A direct proof of the complete Crouzeix inequality

Proof. Expanding the denominator in (2.3) after factoring c\lambda shows that its kth coefficient is a polynomial of degree k with leading term c^{-k}z^k. The constant coefficient is one. In particular, the b_k form a basis of the scalar polynomials.

For the joint extension, consider

Write u=\lambda^{-1}, v=\mu^{-1} and h(\lambda)=c\lambda+g(u), where g is holomorphic on |u|<r^{-1}. Then

The divided difference of g is jointly holomorphic, with diagonal value g'(u), so this formula also covers either or both infinities. Off the diagonal and away from the axes, H is nonzero by injectivity of h. On the diagonal it equals h'(\lambda)\ne0; on either axis it equals c. Thus it has no zeros on the reciprocal bidisk.

Direct differentiation gives s=\lambda\partial_\lambda H/H=-u\partial_uH/H. This function is holomorphic and vanishes on both reciprocal axes. Its Taylor series has precisely the strictly mixed powers in (2.5). For any r<\rho<1, Cauchy estimates on the closed reciprocal bidisk of radius \rho^{-1} give a constant C_\rho with |s_{kj}|\le C_\rho\rho^{k+j}. On a compact subset of the exterior product collar, choose \sigma>r below all finite coordinate moduli, and then r<\rho<\min(1,\sigma). The resulting geometric double series dominates the Laurent series uniformly there; terms vanish on the reciprocal axes. This proves normal convergence throughout the product collar and absolute uniform convergence on the boundary. Comparing coefficients at \lambda=\infty in (2.5), first for fixed finite \mu in the exterior collar, yields (2.4).

For distinct \lambda,\mu\in\T, one has 2\Re(\lambda/(\lambda-\mu))=1, and therefore

by (2.2). Continuity gives the same sign on the diagonal. Every term of s and of its conjugate has nonzero frequency in each variable. Their integrals in either variable vanish, proving both normalizations.□

Proof. For u\in L^2(\T;\mathcal H) with values in a Hilbert space \mathcal H, define

Positivity and the normalization in \lambda imply \|(Tu)(\mu)\|^2\le\int K(\lambda,\mu)\|u(\lambda)\|^2d\tau(\lambda). Integrating in \mu and using the normalization in that variable shows that T is an L^2 contraction.

For u(\lambda)=\sum_{k=1}^N C_k\lambda^k, only the s term contributes to the integral, so

All interchanges are justified by the absolute uniform convergence in Theorem 2.1. In particular,

By (2.4), the boundary value of G is the sum of its constant term C_0, its positive part u, and its negative part Tu. These have disjoint Fourier supports. Parseval’s identity gives

which proves both inequalities.□

Proof. Every eigenvalue of A belongs to W(A), by testing a unit eigenvector. Thus the exterior resolvent is defined on and outside the boundary. Compactness of the circle and strict spectral separation allow a smaller collar |\lambda|>r_A, with r<r_A<1, on which it remains holomorphic, including infinity where R(\infty)=I. Expansion at infinity gives the coefficients b_k(A): this follows either by the geometric resolvent series for large |h(\lambda)| or by substituting A into the polynomial coefficient identities in (2.3). Choose r_A<\theta<1. Cauchy estimates on the reciprocal circle of radius \theta^{-1} give the stated geometric bound. They imply absolute uniform convergence on \T, and its constant coefficient b_0(A)=I gives both integrals.

For B=h(\lambda)I-A, direct congruence gives

For each unit vector v, the quadratic form on the right is 2\Re(\overline{q(\lambda)}(h(\lambda)-v^*Av)), which is nonnegative by (2.2). Invertibility of B proves positivity.□

Proof. Choose h from Lemma A.1 and apply Theorem 2.1 to this coordinate. Put \gamma=\|F[A]\|. There is nothing to prove if \gamma=0. Otherwise choose unit singular vectors x,y\in\C^n\otimes\C^m such that

Write x=\sum_{i=1}^mXe_i\otimes e_i and y=\sum_{i=1}^mYe_i\otimes e_i, where X,Y are n-by-m matrices and (e_i) is the standard basis of \C^m. For the resolvent in Theorem 3.1, define matrix functions

Subscripts k\ge0 denote the coefficients of \lambda^{-k}, so M_k=Y^*b_k(A)X, for example. By Theorem 3.1, all four series converge absolutely and uniformly on \T, and their coefficient sequences are square summable. Since b_0=1,

Set a^2=\sum_{k\ge0}\HS{M_k}^2. We will bound the squared boundary L^2 Hilbert–Schmidt norms of P+P^* and Q+Q^* above by 4a^2/\gamma^2 each, and that of M+L^* below by a^2. Positivity will then compare these quantities and force \gamma\le2.

The off-diagonal coefficient functional. For a polynomial G=\sum_k b_kC_k, direct expansion in the tensor basis gives

Indeed, C_ke_j=\sum_i(C_k)_{ij}e_i in the coefficient factor, leaving (Ye_i)^*b_k(A)Xe_j in the base factor. Equation (3.4) is the bilinear entry pairing of the two coefficient arrays. Cauchy–Schwarz gives

Taking G=F and using (3.2) shows that a>0.

The two ordered tests. Polynomial evaluation respects multiplication in its given order. Explicitly, if F(z)=\sum_iD_iz^i and G(z)=\sum_jE_jz^j, then

and the reversed product has coefficients E_jD_i. Hence

Write the full finite Faber expansions as

The functional in (3.5) uses all M_k, so it applies to these full expansions for every finite degree. Apply it to the two identities in (3.6). The lower coefficient estimate in Proposition 2.2 and the two boundary multiplier bounds give

Here \HS{F(z)G(z)}\le\|F(z)\|\HS{G(z)} and \HS{G(z)F(z)}\le\HS{G(z)}\|F(z)\| are used in their respective orders. The upper coefficient estimate for G now gives

The entries paired with C_k in these two functionals are respectively those of P_k and Q_k, by the calculation in (3.4). The weighted coefficient norm on the right gives reciprocal weights on these two coefficient sequences:

For the first inequality, set d_N=\HS{P_0}^2+\tfrac12\sum_{k=1}^N\HS{P_k}^2 and test (3.7) with C_0=\overline{P_0}, C_k=\overline{P_k}/2 for 1\le k\le N, and C_k=0 otherwise; bars mean entrywise conjugation. The paired value is d_N, and the squared weighted test norm is also d_N. Hence d_N\le(a/\gamma)\sqrt{d_N}, which gives d_N\le a^2/\gamma^2 including when d_N=0. Letting N\to\infty proves the first bound. The same finite tests with Q_k prove the second.

The boundary norms of the three blocks. The compression of R+R^* by [X\ Y] has diagonal blocks P+P^*, Q+Q^* and lower off-diagonal block M+L^*. We first compute their boundary norms. In the diagonal fields, (3.3) makes the constant term twice P_0 or Q_0. For k\ge1, the coefficients P_k and P_k^* occur at frequencies -k and k and have equal Hilbert–Schmidt norms; the same holds for Q_k and Q_k^*. Parseval therefore gives four times the reciprocal weighted sum in (3.8):

and the same identity and bound hold for Q+Q^*.

For the cross field, L_0^*=M_0 makes the constant term 2M_0. Its negative coefficients are M_k for k\ge1 and its positive coefficients are L_k^* for k\ge1. Consequently

The positive block comparison. Theorem 3.1 gives R+R^*\succeq0. Its compression by the n-by-2m matrix [X\ Y] is therefore

For J=\diag(I_m,-I_m), the matrix J\Delta J is also positive, so

Integrating this inequality and using (3.9) and (3.10) yields

Since a>0, this proves \gamma\le2.□

Proof. For 0<\delta<1, choose a finite Euclidean \delta-net D_\delta of unit directions that includes 1,-1,i,-i. Let s_K(\nu)=\max_{z\in K}\Re(\bar\nu z) be the support function. Choose t_\delta>0 with |D_\delta|e^{-t_\delta\delta}<1, and define

Each point of K satisfies \Phi_\delta<1, so K\subset\Omega_\delta. If \Phi_\delta(z)\le1, each positive summand is at most one, and hence

In particular, the four axis directions give

Thus one fixed rectangle contains every closed sublevel for 0<\delta<1.

The Hessian is a sum of positive multiples of \nu\nu^{\mathsf T} after identifying \C with \mathbb R^2. The four axis directions ensure it is positive definite. Thus \Phi_\delta is strictly convex and real analytic.

The gradient cannot vanish at level one: a critical point of a differentiable convex function is a global minimum, whereas every point of K has value less than one. The implicit function theorem therefore makes that level a regular real-analytic curve. The open sublevel set is bounded and convex with nonempty interior. Fix an interior point. Each ray from it crosses level one exactly once; convexity and boundedness give existence and uniqueness. Radial projection from the compact boundary to the circle is a continuous bijection, hence a homeomorphism. The boundary is a Jordan curve, and \overline\Omega_\delta=\{\Phi_\delta\le1\}.

It remains to bound the distance of this sublevel from K. For z\in\overline\Omega_\delta\setminus K, let w\in K be nearest to z, put d=|z-w| and \nu=(z-w)/d, and choose \mu\in D_\delta with |\mu-\nu|\le\delta. Choose v\in K attaining s_K(\mu). Since w+t(v-w)\in K for 0\le t\le1, minimality of w gives \Re(\bar\nu(v-w))\le0. Using this inequality and the support inequality for \mu, we obtain

The first and third inequalities use |\mu-\nu|\le\delta. Points of K have distance zero, so

Taking, for example, \delta_j=1/(j+1) completes the construction.□

Proof of Theorem 1.1. Compactness of W(A) follows from compactness of the unit sphere. We recall a short proof of the Toeplitz–Hausdorff convexity theorem [18, 12]. To join two numerical-range values, compress A to the span of their unit testing vectors. If that span has dimension one there is nothing to prove. In dimension two, its quadratic form on unit vectors (u,v) is a complex-valued real-affine function of

which ranges over the full unit sphere in \mathbb R^3. The linear part of a map from \mathbb R^3 to \mathbb R^2 has a nonzero kernel. Any point of the unit ball can be moved in a kernel direction to the sphere without changing its image. Thus the images of the sphere and ball coincide, and this image is convex. Its segment between the chosen values belongs to the numerical range of the compression and hence to W(A).

Apply Lemma 4.1 with K=W(A). By scaling Theorem 3.2,

If this maximum is zero, each entry of P vanishes on an open set and the polynomial is identically zero, making the inequality immediate. Otherwise divide P by the maximum to invoke that theorem. All \overline\Omega_j lie in one compact set. Uniform continuity of the fixed function z\mapsto\|P(z)\|, the containment of K, and the uniform distance convergence imply convergence of these maxima to \max_K\|P\|. This proves (1.1) without any interior assumption on K.

For sharpness take n=2, m=1, P(z)=z, and

Its norm is two, while |x^*Ax|=2|x_1x_2|\le1 for every unit x, with equality at x_1=x_2=1/\sqrt2. Thus the right-hand maximum in (1.1) is one. No smaller universal constant is possible.□

Proof. Fix z_*\in\Omega and apply z\mapsto1/(z-z_*), with infinity mapped to zero, to the spherical exterior (\mathbb C\setminus\overline\Omega)\cup\{\infty\}. Its image U is a bounded simply connected domain containing zero, with a regular real-analytic Jordan boundary. Indeed, each ray from z_* meets \partial\Omega at a unique positive radius; inversion turns the exterior portions of these rays into a star-shaped domain about zero. The Riemann mapping theorem [2, Chapter 6, Section 1.1, Theorem 1] supplies a conformal bijection g:\mathbb D\to U with g(0)=0. Carathéodory’s Jordan-domain theorem [2, Chapter 6, Section 1.2] extends g to a homeomorphism \overline{\mathbb D}\to\overline U.

We justify the stronger boundary continuation needed below. A regular real-analytic boundary arc has a real-analytic parametrization with nonzero derivative. Its holomorphic continuation and the holomorphic inverse-function theorem provide a local coordinate in which the arc is a real interval and the domain is one side of that interval. Choose such coordinates at \zeta\in\partial\mathbb D and p=g(\zeta)\in\partial U. Continuity of the boundary extension lets us shrink the source neighborhood so that g maps its intersection with \overline{\mathbb D} into the target chart. The composition representing g in these coordinates is holomorphic on a half disk, continuous on its real interval, and real-valued there. Schwarz reflection extends it holomorphically across the interval [2, Chapter 4, Section 6.5; Chapter 6, Sections 1.3–1.4]. Apply the same procedure to g^{-1}. Shrink the neighborhoods so that the reflected maps can be composed. Their compositions equal the identity on the original domain side, and hence throughout a smaller neighborhood by the identity theorem. Thus the continued g has nonzero derivative at each boundary point.

To patch the local continuations, choose around every \zeta\in\partial\mathbb D a disk B(\zeta,R_\zeta) carrying such a continuation, agreeing with g on its intersection with \mathbb D. Select finitely many of the third-radius disks V_i=B(\zeta_i,R_i/3) that cover the circle, and denote the larger disks by U_i=B(\zeta_i,R_i). If V_i\cap V_j\ne\varnothing and R_i\ge R_j, then |\zeta_i-\zeta_j|<(R_i+R_j)/3\le2R_i/3<R_i. The connected lens U_i\cap U_j therefore contains \zeta_j and meets \mathbb D. The two continuations coincide there by the identity theorem. They consequently patch with g to a holomorphic map on \mathbb D\cup\bigcup_iV_i, a neighborhood of the closed disk.

The extension is injective on a smaller neighborhood of the closed disk. Otherwise, distinct colliding pairs arbitrarily close to that compact set would have subsequential limits in the closed disk. Continuity and injectivity on the closed disk identify their limits. Both points of each pair would then eventually lie in one local inverse neighborhood of that common limit, a contradiction. By compactness, this injective neighborhood contains the disk \{w:|w|<R_0\} for some R_0>1. Thus g is univalent on that disk. Its sole zero there is 0, and that zero is simple. Set

This gives all mapping and extension assertions, including the simple pole with coefficient c=1/g'(0)\ne0.

Finally, at \lambda=e^{i\theta} the radial and angular derivatives are q(\lambda) and iq(\lambda), respectively. They are perpendicular and nonzero. Increasing the radial parameter enters the exterior of \overline\Omega, so q(\lambda) points outward. The supporting half-plane property of a smooth convex boundary proves the final inequality.□