The whole theorem goes as follows:
Let $(T_1, \ldots, T_N)$ be a tuple of commuting operators in Hilbert space $H$ satisfying:
$$\exists_{M > 0} \ : \ \forall_{p \in \mathbb{C}[z_1, \ldots, z_N]} \ \| p(T_1, \ldots, T_N) \| \le M \| p \|_{\infty}$$
If we have $M=1$ (which is equal to the von Neumann inequality), then the (joint) absolute continuity of $(T_1, \ldots, T_N)$ is equivalent to $\forall_j \ T_j^n \rightarrow 0$ in A-topology, i.e.:
$$\forall_{x,y \in H} \ \lim_{n \rightarrow \infty} \ \sup_{\| p \|_{\infty} \le 1} | \langle p(T_1, \ldots, T_N) T_j^n x, y \rangle | = 0$$
What I mean by "$(T_1, \ldots, T_N)$ being (joint) absolutely continuous" is:
For any $x, y \in H$, we associate a measure $\mu_{x,y}$ with the functional $p \to \langle p(T_1, \ldots, T_N) x, y \rangle$, such that there exists a representing measure $\mu_z$ for $z \in \mathbb{D}^N$ with the property that $\mu_{x,y}$ is absolutely continuous with respect to $\mu_z$.
A representing measure $\mu_z$ for a point $z \in \mathbb{D}^N$ is such a measure, that $u(z) = \int u d\mu_z$ for any $u \in A(\mathbb{D}^N)$ being the polydisk algebra.
Now the question is: Why is the assumption of the von Neumann inequality important?
The proof of this theorem is in this work, but it seems that we first prove that absolute continuity $\Rightarrow$ convergence in A-topology (see "The necessity follows..."), and then convergence in A-topology $\Rightarrow$ absolute continuity (see "For the sufficiency..."). But nowhere is the assumption of the von Neumann inequality used, at least I don't see it. It just seems that the proof solely relies on the absolute continuity of $(T_1, \ldots, T_N)$ and the convergence of $T_j^n$ to zero in A-topology. It gives me the impression that we simply don't need the assumption of the von Neumann inequality in this proof.
Could anyone help me here to show where exactly the assumption of the von Neumann inequality was used in this proof?