Can $f\in L^2(\Omega)$ imply $\nabla f\in H^{-1}(\Omega):=(H_0^1(\Omega))^*$?

Question

I'm reading a regularity proof in a monograph of PDE. A step of the proof may or may not be implied by the following statement (I don't know if it is true or not):

Suppose $f\in L^2(G_R)$ where $$ G_R=\{x\in\mathbb{R}^n\mid x_n>0,|x|<R\}\quad\textrm{or }\quad G_R=\{x\in\mathbb{R}^n\mid |x|<R\}. $$ Then $\nabla f\in H^{-1}(G_R)$ (component-wise), where $H^{-1}(G_R)$ is defined as the dual of $H_0^1(G_R)$.

Would anybody show me whether this is true or not?

Answer 1

Using Hölder, we deduce that $$\int\nabla f\cdot v\ dx:= \int f\mathrm{div }v\ dx\leq\|f\|_{L^2}\|\mathrm{div }v\|_{L^2} \leq\|f\|_{L^2}\|v\|_{H^1} $$ for $v\in H_0^1(\Omega)$, so $\nabla f\in H^{-1}$.