For all $n \in \mathbf{N}$, let $S_{n} f$ be the function of $\mathcal{C}_{0}$ defined by $$S_{n} f = \sum_{j=0}^{n} \sum_{k \in \mathcal{T}_{j}} c_{j,k}(f) \theta_{j,k}$$ where, for all $(j, k) \in \mathcal{I}$, $$c_{j,k}(f) = f\left(\left(k + \frac{1}{2}\right) 2^{-j}\right) - \frac{f(k 2^{-j}) + f((k+1) 2^{-j})}{2} .$$ Let $n \in \mathbf{N}$. Suppose that for all $\ell \in \mathcal{T}_{n}$, $(S_{n-1} f)(\ell 2^{-n}) = f(\ell 2^{-n})$. Show that we also have that for all $\ell \in \mathcal{T}_{n+1}$, $(S_{n} f)(\ell 2^{-n-1}) = f(\ell 2^{-n-1})$. One may distinguish cases according to the parity of $\ell$.
For all $n \in \mathbf{N}$, let $S_{n} f$ be the function of $\mathcal{C}_{0}$ defined by
$$S_{n} f = \sum_{j=0}^{n} \sum_{k \in \mathcal{T}_{j}} c_{j,k}(f) \theta_{j,k}$$
where, for all $(j, k) \in \mathcal{I}$,
$$c_{j,k}(f) = f\left(\left(k + \frac{1}{2}\right) 2^{-j}\right) - \frac{f(k 2^{-j}) + f((k+1) 2^{-j})}{2} .$$
Let $n \in \mathbf{N}$. Suppose that for all $\ell \in \mathcal{T}_{n}$, $(S_{n-1} f)(\ell 2^{-n}) = f(\ell 2^{-n})$. Show that we also have that for all $\ell \in \mathcal{T}_{n+1}$, $(S_{n} f)(\ell 2^{-n-1}) = f(\ell 2^{-n-1})$.
One may distinguish cases according to the parity of $\ell$.