# 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

$$f_N(x):=\sum_{n=0}^N\sum_{k=1}^{2^n}2^{\frac{n}{2}}\Delta_{n,k}(f)\phi_{n,k}(x)$$

$$f_N(x)-f_{N-1}(x)=\sum_{k=1}^{2^N}\frac{1}{2}\Delta_{N,k}(f)2^{\frac{N}{2}+1}\phi_{N,k}(x)$$

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

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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 $$f_N(x)-f_{N-1}(x)=f(x)-f_{N-1}(x)$$
@ Generic Human: Thanks for your answer. There are still some question around, if you explain these in a more detailed way, or comment, I will accept your answer. 1. Why do we need $f(0)=f(1)=0$? 2. Why is $f_N-F_{N-1}=0$ on $D_N$? 3. Why is $\Delta_{N,2^{N+1}x}(f)=f(x)-f(a)-(f(b)-f(x))$ – user20869 Jun 16 '12 at 14:24
(1) and (2) come from the fact that $\phi_{n,k}$ is 0 on all of $D_n$, as the interior of $J_{k,n}$ does not intersect $D_n$. (2) follows immediately. (1) follows by observing that $f_N(0)=0$, but we want $f(0)=f_N(0)$ so we need $f(0)=0$. – Generic Human Jun 16 '12 at 14:26
(3) should actually be $\Delta_{N,2^N b}(f)$, and it's your exact definition with different notation: $x,a,b$ stand for $a=(k-1)2^{-n}$, $b=k2^{-n}$, $x=(k-1/2)2^{-n}$. – Generic Human Jun 16 '12 at 14:32
Sorry $1)$ and $2)$ are still not clear for me. I thought $J_{k,n} = D_n$ ? Or what is your $D_n$? Why should $\phi_{n,k}$ be zero insight $D_n$? Why is $f_N(0)=0$? Why do we need $f(1)=0$ too? – user20869 Jun 16 '12 at 15:31
I meant $D_n$ to be the $n$-th family of dyadic fractions in $[0,1]$, that is $D_n = \{k2^{-n}: 0\le k\le 2^n,\ k\in\mathbb Z \}$. So both boundaries of $J_{k,n}$ belong to $D_n$, but not the interior. Sorry if that was unclear! – Generic Human Jun 16 '12 at 15:44