The statement $\sup_{R}||\phi u_R||_{{H^1}}<\infty$ holds as well because $u_R\in \dot{H^1}$ and $\phi$ has compact support so their product is in $H^1$.

For $\phi\in\mathcal D$, the map $\phi\mapsto\int\phi u_R\,dx$ IS a linear operator, but how do we know it's continuous with respect to $\dot{H^1}$? I guess what I mean is, if we approximate $\phi\in \dot{H^1}$ by $\phi_k\in\mathcal D$ in the space $\dot{H^1}$, how does it follow that $\int \phi_k u_R\,dx\to\int\phi u_R\,dx$ as well?

Are you viewing $u_R$ as a linear operator from $\dot{H^1}$ to $\mathbb{R}$ ($\phi\mapsto\int\phi u_R\,dx$) or from $\dot{H^1}$ to $H^1$ ($\phi\mapsto \phi u_R$)?

The space $\dot{H^1}$ is the space of distributions $f$ such that $\int_{\mathbb{R}^3}|\xi|^2|\mathcal{F}f(\xi)|^2\,d\xi<\infty$ (here $\mathcal{F}$ is the Fourier transform). Basically, it's the space of distributions whose gradient is in $L^2$.

Right, right, I see. OK looks good

Sorry, I deleted my previous comment after I noticed your edit - looks like we were out of sync. OK but now isn't the issue that $g_n<1$ in the interior? And possibly $g_n\to 0$ in the interior because we're normalizing by factors which increase..