I do not understand a claim from a paper:
Let $b:[0,T] \times \mathbb{R}^n \rightarrow \mathbb{R}$ be a bounded function and let $$b^{n} (t,x) = b(t,x) \ast \psi_n(t) \ast\phi_n(x), $$ where $\psi_n: \mathbb{R} \rightarrow \mathbb{R}$ and $\phi_n:\mathbb{R}^n \rightarrow \mathbb{R}$ are kernels defined in a standard way, i.e. $C^{\infty}$ functions with a compact support, integrated to one, and such that the supports get squeezed to a one-point set. Then it claims that $$\| b^n - b\|_{L^p \big( [0,T] \times B(\mathbf{0}, R) \big)} \rightarrow 0 \quad \text{ as } \quad n \to \infty,$$ for some $R>0$.
My first problem is that by associativity of convolution, if we set $f_{n,1} (t,x) = \psi_n (t)$ and $f_{n,2} (t,x) = \phi_n(x) $, then $$ \begin{eqnarray} (f_{n,1} \ast f_{n,2}) (t,x) & = & \int_{\mathbb{R}^{n+1}} f_{n,1} (t-r,x-y) f_{n,2} (r,y) \,d(r,y) \\ &= & \int_{\mathbb{R}^n} \int_{\mathbb{R}} \psi_n (t-r) \phi_n(y) \,dr \,dy =1. \end{eqnarray} $$ I am not sure if I have misunderstood it. But interpreting the convolution this way, I get that it is equal to the constant function $1$, which is strange.
Secondly, in order to apply the standard result that the convolution of a function with a sequence of kernels converges in $L^p$ to the function itself, we must show that $\{ \psi_n(t) \ast\phi_n(x) \}_{n}$ is indeed a sequence of kernels in $\mathbb{R}^{n+1}$. But is it true and how to show this? Any ideas?