Hörmander's proof of: $ u, f \in C(\Omega) $, $ D_j u = f $ in distribution sense $ \implies $ $ D_j u = f $ classically.

Question

I am looking at the following theorem, which can be found in Hörmander's Linear Partial Differential Operators, and I have two questions about the proof.

Here is Hörmander's Theorem 1.2.1, in case you were wondering:

Question 1: Why is it sufficient to prove the theorem for $ u $ having compact support in $ \Omega $?

Here is what I make of the first two lines of the proof. Let's suppose for a moment that Theorem 1.4.2 is already proven for $ u \in C_{0}^{0}(\Omega) $. If we take $ \tilde{u}, f \in C(\Omega) $ and $ \chi \in C_{0}^{\infty}(\Omega) $ and if we assume $ D_j \tilde{u} = f $ in the distributional sense, then $$ D_j(\chi \tilde{u}) = D_j(\chi) \tilde{u} + \chi f $$ in the distributional sense. Since $ D_j(\chi) \tilde{u} + \chi f $ and $ \chi \tilde{u} $ are continuous and the latter has compact support, the theorem implies that $ D_j(\chi \tilde{u}) = D_j(\chi) \tilde{u} + \chi f $ holds also in the classical sense. Now, if we can show that this implies $ \chi (D_j \tilde{u}) = \chi f $ (in classical sense), then it follows that $ D_j \tilde{u} = f $ in the classical sense because the choice of $ \chi $ was arbitrary; and so it's indeed sufficient to prove the theorem for $ u $ continuous of compact support. But how to show that $$ D_j(\chi \tilde{u}) = D_j(\chi) \tilde{u} + \chi f \:\: \text{in classical sense} \implies \chi (D_j \tilde{u}) = \chi f \:\: \text{in classical sense}? $$ Is this what Hörmander had in mind at all?

By the way, I have also come across this proof $u$ continuous and the weak derivative $Du$ continuous $\Rightarrow$ $u \in C^1$?, which seems to get by without first proving the result for $ u \in C_{0}^{0} (\Omega) $ before treating the general case $ u \in C(\Omega) $. It appears to me that, instead of Hörmander's Thm. 1.2.1, the author of the proof has used the (iii) of the following theorem, which may be found in the appendix of L.C. Evans' PDE book.

Answer 1

Regarding Question 1, Hörmander might have had a partition of unity in mind. There are $\chi_k \in C_0^\infty$ such that $\sum_k \chi_k \equiv 1$ on all of $\Omega$. Then $$ D_j u = D_j ((\sum_k \chi_k) u) = \sum_k D_j (\chi_k u), \quad \quad f = (\sum_k \chi_k) f = \sum_k (\chi_k f). $$ Thus, if we can show that $D_j(\chi_k u) = \chi_k f$ in the distributional sense implies the same in the classical sense, then $D_j u = f$ in the distributional sense implies the same in the classical sense. Also, if we can show this generally when $u$ has compact support then it follows that it is valid specifically when $D_j (\chi_k u) = \chi_k f$.

Answer 2

Answer to Question 1:

Let $ \chi \in C_0^\infty (\Omega) $. If $ \tilde{u}, f \in C(\Omega) $, then $ \chi \tilde{u} $ and $ D_j(\chi) \tilde{u} + \chi f $ are continuous functions with compact support. We recall also that the chain rule for distribution derivatives implies that

$$ D_j(\chi \tilde{u}) = D_j(\chi) \tilde{u} + \chi D_j \tilde{u} \:\:\: \text{(in the distributional sense).} $$ Hence, if Hörmander's Thm. 1.4.2 holds true for continuous functions $u$ with compact support and if $ D_j u = f $ in the distributional sense, then $$ D_j(\chi \tilde{u}) = D_j(\chi) \tilde{u} + \chi f \:\:\: \text{(in the classical sense).} \tag{1}$$

If we assume that $ \text{supp} \chi \neq \emptyset $, then If the product of continuous $f$ and compactly supported smooth $\chi$ is $C^1$, then $f$ is $C^1$. implies that on the interior of $ \text{supp} \chi $ we have $$ D_j(\chi \tilde{u}) = D_j(\chi) \tilde{u} + \chi D_j \tilde{u} \:\:\: \text{(in the classical sense).} \tag{2}$$

Combining (1) and (2), we get $ \chi D_j \tilde{u} = \chi f $ (in the classical sense) on $ \text{int} (\text{supp} \: \chi) $. This is true for any compactly supported smooth $ \chi $ with $ \text{supp} \chi \neq \emptyset $ and in particular for $ \chi_{x_0} \in C_0^\infty(\Omega) $ with $\chi(x_0)=1$, $ x_0 \in \Omega $. (Such a function can be constructed for any $ x_0 \in \Omega $ using the standard mollifier). Therefore we may conclude that $ D_j \tilde{u} = f $ on the whole of $ \Omega $.