Can a function $\varphi:\mathbb{R}^d\to \mathbb{R}$ be evaluated at a point in $\mathbb{R}$?

Question

Context: this paper, the equation right at the top of page 7.

$\varphi_x^n$ is defined as $$2^{nd/2}\varphi(2^n(x-y)),$$

and on the first line of the proof of theorem 2.10 (page 6) it says $\varphi:\mathbb{R}^d\to \mathbb{R}$.

However, part of the equation at the top of page 7 reads

$$(\varphi_x^n)\varphi_x^n(y),$$ where $x\in \mathbb{Z}^d \subset \mathbb{R}^d$ and $y\in \mathbb{R}^d$.

How can this work? When we evaluate $\varphi_x^n(y),$ we end up with something that's in $\mathbb{R}$. How can we then feed this into $\varphi_x^n(y)$ again, when the latter's domain is $\mathbb{R}^d$?

Answer 1

If you read correctly, it's not $(\varphi^n_x)\varphi^n_x(y)$.

It's $(\Pi_xf(x))(\varphi^n_x)$ multiplied by $\varphi^n_x(y)$