Let $\Omega \subset \mathbb{R}^2$ be an open set and $u \in \mathcal{D}'(\Omega)$ a distribution such that $\partial_1 \partial _2 u, \partial _1 \partial _2 u\in L^2$. Thus we can write for every test function $\varphi \in \mathcal{C}^{\infty}_c (\Omega)$ that :
$$\langle \partial _1 \partial _2 u, \varphi \rangle = \langle u , \partial _1 \partial _2 \varphi \rangle = \langle \partial _2 \partial _1 u,\varphi \rangle$$
Then we have : $$ \int_{\Omega} (\partial _1 \partial _2 u(x)-\partial _2 \partial _1 u(x)) \cdot \varphi (x) dx $$
Using the density of $\varphi \in \mathcal{C}^{\infty}_c (\Omega)$ in $L^2(\mathbb{R}^2)$ we get that : $$\partial _1 \partial _2 u = \partial _2 \partial _1 u$$ (almost everywhere)
My questions :
- Is this reasonement correct ? (I do think so ...)
- It appears that it proves that for every function $f : \mathbb{R}^2 \to \mathbb{R}$ such that both $\partial _1 \partial _2 u$ and $\partial _2 \partial _1 u$ exist then $$\partial _1 \partial _2 u = \partial _2 \partial _1 u$$ This statement is clearly incorrect (there is counter examples), where is the lackin my proof ?
Thanks.
Your proof is correct : for every function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $\partial_1\partial_2 f$ and $\partial_2\partial_1 f$ are in $\mathbb{L}_2$, the equality holds, but as distributions. This means that the set on which the two functions are not equal has Lebesgue mesure 0. For instance, in the canonical counter example (https://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Requirement_of_continuity), the two functions only differ at the $(0,0)$ point.