Reference to L^1 error for piecewise linear interpolation of Lipschitz functions Let $f:[0,1]\to\mathbb{R}$ be a Lipschitz function, and $\pi f$ be its piecewise linear interpolant on an equispaced grid with $n$ points.
It should be true (if I am not making mistakes with the constant) that
$$
\int_0^1 |f - \pi f| \leq \frac{1}{4n} \operatorname{Lip}(f).
$$
Do you have a reference that I can cite for this result, without having to re-prove it? All the references I have found by looking around assume better regularity.
 A: Clearly it suffices to prove  that for any $a<b$ we have $\DeclareMathOperator{\Lip}{Lip}$
$$
\int_a^b \big|\; f-L(f)\;\big| dx\leq \frac{b-a}{4}\Lip(f),
$$
where $L(f)$ is the linear function that connects the endpoints of the graph of $f$
$$
x\mapsto f(a)+\frac{f(b)-f(a)}{b-a}x.
$$
After rescaling  and translations we can assume $a=0$, $b=1$, $f(a)=0$ so it suffices to prove
$$
\int_0^1\big|\; f(x) -f(1) x\;\big| dx\leq \frac{1}{4}\Lip(f),
$$ 
$\forall f\in \Lip([0,1])$, $f(0)=0$.
We have
$$
\int_0^1\big|\; f(x) -f(1) x\;\big| dx=\int_0^1\Bigg|\; \int_0^x\big(\;f'(y)-f(1)\;\big)dy\;\Bigg| dx
$$ 
$$
=  \int_0^1 \Bigg|\; \int_0^xf'(y)-f(1)\;dy\;\Bigg|\; dx\leq \int_0^1  \int_0^x \big|\;f'(y)-f(1)\;\big|\;dy\; dx
$$
$$
= \int_0^1 \int_y^1\big|\;f'(y)-f(1)\;\big|dx\; dy= \int_0^1 (1-y)\;\big|\;f'(y)-f(1)\;\big|\;dy
$$
Now observe that $\DeclareMathOperator{\Mean}{Mean}$
$$
f(1)=\Mean(f').
$$
We need to find the  optimal constant $C$ in the inequality
$$
\Big\Vert g-\Mean(g)\Big\Vert_{L^\infty}\leq C\Vert g\Vert_{L^\infty},\;\;g\in L^\infty(0,1).
\tag{1}
$$
It is not very hard to see that $C\in [1,2]$. We deduce
$$
\int_0^1\big|\; f(x) -f(1) x\;\big| dx\leq \frac{C}{2}\Vert f'\Vert_\infty.
$$
Another approach.
$$
\int_0^1 (1-y)\;\big|\;f'(y)-f(1)\;\big|\;dy\leq \left(\int_0^1 (1-y)^2 dy\right)^{1/2}\left(\int_0^1 (f'(y)-f(1))^2dy\right)^{1/2}.
$$
On the other hand, for any $g\in L^2(0,1)$ we have
$$
\int_0^1\Big(\; g(y)-\Mean(g)\;\big)^2  dy\leq \int_0^1 g(y)^2 dy \leq \Vert g\Vert_{L^\infty}^2.
$$
Hence
$$
\int_0^1 (1-y)\;\big|\;f'(y)-f(1)\;\big|\;dy\leq \left(\int_0^1 (1-y)^2 dy\right)^{1/2}\Vert f'\Vert_{L^\infty}=\frac{1}{\sqrt{3}}\Lip(f).
$$
 I don't know yet how to improve the constant $\frac{1}{\sqrt{3}}$ to $\frac{1}{4}$.
A: This is a special form of a Jackson inequality, if I am not mistaken. I fail to find the specific case you are interested in, but a start is https://www.encyclopediaofmath.org/index.php/Jackson_inequality.
Edit: As remarked, Jackson inequalities usually deal with best possible general approximations, but the name also refers to other situations. The closest result to what the OP is looking for I could find is Theorem 15 in 

Sharp Jackson Inequalities for Piecewise Linear Interpolation and Rectangular Formula;
  Oleg L.Vinogradov, Vladimir V.Zhuk; 11th IFAC Workshop on Control Applications of Optimization 2000, St Petersburg, Russia, 3-6 July 2000;
  https://www.sciencedirect.com/science/article/pii/S1474667017396611

Although they formulate their result with the modulus of continuity of $f'$, they only use that the function equal the integral of the derivative and thus, their result and proof also work for the Lipschitz case (where the modulus of continuity is $L$). They also claim to have a sharp inequality for $|f(x) - \pi f(x)|$ and thus, integrating their inequality will give your case.
