$L^1$ norm of derivative of Dirichlet kernel Let $D_n(x)$ denote the $n$-th Dirichlet kernel. We all know that $D_n$ has $L^1$ norm $O(\log n)$ (over a full period). But I'm wondering about the $L^1$ norm of its derivative. Is it $O(n)$ or $O(n \log n)$? Would anyone know a proof? Thanks.
Edit: Here's a proof that it is at least $O(n \log n)$ (I wonder if one could do better):
An application of Abel's summation formula shows that
$$
-iD_n'(x) := \sum_{|m| \leq n}me^{imx} = nD_n(x) - 1 - \int_1^n D_{\lfloor t \rfloor}(x)dt
$$
and so
$$
|D_n'(x)| \leq n|D_n(x)| + 1 + \int_1^n |D_{\lfloor t \rfloor}(x)|dt.
$$
Now integrate the above over $x \in [0,2\pi]$ to get $O(n \log n)$.

By symmetry, it suffices to look at the interval $[0,\pi]$. From
$$D_n(t) = \frac{\sin \bigl(\bigl(n+\frac{1}{2}\bigr)t\bigr)}{\sin \frac{t}{2}}$$
(with the appropriate interpretation if $t = 0$) we can read off the zeros of $D_n$, they are $z_k = \frac{k\pi}{n+\frac{1}{2}}$ for $1 \leqslant k \leqslant n$. Between $z_k$ and $z_{k+1}$, $D_n$ has constant sign, and its modulus attains a maximum $M_k$, so
$$\int_{z_k}^{z_{k+1}} \lvert D_n'(t)\rvert\,dt = 2M_k.$$
Since the denominator is strictly increasing on $[0,\pi]$ and the modulus of the numerator is bounded by $1$, with that bound attained at $w_k = \frac{k+\frac{1}{2}}{n+\frac{1}{2}}\pi$ and nowhere else in $[z_k,z_{k+1}]$, we have
$$\frac{1}{\sin w_k} < M_k < \frac{1}{\sin z_k}.$$
On $\bigl(0,\frac{\pi}{2}\bigr)$ we have
$$0 < \frac{1}{\sin x} - \frac{1}{x} < \sin x < x.$$
The first inequality follows from $\sin x < x$ there, and the second from
\begin{align}
&&x\cos x < x &< \tan x \\
&\implies& \cos^2 x = 1 - \sin^2 x &< \frac{\sin x}{x} \\
&\implies& 1 - \frac{\sin x}{x} &< \sin^2 x \\
&\implies& \frac{1}{\sin x} - \frac{1}{x} &< \sin x.
\end{align}
Thus
$$2\sum_{k = 1}^{n-1} \frac{1}{w_k} < \int_{z_1}^{z_n} \lvert D_n'(t)\rvert\,dt < 2\sum_{k = 1}^{n-1} \biggl(\frac{1}{z_k} + z_k\biggr).$$
Now we can "explicitly" compute those sums,
\begin{align}
\sum_{k = 1}^{n-1} \frac{1}{w_k} &= \frac{2n+1}{\pi} \sum_{k = 1}^{n-1} \frac{1}{2k+1} = \frac{2n+1}{\pi}\biggl(-1 + H_{2n}- \frac{1}{2}H_n\biggr) \\
&= \frac{2n+1}{\pi}\biggl(\frac{1}{2}\log n + \log 2 + \frac{1}{2}\gamma - 1 + O(n^{-1})\biggr), \\
\sum_{k = 1}^{n-1} z_k &= \frac{\pi}{n+\frac{1}{2}}\frac{n(n-1)}{2} = \frac{\pi n(n-1)}{2n+1}, \\
\sum_{k = 1}^{n-1} \frac{1}{z_k} &= \frac{n+\frac{1}{2}}{\pi} H_{n-1} = \frac{n+\frac{1}{2}}{\pi}\bigl(\log n + \gamma + O(n^{-1})\bigr)
\end{align}
and obtain
$$\frac{2n+1}{\pi}\bigl(\log n + \gamma - 2(1-\log 2) + O(n^{-1})\bigr) < \int_{z_1}^{z_n} \lvert D_n'(t)\rvert\,dt < \frac{2n+1}{\pi}\bigl(\log n + \gamma + O(n^{-1})\bigr).$$
With $D_n(0) = 2n+1$ and $D_n(\pi) = (-1)^n$ we have
$$\int_0^{z_1} \lvert D_n'(t)\rvert\,dt = 2n+1\quad\text{and}\quad \int_{z_n}^{\pi} \lvert D_n'(t)\rvert\,dt = 1,$$
so
$$\lVert D_n'\rVert_{L^1([-\pi,\pi])} = 2\int_0^{\pi} \lvert D_n'(t)\rvert\,dt = \frac{4}{\pi}n\log n + O(n).$$
