Suppose we have a function $f \in \mathrm{L}^p(\Omega)$ for an open set $\Omega \subset \mathbb{R}^n$, such that the distributional derivatives for all multi-indices $\alpha$ up to degree $m$ are again in $\mathrm{L}^p$.
While going through a calculation, it was used that for a given test function $\varphi \in {D}(\Omega)$, we have $$ \int\limits_{\Omega} \partial^\alpha f \varphi \mathrm{d}x \, = \, (-1)^{|\alpha|}\int\limits_{\Omega} f \partial^\alpha\varphi \mathrm{d}x \:. $$ I do not understand how one can derive this equality. If the function $f$ was continuously differentiable up to degree $\alpha$, I would understand how this works, but here $f$ is only in $\mathrm{L}^p$, so I do not know how I can "transfer" the derivative onto $f$. Thanks in advance, any help is greatly appreciated.
It is simply a matter of meaning of definition, as stated by stange in their comment: saying that the distributional derivatives of $f$ of multiindex order $\alpha$, $|\alpha|\le m\in \Bbb N$ are in $\mathrm{L}_p$ means the there is a finite set of functions $\{g_\alpha\in\mathrm{L}_p\mid |\alpha|\le m\}$ such that $$ (-1)^{|\alpha|}\int\limits_{\Omega} f \partial^\alpha\varphi \mathrm{d}x\, =\, \int\limits_{\Omega} g_\alpha \varphi \mathrm{d}x\quad\forall\alpha\text{ s.t. } |\alpha|\le m\; \text{ and }\; \forall\varphi \in \mathscr{D}(\Omega). $$ Then it is conceptually simple to define $$ \partial^\alpha f\triangleq g_\alpha \quad\forall\alpha\text{ s.t. } |\alpha|\le m $$ In short this simply means that, despite the fact that $f$ could not be differentiable on a dense subset of $\Omega$, it can nevertheless have a generalized $\mathrm{L}_p$-derivative as long as this set of nondifferentiability has zero measure.