Does a constant 4-acceleration implies constant acceleration? Before asking the question I explain here my line of thoughts:
I'm almost sure that a constant acceleration $\bf{a}$ implies constant 4-acceleration $w^\mu$. So I assume constant acceleration, then, in the rest frame of the particle where its velocity is zero we have $w^\mu=(0,\bf{a})$, thus, if $\bf a$ is constant, the quantity  $w^\mu w_\mu$ will be constant aswell and Lorentz-invariant. If there's a mistake you're free to point it out. 
Anyway, assuming what I've said true I ask the opposite question: does constant four acceleration implies constant acceleration? 
I didn't come up with a satisfying yes, I tried to think in analogy of what I've written above, but I only obtained that just in the comoving reference frame i have constant acceleration, but then in another reference frame where $w^\mu$ is function both of the velocity and the acceleration of the particle, how can I know they don't modify themselves in order to keep the initial assumption of constant 4-acceleration valid?
 A: The norm of the four-acceleration (i.e. the proper acceleration) defined by:
$$ A^2 = g_{\alpha\beta} a^\alpha a^\beta $$
is a Lorentz scalar and therefore the same for all observers. However there is nothing to stop me from choosing some bizarre coordinates in which the four acceleration is time dependent but this is cancelled out by the time dependence in the metric to give a constant proper acceleration. So in this sense a constant proper acceleration does not imply a constant four acceleration or vice versa.
A: No. Suppose a relativistic mass accelerates at constant proper acceleration $a$. Its rapidity after a proper time $\tau$ is $\phi=\phi_0+\frac{a\tau}{c}$. The speed is $v=c\tanh\phi$, so  $\frac{dv}{d\tau}=a\text{sech}^2\phi$ falls over time, as does $\frac{dv}{dt}=\gamma^-1\frac{dv}{d\tau}=a\text{sech}^3\phi$.
A: If 
\begin{align}
\mathbf{v} & =\textrm{velocity 3-vector}
\tag{01a}\\
\mathbf{a} & =\textrm{acceleration 3-vector}
\tag{01b}\\
\mathbf{U} &=\textrm{velocity 4-vector}
\tag{01c}\\
\mathbf{A} &=\textrm{acceleration 4-vector}
\tag{01d}
\end{align}
then
\begin{align}
\mathbf{A} & = \gamma_{v}\left(\gamma_{v}\,\mathbf{a}+\dfrac{\mathrm{d}\gamma_{v}}{\mathrm{d}t}\mathbf{v},\dfrac{\mathrm{d}\gamma_{v}}{\mathrm{d}t}c \right)
\tag{02a}\\
\Vert\mathbf{A}\Vert^{2} & = -\gamma_{v}^{4}\left[\Vert\mathbf{a}\Vert^{2}+\left(\gamma_{v}^{2}-1\right) \left(\dfrac{\mathrm{d}v}{\mathrm{d}t}\right) ^{2}     \right]
\tag{02b}
\end{align}
You have all answers from equations (02).
Note that in the rest frame of the particle $\:(\mathbf{v}_{0}\equiv \boldsymbol{0}, \gamma_{v}=1,\mathbf{a}=\mathbf{a}_{0})\:$ 
\begin{align}
\mathbf{A}_{0} & = \left(\mathbf{a}_{0},0 \right)
\tag{03a}\\
\Vert\mathbf{A}_{0}\Vert^{2} & = -\Vert\mathbf{a}_{0}\Vert^{2}
\tag{03b}
\end{align} 
