electromagnetism - Question about physical degree of freedom in Maxwell Theory: Why Coulomb gauge can fix all redundant degree of freedom

Given $4$-potential $A^\mu(x)=(\phi(x),\mathbf{A}(x))$, the vacuum Maxwell equations: $$\nabla^2\phi+\frac{\partial}{\partial t}(\nabla\cdot \mathbf{A} )=0$$ $$\nabla^2 \mathbf{A} -\frac{\partial^2 \mathbf{A} }{\partial t^2} - \nabla(\nabla\cdot\mathbf{A} + \frac{\partial \phi}{\partial t})=0$$

There is redundant degree of freedom(d.o.f) in $A^\mu(x)$: $$A^\mu(x)\rightarrow A^\mu(x) +\partial^\mu \lambda(x) $$

In Coulomb gauge: $$\nabla\cdot \mathbf{A}(x)=0 \tag{1}$$ Vacuum Maxweel equation becomes: $$\nabla^2\phi =0 \tag{2}$$ $$\nabla^2 \mathbf{A} -\frac{\partial^2 \mathbf{A} }{\partial t^2} = \nabla( \frac{\partial \phi}{\partial t}) \tag{3}$$

Then we can always choose $\phi(x)=0$. So question becomes that physical degree of freedom is $A^\mu(x)= (0,\mathbf{A}(x))$ with one constraint $\nabla\cdot \mathbf{A}(x)=0$. Every textbook then says the physical degree of freedom is $2$. But it seems there are still redundant d.o.f , we can always make $$\mathbf{A}(\mathbf{x},t)\rightarrow \mathbf{A}(\mathbf{x},t)+\nabla \Lambda(\mathbf{x})$$ such that $$\nabla^2 \Lambda(\mathbf{x}) =0\tag{4}$$ But above equation is Laplace equation that has nontrivial solutions, harmonic function. For example, $\Lambda(\mathbf{x}) = xyz $.

My questions

1. 
   

   Using $\phi(x) =0 $ and $\nabla\cdot \mathbf{A}(\mathbf{x},t)=0$, I have substract two redundant d.o.f. , why fixing $\Lambda(\mathbf{x})$ further can not substract more redundant d.o.f. ?

   
   



2. 
   

   Many textbooks will argue that $A^\mu(x)$ should vanish at spacial infinity, so Laplace equation $(4)$ with zero boundary condition in infinity has only trivial solution. But why do we have to require $A^{\mu}$ vanish at spacial infinity? For example, a uniform magnetic field has $\mathbf{A}(x) = \mathbf{B}\times \mathbf{r}/2$ which does not vanish in infinity. If you require that $A^\mu$ vanish at spacial infinity, you even cannot get constant electrical or magnetic field solutions from Vacuum Maxwell equations. And even the elctromagnetic wave solution $e^{i k(t -x)}$ is also nonvanishing at spacial infinity. This question has some relation with Coulomb gauge fixing and “normalizability”

   
   


3. 
   

   Why textbook says "Lorentz gauge is Lorentz invariant but cannot fix all redundant d.o.f. Coulomb gauge can fix all redundant d.o.f but is not Lorentz invariant. "? But it's obvious that only Coulomb gauge $(1)$ also cannot fix all redundant d.o.f. We can see that the gauge fixing $\phi =0 $ is not a consequence of $(2)$. It's forced artificially and $\phi=0$ is independent from Coulomb gauge. For example, vacuum Maxwell equations $(2),(3)$ can have uniform electrical field solution $\phi(x)=-\mathbf{E}\cdot \mathbf{r}$, $\mathbf{A}=0$ satisfying only Coulomb gauge $(1)$ but $\phi(x)\neq 0$. If we requires $\phi=0$ and $\nabla \cdot \mathbf{A}=0$, the solution becomes $\phi=0$ and $\mathbf{A}= \mathbf{E} t$.