The bound on a smooth convex domain

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

This lesson proves the complete Crouzeix inequality, Theorem A of The numerical range and the Crouzeix inequality. For a matrix \(A\) whose numerical range lies in a smooth convex domain \(\Omega\), substituting \(A\) into the Faber expansion of Faber polynomials and a positive kernel gives the matrix function \[ \begin{aligned} R(\lambda)&=q(\lambda)\bigl(h(\lambda)I-A\bigr)^{-1}\\ &=\sum_{k\ge0}b_k(A)\lambda^{-k} \end{aligned} \] on the unit circle, and its Hermitian part \(R+R^*\) is positive semidefinite (Theorem 5.1). Compressing this positive field with the singular vectors of \(F[A]\), and applying the coefficient estimate to the products \(FG\) and \(GF\) in both orders, gives \(\|F[A]\|\le2\) when \(\|F\|\le1\) on \(\partial\Omega\) (Theorem B). Theorem A follows by shrinking the domains of Smooth convex neighbourhoods to the numerical range. The argument is OpenAI's [OpenAI-DC, Sections 3 and 4]. The field \(R+R^*\) is the double-layer kernel of the numerical range used by Delyon and Delyon [DD] and by Crouzeix and Palencia [CP].

Setting. \(\Omega\), \(h\), \(c\), \(\kappa\), \(r_0\), \(q\), \(b_k\) and \(s_{kj}\) are as in Faber polynomials and a positive kernel, and \(A\in M_n(\mathbf C)\) satisfies \(W(A)\subset\Omega\). For a polynomial \(P(z)=\sum_kB_kz^k\) with \(B_k\in M_m(\mathbf C)\), \(P[A]=\sum_kA^k\otimes B_k\) as in the first lesson. \(\|\cdot\|\) is the operator norm and \(\|\cdot\|_2\) the Hilbert–Schmidt norm, \(d\tau(e^{it})=dt/2\pi\), and for a matrix polynomial \(H\) \[ \|H\|_\partial^2=\int_{\mathbf T}\|H(h(\mu))\|_2^2\,d\tau(\mu). \]

1. The resolvent

Theorem 5.1 (OpenAI). For \(|\lambda|\ge1\), \(h(\lambda)I-A\) is invertible. The function \(R(\lambda)=q(\lambda)(h(\lambda)I-A)^{-1}\) satisfies, for \(\lambda\in\mathbf T\), \[ \begin{aligned} R(\lambda)&=\sum_{k\ge0}b_k(A)\lambda^{-k},\\ R(\lambda)+R(\lambda)^*&\succeq0, \end{aligned} \] and there are \(C<\infty\) and \(0<\theta<1\) with \(\|b_k(A)\|\le C\theta^k\) for all \(k\). In particular \[ \int_{\mathbf T}R\,d\tau=I,\qquad\int_{\mathbf T}(R+R^*)\,d\tau=2I. \]

Proof. Every eigenvalue of \(A\) lies in \(W(A)\subset\Omega\), while \(h\) maps \(\{|\lambda|\ge1\}\) into \(\mathbf C\setminus\Omega\) (Theorem 3.7 of An exterior conformal coordinate). Since \(h\) is injective, at most \(n\) points \(\lambda\) of the collar \(\{|\lambda|>r_0\}\) have \(h(\lambda)\) in the spectrum, all with \(|\lambda|<1\). Choose \(r_A\in(r_0,1)\) larger than their moduli. Then \(\det(h(\lambda)I-A)\ne0\) for \(|\lambda|>r_A\).

In the variable \(u=1/\lambda\), \(h=(c+u\kappa(u))/u\) and \(q=(c-u^2\kappa'(u))/u\), so \[ R(1/u)=\bigl(c-u^2\kappa'(u)\bigr)D_A(u)^{-1}, \] where \[ \begin{aligned} D_A(u)&=cI+u\bigl(\kappa(u)I-A\bigr)\\ &=cI+(\kappa_0I-A)u\\ &\quad+\sum_{j\ge1}\kappa_ju^{j+1}I. \end{aligned} \] Since \(\det D_A(u)=u^n\det(h(1/u)I-A)\) for \(u\ne0\) and \(\det D_A(0)=c^n\), Cramer's rule shows that \(R(1/u)\) is holomorphic, entry by entry, on \(\{|u|<1/r_A\}\), including \(u=0\), where it equals \(I\). Let \(\sum_kB_ku^k\) be its Taylor expansion. Multiplying out \[ D_A(u)\sum_{k\ge0}B_ku^k=\bigl(c-u^2\kappa'(u)\bigr)I \] and comparing coefficients gives \(B_0=I\) and the recursion of Proposition 4.1 of Faber polynomials and a positive kernel with \(z\) replaced by \(A\). So \(B_k=b_k(A)\) by induction.

Choose \(\theta\in(r_A,1)\) and let \(C=\max_{|u|=1/\theta}\|R(1/u)\|\). For unit vectors \(x,y\), the Taylor coefficients of \(x^*R(1/u)y\) are \(x^*b_k(A)y\), and the Cauchy estimates (Theorem 3.3.8 of Lebl's guide in the core course Complex Analysis) give \(|x^*b_k(A)y|\le C\theta^k\). So \(\|b_k(A)\|\le C\theta^k\), the series converges absolutely and uniformly on \(\mathbf T\), and integrating term by term gives \(\int R\,d\tau=b_0(A)=I\).

For \(\lambda\in\mathbf T\) put \(B=h(\lambda)I-A\), so \(R=qB^{-1}\). Then \[ B^*(R+R^*)B=qB^*+\bar qB, \] and for every unit vector \(v\), \[ \begin{aligned} &v^*(qB^*+\bar qB)v=2\operatorname{Re}\bigl(\bar q\,v^*Bv\bigr)\\ &\quad=2\operatorname{Re}\Bigl(\overline{q(\lambda)}\bigl(h(\lambda)-v^*Av\bigr)\Bigr)\ge0 \end{aligned} \] by the support inequality, because \(v^*Av\in W(A)\subset\overline\Omega\). As \(B\) is invertible, \(R+R^*\succeq0\). \(\square\)

2. The bound on a smooth convex domain

Theorem B (OpenAI). Let \(\Omega\) be a smooth convex domain and \(A\in M_n(\mathbf C)\) with \(W(A)\subset\Omega\). For every polynomial \(F\) with coefficients in \(M_m(\mathbf C)\) such that \(\|F(z)\|\le1\) for \(z\in\partial\Omega\), \[ \|F[A]\|\le2 . \]

Proof. Put \(\gamma=\|F[A]\|\) and suppose \(\gamma>0\). Choose a unit vector \(x\in\mathbf C^n\otimes\mathbf C^m\) with \(\|F[A]x\|=\gamma\) and put \(y=F[A]x/\gamma\). Then \(x\) maximizes \(v\mapsto v^*F[A]^*F[A]v\) on unit vectors, so it is an eigenvector of \(F[A]^*F[A]\) for the eigenvalue \(\gamma^2\), and \[ F[A]x=\gamma y,\qquad F[A]^*y=\gamma x . \] Let \(e_1,\dots,e_m\) be the standard basis of \(\mathbf C^m\), and write \(x=\sum_iXe_i\otimes e_i\) and \(y=\sum_iYe_i\otimes e_i\) with \(n\times m\) matrices \(X,Y\); then \(\|X\|_2=\|Y\|_2=1\). Define \(m\times m\) matrix functions on \(\mathbf T\): \[ \begin{aligned} P&=X^*RX,&Q&=Y^*RY,\\ M&=Y^*RX,&L&=X^*RY. \end{aligned} \] Their coefficients of \(\lambda^{-k}\) are \(P_k=X^*b_k(A)X\), and similarly \(Q_k\), \(M_k\), \(L_k\). Since \(b_0(A)=I\), \(P_0=X^*X\) and \(Q_0=Y^*Y\) are Hermitian and \(L_0^*=M_0\). Put \[ a^2=\sum_{k\ge0}\|M_k\|_2^2 , \] which is finite because \(\|M_k\|_2\le\|Y\|\,\|b_k(A)\|\,\|X\|_2\le C\theta^k\).

Step 1: pairings. Let \(G=\sum_{k=0}^Nb_kC_k\) with \(C_k\in M_m(\mathbf C)\). Then \(G[A]=\sum_kb_k(A)\otimes C_k\). With the columns \(x_j=Xe_j\) and \(y_i=Ye_i\), so that \(x=\sum_jx_j\otimes e_j\) and \(y=\sum_iy_i\otimes e_i\), \[ \begin{aligned} &(y_i\otimes e_i)^*\bigl(b_k(A)\otimes C_k\bigr)(x_j\otimes e_j)\\ &\quad=\bigl(y_i^*b_k(A)x_j\bigr)\bigl(e_i^*C_ke_j\bigr)\\ &\quad=(M_k)_{ij}(C_k)_{ij}, \end{aligned} \] so \[ \begin{aligned} y^*G[A]x&=\sum_k\sum_{i,j}(M_k)_{ij}(C_k)_{ij},\\ x^*G[A]x&=\sum_k\sum_{i,j}(P_k)_{ij}(C_k)_{ij},\\ y^*G[A]y&=\sum_k\sum_{i,j}(Q_k)_{ij}(C_k)_{ij}. \end{aligned} \] By the Cauchy–Schwarz inequality, \[ |y^*G[A]x|\le a\Bigl(\sum_k\|C_k\|_2^2\Bigr)^{1/2}. \] For \(G=F\) the left side is \(y^*F[A]x=\gamma>0\), so \(a>0\).

Step 2: the two ordered tests. For a polynomial \(G\) with coefficients in \(M_m(\mathbf C)\), \((FG)[A]=F[A]G[A]\) and \((GF)[A]=G[A]F[A]\) (Section 2 of the first lesson), so \[ \begin{aligned} y^*(FG)[A]x&=\gamma\,x^*G[A]x,\\ y^*(GF)[A]x&=\gamma\,y^*G[A]y . \end{aligned} \] Expand \(FG=\sum_kb_kU_k\) and \(GF=\sum_kb_kV_k\) (Proposition 4.1). On \(\partial\Omega=h(\mathbf T)\), \(\|F(z)G(z)\|_2\le\|F(z)\|\,\|G(z)\|_2\), which is at most \(\|G(z)\|_2\). With Step 1 and the lower inequality of the coefficient estimate (Proposition 4.6), this gives \[ \begin{aligned} \gamma|x^*G[A]x|&\le a\Bigl(\sum_k\|U_k\|_2^2\Bigr)^{1/2}\\ &\le a\|FG\|_\partial\le a\|G\|_\partial, \end{aligned} \] and in the same way, with \(\|G(z)F(z)\|_2\le\|G(z)\|_2\|F(z)\|\), \(\gamma|y^*G[A]y|\le a\|G\|_\partial\). For \(G=\sum_kb_kC_k\) put \[ \mathcal N(G)^2=\|C_0\|_2^2+2\sum_{k\ge1}\|C_k\|_2^2 . \] The upper inequality of Proposition 4.6 gives \[ \max\bigl(|x^*G[A]x|,|y^*G[A]y|\bigr)\le\frac a\gamma\,\mathcal N(G). \]

Step 3: weighted bounds for \(P\) and \(Q\). For \(N\ge1\) put \(d_N=\|P_0\|_2^2+\frac12\sum_{k=1}^N\|P_k\|_2^2\), and test Step 2 with \(C_0=\overline{P_0}\), \(C_k=\overline{P_k}/2\) for \(1\le k\le N\) (entrywise conjugates) and \(C_k=0\) otherwise. By Step 1, \(x^*G[A]x=d_N\), and \(\mathcal N(G)^2=\|C_0\|_2^2+2\sum_k\|C_k\|_2^2=d_N\). So \(d_N\le(a/\gamma)\sqrt{d_N}\), that is \(d_N\le a^2/\gamma^2\). Letting \(N\to\infty\), and arguing in the same way with \(Q\), \[ \begin{aligned} \|P_0\|_2^2+\tfrac12\sum_{k\ge1}\|P_k\|_2^2&\le a^2/\gamma^2,\\ \|Q_0\|_2^2+\tfrac12\sum_{k\ge1}\|Q_k\|_2^2&\le a^2/\gamma^2. \end{aligned} \]

Step 4: boundary norms. On \(\mathbf T\), \(P^*=\sum_kP_k^*\lambda^k\), so \(P+P^*\) has constant term \(2P_0\) and the coefficients \(P_k\), \(P_k^*\) at \(\lambda^{-k}\), \(\lambda^k\) for \(k\ge1\). Parseval's identity (Lemma 4.5) and Step 3 give \[ \begin{aligned} &\int_{\mathbf T}\|P+P^*\|_2^2\,d\tau\\ &\quad=4\|P_0\|_2^2+2\sum_{k\ge1}\|P_k\|_2^2\le\frac{4a^2}{\gamma^2}, \end{aligned} \] and the same bound for \(Q+Q^*\). The function \(M+L^*\) has constant term \(M_0+L_0^*=2M_0\) and the coefficients \(M_k\), \(L_k^*\) at \(\lambda^{-k}\), \(\lambda^k\), so \[ \begin{aligned} &\int_{\mathbf T}\|M+L^*\|_2^2\,d\tau\\ &\quad=4\|M_0\|_2^2+\sum_{k\ge1}\bigl(\|M_k\|_2^2+\|L_k\|_2^2\bigr)\\ &\quad\ge a^2 . \end{aligned} \]

Step 5: a positive block matrix. By Theorem 5.1, the \(2m\times2m\) matrix \[ \begin{aligned} \Delta&=\begin{pmatrix}X&Y\end{pmatrix}^*(R+R^*)\begin{pmatrix}X&Y\end{pmatrix}\\ &=\begin{pmatrix}P+P^*&L+M^*\\M+L^*&Q+Q^*\end{pmatrix} \end{aligned} \] is positive semidefinite at every point of \(\mathbf T\), and so is \(J\Delta J\) for \(J=\operatorname{diag}(I_m,-I_m)\). Hence \[ 0\le\operatorname{tr}\bigl(\Delta^{1/2}J\Delta J\Delta^{1/2}\bigr)=\operatorname{tr}(\Delta J\Delta J). \] For a Hermitian block matrix with diagonal blocks \(D_1,D_2\) and lower block \(E\), \(\operatorname{tr}(\Delta J\Delta J)=\operatorname{tr}D_1^2+\operatorname{tr}D_2^2-2\operatorname{tr}(E^*E)\). So, pointwise on \(\mathbf T\), \[ \begin{aligned} 2\|M+L^*\|_2^2&\le\|P+P^*\|_2^2\\ &\quad+\|Q+Q^*\|_2^2 . \end{aligned} \] Integrating and using Step 4, \[ 2a^2\le\frac{8a^2}{\gamma^2}. \] Since \(a>0\), \(\gamma\le2\). \(\square\)

The proof uses the two products \(FG\) and \(GF\) and never commutes coefficients of \(F\) and \(G\); this is why the bound holds for matrix coefficients of every size \(m\). Convexity of \(\Omega\) enters twice, both times through the support inequality: in the positivity of the kernel behind Proposition 4.6, and in \(R+R^*\succeq0\).

3. Proof of Theorem A

Let \(A\in M_n(\mathbf C)\) and let \(P\) be a polynomial with coefficients in \(M_m(\mathbf C)\). The numerical range \(K=W(A)\) is compact and convex (Section 1 of the first lesson). For \(0<\delta<1\), the domain \(\Omega_\delta\) of Smooth convex neighbourhoods is a smooth convex domain containing \(K\) (Lemma 2.1 and Proposition 2.3 there). If \(P\) vanishes on the infinite set \(\partial\Omega_\delta\), every entry of \(P\) is the zero polynomial and there is nothing to prove. Otherwise Theorem B, applied to \(P\) divided by \(\max_{\partial\Omega_\delta}\|P\|\), gives \[ \|P[A]\|\le2\max_{\partial\Omega_\delta}\|P\|\le2\max_{\overline\Omega_\delta}\|P\|. \] All the sets \(\overline\Omega_\delta\) lie in one rectangle \(Q\) (Lemma 2.1(1)), on which \(z\mapsto\|P(z)\|\) is uniformly continuous, with modulus of continuity \(\omega\). Every point of \(\overline\Omega_\delta\) has distance at most \(\varepsilon_\delta=\delta(1+\operatorname{diam}K)/(1-\delta)\) from \(K\) (Lemma 2.1(4)), so \[ \max_{\overline\Omega_\delta}\|P\|\le\max_K\|P\|+\omega(\varepsilon_\delta), \] and \(\omega(\varepsilon_\delta)\to0\) as \(\delta\to0\). Hence \(\|P[A]\|\le2\max_{W(A)}\|P\|\). The example of Section 3 of the first lesson shows that the constant \(2\) cannot be lowered. \(\square\)

4. Exercises

5.1 (easy). Show that the hypothesis \(W(A)\subset\Omega\) in Theorem B cannot be replaced by the condition that the spectrum of \(A\) lies in \(\Omega\).

5.2 (easy). Show that \(\|b_k(A)\|\le2\) for every \(k\ge1\) when \(W(A)\subset\Omega\). (This bound is due to Beckermann; see [BC].)

5.3 (medium). For scalar coefficients \(c_0,\dots,c_N\), put \(G=\sum_kc_kb_k\) and \(g(\lambda)=\sum_kc_k\lambda^k\). Show that \(G(A)=\int_{\mathbf T}(R+R^*)g\,d\tau-c_0I\), and deduce \(\|G(A)+c_0I\|\le2\max_{\mathbf T}|g|\).

5.4 (medium). For the ellipse \(h(\lambda)=\lambda+a/\lambda\) with \(0<a<1\) and \(A=0\in M_1(\mathbf C)\), compute \(R(\lambda)\), check \(R+R^*\ge0\) on \(\mathbf T\) directly, and deduce \(b_{2j}(0)=2(-a)^j\) and \(b_{2j+1}(0)=0\) for \(j\ge1\).

5. Solutions

5.1. For \(t>2\), \(A=\bigl(\begin{smallmatrix}0&t\\0&0\end{smallmatrix}\bigr)\) has spectrum \(\{0\}\subset\mathbf D\), and \(F(z)=z\) satisfies \(|F|=1\) on \(\partial\mathbf D\), but \(\|F[A]\|=t>2\). (Its numerical range is the disc of radius \(t/2>1\), Exercise 1.1 of the first lesson.)

5.2. For \(k\ge1\), \(\int_{\mathbf T}R\lambda^kd\tau=b_k(A)\) and \(\int_{\mathbf T}R^*\lambda^kd\tau=\sum_jb_j(A)^*\int\lambda^{j+k}d\tau=0\), so \(b_k(A)=\int(R+R^*)\lambda^kd\tau\). For unit vectors \(x,y\) put \(f_x=x^*(R+R^*)x\) and \(f_y=y^*(R+R^*)y\); these are nonnegative functions on \(\mathbf T\) with integral \(2\), because \(\int(R+R^*)\,d\tau=2I\). The Cauchy–Schwarz inequality for the positive semidefinite form \(R+R^*\) gives \[ |y^*b_k(A)x|\le\int_{\mathbf T}f_y^{1/2}f_x^{1/2}\,d\tau, \] and by the Cauchy–Schwarz inequality for the integral, the right side is at most \((\int f_y\,d\tau)^{1/2}(\int f_x\,d\tau)^{1/2}=2\).

5.3. As in 5.2, \(\int R\lambda^kd\tau=b_k(A)\) for \(k\ge0\), \(\int R^*\lambda^kd\tau=0\) for \(k\ge1\) and \(\int R^*d\tau=I\). So \(\int(R+R^*)g\,d\tau=G(A)+c_0I\). For unit \(x,y\), with \(f_x\) and \(f_y\) as in 5.2, \[ \begin{aligned} &|y^*(G(A)+c_0I)x|\\ &\quad\le\max_{\mathbf T}|g|\int_{\mathbf T}f_y^{1/2}f_x^{1/2}\,d\tau\le2\max_{\mathbf T}|g| . \end{aligned} \]

5.4. Here \(0\in\Omega\), \(W(A)=\{0\}\), and \[ R(\lambda)=\frac{q(\lambda)}{h(\lambda)}=\frac{\lambda-a/\lambda}{\lambda+a/\lambda}=\frac{\lambda^2-a}{\lambda^2+a}. \] With \(w=\lambda^2\in\mathbf T\), \[ \begin{aligned} \operatorname{Re}\frac{w-a}{w+a}&=\frac{\operatorname{Re}\bigl((w-a)(\bar w+a)\bigr)}{|w+a|^2}\\ &=\frac{1-a^2}{|w+a|^2}>0 . \end{aligned} \] Expanding in \(\lambda^{-1}\), \[ \frac{1-a\lambda^{-2}}{1+a\lambda^{-2}}=1+2\sum_{j\ge1}(-a)^j\lambda^{-2j}, \] and the coefficients are \(b_k(0)\) by Theorem 5.1. For example \(b_2(0)=-2a\), in agreement with \(b_2(z)=z^2-2a\).

References