Let $h\in C_c^\infty(\mathbb R^n)$ and consider the equation $$(-\Delta)^s u = h,$$ where $(-\Delta)^s u$ denotes the fractional Laplacian of $u$. Let $\Gamma$ be its fundamental solution, i.e. $(-\Delta)^s \Gamma =\delta$, where $\delta $ stands for the Dirac delta at $0$.
At class (but it is written also in many examples on my textbook) it is written that, under these conditions, setting $\bar u = u-\Gamma\ast h$, follows that $\bar u$ solves the problem $$(\ast)\quad (-\Delta)^s \bar u = 0.$$
Why is that true? I mean, I can understand on my own that $$(-\Delta)^s u = h,\quad (-\Delta)^s (\Gamma\ast h) =\delta\ast h = h,$$ but why is $(\ast)$ true?