The proof of commutativity of weak partial derivatives is given in Evans PDE textbook but I really don't follow any of it. If someone could walk me through it I would be extremely grateful. Sorry that I can't comment with what I've tried as I really haven't got very far. Thanks
Assume $u,v \in W^{k,p}(U)$, $|\alpha|\leq k$. Then $D^\alpha u \in W^{k-|\alpha|,p}(U)$ and $D^\beta (D^\alpha u)=D^\alpha (D^\beta u)=D^{\alpha+\beta} u$ for all multiindices $\alpha, \beta$ with $|\alpha|+|\beta|\leq k$
Proof
Fix $\phi \in C^{\infty}_c(U)$. Then $D^\beta \phi \in C^{\infty}_c(U)$ and so \begin{align} \int_U D^\alpha u D^\beta \phi\;dx&=(-1)^{|\alpha|}\int_U uD^{\alpha+\beta}\phi\;dx \\ &=(-1)^{|\alpha|}(-1)^{|\alpha+\beta|}\int_U D^{\alpha+\beta}u\phi\;dx \\ &=(-1)^{|\beta|}\int_U D^{\alpha+\beta}u\phi\;dx \end{align}
Many thanks.
First of all, Evans' primary goal here is not to prove that weak derivatives commute, but rather, to prove that there exists a $\beta$th weak derivative for $D^\alpha u$, namely: $$ D^\beta(D^\alpha u) = D^{\alpha + \beta} u.$$ [If $|\alpha| + |\beta| < k$, then the $\alpha$th and $(\alpha + \beta)$th weak derivatives of $u$ exist by assumption, and they are denoted by $D^\alpha u$ and $D^{\alpha + \beta}u$ respectively; our task is to prove that $D^\alpha u$ has a $\beta$th weak derivative, equal to $D^{\alpha + \beta} u$.]
However, once Evans has proved this, the commutativity of weak derivatives follows immediately, from the symmetry between $\beta$ and $\alpha$.
So what does it mean to say that the $\beta$th weak derivative of a locally summable function $v$ (say) is equal to a locally summable function $w$ (say)? It means that, for all test functions $\phi \in C_c^\infty(U)$, $$ \int_U vD^\beta \phi = (-1)^{|\beta|} \int_Uw \phi. $$ In our case, $v=D^\alpha u$ and $w = D^{\alpha + \beta} u$, so our task is to prove that $$ \int_U D^\alpha u D^\beta \phi = (-1)^{|\beta|} \int_U D^{\alpha + \beta}u \phi \ \ \ \ (\ast)$$ for all $\phi \in C_c^\infty(U)$.
To prove this, first recall that $D^{\alpha + \beta}u$ is by definition the $(\alpha + \beta)$th weak derivative of $u$. Therefore, for all $\phi \in C_c^\infty(U)$: $$ \int_U D^{\alpha + \beta} u \phi = (-1)^{|\alpha| + |\beta|} \int_U u D^{\alpha + \beta} \phi$$ Also, $D^{\alpha + \beta} \phi$ is equal to $D^\alpha ( D^\beta \phi)$. (Remember, these are genuine, not weak, derivatives of $\phi$!). So: $$ \int_U u D^{\alpha + \beta} \phi = \int_U u D^{\alpha } ( D^\beta \phi) $$ Next, observe that $D^\beta \phi$ is in $C_c^\infty(U)$! So applying the statement that $D^\alpha u$ is the $\alpha$th weak derivative of $u$, with $D^\beta\phi$ as the "test function" instead of $\phi$, we learn that $$ \int_U u D^{\alpha}( D^\beta \phi ) = (-1)^{|\alpha|} \int_U D^\alpha u D^\beta \phi $$ Combining the two equations above gives the desired result $(\ast)$.
Finally, we can say a little more about $D^\alpha u$. Its $\beta$th weak derivative, being equal to $D^{\alpha + \beta}u$, is in $L^p(U)$. This is true for any $\beta$ such that $|\beta | < k - |\alpha|$. Hence $D^\alpha u$ is in $W^{k - |\alpha|, p}(U)$.