Assume $u \in L^1(\mathbb{R}^n)$ and $\mathrm{ess\,supp}(u) \subset U,$ where $U$ is a bounded open set. Now we compute the convolution of $u$ with a function $\eta \in C(\mathbb{R}^n)$ with $\mathrm{ess\,supp}(\eta) \subset B(0,h).$ Does this mean then that $\forall x \in \mathrm{ess\,supp}(u * \eta): dist(x,U)\le h$? Somehow this convolution confuses me completely, so I am not sure if this holds, although it sounds natural
By support I mean for an $L^p$ function that $u$ is zero outside $\mathrm{ess\,supp}(u)$ a.e.
Yes. In general, if $f$ and $g$ are two integrable functions defined on $\Bbb R^n$, then $$supp(f*g) \subset supp(f)+supp(g),$$ where the right-hand side is the sum of the two sets: $S+T = \{s+t\colon s\in S,\, t\in T\}$. This can be proved directly from the definition of convolution, and it implies the statement you wrote.