How to show that $[L_i, v_j]=i\hbar\sum_k \epsilon_{ijk}v_k$ for any operator $\textbf{v}$ constructed from $\textbf{x}$ and/or $\nabla$? In Weinberg's Lectures on Quantum Mechanics (pg 31), he said that the commutator relation
$$[L_i, v_j]=i\hbar\sum_k \epsilon_{ijk}v_k$$
is true for any vector operator $\textbf{v}$ constructed from $\textbf{x}$ and/or $\nabla$, where $\textbf{L}$ is the angular momentum operator given by $\textbf{L}=-i\hbar\textbf{x} \times \nabla$.
An example for vector $\textbf{v}$ is the angular momentum $\textbf{L}$ itself: $$[L_i,L_j] = i\hbar \sum_k \epsilon_{ijk} L_k.$$
Other examples are $\textbf{v}=\textbf{x}$ and $\textbf{v}=\nabla: $ $$[L_i,x_j] = i\hbar \sum_k \epsilon_{ijk} x_k,$$ $$[L_i,\frac{\partial}{\partial x_j}] = i\hbar \sum_k \epsilon_{ijk} \frac{\partial}{\partial x_k}.$$
How can it be shown that the commutator relation $[L_i, v_j]=i\hbar\sum_k \epsilon_{ijk}v_k$ is indeed true for any vector operator $\textbf{v}$ constructed from $\textbf{x}$ and/or $\nabla$?
 A: Recall
\begin{align}
L_i = -i\hbar \sum_{j, k=1}^3\epsilon_{ijk}x_j\frac{\partial}{\partial x_k} = -i\hbar\epsilon_{ijk}x_j\frac{\partial}{\partial x_k}
\end{align}
then we see that
\begin{align}
[L_i, x_\ell]f = L_ix_\ell f-x_\ell L_if. 
\end{align}
Here, we see that
\begin{align}
L_iv_\ell f =&\ -i\hbar \epsilon_{ijk}x_j\frac{\partial}{\partial x_k}(x_\ell f)\\
=&\ -i\hbar \epsilon_{ijk}x_j\frac{\partial x_\ell}{\partial x_k}f-i\hbar \epsilon_{ijk}x_jv_\ell\frac{\partial f}{\partial x_k}
\end{align}
and
\begin{align}
v_\ell L_i f = -i\hbar \epsilon_{ijk}x_jx_\ell\frac{\partial f}{\partial x_k}.
\end{align}
Finally, we have that
\begin{align}
[L_i, v_\ell]f = -i\hbar \epsilon_{ijk}x_j\frac{\partial x_\ell}{\partial x_k}f
\end{align}
or equivalently
\begin{align}
[L_i, v_\ell]= -i\hbar \epsilon_{ijk}x_j\frac{\partial x_\ell}{\partial x_k}=-i\hbar \epsilon_{ijk}x_j\delta_{\ell, k} = -i\hbar \epsilon_{ij\ell}x_j = i\hbar \epsilon_{i\ell j}x_j.
\end{align}
The proof for $v_i = \partial_i$ is similar.
