Why is a Harmonic distribution a function? I want to know how one proves: 
``If for a distribution $T$ it holds that $\Delta T$=0, then $T$ is an Harmonic function (see Page in the book of Donoghue with the Theorem)''. 
In the proof in this book I do not understand the last sentence, can someone explain this: ``It follows that $S$ is a $C^\infty$-function in the ball...'' 
(How does it follow from the convolution $E_1 * \Delta S =0$ in that ball that $S$ is a $C^\infty$ function in that ball?) 
With a distribution in this case is meant a linear functional on $C^\infty(\mathbb{R}^d)$ (e.g., not a tempered distribution). 
 A: Since differential operators are "local", even on distributions, we can multiply a given harmonic distribution by a smooth cut-off function so that it is compactly supported, hence tempered, and satisfies $\Delta u=f$ with $f$ another compactly-supported distribution, identically $0$ on a large ball. Then everything in sight is a tempered distribution, so Fourier transforms can be used...
A: Short answer: As is commented by Vobo, actually there is no problem (I just overlooked it). The claim follows from the equality $S= E_1 * \Delta S + E_2 * \Delta S$. 
Longer comment on the answer: So basically there are multiple points of view on the statement. One can prove that a Harmonic distribution is a Harmonic function without any knowledge of Tempered distributions or Fourier transforms. However, with this knowledge - as is commented by Paul Garrett - one can prove a little bit more. In case $T$ itself is tempered one can actually prove that $T$ is a polynomial (see e.g. Show that the only tempered distributions which are harmonic are the the harmonic polynomials). 
Indeed, one proves that the support of $\mathcal F(T)$ (the Fourier transform of $T$) is $\{0\}$, which then implies that $\mathcal F(T)= \sum_{\alpha : |\alpha| \le k} c_\alpha \partial^\alpha \delta_0$. Applying the Fourier inverse we find that 
$T= \sum_{\alpha :|\alpha|\le k} d_\alpha x^\alpha $ 
(as $\mathcal F \mathbb 1 = \delta_0$, and $\mathcal F x^\alpha = (-1)^\alpha \delta^\alpha$), where $d_\alpha = (-1)^\alpha c_\alpha$. 
