Remaining Gauge Freedom in Lorenz gauge If one fixes the gauge in Electrodynamics to fulfill the Lorenz gauge $\partial_\mu A^\mu=0$, then the gauge scalar field $\chi$ has to fulfill (eq 3 page 5):
\begin{equation}
\partial^\mu\partial_\mu \chi=-\partial_\mu A^\mu~~~~~~~~~~~~~~(1)
\end{equation}
If the 4 potential transforms as $A^\mu \longrightarrow A^\mu+\partial^\mu \chi$. 
I think I got that point. There exists a solution to (1) for $\chi$, since it is the solution for an inhomogeneous wave equation with well behaved source fields. Is that correct?
I have read that the gauge is not fully fixed by the Lorenz gauge. Does that mean that (1) does not determine $\chi$ completely? If yes what is the fully left freedom? And where does this condition
\begin{equation}
\partial^\mu\partial_\mu \chi=0~~~~~~~~~~~~~~(2)
\end{equation}
comes into play, which can be found here (on page 6). Right now it seems to me that $\chi$ then has to fulfill (1) and (2) which would be a bit weird..?
So my clear question is:


*

*How is it possible that a gauge freedom of the type (2) is still left by demanding (1)?

*And if this should be wrong and one only demands (1) what is the left
gauge freedom then?

 A: Lorenz gauge reads:
$$
\partial_\mu A^\mu = 0.
$$
In slide 5 what they want to say is that in general $A_\mu$ does not satisfy Lorenz gauge fixing, but you can always choose a $\chi$ such that $A'_\mu = A_\mu + \partial_\mu \chi$ does:
$$
0 = \partial_\mu A'^\mu = \partial_\mu A^\mu + \partial_\mu\partial^\mu \chi \qquad (1)
$$
is indeed an equation for $\chi$ given some $A^\mu$.
This condition does not uniquely determine the function $\chi$. Consider some function $\xi$ such that $ \partial_\mu\partial^\mu \xi =0$. A gauge transformation generated by $\xi$ does not change the 4-divergence of $A_\mu$:
$$
\partial^\mu A'_\mu = \partial^\mu (A_\mu + \partial_\mu \xi) = \partial^\mu A_\mu
$$
This is called a 'residual gauge freedom'.
So, if you start with a generic potential $A_\mu$, you can perform a gauge transformation so that it satisfies Lorenz gauge by solving (1) and finding $\chi$; but this is just up to a function $\xi$:
$$
\partial_\mu A'^\mu = \partial_\mu ( A^\mu + \partial^\mu \chi + \partial^\mu \xi) = \partial_\mu ( A^\mu + \partial^\mu \chi ) + \partial^\mu \partial_\mu\xi = \partial^\mu \partial_\mu \xi = 0
$$
In other words if you pick a solution $\chi$ of (1) and you add a function $\xi$, the sum will still solve (1).

Indeed in those slides they are not saying that the gauge transformation function satisfies both equations. They are considering two gauge transformations; the first fixing the divergence as in my (1), the second leaving that equation invariant. 
