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The definition of weak derivative in the book Partial Differential Equations by Evans is stated as follows:

Suppose $u,v \in L_{loc}^1(U)$, and $\alpha$ is a multiindex. We say that $v$ is the $\alpha$th-weak partial derivative of $u$, written \begin{equation*} D^\alpha u=v, \end{equation*} provided \begin{equation*} \int_U u D^\alpha \phi d x=(-1)^{|\alpha|} \int_U v \phi d x \end{equation*} for all test functions $\phi \in C_c^\infty(U)$.

I have tried to prove that $u$ has the $\beta$th-weak partial derivative for any $\beta \leq \alpha$ if it has the $\alpha$th-weak partial derivative but failed. The reason why I tried to do this is that the text followed in this book seems to assume it holds, though not that obvious for me at the first glance.

PS: Suppose $\alpha=(\alpha_1,\dots,\alpha_n)$, $\beta=(\beta_1,\dots,\beta_n)$ are multiindices. By $\beta \leq \alpha$ we mean $\beta_i \leq \alpha_i$, $i=1,\dots,n$.

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    $\begingroup$ It is not true. Consider functions $u(x,y)$ which are independent of $x$. Then $u_{xy}=0$. But this does not tell you anything about the regularity of $u_y$. $\endgroup$ Commented Mar 17, 2020 at 16:56
  • $\begingroup$ @MichaelRenardy Thank you very much! $\endgroup$
    – Wentao Hu
    Commented Mar 17, 2020 at 17:32
  • $\begingroup$ There seems to be some misunderstanding here—we are not talking about regularity in the classical sense. $u$, as a locally integrable function, is a distribution and so has (distributional) derivatives of all orders. The point of the given definition is not to specify when a derivative exists but when this derivative is not just a distribution but even a (specific) locally integrable function, namely $v$. If $u_{xy}=0$, then $u$ is a sum $u_1(x)+u_2(y)$ (under suitable conditions on the domain) where the two summands are distributions of a single variable.. $\endgroup$
    – user131781
    Commented Mar 17, 2020 at 17:52
  • $\begingroup$ @user131781: And it is possible that $u_2$ is a locally integrable function, but $u_2'$ is not. $\endgroup$ Commented Mar 17, 2020 at 17:58
  • $\begingroup$ Of course (plus 5). $\endgroup$
    – user131781
    Commented Mar 17, 2020 at 18:00

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