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