Multiplier boundary term and hidden normal trace

The Dirichlet realization is self-adjoint on the nonzero complex $L^2(\mathrm{\Omega })$ space. The mathematical flow supplier, used with generator ${\mathrm{\Delta }}_D$, gives the same spectral evolution $v=e^{it{\mathrm{\Delta }}_D}{v}_0$ as in the localized trace theorem. The hypotheses of its self-adjoint calculus hold under the assumed Choice axiom, which also supplies the choice assumptions of the Sobolev and bounded-domain suppliers.

Write $E_A$ for the spectral measure of $A$, set $v_{0,N}=E_A([1,N])v_0$, and put $w_N(t)=e^{it{\mathrm{\Delta }}_D}{v}_{0,N}$. The spectral support of $A$ is in $[1,\mathrm{\infty })$. The realization's form identity gives $\mathrm{\parallel }{v}_0-v_{0,N}{\mathrm{\parallel }}_{H^1}^2={\int }_{(N,\mathrm{\infty })}\lambda d(E_A{)}_{v_0}(\lambda )\to 0$ by dominated convergence. Multiplication by the spectral exponential has modulus one, so $w_N\to v$ in $C([0,T];H^1)$ and $\mathrm{\parallel }{w}_N(t){\mathrm{\parallel }}_{H^1}=\mathrm{\parallel }{v}_{0,N}{\mathrm{\parallel }}_{H^1}$. Each $w_N$ has bounded spectral support and hence lies in every integer power domain of $A$. The flow derivative is ${\mathrm{\partial }}_t{w}_N=i{\mathrm{\Delta }}_D{w}_N$; commuting this identity with $A$ shows differentiability also in its graph norm, since all relevant powers of $A$ are bounded on that spectral subspace. The graph norm equivalence therefore gives $w_N\in C^1([0,T];H^2(\mathrm{\Omega }))$. Each $w_N$ and its time derivative belong to $H_0^1(\mathrm{\Omega })$.

Apply the localized normal-trace theorem to $v$. Its unique trace is the spacetime $L^2$ limit of the normal traces of these spectral approximants, and it gives the displayed estimate. Its compact field $Q=\nu$ has $Q\cdot \nu =1$ on the boundary; the theorem controls its remaining multiplier terms by the energy norm. This proves trace existence and the bound under smooth compact-boundary geometry, independently of a radial sign condition.

Fix an arbitrary real smooth compactly supported field $q$. Choose $R$ so large that both $\mathrm{\Theta }$ and $\mathrm{supp}q$ lie strictly inside $B_R(0)$. Then $D=B_R(0)\setminus \mathrm{\Theta }$ is bounded with smooth boundary, whose inner normal is the exterior normal $\nu$; $q$ vanishes near its artificial outer sphere. For $w=w_N$, the restriction to $D$ is $H^2$. Its trace on $\mathrm{\partial }\mathrm{\Omega }$ is zero: approximate $w\in H_0^1(\mathrm{\Omega })$ in $H^1$ by interior test functions and use local trace continuity. Moreover $T\mathrm{\nabla }w=\nu {\mathrm{\partial }}_{\nu }w$ there. To justify the latter equality, approximate $w{\mathrm{\mid }}_D$ in $H^2(D)$ by ambient smooth restrictions. In a boundary graph chart $x=(y,h(y))$, their boundary derivatives satisfy ${\mathrm{\partial }}_{y_a}(w{\mathrm{\mid }}_{\mathrm{\partial }D})=T({\mathrm{\partial }}_{a}w)+({\mathrm{\partial }}_{a}h)T({\mathrm{\partial }}_{n}w)$ before taking limits. Both boundary restrictions and first-derivative traces converge in local boundary $L^2$ by the trace theorem, so the equality passes to distributions in $y$. The limiting restriction is zero on the inner boundary; thus every tangential component of $T\mathrm{\nabla }w$ vanishes. Resolving the vector into its normal and tangential components gives the asserted equality, with ${\mathrm{\partial }}_{\nu }w=\nu \cdot T\mathrm{\nabla }w$. All normal and product boundary terms below are consequently $L^2$ and $L^1$ traces, respectively.

For this regular solution put $d=\mathrm{div}q$ and $M_q[w](t)=\mathrm{Im}{\int }_D\bar{w}q\cdot \mathrm{\nabla }w$. Differentiate using $w_t=i\mathrm{\Delta }w$, then move $q\cdot \mathrm{\nabla }$ off $\mathrm{\Delta }w$ in the second term. The result is $M_q[w{]}^{\mathrm{\prime }}=-2\mathrm{Re}{\int }_D\mathrm{\Delta }\bar{w}q\cdot \mathrm{\nabla }w-\mathrm{Re}{\int }_{D}d\bar{w}\mathrm{\Delta }w$. Indeed, the differentiated terms are $-\mathrm{Re}{\int }_D\mathrm{\Delta }\bar{w}q\cdot \mathrm{\nabla }w+\mathrm{Re}{\int }_D\bar{w}q\cdot \mathrm{\nabla }\mathrm{\Delta }w$; integration by parts changes the latter to $-\mathrm{Re}{\int }_D(q\cdot \mathrm{\nabla }\bar{w}+d\bar{w})\mathrm{\Delta }w$. Its boundary contribution vanishes because $Tw=0$ on the inner boundary and $q=0$ near the outer sphere. Here $\mathrm{\Delta }w=w_t\mathrm{/}i\in H^2$, so this first integration by parts uses $H^1$ factors.

Define $B_q[w]={\int }_{\mathrm{\partial }\mathrm{\Omega }}(q\cdot \nu )\mathrm{\mid }{\mathrm{\partial }}_{\nu }w{\mathrm{\mid }}^2$, $I_q[w]=\mathrm{Re}{\int }_D{\sum }_{j,k}({\mathrm{\partial }}_j{q}_k)({\mathrm{\partial }}_j\bar{w})({\mathrm{\partial }}_{k}w)$, and $J_q[w]=\mathrm{Re}{\int }_D\bar{w}\mathrm{\nabla }d\cdot \mathrm{\nabla }w$. Integrating the first term in step 4.1 once in each $j$ gives $-2\mathrm{Re}{\int }_D\mathrm{\Delta }\bar{w}q\cdot \mathrm{\nabla }w=-2B_q[w]+2I_q[w]+{\int }_{D}q\cdot \mathrm{\nabla }(\mathrm{\mid }\mathrm{\nabla }w{\mathrm{\mid }}^2)$. In this formula the last integral denotes the sum of the integrable first-derivative products. Applying coordinate Gauss–Green to the $H^1$ factors ${\mathrm{\partial }}_j\bar{w}$ and $q_k{\mathrm{\partial }}_{j}w$, and summing real parts, makes it $B_q[w]-{\int }_{D}d\mathrm{\mid }\mathrm{\nabla }w{\mathrm{\mid }}^2$, since $\mathrm{\mid }T\mathrm{\nabla }w{\mathrm{\mid }}^2=\mathrm{\mid }{\mathrm{\partial }}_{\nu }w{\mathrm{\mid }}^2$ on the inner boundary. The second term in step 4.1 is $-\mathrm{Re}{\int }_{D}d\bar{w}\mathrm{\Delta }w={\int }_{D}d\mathrm{\mid }\mathrm{\nabla }w{\mathrm{\mid }}^2+J_q[w]$, with zero boundary contribution because $Tw=0$ where $q$ or its derivatives meet the boundary. Smooth approximation proves the product expansions for the $H^2$ functions, and trace continuity and the $2,2$ product bound pass their boundary products to the limit. Combining these equalities cancels the two terms with $d\mathrm{\mid }\mathrm{\nabla }w{\mathrm{\mid }}^2$ and proves $M_q[w{]}^{\mathrm{\prime }}=-B_q[w]+2I_q[w]+J_q[w]$ for arbitrary $q$. No equality $q=\nu$ was used.

Integrate the locally proved identity to obtain $\frac{1}{2}{\int }_0^T{B}_q[w_N](t)dt=\frac{1}{2}(M_q[w_N](0)-M_q[w_N](T))+{\int }_0^T{I}_q[w_N](t)dt+\frac{1}{2}{\int }_0^T{J}_q[w_N](t)dt$. Since $q$ vanishes outside $D$, its volume integrals are the corresponding integrals on $\mathrm{\Omega }$. The convergence $w_N\to v$ in $C_t{H}^1$ passes every momentum and volume term to the limit by boundedness of $q,Dq,D^{2}q$ and the $2,2$ product bound. If $g_N={\mathrm{\partial }}_{\nu }{w}_N$ and $g={\mathrm{\partial }}_{\nu }v$, the trace theorem used in step 3.1 gives $g_N\to g$ in boundary spacetime $L^2$ and $\mathrm{\parallel }\mathrm{\mid }{g}_N{\mathrm{\mid }}^2-\mathrm{\mid }g{\mathrm{\mid }}^2{\mathrm{\parallel }}_1\le (\mathrm{\parallel }{g}_N{\mathrm{\parallel }}_2+\mathrm{\parallel }g{\mathrm{\parallel }}_2)\mathrm{\parallel }{g}_N-g{\mathrm{\parallel }}_2\to 0$. Since $q\cdot \nu$ is bounded on the compact boundary, the weighted boundary integrals converge too. This proves the energy-space identity with exactly the boundary term stated.

For the radial boundary expression take $q=\rho x$, with $\rho =1$ near $\mathrm{\Theta }$. On $\mathrm{\partial }\mathrm{\Omega }$ this equals $x$, hence its boundary term is the one of the usual radial multiplier. The cutoff allows the energy identity just proved without assuming that $xv$ is globally square integrable. For a star-shaped obstacle use instead $\rho (x-x_0)$; its boundary weight agrees with the translated radial weight. The supplied geometric calculation gives the nonpositive obstacle sign and, on the bounded truncated geometry with the uncut radial field, the positive enclosing-sphere sign. The compact cutoff identity itself has zero weight on a sphere outside its support. The hidden trace estimate from the second phase used neither radial sign. This establishes the claimed distinction between the radial sign condition and existence of the trace.