\otimes {\operatorname{id}}+ {\operatorname{id}}\otimes \partial_i) (Y_1 \otimes Y_2)\\
 &= Y_1 \xi_i Y_2 - ({\operatorname{id}}\otimes\tau)(\partial_i Y_1) Y_2 - Y_1 (\tau\otimes{\operatorname{id}})(\partial_i Y_2)\\
 &= \partial_i^\ast(Y_1 \otimes 1) Y_2 - Y_1 (\tau\otimes{\operatorname{id}})(\partial_i Y_2)\\
 &= \partial_i^\ast(Y_1 \otimes 1) Y_2 - Y_1 ({\operatorname{id}}\otimes\tau)(\partial_i Y_2^\ast)^\ast\end{aligned}$$ and thus, by applying the estimates and , that $$\begin{aligned}
\|\partial_i^\ast(Y_1 \otimes Y_2)\|_2
 &\leq \|\partial_i^\ast(Y_1 \otimes 1)\|_2 \|Y_2\| + \|Y_1\| \|({\operatorname{id}}\otimes\tau)(\partial_i Y_2^\ast)\|_2\\
 &\leq 3 \|\xi_i\|_2 \|Y_1\| \|Y_2\|.\end{aligned}$$ This shows the validity of (a). For proving (b), we first follow the idea of “integration by parts” in order to obtain $$\begin{aligned}
({\operatorname{id}}\otimes\tau)((\partial_i Y_1) 1 \otimes Y_2)
 &= ({\operatorname{id}}\otimes\tau)(\partial_i(Y_1 Y_2)) - ({\operatorname{id}}\otimes\tau)(Y_1 \otimes 1 (\partial_i Y_2))\\
 &= ({\operatorname{id}}\otimes\tau)(\partial_i(Y_1 Y_2)) - Y_1 ({\operatorname{id}}\otimes\tau)(\partial_i Y_2)\end{aligned}$$ for arbitrary $Y_1,Y_2\in{\mathbb{C}}\langle X_1,\dots,X_n\rangle$. From this, we can easily deduce by using that $$\begin{aligned}
\|({\operatorname{id}}\otimes\tau)(&(\partial_i Y_1) 1 \otimes Y_2)\|_2\\
 &\leq \|({\operatorname{id}}\otimes\tau)(\partial_i(Y_1 Y_2))\|_2 + \|Y_1\| \|({\operatorname{id}}\otimes\tau)(\partial_i Y_2)\|_2\\
 &\leq 4 \|\xi_i\|_2 \|Y_1\| \|Y_2\|\end{aligned}$$ which is the first inequality. The second inequality can be proven similarly.

We will also need the following easy application of Kaplansky’s density theorem.

\[approximation\] For any $w=w^\ast\in{\operatorname{vN}}(X_1,\dots,X_n)$, there exists a sequence $(w_k)_{k\in{\mathbb{N}}}$ of elements in ${\mathbb{C}}\langle X_1,\dots,X_n\rangle$ such that:

-   $w_k = w_k^\ast$

-   $\displaystyle{\sup_{k\in{\mathbb{N}}} \|w_k\| \leq \|w\|}$

-   $\|w_k-w\|_2 \rightarrow 0$ for $k\rightarrow\infty$

First of all, we note that 