Boundary trace quotient and chart norm equivalence

If $u-v\in H_0^1(\mathrm{\Omega })$, then [F2] gives $Tu=Tv$, so $\bar{T}[u]:=Tu$ is well-defined on $\mathcal{D}(\mathrm{\Omega })$. By [F1], $\mathrm{\parallel }\bar{T}[u]{\mathrm{\parallel }}_{W^{1\mathrm{/}2,2}(\mathrm{\partial }\mathrm{\Omega })}\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })}$ for every representative $u$; taking the infimum over its coset gives $\mathrm{\parallel }\bar{T}[u]\mathrm{\parallel }\le C\mathrm{\parallel }[u]{\mathrm{\parallel }}_{\mathcal{D}}$. If $\bar{T}[u]=0$, [F2] gives $u\in H_0^1$, hence $[u]=0$; thus the induced map is injective.

By [F1], every $f\in W^{1\mathrm{/}2,2}(\mathrm{\partial }\mathrm{\Omega })$ is $Tu$ for some $u\in H^1(\mathrm{\Omega })$, so $\bar{T}$ is onto. More quantitatively, choose the lift $Rf$ from [F3]. Then $\bar{T}[Rf]=f$ and $\mathrm{\parallel }[Rf]{\mathrm{\parallel }}_{\mathcal{D}}\le \mathrm{\parallel }Rf{\mathrm{\parallel }}_{H^1}\le C\mathrm{\parallel }f{\mathrm{\parallel }}_{W^{1\mathrm{/}2,2}}$, proving that the inverse is bounded.

The trace of a constant function is that same constant, so $\bar{T}$ maps the closed constant subspace $\mathbb{K}\mathbf{1}\subset \mathcal{D}(\mathrm{\Omega })$ onto the constant subspace of $W^{1\mathrm{/}2,2}(\mathrm{\partial }\mathrm{\Omega })$. The bounded isomorphism and its bounded inverse from steps 1.1–1.2 therefore induce a bounded isomorphism on the two quotient spaces, with equivalent quotient norms.

If $\mathrm{\ell }$ is a continuous linear functional on $W^{1\mathrm{/}2,2}(\mathrm{\partial }\mathrm{\Omega })$ that annihilates constants, then $\dot{\mathrm{\ell }}([f]):=\mathrm{\ell }(f)$ is independent of the representative because $\mathrm{\ell }(f+c)=\mathrm{\ell }(f)$ for every constant $c$. By the quotient-norm definition, $\mathrm{\mid }\dot{\mathrm{\ell }}([f])\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }{\inf }_c\mathrm{\parallel }f+c{\mathrm{\parallel }}_{W^{1\mathrm{/}2,2}}$, so $\dot{\mathrm{\ell }}$ is continuous with quotient dual norm at most $\mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }$. Conversely, any functional that factors as $\mathrm{\ell }=\phi \circ q$ vanishes on constants because $q(c)=0$; and $\mathrm{\mid }\phi (qf)\mathrm{\mid }\le \mathrm{\parallel }\phi \mathrm{\parallel }\mathrm{\parallel }qf\mathrm{\parallel }\le \mathrm{\parallel }\phi \mathrm{\parallel }\mathrm{\parallel }f\mathrm{\parallel }$ because the quotient map $q$ has norm at most one. Hence $\mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }\le \mathrm{\parallel }\phi \mathrm{\parallel }$, while the first inequality gives $\mathrm{\parallel }\phi \mathrm{\parallel }\le \mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }$ for the induced functional. This proves the equivalence and equality of dual norms.