What is the optimal Fourier series convergence rate estimate for $|x|$? What is the known best estimate of the rate of convergence in $\|\cdot\|_\infty$ (or maximal absolute value) of the Fourier series of $|x|,\, x\in[-1,1]$? If I look at the coefficients of the Fourier series, it is $\sim \frac1{n^2}$ so the error should be $O\big(\frac1n\big)$ where $n$ is the highest power of the partial sum of the Fourier series. Is the convergence rate higher than $\frac1n$?
 A: The convergence rate is indeed not significantly higher. If it was $\mathcal{O}(n^{-1-\epsilon})$ for some $\epsilon$, then Bernstein's theorem would imply that $|x|$ is in the H$\ddot{\text o}$lder space $C^{1+\epsilon}$ which it isn't.
A: Here is my corrected answer. Intially, I have misread the function. Of course $f(x) =|x|$ defined on $[-1,1]$ can be periodically extended as a continuous function. It is somesort of a 'triangle-function'.

We are going to show that the convergence rate is exactly $1/n$. 

First note, that
$$\widehat{f}(k) = \frac{1}{2} \int_{-1}^1 |x| \exp(-\pi \mathrm{i} xk) \, \mathrm{d} x = \frac{ k \sin(-\pi k) + \cos(-\pi k) -1 }{k^2} = \frac{\cos( \pi k) -1}{k^2}.$$
In order to obtain a statement about the convergence rate, we rewrite the partial sums as the convolution of the function with the Dirichlet kernel. The kernel is given by
$$D_n(t):=\sum_{k=-n}^n \exp(\pi \mathrm{i} kt) = \frac{\sin(\pi(1/2+n)t)}{\sin(\pi t/2)}$$
and thus we get
$$\sum_{k=-n}^n \widehat{f}(k) e^{\pi\mathrm{i} k x} = \frac{1}{2} \int_{-1}^1 f(x-y) D_n(y) \, \mathrm{d} y.$$
Hence, we can write
$$f(x)-\sum_{k=-n}^n \widehat{f}(k) e^{\pi\mathrm{i} k x} = \frac{1}{2} \int_{-1}^1 (f(x)-f(x-y)) D_n(y) \, \mathrm{d} y $$
Using the definition of $f$, whereby we should recall that $f$ is a periodic extension, the last integral can be written as
\begin{align}
\frac{1}{2} \int_{-1}^{\max(-1,x-1)} (|x| -|x-y-2|) D_n(y) \mathrm{d} y&+\frac{1}{2} \int_{\min(1,x+1)}^{1} (|x| -|x-y+2|) D_n(y) \mathrm{d} y \\
&+ \frac{1}{2} \int^{\min(1,x+1)}_{\max(-1,x-1)} (|x|-|x-y|) D_n(y) \mathrm{d} y \tag1
\end{align}
This formula is a bit messy, but we need to distinguish if $|x-y| \leq 1$, $x-y >1$ or $x-y < -1$. And this three terms correspond to these cases. We need to to handle all three integrals in (1) separately.
For simplicity let us assume that $x \geq 0$. In fact, the other case are similar. Note that the critical points are $x=-1,0,1$ because $f$ is not differentiable in this points. In this case, the middle-integral in (1) vanishes, because $\min(1,x+1) =1$, and if $x>0$ the first integral is equal to
$$\frac{1}{2} \int_{-1}^{x-1} (x-(2+y-x)) D_n(y) \mathrm{d} y.$$
If $x=0$ this integral vanishes too. The function 
$$\tag{g}g(y):=y/\sin(y \pi/2)$$ is smooth on $[-1,1]$, because we have a removable singularity in $y=0$. By partial integration, we find
$$\frac{1}{2} \int_{-1}^{x-1} y D_n(y) \mathrm{d}y=\mathcal{O}(1/n) \tag2$$
uniformly in $x \in [0,1]$. Of course, we can do this step in detail: By applying two times partial integration, we may show that
$$\frac{1}{2} \int_{-1}^{x-1} y D_n(y) \mathrm{d}y = g(x-1) \frac{\cos(\pi(1/2+n)(x-1))}{\pi(1+2n)} + \mathcal{O}(1/n^2)\tag{3.1}.$$
Thus, the critical part is
\begin{align}
(x-1)\int_{-1}^{x-1} D_n(y) \mathrm{d} y = (x-1)& \left. \frac{\cos(\pi (1/2+n)y)}{\pi(1/2+n)} \frac{1}{\sin(\pi y/2)} \right|_{y=-1}^{x-1} \\
& + \frac{\pi}{2} (x-1) \int_{-1}^{x-1} \frac{\cos(\pi (1/2+n)y)}{\pi(1/2+n)} \frac{\cos(\pi y/2)}{\sin(\pi y/2)^2} \mathrm{d} y \tag{3.2}
\end{align}
Note that in (3.2) the first term is of order $\mathcal{O}(1/n)$. Additionally, using $|\sin(y\pi/2)| \geq |y|$ we get that the last term is also $\mathcal{O}(1/n)$ uniformly in $x \in (0,1]$. 
We can use a similar argument to show that the third term in (1) is also of order $\mathcal{O}(1/n)$. In fact, this term is equal to
$$ \tag{5} \frac{1}{2} \int_{x-1}^x y D_n(y) \mathrm{d} y + \frac{1}{2} \int_x^1 (2x-y) D_n(y) \mathrm{d} y.$$
The same argument as before (using that $g$ is smooth) shows that the critcal part is
$$\tag{6} x \int_x^1 D_n(y) \mathrm{d} y$$
and this can be handled as in the previous step.
We have already shown that the error-term is $\mathcal{O}(1/n)$ uniformly in $x \in \mathbb{R}$. Calling the error term $R_n(x)$, that is (1), we are going to show that $R_{n_k}(x) n_k \not\rightarrow 0$ for some special subsequence depending on special $x \in (0,1)$. This implies that error-term cannot be better than $\mathcal{O}(1/n)$.
We begin with adding (3.1) and the Integrands with $y$ in (5), that is
$$\frac{1}{2}\int_{-1}^x y D_n(y) \mathrm{d}y - \frac{1}{2} \int_x^1 yD_n(y) \mathrm{d}y.$$
Using the same argument as in (3.1) we can easily verify that the this term is
$$g(x) \frac{\cos(\pi(1/2+n)x)}{\pi(1/2+n)} + \mathcal{O}(1/n^2).$$
In (3.2) the third term is $o(1/n)$, depending on $x \in (0,1)$ now, by the Riemann-Lebesgue-Lemma. Moreover, we do the same for (6) as in (3.2) to get
$$x\int_x^1 D_n(y) \mathrm{d} y = -g(x) \frac{\cos(\pi (1/2+n)x)}{\pi(1/2+n)} +o_x(1/n).$$
Thus, adding all together we see that
$$R_{n}(x)  = g(x-1) \frac{\cos(\pi(1/2+n)(x-1))}{\pi(1/2+n)} + o_x(1/n)$$
If we now take $x \in (0,1)$ irrational, then there exists a subsequence $(n_k)_{k \in \mathbb{N}}$ such that $\cos(\pi (1/2+n_k) (1-x)) \rightarrow 1$. Thus $R_{n_k}(x) \pi (1/2+n_k) \rightarrow g(x-1) \neq 0$. (Note that $g$ has no zeros!)
