}. \end{equation} Hence, integrating by parts, \begin{align} \pi s_N(x)=\left[g(x,y)\int_x^yD_N\right]_{y=0}^{y=\pi} -\int_0^\pi\frac{\sin x\int_x^yD_N}{\cos x-\cos y}\text dy. \end{align} The boundary terms vanish since $g(x,y)$ vanishes when $y=0,\pi$, so if we write $\int_x^yD_N=K_N(x,y)$ then \begin{equation} \pi s_N(x)=-\int_0^\pi K_N(x,y)\partial_yg(x,y)\text dy. \end{equation}

Observe that $\partial_yg(x,y)$ is singular as $y\rightarrow x$, and indeed, by Taylor-expanding $\cos y$ around $x$, behaves like \begin{equation} \frac{1}{y-x}\big(1+O(y-x)\big). \end{equation} Clearly, then, one needs to prove that $K_N(x,y)$ will "kill" $(y-x)^{-1}$ in some uniform fashion as $y\rightarrow x$ (and $N\rightarrow\infty$!). Unfortunately using the $\sin$-representation of $D_N$ to Taylor-expand $K_N$ around $y=x$ gives \begin{equation} K_N(x,y)=D_N(x)\int_x^y\big(1+N\cdot O(t-x)\big)\text dt= D_N(x)(y-x)\big(1+N\cdot O(y-x)\big), \end{equation} where I have left out the factor $N$ from the $O$-term to illustrate the non-uniformity of the convergence.

There are a couple of other failed attempts I made in a similar vein (for example using the $\cos$-representation of $D_N$), but for fear of making this post too long, I will leave them out. Any ideas on how to proceed would be greatly appreciated, though I would prefer them left as ideas, and not fully fleshed-out answers. Thanks in advance!

Answer 1

As requested, here is the sketch on an idea only.

Step $1$

Start with the partial sums

$$s_N(x) =\sum_{n=1}^N \frac{\cos nx}{n}$$

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Step $2$

Differentiate $s_N$ term by term to arrive at

$s_N'(x) =-\sum_{n=1}^N \sin nx= -\csc (x/2) \sin(Nx/2)\sin((N+1)x/2)$

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Step $3$

Using the closed-form expression for $s_N'(x)$, express $s_N(x)$ as an integral of $s_N'$ as

$$s_N(x)=s_N(\pi)+\int_{\pi}^x s_N'(x')dx'$$

where $-1\ge s_N(\pi)<-1/2$.

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Step $4$

Find a bound for the integral of $s_N'$.

$$\begin{align}\left|\int_{\pi}^x s_N'(x')dx'\right|&= \left|\,\int_{\pi}^x \left(-\csc (x'/2) \sin(Nx'/2)\sin((N+1)x'/2)\right)\,dx'\,\right|\\\\ &\le \int_{x}^{\pi} \left|-\csc (x'/2) \sin(Nx'/2)\sin((N+1)x'/2)\right|\,dx'\\\\ &\le\int_x^{\pi} \csc (x'/2) \,dx'\\\\ &=2\log\left(\cot\left(\frac{x}{4}\right)\right) \end{align}$$

So, choose $h(x) =|s_N(\pi)|+2\log\left(\cot\left(\frac{x}{4}\right)\right)$, which is an $L^1$ function on $(0,\pi)$ since $h \sim 2\log(x)$ for small $x$.

Note: $2\log\left(\cot\left(\frac{x}{4}\right)\right)\ge 0$ on $(0,\pi)$.

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Hints for Other Possible Ways Forward:

$$\sin(Nx/2)\sin((N+1)x/2)=\frac12\left(\cos(x/2)-\cos((N+1/2)x)\right)$$

Answer 2

Young's inequality (see Alzer, for instance) gives: $$ s_N(x)\geq -1 \tag{1}$$ and summation by parts gives: $$ s_N(x) = \frac{1}{N}\left(D_N(x)-\frac{1}{2}\right)+\sum_{n=1}^{N-1}\frac{D_n(x)-\frac{1}{2}}{n(n+1)}\tag{2} $$ hence the claim follows from the fact that: $$ \| D_N \|_{L^1}=O(\log n). \tag{3}$$ The trick is just to locate the real roots of $D_N$ to be able to state: $$ \left|D_N(x)\right|\leq \min\left(N,\frac{C}{x}\right).\tag{4}$$

Answer 3

You might want to consider complex variable technique and Abel summable method.

Starting with $f(z) = Log(1 - z) = - \sum\limits_{n = 1}^\infty {{{\frac{z}{n}}^n}} $

we get

$f(r{e^{i\theta }}) = - \sum\limits_{n = 1}^\infty {\frac{{\cos (n\theta )}}{n}{r^n}} - i\sum\limits_{n = 1}^\infty {\frac{{\sin (n\theta )}}{n}{r^n}} $

and so

${\mathop{\rm Re}\nolimits} Log(1 - r{e^{i\theta }}) = - \sum\limits_{n = 1}^\infty {\frac{{\cos (n\theta )}}{n}{r^n}} $.

Since $ - \sum\limits_{n = 1}^\infty {\frac{{\cos (n\theta )}}{n}} $ is convergent for $\theta$ not a multiple of 2 $\pi$, by Abel’s theorem,

$\sum\limits_{n = 1}^\infty {\frac{{\cos (n\theta )}}{n}} = - \mathop {\lim }\limits_{r \to {1^ - }} {\mathop{\rm Re}\nolimits} Log(1 - r{e^{i\theta }})$ .

But

${\mathop{\rm Re}\nolimits} Log(1 - r{e^{i\theta }}) = \ln (|1 - r{e^{i\theta }}|) = \frac{1}{2}\ln (1 + {r^2} - 2r\cos (\theta ))$

and so

$\mathop {\lim }\limits_{r \to {1^ - }} {\mathop{\rm Re}\nolimits} Log(1 - r{e^{i\theta }}) = \mathop {\lim }\limits_{r \to {1^ - }} \frac{1}{2}\ln (1 + {r^2} - 2r\cos (\theta )) = \frac{1}{2}\ln (2 - 2\cos (\theta )) = \frac{1}{2}\ln (4{\sin ^2}({\textstyle{\theta \over 2}})) = \ln (2\sin ({\textstyle{\theta \over 2}}))$

Thus $\sum\limits_{n = 1}^\infty {\frac{{\cos (n\theta )}}{n}} = - \ln (2\sin ({\textstyle{\theta \over 2}})) = \ln \left( {\frac{1}{{2\sin ({\textstyle{\theta \over 2}})}}} \right)$ .

To show that the convergence is dominated by a integrable function, you can use the Fejer sum. For the detail please see Theorem 11 of

Riemann summable everywhere series, Two special cosine series and Abel summable series https://037598a680dc5e00a4d1feafd699642badaa7a8c.googledrive.com/host/0B4HffVs7117IbmZ2OTdKSVBZLVk/Fourier%20Series/R-Summability%20two%20cosine%20series.pdf

Indeed

$\left| {{t_n}(\theta )} \right| = \left| {\sum\limits_{k = 1}^n {\frac{{\cos (k\theta )}}{k}} } \right| \le \ln \left( {\frac{1}{{2\sin \left( {{\textstyle{\theta \over 2}}} \right)}}} \right) + \frac{9}{2}$ .

All you have to do is just to add the constant 9/2 to the function and this is the required function. The details are in the above article.