How to justify the Coulomb gauge fixing condition choice with $$ A_0=0, \quad \nabla \cdot \vec{A} =0? $$
Below in the text image, I find a text explaining that imposing $A_0=0$ is always possible because I think we set $$ A_\mu \mapsto A_\mu' = A_\mu + \partial_\mu \omega. $$
We can choose $A_0 \mapsto A_0' = A_0 + \partial_0 \omega =0$. This means finding $\omega$ such that $A_0 + \partial_0 \omega =0$.
Then we impose the equation of motion of $A_0$, which seems to be, at least for pure Maxwell theory, to $$ \partial_j(\partial_j A^0 - \partial_0 A^j )=0. $$
Even if $A^0 =0$, we have still $$ \partial_j( \partial_0 A^j )=0. \tag{1} $$
This seems not implying that the Coulomb gauge condition requires $$ \partial_j A^j =\nabla \cdot \vec{A} =0. \tag{2} $$
How to show that (1) can deduce (2)?
p.s. See also a different condition is discussed in https://physics.stackexchange.com/a/33140/310987
