Global conservation + Lorentz invariance = local conservation? On the page 83 of "Quantum Field Theory Lectures of Sidney Coleman", Coleman showed an interesting example:

It seems that global conservation law and local conservation law can be related. Can such a relation be made more clear? For example, if I have a global charge conservation law,
$$ \frac{d}{dt} \int J^{0} d^{3}\vec{x} = 0 $$
by considering Lorentz invariance, can I derive the corresponding local charge conservation law
$$ \frac{\partial}{\partial t}J^{0} + \nabla \cdot \vec{J} = 0 $$
and the explicit expression for $ J^{i}  $,$i = 1,2,3$?
 A: One line of reasoning goes as follows:

*

*We assume there exists a notion of a local charge density $\rho({\bf r},t)$.


*By Lorentz symmetry, we assume that there exists a 4-vector current density $J^{\mu}$ such that the 0-component $J^0=\rho$ is the charge density.


*The 4-divergence
$$s~:=~d_{\mu}J^{\mu}\tag{1}$$
is then a Lorentz scalar.


*The global conservation law (=conservation law in integral form) states that the charge $$Q(t)~:=~\int_{\mathbb{R^3}}\! \mathrm{d}^3r~\rho({\bf r},t)\tag{2}$$ is conserved: $$\frac{dQ}{dt}~=~0.\tag{3}$$


*Lorentz symmetry then implies that the Radon transform
$$(Rs)(\Sigma)~:=~ \int_{\Sigma}\! \mathrm{d}^3r~s~=~0\tag{4}$$ vanishes, where $\Sigma$ is an arbitrary space-like affine hyperplane $\subseteq \mathbb{R}^{3,1}$.


*The projection/central/Fourier slice theorem$^1$ then implies that the Fourier transform $\hat{s}$ vanishes for time-like 4-wave-vectors:
$$|{\bf k}|~<~|\omega| \quad\Rightarrow\quad \hat{s}({\bf k},\omega)~=~0. \tag{5}$$


*On the other hand, we assume that the matter $J^{\mu}$ obeys causality, i.e. that the Fourier transform $\hat{J}^{\mu}$ only has support inside the time-like light-cone $|{\bf k}|<|\omega|$.


*Altogether, this implies that $\hat{s}=0$ vanishes identically. By an inverse Fourier transformation, we get the continuity equation in differential form:
$$ s({\bf r},t)~=~0, \tag{6}$$
i.e. the local conservation law. $\Box$
References:

*

*Sidney Coleman, QFT Lectures, p. 83.


*Feynman lectures, vol. II sec. 27-1.
--
$^1$ It is useful to first work out the Radon correspondence in 1+1D where the formulas simplify significantly.
A: After reading Qmechanic's answer, I feel I get an understanding of the spirit of the answer and am writing my understanding, from a -not quite rigorous- point-charge viewpoint.
I shall assume a finite many of point charges. In this picture, to prove local current conservation, it is sufficient to prove that, if Lorentz invariance and global charge conservation is assumed, no isolated charge should ever come into or out of existence, or more generally change its charge by any amount.
The proof goes by contradiction. Assuming the space-time coordinates of the events where any charge changes its charge by any amount to be {$t_i$, $\vec{r}_i$}. Without loss of generality, I shall focus on an isolated event with space-time coordinate $t_0=0$, $\vec{r}_0=(0,0,0)$.
Under a Lorentz transformation characterized by velocity $\vec{v}$, the transformed time $t'=\frac{1}{1-\frac{v^2}{c^2}} (t- \frac{\vec{v} \cdot \vec{r}}{c^2})$. Since there are only a finite many of {$t_i$, $\vec{r}_i$}, and since we assumed an isolated event, all other $t_i$, $\vec{r}_i$ differ from $t_0=0$, $\vec{r}_0=(0,0,0)$ by a finite amount (differing either in space or time or both). It is fairly easy to convince oneself that, there is always a Lorentz frame in which $t'_0$ differs from all other $t'_i$ by an finite amount.
So in this frame at time $t'_0$, there is only one change-of-charge event in the whole of space, which is at $t'_0$ and $\vec{r}'_0$. Thus in this frame the total charge before and after $t'_0$ is not conserved, in contradiction with global charge conservation.
Since there are nothing special about $t_0$ and $\vec{r}_0$, this means no isolated change of charge can ever occur, which as mentioned in the beginning of the answer, is equivalent to local current conservation in the point-charge picuture.
