$L^1$ 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 $O(n \log n)$.
 A: Rough back-of-envelope estimate: Pretend that $D_n(t) = \sin(\pi nt/2)/t$ for $-1\le t\le 1$. That should be close enough to determine the order of magnitude of the total variation; if you want you can easily convert what I write here to actual inequalities for the actual Dirichlet kernel.
We have $D_n(2k/n) = -n/(2k)$ and $D_n((2k+1)/n)=n/(2k+1)$. So $|D_n(2k/n)-D_n((2k+1)/n)|$ is roughly $n/k$; hence the total variation is something like $n\sum_{k=1}^{n/2}1/k$, or $n\log(n)$.
If this is homework you better clean it up before handing it in...
A: 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
$$\lVert D_n'\rVert_{L^1([-\pi,\pi])} = 2\int_0^{\pi} \lvert D_n'(t)\rvert\,dt = \frac{4}{\pi}n\log n + O(n).$$
A: Staring back at this after some time I realized that the Abel summation trick does it as well. Only one more step was needed. I give the proof for the general case of high derivatives since it seems difficult to find these online for some reason. 
Proposition: Let $k,N \geq 1$. Then 
$$
\left|\|D_N^{(k)}\|_{L^1([-\pi,\pi])} - N^{k}\|D_N\|_{L^1([-\pi,\pi])}\right| \leq 6\pi(2k-1)N^k + 4\pi k N^{k-1}.
$$
Proof: An application of Abel's summation formula  https://en.wikipedia.org/wiki/Abel%27s_summation_formula 
gives
$$
i^{-k}D_N^{(k)}(x) = N^k D_N(x) - k I(x,k,N),
$$
where
$$
I(x,k,N) = \int_0^N t^{k-1}D_{\lfloor t \rfloor }(x)dt.
$$
From the triangle inequalities (the regular one and the reversed one) we can see that $|A + B| = |A| + O^*(B)$ for any complex numbers $A,B$, where $O^*(B)$ denotes a quantity that is at most $|B|$ in absolute value. Hence 
$$
|D_N^{(k)}(x)| = N^k|D_N(x)| + O^*(kI(x,k,N)).
$$
Thus to conclude we only need to prove that $\int_{-\pi}^\pi|I(x,k,N)|dx \leq 6\pi \frac{2k-1}{k}N^k + 4\pi N^{k-1}$.
An integration by parts shows that
$$
I(x,k,N) = N^{k-1}G(x,N) - (k-1)\int_0^N y^{k-2}G(x,y)dy,
$$
where
$$
G(x,y) = \int_0^y D_{\lfloor t \rfloor}(x)dt = \sum_{0 \leq m < \lfloor y \rfloor}D_{m}(x) + \{y\}D_{\lfloor y \rfloor}(x)
$$
with $\{y\} := y - \lfloor y \rfloor$. We recognize the sum above as $\lfloor y \rfloor F_{\lfloor y \rfloor}(x)$ with $F_m(x)$ the Fejer kernel. The trivial bound gives $|D_{\lfloor y \rfloor}(x)| \leq 2y+1$ (one could do better in $L^1$ but this suffices for our purposes). Hence $|G(x,y)| \leq yF_{\lfloor y \rfloor }(x) + 2y+1$. Thus
$$
|I(x,k,N)| \leq N^{k-1}(NF_N(x) + 2N + 1) + (k-1)\int_0^N y^{k-2}(yF_{\lfloor y \rfloor }(x) + 2y + 1)dy.
$$
Now the result follows when we integrate over $x \in [-\pi,\pi]$ and use the fact that the Fejer kernel integrates to $2\pi$ there.
Remark: Note from the proof above that the main contribution to the $L^1$ norm of $D_N^{(k)}(x)$ comes from the ''endpoint'' of the Abel summation by parts, i.e., the term $N^k|D_N(x)|$. This is a typical problem with the rough indicator functions $1_{[-N,N]}(x)$ (we are summing $D_N(x) = \sum_{n \in \mathbb{Z}}1_{[-N,N]}(n)e^{inx}$). If we instead had a sum like $D_{N,\varphi}(x) = \sum_{n \in \mathbb{Z}}\varphi(n/N)e^{inx}$ with say $\varphi$ even of class $C_0^{2}(\mathbb{R})$ supported on $[-1,1]$ with $\|\varphi^{(j)}\|_\infty = O(1)$ for $j=0,1,2$, then the end-points go away and arguments similar to those above show that $\|D_{N,\varphi}^{(k)}\|_{L^1([-\pi,\pi])} = O_k(N^k)$; i.e., we save a log.
