Modal trace bound and continuity of the exact-DtN variational form

Each $H^1((a,R);{\mathcal{H}}_{\mathrm{\ell }})$ is complete because ${\mathcal{H}}_{\mathrm{\ell }}$ is finite dimensional. A Cauchy sequence in the weighted square-sum norm is Cauchy in each coordinate; take its coordinatewise $H^1$ limits. For every finite number of coordinates the squared norm of the limit is bounded by the liminf of the sequence norms, so the coordinatewise limit has finite total norm; applying the same argument to differences shows convergence in the sum norm. Tails of the defining convergent nonnegative series tend to zero, so finite-degree truncations are dense. For any scalar or finite-dimensional Hilbert-valued $w\in H^1(a,R)$, set $A=\mathrm{\parallel }w{\mathrm{\parallel }}_{L^2(a,R)}$ and $B=\mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(a,R)}$. Absolute continuity and $\mathrm{\mid }w(R){\mathrm{\mid }}^2-\mathrm{\mid }w(t){\mathrm{\mid }}^2=2\mathrm{Re}{\int }_t^R⟨w^{\mathrm{\prime }}(s),w(s)⟩ds$ give, after averaging $t$ over $[a,R]$ and applying Cauchy–Schwarz, $\mathrm{\parallel }w(R){\mathrm{\parallel }}^2\le (R-a{)}^{-1}{A}^2+2AB$. Multiplying by $1+\mathrm{\ell }$ and using $2(1+\mathrm{\ell })AB\le (1+\mathrm{\ell }{)}^2{A}^2+B^2$, $(1+\mathrm{\ell }{)}^2\le 2(1+\mathrm{\ell }(\mathrm{\ell }+1))$, and $B^2\le a^{-2}{\int }_a^R{r}^2\mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}^{2}dr$ gives $(1+\mathrm{\ell })\mathrm{\parallel }w(R){\mathrm{\parallel }}^2\le C_{a,R}{\int }_a^R(r^2\mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}^2+(1+\mathrm{\ell }(\mathrm{\ell }+1))\mathrm{\parallel }w{\mathrm{\parallel }}^2)dr$. Replacing $r$ by $a+R-r$ proves the same inequality at the inner endpoint. Summing with $w=u_{\mathrm{\ell }}$ proves the weighted trace estimate.

The volume terms of $\mathfrak{a}$ are bounded by a constant times $\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathcal{V}}\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathcal{V}}$ because $r$ lies in $[a,R]$ and $k$ is fixed. By [F2], $\mathrm{\mid }{\tau }_{\mathrm{\ell }}\mathrm{\mid }\le C_{k,R}(1+\mathrm{\ell })$. Cauchy–Schwarz over degrees and the trace estimate from step 1.1 give ${\sum }_{\mathrm{\ell }}\mathrm{\mid }{\tau }_{\mathrm{\ell }}⟨u_{\mathrm{\ell }}(R),v_{\mathrm{\ell }}(R)⟩\mathrm{\mid }\le C{({\sum }_{\mathrm{\ell }}(1+\mathrm{\ell })\mathrm{\parallel }{u}_{\mathrm{\ell }}(R){\mathrm{\parallel }}_2^2)}^{1\mathrm{/}2}{({\sum }_{\mathrm{\ell }}(1+\mathrm{\ell })\mathrm{\parallel }{v}_{\mathrm{\ell }}(R){\mathrm{\parallel }}_2^2)}^{1\mathrm{/}2}\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathcal{V}}\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathcal{V}}\mathrm{.}$ Thus the boundary series is absolutely convergent and the whole form is continuous.