The complete Crouzeix theorem: optimal similarity and a common positive boundary representation

Proof. For |\zeta|=1, E_\zeta=I-D/\zeta is invertible and

Thus \Phi_D is positive and continuous. The strict spectral-radius bound, rather than a strict norm bound, implies \sum_{j\ge0}\|D^j\|<\infty. Consequently the following series is absolutely uniformly convergent on the circle:

For any m and any M_m-valued h holomorphic near \overline\D, integration of its uniformly convergent Taylor series gives

Apply this with h=v\circ f^{-1}. The conformal collar makes h holomorphic near \overline\D, and scalar composition gives h[D]=v[A']. Locally writing \psi(e^{i\theta})=e^{i\eta(\theta)} shows that j(e^{i\theta})=\eta'(\theta)>0. Changing variables \zeta=\psi(t) proves the first identity in (1.2). The zeroth moment gives \int\Lambda\,d\sigma=I.

For each m, define

Normalization makes V_m an isometry. If M_v multiplies by I_n\otimes v(G(t)), then

Similarity on the first factor gives the remaining estimate with constant \kappa. Neither the definition of S nor that of \Lambda depends on m or v.□

Proof. We first obtain positive matrices X_0 and Y_0 supported on the \kappa^2 and 1 eigenspaces of H, respectively. A third positive matrix Z_0 will express their difference as Z_0-TZ_0T^*; this balance will provide the equal Gram matrices needed for the unitary realization.

Separation, including a possible contact point. In the real vector space \R\oplus\operatorname{Herm}_n^3, the affine set

is disjoint from the open convex set

Here is the finite-dimensional separation argument used for these sets; the general convex-separation framework is treated in [22, Section 11, Theorem 11.2]. Write \mathcal A_T=a_0+L, and let \pi project orthogonally onto L^\perp. The set \mathcal K=\pi(\mathcal O-a_0) is nonempty, open in L^\perp, and convex, and it does not contain zero. If 0\notin\overline{\mathcal K}, a nearest point z_0 in the closed convex set \overline{\mathcal K} satisfies \langle z_0,c\rangle\ge\|z_0\|^2>0 for every c\in\mathcal K.

If 0\in\overline{\mathcal K}, fix c_*\in\mathcal K and set z_j=-c_*/j. Each z_j is outside \overline{\mathcal K}: otherwise zero would lie on the open segment joining an interior point c_* to a point of the closure, and hence would belong to \mathcal K. To verify that segment fact, take a ball about c_* inside \mathcal K, approximate the other endpoint by points of \mathcal K, and use convexity; a ball about each fixed interior point of the limiting segment is still contained in \mathcal K. Let y_j be nearest to z_j in \overline{\mathcal K}. Since zero belongs to this closed set, \|y_j-z_j\|\le\|z_j\|, so y_j\to0. A subsequence of \nu_j=(y_j-z_j)/\|y_j-z_j\| tends to a unit vector \nu. The nearest-point inequality gives \langle\nu_j,c-y_j\rangle\ge0, hence \langle\nu,c\rangle\ge0 for every c\in\mathcal K. In the first case use \nu=z_0. In either case \ell(v)=-\langle\nu,\pi v\rangle is nonzero and \ell(a)\ge\ell(o) for all a\in\mathcal A_T, o\in\mathcal O.

Because \mathcal A_T is affine, \ell is constant on it. Boundedness above on \mathcal O forces the signs

If \lambda=0, strict feasibility produces a negative value on \mathcal A_T, while separation requires a value at least \sup_{\mathcal O}\ell=0. Thus \lambda>0; scale it to one. Constancy on \mathcal A_T then gives

The constant value is \tr Y_0\ge\kappa^2. At (\kappa^2,H) it is

Each subtracted trace is nonnegative, so all vanish and \tr Y_0=\kappa^2.

Endpoint support and balance. Write \|B\|_{\HS}=(\tr(B^*B))^{1/2} for the unnormalized Hilbert–Schmidt norm. For positive matrices B,C, \tr(BC)=\|C^{1/2}B^{1/2}\|_{\HS}^2; vanishing implies that the range of B lies in the kernel of C. Therefore X_0 is supported on the \kappa^2 eigenspace of H, and Y_0 on its 1 eigenspace.

Choose square factors and a positive matrix Q by

The support and trace relations give the asserted norms and endpoint identities. Their column ranges are orthogonal because \kappa>1, so X^*Y=0 as a matrix. Congruence of (2.4) gives

Unitary realization and transpose. The balance-to-isometry step is the classical interpolation mechanism of [15, Section 3 of the versioned preprint]; its particular endpoint data and finite-dimensional realization are established here. Put L=Q^{1/2}. The rows R_1=[L\ Y] and R_2=[DL\ X] have equal products with their adjoints. The map R_2^*v\mapsto R_1^*v is therefore a well-defined isometry of their ranges. Extend it to a unitary using equal-dimensional orthogonal complements, then take its adjoint. This yields

Thus L=DLa+Xc and Y=DLb+Xd. Since \|a\|\le1 and \rho(D)<1, iteration gives the norm-convergent identity

The transfer-function norm identity is the unitary-colligation identity of [8, Section 3.1, equations (3.1)–(3.4)]. Define, for |\zeta|<1,

For e\in\C^n and v=(I-\zeta a)^{-1}be, unitarity gives

Hence R is contractive. Ordinary transpose preserves its norm and holomorphy. Its rational entries are bounded in \D, so every apparent pole on \T is removable: a genuine pole would diverge on an interior radial approach. After cancellation, finitely many remaining poles stay away from \overline\D. This supplies a common neighborhood of that closed disk on which F_\D is holomorphic; continuity retains contractivity on the circle.

Finally,

The Taylor coefficients of F_\D are the transposes of d and ca^jb. Substitution in (2.6) proves F_\D[D]x=y. The conformal collar and scalar composition give the claims for F_\Omega. Rationality is asserted in the disk variable; F_\Omega is only required to be holomorphic near \overline\Omega.□

Proof. Expand (3.1) 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

Write u=t^{-1}, v=w^{-1}, and G(t)=at+g(u), with g holomorphic for |u|<r^{-1}. Then

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

is holomorphic on the reciprocal bidisk and vanishes on both axes. Its Taylor series has only the strictly mixed powers in (3.2). 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 \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\T, the circle identity 2\operatorname{Re}(t/(t-w))=1 gives

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 (3.1). Cauchy estimates in the reciprocal disk give geometric decay and absolute uniform convergence on \T.

Finally put R_t=G(t)I-A. Direct congruence gives

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

The mixed expansion in (3.2) gives, for j>0,

Both marginal identities say that \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,

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 (3.2) and (3.4) give \cC u=u_0+u_++\M u_+ for finite Laurent data. The three terms are orthogonal, so \|\cC u\|_2^2\le2\|u\|_2^2. Density defines the bounded extension, and the same formula gives \cC^2=\cC. Taking its adjoint and using the strict Fourier exchange proves (3.5); conjugation by the antiunitary map u\mapsto u^* gives the second formula. The weighted estimate follows from

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 \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

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 \cC v. Cauchy’s formula gives c_v=v_\Omega in \Omega, and thus \cC v=v on \T. For Laurent data u, both F_\Omega c_u and c_uF_\Omega are holomorphic near \overline\Omega by (3.4). Their traces are fixed by \cC, giving both identities in (3.7) for finite Laurent u. Left and right multiplication by the bounded F are bounded on \cH_k; Laurent approximation extends the two identities to arbitrary u\in\cH_k.

Lastly, integrating (3.3) against a finite Laurent polynomial gives

by (3.4) and the prescribed base-first evaluation. Applying the same formula to u^* and taking the full matrix adjoint proves (3.8). The constant case gives mass 2I_n.□

Proof. Put g=\cC^\dagger r, P_1=\cC^\dagger p, P_2=\cC^\dagger q. On \cH_k, the adjoint of left multiplication by F is left multiplication by F^*; the adjoint of right multiplication is right multiplication by F^*, because

Taking the Hilbert adjoints of both identities in (3.7) therefore gives, with the factors on their original sides,

If g=0, the first identity makes P_1=0, contradicting mean preservation and \tr(p_0)>0. Thus g\ne0.

Each P_j has only nonnegative Fourier powers, and its mean is Hermitian. Fourier orthogonality gives

Since \|F\|\le1, both F^*g and gF^* have Hilbert–Schmidt L^2 norm at most \|g\|_2. Apply (3.6) separately to the two inputs in (5.1). Using the star relation in (3.5) and p=p^*, q=q^*, we obtain

Put h=\widetilde\cC^\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

Finally, set J=\diag(I_k,-I_k). Positivity of both \Delta and J\Delta J gives

The entries are in L^2, so this inequality is integrable. Combining it with (5.2) and (5.3) yields

Since g\ne0, division proves \kappa\le2.□

Completion of Theorems 1.2 and 1.3. If \kappa=1, the required bound already holds. Otherwise Proposition 2.2 and Section 4 supply all the hypotheses of Lemma 5.1 at auxiliary size k=n: (4.4) gives both multiplication orders, (4.6) gives the zero matrix mean and positive trace, (4.8) supplies positivity, and (4.10) identifies its entries. Hence \kappa\le2. Proposition 2.1 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 \R^2 and put

The Lipschitz bound gives the uniform estimate

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 6.1. 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

Taking their trace against the compression is a real-affine map into \C\simeq\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 6.2 to K=W(A). For each fixed polynomial P(z)=\sum_{j=0}^dB_jz^j, with arbitrary B_j\in M_m(\C), Theorems 1.2 and 1.3 give

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=\left(\begin{smallmatrix}0&2\\0&0\end{smallmatrix}\right). Its norm is two, whereas, for every unit x\in\C^2,

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 \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

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\C^m,

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 6.1. Thus W(A_E)\subset W(A). Applying Corollary 6.1 in an orthonormal basis of E yields

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 6.1 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 (6.1). To check spectral containment, let \lambda\notin K and \delta=\dist(\lambda,K)>0. For every x\in\mathcal H, Cauchy–Schwarz gives

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 \spec(A)\subset K.

Now fix U and F as in the statement. Choose \epsilon>0 such that the compact convex neighborhood

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, pp. 235–238], gives matrix polynomials P_\ell with

Indeed, apply the scalar theorem to each of the finitely many entries; entrywise uniform convergence implies convergence in matrix operator norm. By Lemma 6.2, 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

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

The maxima of \|P_\ell\| on K also converge to that of \|F\|. Applying (6.1) to P_\ell and taking the limit proves (6.2). 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 6.1 is already a bounded Hilbert-space operator and is both rational and holomorphic. It proves the stated uniform sharpness.□

Proof of Lemma 1.1. A disk coordinate for a bounded simply connected domain. Let U\subset\C be a bounded simply connected domain and fix z_*\in U. Consider the injective holomorphic maps \phi:U\to\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\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\D\setminus\phi(U). Necessarily w\ne0. The function

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

vanishes at z_* and has derivative magnitude

This is impossible. Thus \phi maps U conformally onto \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\D) is a compact subset of U, because the conformal inverse is continuous on \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

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

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

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

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

an open neighborhood of \overline U. We continue to call it \phi.

The extended map takes \partial U into \T. Its nonzero derivative makes the restriction to the boundary a local diffeomorphism onto an arc of \T. The boundary image is consequently open in \T, and it is also nonempty and compact, so it is all of \T. There cannot be two distinct boundary points with the same image \eta\in\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 \T, and the interior side is mapped into \D. Points in \D sufficiently close to \eta would then have two preimages in U, contradicting interior injectivity. Since interior and boundary images lie in \D and \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\D. Its holomorphic inverse gives a univalent extension of \phi^{-1} to a neighborhood of \overline\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

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

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\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

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

In particular its restriction to \T is a smooth diffeomorphism onto \partial\Omega.

Orientation and normal direction. For t=e^{i\theta}, the radial and angular derivatives are

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 \T, and its derivative preserves orientation in the plane. Thus f\circ G:\T\to\T is a smooth orientation-preserving diffeomorphism. This proves every assertion of the lemma.□