Let $\phi: \mathbb{R} \rightarrow \mathbb{R}$ be:
- A smooth non-negative function compactly supported on $(-1, 1)$
- Identically equal to 1 on $(-\frac12, \frac12)$
- $\sum_{n \in \mathbb{Z}} \phi^2(x - n) = 1.$
I would like to extend this function so that the domain is now on $\mathbb{R}^n$. Is there a standard way to do this? One construction I came across is the following for a finite domain $\Omega \in \mathbb{R}^n$ which resembles a sort of tensor product.
For simplicity take $\Omega$ to be a hypercube. Partition the cube into equally spaced points $\{y\}$ with spacing $\Delta$. Then define for each $y$, $$\phi_y(x) = \prod_{i=1}^n \phi\left(\frac{x_i - y_i}{1.5\Delta}\right)$$ where $x = \{x_i\}$ and similarly for $y$ ($y_i$ is the $i$th component of $y$). With the choice of $1.5 \cdot \Delta$, the product of $\phi_y$ and $\phi_{y'}$ is nonzero only if $y, y'$ are adjacent points on the grid.
Is this enough to conclude that $$\sum_y \phi_y^2(x) = 1$$ for all $x \in \Omega$?
