Smoothness in Haar basis The rate of decay of fourier co-efficients of a function $f$ determines the order of differentiability of $f$. Is there an equivalent result for the case when Fourier basis is replaced by Haar wavelet basis?
 A: Consider $V_{0},$ the space of piecewise constant functions with knots at the integers. Then if we define $V_{j}=\{f\in L_{2}(\mathbb{R}^{n}):f(2^{j}x)\in V_{0}\}$ for all $j\in\mathbb{Z},$ the orthogonal complement of $V_{j}$ in $V_{j-1}$ is $W_{j-1}=\{\psi_{j,k}:k\in\mathbb{Z}\},$ where $\psi_{j,k}=2^{j/2}\psi(2^{j}x-k),$ and $\psi(x)$ is the Haar mother wavelet, equal to 1 on $(0,1/2],$ and equal to $-1$ on $(1/2,1].$
Then if $f\in V_{0},$ the projection of $f$ on $W_{j-1}$, for all $j\leq0$, is simply $0,$ since $\psi_{j,k}$ is supported on the interval $(k/2^{j},(k+1)/2^{j}],$ which always lies inside some interval $(k',k'+1],$ and this means that $\langle f,\psi_{j,k}\rangle=0$ for all $k.$ In other words, $$f\in V_{0}\text{ has the most rapid decay of its Haar wavelet coefficients possible.}$$
But if we take any nonzero $f\in V_{0},$ it must be discontinuous. Note that we may write $f(x)=\sum_{k\in\mathbb{Z}}c_{k}\phi(x-k),$ where $\phi(x)=1$ on $(0,1]$ and is zero elsewhere. Then if $\lim_{x\rightarrow 1^{+}}f(k+x)=\lim_{x\rightarrow 1^{-}}f(k+x),$ this forces $c_{k+1}=c_{k},$ and thus if $f$ is continuous on $\mathbb{R},$ all of the $c_{k}$ must be equal, which forces $f$ to be zero, since it is square-integrable and constant.
