Proof. Expand (9) at infinity after factoring \(at\) from the denominator. The constant term is one; the coefficient of \(t^{-k}\) is a polynomial of degree \(k\) with leading term \(a^{-k}z^k\).

For the joint continuation, put \[Q(w,t)=\frac{G(t)-G(w)}{t-w}.\] Write \(u=t^{-1}\), \(v=w^{-1}\), and \(G(t)=at+g(u)\), with \(g\) holomorphic for \(|u|<r^{-1}\). Then \[Q=a-uv\frac{g(u)-g(v)}{u-v}.\] The divided difference is jointly holomorphic, with its derivative value on the diagonal. Off the finite diagonal, \(Q\ne0\) by univalence of \(G\); on that diagonal \(Q=G'(t)\ne0\); on either reciprocal axis, including their intersection, \(Q=a\ne0\). Consequently \[B(w,t)=t\frac{\partial_tQ(w,t)}{Q(w,t)} =-u\frac{\partial_uQ}{Q}\] is holomorphic on the reciprocal bidisk and vanishes on both axes. Its Taylor series has only the strictly mixed powers in (10). Cauchy estimates on the closed reciprocal bidisk of radius \(\rho^{-1}\) give \(|\beta_{ij}|\le C_\rho\rho^{i+j}\) for every \(r<\rho<1\). For a compact subset of the product collar, choose \(\rho\) above \(r\) but smaller than one and than the moduli of both coordinates on that compact set. The estimate then dominates the double series by a product of convergent geometric series. This proves normal convergence throughout the product collar, including at infinity, and absolute uniform convergence on \(\mathbb T^2\). Comparing coefficients at \(t=\infty\) in the defining identity for \(B\), first with fixed finite \(w\) in the collar, gives the displayed boundary formula for \(b_j\circ G\).

For distinct \(w,t\in\mathbb T\), the circle identity \(2\operatorname{Re}(t/(t-w))=1\) gives \[m(w,t)=2\operatorname{Re}\frac{q_G(t)}{G(t)-G(w)}\ge0.\] Indeed \(q_G(t)\) is the outward normal, so convexity gives \(\operatorname{Re}(\overline{q_G(t)}(G(t)-G(w)))\ge0\). Continuity of the mixed-series expression supplies the same sign on the diagonal. Every term in \(B\) and its conjugate has nonzero frequency in each variable, so integration in either variable leaves exactly one.

For the matrix resolvent, every eigenvalue of \(A\) lies in \(W(A)\) by testing a unit eigenvector. Thus \(G(t)I-A\) is invertible on and outside the circle. Compactness of the circle permits a smaller exterior collar on which the resolvent remains holomorphic, including infinity where \(K_A(\infty)=I\). Expanding for large \(|t|\) gives the coefficients \(b_k(A)\) by the same polynomial identities as (9). Cauchy estimates in the reciprocal disk give geometric decay and absolute uniform convergence on \(\mathbb T\).

Finally put \(R_t=G(t)I-A\). Direct congruence gives \[R_t^*(K_A+K_A^*)R_t=q_G(t)R_t^*+\overline{q_G(t)}R_t.\] At a unit vector \(x\), its quadratic form is \(2\operatorname{Re}(\overline{q_G(t)}(G(t)-x^*Ax))\ge0\), by the supporting half-plane and \(W(A)\subset\Omega\). Invertibility of \(R_t\) proves \(E_A\succeq0\). ◻

Proof. Positivity and the two marginal identities give the contraction locally: Jensen’s inequality at each \(w\) and then integration imply \[\|(\mathsf Mu)(w)\|_{\mathrm{HS}}^2 \le\int_\mathbb Tm(w,t)\|u(t)\|_{\mathrm{HS}}^2\,d\sigma(t), \qquad \|\mathsf Mu\|_2\le\|u\|_2.\] The mixed expansion in (10) gives, for \(j>0\), \[\mathsf M(t^j)(w)=\sum_{i\ge1}\beta_{ij}w^{-i},\qquad \mathsf M(t^{-j})(w)=\sum_{i\ge1}\overline{\beta_{ij}}w^i.\] Both marginal identities say that \(\mathsf M\) fixes constants and preserves means. The real kernel makes it commute with pointwise matrix adjoint. Its actual adjoint has kernel \(m(t,w)\), and hence, for \(i>0\), \[\mathsf M^\dagger(t^{-i})(w)=\sum_{j\ge1}\overline{\beta_{ij}}w^j, \qquad \mathsf M^\dagger(t^i)(w)=\sum_{j\ge1}\beta_{ij}w^{-j}.\] It is a contraction, fixes constants, preserves means, and commutes with pointwise adjoint. No symmetry of \(m\) is used. By boundedness and density of Fourier polynomials, the displayed actions extend to the closed strictly positive and strictly negative Fourier subspaces.

Equations (10) and (12) give \(\mathcal Cu=u_0+u_++\mathsf Mu_+\) for finite Laurent data. The three terms are orthogonal, so \(\|\mathcal Cu\|_2^2\le2\|u\|_2^2\). Density defines the bounded extension, and the same formula gives \(\mathcal C^2=\mathcal C\). Taking its adjoint and using the strict Fourier exchange proves (13); conjugation by the antiunitary map \(u\mapsto u^*\) gives the second formula. The weighted estimate follows from \[\|\mathcal C^\dagger u\|_2^2+\|(\mathcal C^\dagger u)_0\|_2^2 =2\|u_0\|_2^2+\|u_++\mathsf M^\dagger u_-\|_2^2 \le2\|u\|_2^2.\]

We need the physical interpretation also for analytic boundary traces. If \(v=v_\Omega\circ G\) and \(v_\Omega\) is holomorphic near \(\overline\Omega\), then \(v\) is holomorphic on an annulus about \(\mathbb T\). On a smaller closed annulus its Laurent series converges absolutely uniformly. If \(r<|w|<1\), then \(G(w)\in\Omega\): an exterior or boundary value would also have a preimage with modulus at least one, contradicting univalence on the collar. For \(|w|<1\) sufficiently close to one, expansion of \(t/(t-w)+B(w,t)\) in the Cauchy integral gives \[c_v(G(w))=\sum_{j\ge0}v_jw^j+ \sum_{i,j\ge1}\beta_{ij}w^{-i}v_j.\] The series converge uniformly after shrinking the annulus again, by the geometric Laurent bounds for \(v_j\) and the normally convergent mixed expansion. At the boundary the expression is \(\mathcal Cv\). Cauchy’s formula gives \(c_v=v_\Omega\) in \(\Omega\), and thus \(\mathcal Cv=v\) on \(\mathbb T\). For Laurent data \(u\), both \(F_\Omega c_u\) and \(c_uF_\Omega\) are holomorphic near \(\overline\Omega\) by (12). Their traces are fixed by \(\mathcal C\), giving both identities in (15) for finite Laurent \(u\). Left and right multiplication by the bounded \(F\) are bounded on \(\mathcal H_k\); Laurent approximation extends the two identities to arbitrary \(u\in\mathcal H_k\).

Lastly, integrating (11) against a finite Laurent polynomial gives \[\int_\mathbb TK_A\otimes u\,d\sigma =\sum_{j\ge0}b_j(A)\otimes U_j=c_u[A]\] by (12) and the prescribed base-first evaluation. Applying the same formula to \(u^*\) and taking the full matrix adjoint proves (16). The constant case gives mass \(2I_n\). ◻

Proof. Put \(g=\mathcal C^\dagger r\), \(P_1=\mathcal C^\dagger p\), \(P_2=\mathcal C^\dagger q\). On \(\mathcal H_k\), the adjoint of left multiplication by \(F\) is left multiplication by \(F^*\); the adjoint of right multiplication is right multiplication by \(F^*\), because \[\mathop{\mathrm{tr}}((uF)^*v)=\mathop{\mathrm{tr}}(F^*u^*v)=\mathop{\mathrm{tr}}(u^*vF^*).\] Taking the Hilbert adjoints of both identities in (15) therefore gives, with the factors on their original sides, \[ P_1=\mathcal C^\dagger(F^*g),\qquad P_2=\mathcal C^\dagger(gF^*). \tag{27}\] If \(g=0\), the first identity makes \(P_1=0\), contradicting mean preservation and \(\mathop{\mathrm{tr}}(p_0)>0\). Thus \(g\ne0\).

Each \(P_j\) has only nonnegative Fourier powers, and its mean is Hermitian. Fourier orthogonality gives \[\|P_j+P_j^*\|_2^2 =4\|(P_j)_0\|_2^2+2\|(P_j)_+\|_2^2 =2\bigl(\|P_j\|_2^2+\|(P_j)_0\|_2^2\bigr).\] Since \(\|F\|\le1\), both \(F^*g\) and \(gF^*\) have Hilbert–Schmidt \(L^2\) norm at most \(\|g\|_2\). Apply (14) separately to the two inputs in (27). Using the star relation in (13) and \(p=p^*\), \(q=q^*\), we obtain \[ \|p_e\|_2^2\le4\|g\|_2^2,\qquad \|q_e\|_2^2\le4\|g\|_2^2. \tag{28}\]

Put \(h=\widetilde\mathcal C^\dagger r\). Mean preservation and the whole matrix equality \(r_0=0\) put \(g\) in the strictly positive Fourier subspace and \(h\) in the strictly negative one. Therefore \[ \|r_e\|_2^2=\kappa^2\|g\|_2^2+\kappa^{-2}\|h\|_2^2 \ge\kappa^2\|g\|_2^2. \tag{29}\] Finally, set \(J=\mathop{\mathrm{diag}}(I_k,-I_k)\). Positivity of both \(\Delta\) and \(J\Delta J\) gives \[0\le\mathop{\mathrm{tr}}\bigl(\Delta^{1/2}(J\Delta J)\Delta^{1/2}\bigr) =\mathop{\mathrm{tr}}(\Delta J\Delta J) =\|p_e\|_{\mathrm{HS}}^2+\|q_e\|_{\mathrm{HS}}^2-2\|r_e\|_{\mathrm{HS}}^2.\] The entries are in \(L^2\), so this inequality is integrable. Combining it with (28) and (29) yields \[2\kappa^2\|g\|_2^2\le2\|r_e\|_2^2 \le\|p_e\|_2^2+\|q_e\|_2^2\le8\|g\|_2^2.\] Since \(g\ne0\), division proves \(\kappa\le2\). ◻

Completion of Theorems 2 and 3. If \(\kappa=1\), the required bound already holds. Otherwise Proposition 5 and Section 4 supply all the hypotheses of Lemma 9 at auxiliary size \(k=n\): (20) gives both multiplication orders, (22) gives the zero matrix mean and positive trace, (24) supplies positivity, and (26) identifies its entries. Hence \(\kappa\le2\). Proposition 4 already used this same metric and density for every independently chosen coefficient size \(m\), which proves all statements of the two theorems. ◻

Proof. The distance function \(d_K\) is convex and \(1\)-Lipschitz. For convexity, choose nearest points in \(K\) to two given points, use their convex combination as a competitor, and apply the triangle inequality. Let \(Z\) be a standard Gaussian in \(\mathbb R^2\) and put \[h_\delta(x)=\mathbb E\,d_K(x-\delta Z)+\delta|x|^2, \qquad c=\mathbb E|Z|.\] The Lipschitz bound gives the uniform estimate \[\big|\mathbb E\,d_K(x-\delta Z)-d_K(x)\big|\le c\delta.\] The convolution is real analytic: write it against the translated Gaussian density, complexify both coordinates of \(x\), and use Gaussian domination on every compact set of complex coordinates. The factor \(d_K(y)\) has at most linear growth, so the integral defines a holomorphic function of those coordinates. Convexity passes through convolution; therefore \(h_\delta\) is proper and strictly convex, with Hessian at least \(2\delta I\).

Put \(R_K=\max_{x\in K}|x|\), and choose \(\delta>0\) with \(c\delta+\delta R_K^2<\epsilon/2\) and \(c\delta<\epsilon/2\). Then \(h_\delta<\epsilon/2\) on \(K\), while \(h_\delta\le\epsilon/2\) implies \(d_K\le h_\delta+c\delta<\epsilon\). Thus \(\Omega_\epsilon=\{h_\delta<\epsilon/2\}\) has the desired containments and is bounded, open, and convex. The chosen level is above the minimum. A critical point of a differentiable convex function is a global minimum, so the gradient is nonzero on this level. The implicit function theorem gives a regular real-analytic boundary. From an interior point, each ray meets this compact boundary exactly once. Radial projection is a continuous bijection onto the circle and hence a homeomorphism. The boundary is therefore a Jordan curve. ◻

Proof of Corollary 10. For completeness, the Toeplitz–Hausdorff convexity theorem [26, 13] follows here from two-dimensional compression. Compress to the span of unit vectors realizing any two chosen values. In dimension two, the rank-one orthogonal projections are \[\frac12\begin{pmatrix}1+u&v+iw\\v-iw&1-u\end{pmatrix}, \qquad u^2+v^2+w^2=1,\quad u,v,w\in\mathbb R.\] Taking their trace against the compression is a real-affine map into \(\mathbb C\simeq\mathbb R^2\). Its linear part has a nonzero kernel. Each point of the unit ball can be moved in a kernel direction to the sphere without changing its image, so the images of the sphere and ball coincide. That image is convex. The one-dimensional compression case is immediate. Compactness of \(W(A)\) follows from compactness of the unit sphere.

Apply Lemma 11 to \(K=W(A)\). For each fixed polynomial \(P(z)=\sum_{j=0}^dB_jz^j\), with arbitrary \(B_j\in M_m(\mathbb C)\), Theorems 2 and 3 give \[\left\|\sum_{j=0}^dA^j\otimes B_j\right\| \le2\max_{\overline\Omega_\epsilon}\|P(z)\|.\] Uniform continuity on a fixed compact neighborhood of \(K\) makes the maxima on the right tend to \(\max_K\|P\|\). This includes point and segment numerical ranges. It is a limit of scalar bounds for a fixed polynomial; no limiting similarity or boundary density on \(K\) is asserted. For sharpness take \(n=2\), \(m=1\), \(P(z)=z\), and \(A=\begin{psmallmatrix}0&2\\0&0\end{psmallmatrix}\). Its norm is two, whereas, for every unit \(x\in\mathbb C^2\), \[|x^*Ax|=2|x_1x_2|\le |x_1|^2+|x_2|^2=1,\] with equality when \(x_1=x_2=1/\sqrt2\). Hence \(\max_{W(A)}|P|=1\), proving optimality uniformly over all finite base sizes, coefficient sizes, and degrees. ◻

Proof. Assume \(\mathcal H\ne\{0\}\). We first prove the polynomial bound. With \(e_1,\ldots,e_m\) the standard basis of \(\mathbb C^m\), every vector in the Hilbert tensor product has the exact form \(\xi=\sum_{j=1}^m\xi_j\otimes e_j\), with \(\xi_j\in\mathcal H\). For \(\xi\ne0\), define the nonzero finite-dimensional subspace \[E=\operatorname{span}\{A^k\xi_j:0\le k\le d,\ 1\le j\le m\}, \qquad A_E=Q_EA|_E,\] where \(Q_E\) is the orthogonal projection onto \(E\). Induction gives \(A_E^k\xi_j=A^k\xi_j\) for \(0\le k\le d\): at each step the next iterate already belongs to \(E\). Hence, as vectors in \(E\otimes\mathbb C^m\), \[P[A]\xi=\sum_{k=0}^d\sum_{j=1}^m A^k\xi_j\otimes B_ke_j =\sum_{k=0}^d\sum_{j=1}^m A_E^k\xi_j\otimes B_ke_j =P[A_E]\xi.\] For each unit \(x\in E\), orthogonality of \(Q_E\) gives \(\langle x,A_Ex\rangle=\langle x,Q_EAx\rangle=\langle x,Ax\rangle\). With the conjugate-linear-first convention, these are exactly the numerical-range values used in Corollary 10. Thus \(W(A_E)\subset W(A)\). Applying Corollary 10 in an orthonormal basis of \(E\) yields \[\|P[A]\xi\|\le2\max_{z\in W(A_E)}\|P(z)\|\,\|\xi\| \le2\sup_{z\in W(A)}\|P(z)\|\,\|\xi\|.\] Taking the supremum over unit \(\xi\) proves the inequality. This argument uses no sequence of subspaces exhausting \(\mathcal H\).

The same two-vector compression argument used in the proof of Corollary 10 shows that \(W(A)\) is convex. It is bounded by \(\|A\|\), so \(K\) is compact and convex. Continuity of \(P\) and density of \(W(A)\) in \(K\) give the equality of the supremum and maximum in (30). To check spectral containment, let \(\lambda\notin K\) and \(\delta=\mathop{\mathrm{dist}}(\lambda,K)>0\). For every \(x\in\mathcal H\), Cauchy–Schwarz gives \[\|(\lambda I-A)x\|\,\|x\| \ge |\langle x,(\lambda I-A)x\rangle| \ge\delta\|x\|^2.\] Since \(W(A^*)=\{\overline z:z\in W(A)\}\), the same argument gives \(\|(\overline\lambda I-A^*)y\|\ge\delta\|y\|\) for every \(y\). The first lower bound makes \(\lambda I-A\) injective with closed range, and the second makes it dense because its orthogonal complement is the kernel of \(\overline\lambda I-A^*\). Thus \(\lambda I-A\) is invertible, proving \(\mathop{\mathrm{spec}}(A)\subset K\).

Now fix \(U\) and \(F\) as in the statement. Choose \(\epsilon>0\) such that the compact convex neighborhood \[K_\epsilon=\{z\in\mathbb C:\mathop{\mathrm{dist}}(z,K)\le\epsilon\}\] is contained in \(U\). Its complement is connected. Runge’s polynomial approximation theorem, obtained by rational approximation and movement of poles to infinity in the connected complement [23], gives matrix polynomials \(P_\ell\) with \[\max_{z\in K_\epsilon}\|P_\ell(z)-F(z)\|\longrightarrow0.\] Indeed, apply the scalar theorem to each of the finitely many entries; entrywise uniform convergence implies convergence in matrix operator norm. By Lemma 11, choose an admissible \(\Omega\) with \(K\subset\Omega\) and \(\overline\Omega\subset K_\epsilon\), and let \(\Gamma=\partial\Omega\) have positive orientation. The holomorphic functional calculus, entry by entry, reads \[F[A]=\frac{1}{2\pi i}\int_\Gamma (zI-A)^{-1}\otimes F(z)\,dz.\] The integral converges in operator norm. For a polynomial, Cauchy’s theorem moves the contour to a large circle, where the resolvent’s Neumann expansion gives the prescribed base-first evaluation. Consequently \[\|F[A]-P_\ell[A]\| \le\frac{\operatorname{length}(\Gamma)}{2\pi} \max_{z\in\Gamma}\|(zI-A)^{-1}\| \max_{z\in\Gamma}\|F(z)-P_\ell(z)\| \longrightarrow0.\] The maxima of \(\|P_\ell\|\) on \(K\) also converge to that of \(\|F\|\). Applying (30) to \(P_\ell\) and taking the limit proves (31). The larger compact neighborhood is used to obtain convergence on the fixed contour, not just on \(K\).

A matrix-valued rational function whose poles lie outside \(K\) is holomorphic on a neighborhood \(U\) of \(K\) that avoids all its poles; the preceding choice then puts every pole outside \(\overline\Omega\). Spectral containment makes its rational evaluation well-defined, and this agrees with the displayed contour evaluation. For example, the resolvent identity identifies the contour evaluation of \((z-a)^{-1}\) with \((A-aI)^{-1}\) when \(a\notin\overline\Omega\); differentiation in \(a\) gives the higher pole orders, and partial fractions give every rational function. This proves the complete spectral-set assertion. Finally, the \(2\)-by-\(2\) scalar-coefficient example in the proof of Corollary 10 is already a bounded Hilbert-space operator and is both rational and holomorphic. It proves the stated uniform sharpness. ◻

Proof of Lemma 1. A disk coordinate for a bounded simply connected domain. Let \(U\subset\mathbb C\) be a bounded simply connected domain and fix \(z_*\in U\). Consider the injective holomorphic maps \(\phi:U\to\mathbb D\) satisfying \(\phi(z_*)=0\). This family is nonempty, since a sufficiently small multiple of \(z-z_*\) belongs to it. The supremum \(\alpha\) of \(|\phi'(z_*)|\) over the family is positive and finite: a disk centered at \(z_*\) with closure in \(U\) gives a uniform upper bound by Cauchy’s derivative estimate.

Choose a sequence approaching this supremum. Cauchy’s estimates and compact exhaustion give a subsequence converging locally uniformly to a holomorphic function \(\phi\). Its derivative at \(z_*\) has modulus \(\alpha\), so it is nonconstant; the maximum principle then gives \(\phi(U)\subset\mathbb D\). The limit is injective. Indeed, if two distinct points had the same image, choose disjoint small closed disks about them on whose boundaries the limit minus this common image does not vanish. The argument principle preserves at least one zero in each disk for every sufficiently late member of the sequence, contradicting its injectivity.

Suppose that \(w\in\mathbb D\setminus\phi(U)\). Necessarily \(w\ne0\). The function \[q(z)=\frac{\phi(z)-w}{1-\overline w\phi(z)}\] is holomorphic and nonvanishing on the simply connected domain \(U\), and therefore has a holomorphic square root \(h\). It satisfies \(|h|<1\) and is injective, since \(h^2=q\) is injective. Set \(\beta=h(z_*)\), so that \(|\beta|=\sqrt{|w|}\). The injective disk-valued function \[\psi(z)=\frac{h(z)-\beta}{1-\overline\beta h(z)}\] vanishes at \(z_*\) and has derivative magnitude \[|\psi'(z_*)| =\frac{1-|w|^2}{2\sqrt{|w|}(1-|w|)}\,|\phi'(z_*)| =\frac{1+|w|}{2\sqrt{|w|}}\,\alpha >\alpha.\] This is impossible. Thus \(\phi\) maps \(U\) conformally onto \(\mathbb D\).

Reflection at a regular analytic boundary. Suppose now that \(\partial U\) is a regular real-analytic Jordan curve. For each \(0<r<1\), the set \(\phi^{-1}(r\overline\mathbb D)\) is a compact subset of \(U\), because the conformal inverse is continuous on \(\mathbb D\). Consequently \(|\phi(z)|\to1\) as \(z\) approaches \(\partial U\) from inside.

Fix \(p\in\partial U\). A regular real-analytic parametrization of the boundary near \(p\) extends holomorphically with nonzero derivative. Its local holomorphic inverse straightens the boundary. After shrinking and choosing its sign, we therefore have a holomorphic coordinate \(z=\chi(\xi)\) taking \(0\) to \(p\), a real diameter to the boundary arc, and the upper half of a small disk to the part of \(U\) in this coordinate neighborhood. Choose the neighborhood disjoint from \(z_*\). Then \[u(\xi)=\log|\phi(\chi(\xi))|\] is harmonic and strictly negative on the upper half disk, and is continuous with value zero on its real diameter.

Here the harmonic reflection can be obtained directly. On the boundary of a smaller full disk, prescribe \(u\) on the upper semicircle and the negative of its reflected values on the lower semicircle. These boundary values are continuous, including at the endpoints of the diameter. Their Poisson extension is odd under reflection in the real axis, hence vanishes on the diameter. Uniqueness for the Dirichlet problem on the upper half disk shows that it agrees there with \(u\). Denote the resulting harmonic extension by \(\widehat u\).

The gradient of \(\widehat u\) at zero does not vanish. To see this, its first nonzero homogeneous Taylor term has the form \[c\,\operatorname{Im}(\xi^k),\qquad c\in\mathbb R\setminus\{0\},\quad k\ge1,\] because \(\widehat u\) is harmonic and vanishes on the real diameter. Such a term exists since \(u<0\) on the upper half disk. If \(k\ge2\), \(\sin(k\theta)\) takes both signs for \(0<\theta<\pi\); along two corresponding rays the leading term would force \(u\) to take both signs near zero. Thus \(k=1\).

Take a harmonic conjugate \(\widehat v\) on the full disk. The function \(\exp(\widehat u+i\widehat v)\) is holomorphic and nonvanishing, with nonzero derivative at zero. On the upper half disk it has the same modulus as \(\phi\circ\chi\), so their quotient is a constant of modulus one. Adjusting that constant and returning to the physical coordinate extends \(\phi\) holomorphically across \(p\), with nonzero derivative there.

Patching and univalence on a neighborhood of the closure. We spell out the passage from these local extensions to one extension, including overlaps that may lie outside \(U\). For each boundary point \(p\), choose an extension disk \(B(p,R_p)\) contained in its reflected coordinate neighborhood. Its extension agrees with \(\phi\) throughout \(U\cap B(p,R_p)\); shrink \(R_p\) so that its derivative is nonzero on the disk. Select finitely many points \(p_i\) such that the smaller disks \[V_i=B(p_i,R_i/3),\qquad R_i=R_{p_i},\] cover \(\partial U\), and let \(\phi_i\) denote the extension on \(U_i=B(p_i,R_i)\). If \(V_i\cap V_j\ne\varnothing\) and \(R_i\ge R_j\), then \[|p_i-p_j|<\frac{R_i+R_j}{3}\le\frac{2R_i}{3}<R_i.\] Thus the connected open lens \(U_i\cap U_j\) contains the boundary point \(p_j\) and meets \(U\) in a nonempty open set. Both extensions agree with \(\phi\) there, so the identity theorem makes them equal on the entire lens. Their restrictions therefore patch with \(\phi\) to a holomorphic function on \[U\,\cup\,\bigcup_i V_i,\] an open neighborhood of \(\overline U\). We continue to call it \(\phi\).

The extended map takes \(\partial U\) into \(\mathbb T\). Its nonzero derivative makes the restriction to the boundary a local diffeomorphism onto an arc of \(\mathbb T\). The boundary image is consequently open in \(\mathbb T\), and it is also nonempty and compact, so it is all of \(\mathbb T\). There cannot be two distinct boundary points with the same image \(\eta\in\mathbb T\). Indeed, choose disjoint local inverse neighborhoods about them. Each inverse, restricted to a sufficiently small neighborhood of \(\eta\), takes its disk side into \(U\): the boundary arc is mapped to \(\mathbb T\), and the interior side is mapped into \(\mathbb D\). Points in \(\mathbb D\) sufficiently close to \(\eta\) would then have two preimages in \(U\), contradicting interior injectivity. Since interior and boundary images lie in \(\mathbb D\) and \(\mathbb T\), respectively, \(\phi\) is injective on \(\overline U\).

It is in fact injective on an open neighborhood of \(\overline U\). Otherwise there would be distinct pairs \(z_j,w_j\), approaching the compact set \(\overline U\), with \(\phi(z_j)=\phi(w_j)\). Subsequence limits \(z,w\in\overline U\) have the same image, hence \(z=w\). Both sequences would eventually lie in one local inverse neighborhood of \(z\), a contradiction. The image of a sufficiently small such neighborhood is open and contains \(\overline\mathbb D\). Its holomorphic inverse gives a univalent extension of \(\phi^{-1}\) to a neighborhood of \(\overline\mathbb D\).

Applying these conclusions to the admissible domain \(\Omega\) gives the interior coordinate \(f\) and both of its asserted extensions.

The exterior coordinate. Fix \(z_0\in\Omega\) and apply the inversion \[\iota(z)=\frac{1}{z-z_0},\qquad \iota(\infty)=0,\] to the exterior of \(\overline\Omega\), including infinity. Its image \(U_{\mathrm{ext}}\) is a bounded domain containing zero. More explicitly, convexity and \(z_0\in\Omega\) give a finite positive boundary radius \(R(\theta)\) on each ray \(z_0+r e^{i\theta}\), and \[U_{\mathrm{ext}} =\{0\}\,\cup\, \bigl\{s e^{-i\theta}:0<s<R(\theta)^{-1}\bigr\}.\] This domain is star-shaped about zero, hence simply connected. Boundedness follows also from a positive disk about \(z_0\) contained in \(\Omega\); a neighborhood of zero is in the image because \(\Omega\) is bounded. Inversion preserves the regular real-analytic Jordan boundary, since its pole \(z_0\) is not on that boundary.

By the preceding arguments there is a conformal map \(\phi_{\mathrm{ext}}:U_{\mathrm{ext}}\to\mathbb D\) with \(\phi_{\mathrm{ext}}(0)=0\), whose inverse \(h\) extends univalently to a disk \(\{\zeta:|\zeta|<\rho\}\) for some \(\rho>1\). Here \(h(0)=0\) is its only zero on this larger disk, and it is simple. Define \[G(t)=z_0+\frac{1}{h(1/t)},\qquad |t|>r_0:=\rho^{-1}.\] The function \(G\) is univalent on this region. It maps \(|t|>1\) onto the exterior of \(\overline\Omega\), takes infinity to infinity, and has a simple pole there. Its convergent Laurent expansion is therefore \[G(t)=at+b+\sum_{\ell\ge1}c_\ell t^{-\ell}, \qquad a=h'(0)^{-1}\ne0.\] In particular its restriction to \(\mathbb T\) is a smooth diffeomorphism onto \(\partial\Omega\).

Orientation and normal direction. For \(t=e^{i\theta}\), the radial and angular derivatives are \[\left.\frac{\partial}{\partial r}G(re^{i\theta})\right|_{r=1} =tG'(t), \qquad \frac{d}{d\theta}G(e^{i\theta})=itG'(t).\] They are nonzero and perpendicular. Increasing \(r\) enters the exterior of \(\overline\Omega\), so \(tG'(t)\) points in the outward normal direction. The tangent \(itG'(t)\) is consequently the positive boundary tangent. The interior conformal map \(f\) preserves boundary orientation: locally it maps the interior side of \(\partial\Omega\) to the interior side of \(\mathbb T\), and its derivative preserves orientation in the plane. Thus \(f\circ G:\mathbb T\to\mathbb T\) is a smooth orientation-preserving diffeomorphism. This proves every assertion of the lemma. ◻