Let $\Omega$ be a bounded open set in $\mathbb{R}^n$ and $\rho\in C^1(\overline{\Omega})$ . Show that the mapping $v\to\rho v$ defines a continuous linear operator of $H^1(\Omega)$ into itself . If $\rho>0$ on $\overline{\Omega}$ show that this operator is an isomorphism of $H^1(\Omega)$ onto itself .
In other words , it is enough to show $$||\rho v||_{H^1}\leq C||v||_{H^1}$$ for some $C>0$ . We know by standard definitions of norm $$||\rho v||^2_{H^1(\Omega)}=\int_\Omega|\rho v|^2+\int_\Omega|\nabla(\rho v)|^2=||\rho v||_2^2+||\nabla(\rho v)||_2^2$$ Now since $\rho\in C^1(\overline{\Omega})$ , one has $||\rho v||_2^2\leq||\rho||_\infty^2||v||^2_2$ . Now if we somehow able to show that $$\int_\Omega|\nabla(\rho v)|^2\leq||\rho||^2_\infty||\nabla v||_2^2$$ then we are done . But the product rule is not applicable here since we don't have that $v\in H^1(\Omega)\color{blue}{\cap L^\infty(\Omega)}$ . I suspect that the boundedness of $\Omega$ has to do something here (Poincaré 's inequality won't help since that works only in $H_0^1(\Omega)$) . But how will that help I can't understand .
Also some help in the second part is appreciated . Regards .
You can just write $$ \|v\nabla \rho\|_2 \leq \|v\|_2 \|\nabla \rho\|_\infty \\ \|\rho\nabla v\|_2 \leq \|\rho\|_\infty \|\nabla v\|_2. $$ Is it sufficient for you to finish the question?