Consider the Peetre theorem, as stated here on wikipedia. Isn't the assumption that $D$ not increase supports reduntant? I would think that the fact that $D$ is a morphism of sheaves already implies this.
Let us consider the simplest example, $D$ is an endmorphism of the (vector space valued) sheaf of locally-defined smooth functions $\mathbb{R} \to \mathbb{R}$. Now, there are two ways to interpret the assumption that $D$ does not increase supports, corresponding to the two common defintions of the support of a smooth map $f : U \to \mathbb{R}$:
- Sometimes, the support of $f$ is just $\{x \in U : f(x) \neq 0\}$.
- Sometimes, the support is the closure of the above set.
I think it is safe to assume that we are not using the first definition, since even the ordinary derivative increases supports if they are defined this way. Consider $f(x) = x$ on $\mathbb{R}$. Then, the support of $f$ is $\mathbb{R}\setminus \{0\}$, but the derivative $f'(x)=1$ is supported on the whole line.... so in the sense the support has increased.
OK, so we may assume that when we say $D$ does not increase supports that we are talking about the closed supports. But now, let $f : U \to \mathbb{R}$ be a locally defined smooth function. I claim that $D$ does not increase the support of $f$ simply by virtue of being a sheaf morphism. Indeed, let $x$ belong to $U \setminus \overline{\mathrm{supp}}(f)$. Then, there is a small open set $V \subset U$ with $x \in V$ and $V \cap \overline{\mathrm{supp}}(f) = \varnothing$. Since $D$ is a sheaf morphisms, we have that $$(Df)|_V = D(f|_V) = D( \text{the zero function on }V) = \text{the zero function on }V,$$ by linearity. Thus, $x$ is also not in the support of $Df$, and the support is not increased.