Let $f\in L^1_{loc}(\Omega)$ of some open subset of $\Omega \subset \mathbb{R}^n$. It is easy to prove that if $f$ is $|\alpha|$ times continuously differentiable, $\partial^\alpha f$ is also in $L^1_{loc}(\Omega)$ and defines a distribution which coincides with $\partial^\alpha T_f$. But if $f$ is only assumed to be locally integrable, are its weak derivatives also locally integrable ?
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Usually the fact that assumes $f \in L^{1}_{loc}(\Omega)$ and $|\alpha|$-continuous differentiable is because it justifies the definition of distributional derivative, as a generalization of integration by parts, that is if $u \in \mathcal{D}'(\Omega)$ then $\langle D^\alpha u , \varphi \rangle := (-1)^{|\alpha|} \langle u, D^\alpha \varphi \rangle$ and it occurs that the distributional derivative is still a distribution, basically the operator $D^\alpha : \mathcal{D}'(\Omega) \longrightarrow \mathcal{D}'(\Omega)$ is "closed", where $D^\alpha$ is distributional derivative.
Formally the linear operator $D^\alpha : L^{1}_{loc} ( \Omega) \longrightarrow \mathcal{D}'(\Omega)$ is restriction of the operator $D^\alpha : \mathcal{D}'(\Omega) \longrightarrow \mathcal{D}'(\Omega)$ and in this case we place$D^\alpha f = D^\alpha u_f$ to indicate the weak derivative (also distributional derivative) of a function $f \in L^{1}_{loc}(\Omega)$ and therefore it is characterized by the identity $$ \displaystyle \langle \varphi, D^\alpha f \rangle = (-1)^{|\alpha|} \int_{\Omega} f D^\alpha \varphi dx $$ for all $\varphi \in \mathcal{D}(\Omega)$. This means precisely that $f \in L^1_{loc}(\Omega)$ admits weak derivative $D^\alpha u_f=D^\alpha f$ if there is a function $u_g=g \in L^{1}_{loc}(\Omega)$ such that $\langle \varphi, D^\alpha f \rangle = \langle \varphi, g \rangle$, that is if $$ \displaystyle (-1)^{|\alpha|} \int_{\Omega} f D^\alpha \varphi dx = \int_{\Omega} g \varphi dx \tag{$\star$} $$ for all $\varphi \in \mathcal{D}(\Omega)$. In other words for a function $f$ locally integrable that admits weak derivatives, then by definition also its weak derivatives is locally integrable.
answered Oct 12, 2016 at 18:39
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$\begingroup$ Having re-read the last statement a few times, with the integral, I'm not quite sure where you get that the $g$ function always exists when $f$ is locally integrable. $\endgroup$ Commented Oct 13, 2016 at 12:14
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$\begingroup$ If $f$ is locally integrable and admits weak derivatives then also $D^\alpha u_f$ is locally integrable by definition. In general, if $f$ is locally intragrable not always admits weak derivative. $\endgroup$ Commented Oct 13, 2016 at 16:04
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$\begingroup$ Hey hang on there I apologise I wasn't clear with the definitions ! so if such a $g$ function exists, when you say it is locally integrable by definition it's because if it weren't locally integrable then there would be a compact on which it wasn't integrable in which case one could always find some test function which was not integrable against $g$ on that compact, is that right ? $\endgroup$ Commented Oct 14, 2016 at 6:38
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$\begingroup$ Locally integrable means that $|| f||_{L^1(K)} < \infty$ for all compact sets $K \subset \Omega$. The test functions are always locally integrable, and the functions of $L^1_{loc}(\Omega)$, as distributions, are defined by $u_f(\varphi)=\int_{\Omega} f(x) \varphi(x) dx$ for all $\varphi \in C^{\infty}_c(\Omega)$. In particular ($\star$) is finite. $\endgroup$ Commented Oct 14, 2016 at 9:49