If $f\in L^2(\Omega)$, where $\Omega$ is a domain in $\mathbb{R}^n$, why is it that the distributional of $f$, say with respect to $x_1$, is in $H^{-1}(\Omega)$, the dual space of $H^1_{0}(\Omega)$?
I tried to make sense of it by writing out the definition of distributional derivative, but I could not see why $\partial x_1 f $ is in $H^{-1}$
For $\varphi \in C^{\infty}_c$ one defines $\langle\partial_{x_1}f,\varphi\rangle = -\int_{\Omega}f\partial_{x_1}\varphi$. Extending this definition to $H^1_0(\Omega)$ we obtain a linear functional on $H^1_0(\Omega)$. Continuity, i.e. boundedness, follows from Holder's inequality (together with Poincare's inequality, depending on the norm you define on $H^1_0$): $$\Big|\int f\partial_{x_1}\varphi\Big|\le \|f\|_{L^2}\|\nabla\varphi\|_{L^2} \Big(\le C\|f\|_{L^2}\|\varphi\|_{H^1_0}\Big).$$