Q:

Why is this a linear interpolation? 

Let $J_{k,n}$ be the dyadic partition of $[0,1]$, i.e. $n\in \mathbb{N}_0,k=1,\dots,2^n$, $J_{k,n}:=((k-1)2^{-n},k2^{-n}]$ and we denote with $\phi_{n,k}$ the Schauder functions over $J_{k,n}$, i.e. the triangle function with peak of height $2^{-\frac{n}{2}-1}$ at the middle point $(2k-1)2^{-(n+1)}$.
Furthermore, we define for a function $f\in C[0,1]$ the set $\Delta_{n,k}(f):=(f((2k-1)2^{-(n+1)})-f((k-1)2^{-n}))-(f(k2^{-n})-f((2k-1)2^{-(n+1)}))$. Then put 
$$f_N(x):=\sum_{n=0}^N\sum_{k=1}^{2^n}2^{\frac{n}{2}}\Delta_{n,k}(f)\phi_{n,k}(x)$$
A calculation leads to
$$f_N(x)-f_{N-1}(x)=\sum_{k=1}^{2^N}\frac{1}{2}\Delta_{N,k}(f)2^{\frac{N}{2}+1}\phi_{N,k}(x)$$
Why does this last sum imply that $f_N$ is the piecewise linear interpolation of $f$ along the $N$-the dyadic partition.  Sorry I do not see this. 
Thank you for your help
hulik

A:

You need the additional constraint $f(0)=f(1)=0$.
$f_N$ is actually the piecewise linear interpolation of $f$ along the $(N+1)$-th dyadic partition $D_{N+1} = \{k2^{-N-1}: 0\le k\le 2^{N+1},\ k\in\mathbb Z \}$.
Suppose $f_{N-1}$ satisfies this property.  Since $f_N-f_{N-1}$ is 0 on $D_N$, you only need to check that $f_N(x)=f(x)$ for $x\in D_{N+1}\setminus D_N$, that is $x=(a+b)/2$ for $a,b$ consecutive in $D_N$.
$$f_{N-1}(x)=\frac{f(a)+f(b)}{2}$$
$$\Delta_{N,2^N b}(f)=f(x)-f(a)-(f(b)-f(x))\\
=2(f(x)-f_{N-1}(x))$$
proving that
$$f_N(x)-f_{N-1}(x)=f(x)-f_{N-1}(x)$$

