If a vector function $\bf A$ is harmonic (that is $\nabla^2 {\bf A} = 0$) on a region $V$ then in Cartesian coordinates each component of $\bf A$ is also harmonic on $V$. That is $\nabla^2 A_i = 0$ where $i=x,y$ or $z$. It then follows from the extremum principle for harmonic functions that no Cartesian component of $\bf A$ can have a true minimum or maximum in $V$.
If on the other hand $\bf A$ is written in spherical coordinates as ${\bf A} = A_r {\hat{\bf r}} + A_{\theta} {\hat{\bf \theta}} + A_{\phi} {\hat{\bf \phi}}$ then it can shown that $\nabla^2 A_i \ne 0$ where $i=r,\theta$ or $\phi$. In fact the following equations would be obeyed:
0 = $\nabla^2 A_r - \frac{2}{r^2}A_r - \frac{2}{r^2} \frac{\partial A_{\theta}} {\partial \theta} - \frac{2\cos \theta}{r^2 \sin \theta} A_{\theta} - \frac{2}{r^2 \sin \theta} \frac{\partial A_{\phi}} {\partial \phi}$
0 = $\nabla^2 A_{\theta} - \frac{1}{r^2 \sin^2 \theta} A_{\theta} + \frac{2}{r^2} \frac{\partial A_r} {\partial \theta} - \frac{2 \cos \theta}{r^2 \sin^2 \theta} \frac{\partial A_{\phi}} {\partial \phi}$
0 = $\nabla^2 A_{\phi} - \frac{1}{r^2 \sin^2 \theta} A_{\phi} + \frac{2}{r^2 \sin \theta} \frac{\partial A_r} {\partial \phi} + \frac{2 \cos \theta}{r^2 \sin^2 \theta} \frac{\partial A_{\theta}} {\partial \phi}$
Does this mean that $A_r$, $A_{\theta}$ or $A_{\phi}$ can have extrema in $V$? That doesn't seem right. What am I missing?