Cascade algorithm

Remarks

The condition $\sum_n h_n = \sqrt{2}$ keeps the integral fixed: $\int \varphi_{i+1} = \sqrt{2}\sum_n h_n \cdot \frac{1}{2}\int\varphi_i = \int \varphi_i$. The values of $\varphi_i$ at the points $k 2^{-i}$ can be computed by repeated upsampling by $2$ and convolution with $\sqrt{2}\,h$. This is the inverse fast wavelet transform applied to a single coefficient.

Example: for the Haar filter $h_0 = h_1 = 1/\sqrt{2}$ and $\varphi_0 = \mathbf{1}_{[0,1)}$, one step gives $\varphi_1(x) = \mathbf{1}_{[0,1)}(2x) + \mathbf{1}_{[0,1)}(2x - 1) = \mathbf{1}_{[0,1)}(x)$. So the iteration is constant and converges at once to the Haar scaling function.

Non-example: for the filter $h_0 = h_1 = 1$, the sum is $2 \ne \sqrt{2}$. From $\varphi_0 = \mathbf{1}_{[0,1)}$ one gets $\varphi_i = 2^{i/2}\mathbf{1}_{[0,1)}$, which does not converge.

For the filters of the Daubechies wavelets, the iterates converge to the scaling function in $L^2(\R)$. For some filters that satisfy the sum condition, the iteration still fails to converge, so convergence depends on further properties of the filter.