The Sharp Numerical-Range Spectral-Set Inequality

The theorem below is the sharp constant-2 assertion conjectured by Crouzeix [1]. Crouzeix later obtained the dimension-free bound $11.08$ [2], and the previously published general bound is $1+\sqrt{2}$, due to Crouzeix and Palencia [5]. Two contemporaneous preprints also claim the sharp bound, by Jin [14] and by Lorist and Schwenninger [15]. The present proof is stated only for scalar-valued functions and finite-dimensional matrices.

Theorem 1. 

Let $A\in {\mathbb{C}}^{n\times n}$, and let

$$W\left(A\right)=\{x^{\ast }Ax:x\in {\mathbb{C}}^n,\parallel x{\parallel }_2=1\}.$$

Then every complex polynomial p satisfies

$${\parallel p\left(A\right)\parallel }_2\le 2\underset{z\in W\left(A\right)}{max}\left|p\left(z\right)\right|.$$

The same inequality holds for every rational function whose poles lie off $W\left(A\right)$. Thus $W\left(A\right)$, which is compact in finite dimensions, is a 2-spectral set for A.

Proof. 

We first prove the inequality with $W\left(A\right)$ replaced by a particularly regular convex neighborhood. Let $\Omega$ be a bounded convex domain with real-analytic strictly convex boundary, and assume that $W\left(A\right)⋐\Omega$. We shall prove

$${\parallel h\left(A\right)\parallel }_2\le 2{\parallel h\parallel }_{\infty ,\Omega }(h\in H^{\infty }(\Omega )).$$

(1)

The conjugate-Cauchy and positive double-layer framework used below originates in Delyon and Delyon [3] and was developed for convex spectral sets and numerical-range functional calculus in [2,4,5]. For a clarification of the symmetrized mechanism and a recent systematic treatment of the double-layer potential, see [6,7]. The finite-fiber kernel and the all-powers recurrence below are the forms used in the present argument.

We use counterclockwise orientation on $\partial \Omega$. At $\zeta \in \partial \Omega$, let $n\left(\zeta \right)$ be the outward unit normal, regarded as a complex number, and let $ds$ denote arclength. Thus

$$d\zeta =in\left(\zeta \right)ds.$$

(2)

For a function u analytic on a neighborhood of $\overline{\Omega }$, define its conjugate Cauchy transform by

$$\left(\Gamma u\right)\left(z\right)={\displaystyle \frac{1}{2\pi i}}{\int }_{\partial \Omega }{\displaystyle \frac{\overline{u\left(\zeta \right)}}{\zeta -z}}d\zeta ,z\in \Omega .$$

(3)

We next establish the boundary kernel that drives the proof. Let $\varphi :\Omega \to \mathbb{D}$ be a conformal map. Real-analyticity of the boundary and Schwarz reflection imply that $\varphi$ extends conformally across $\partial \Omega$ [8]. Let

$$f=B\circ \varphi ,$$

(4)

where B is a nonconstant finite Blaschke product. Thus f is analytic on a neighborhood of $\overline{\Omega }$ and $\left|f\right|=1$ on $\partial \Omega$. If $\zeta =\zeta \left(s\right)$ is an arclength parametrization and $f\left(\zeta \left(s\right)\right)=e^{it\left(s\right)}$, then

$$\kappa \left(\zeta \right):=t^{\prime }\left(s\right)>0,f^{\prime }\left(\zeta \right)n\left(\zeta \right)=f\left(\zeta \right)\kappa \left(\zeta \right).$$

(5)

Indeed, a finite Blaschke product restricts to an orientation-preserving covering of the unit circle. In particular, all boundary solutions of $f\left(\zeta \right)=\alpha$, $\alpha \in \mathbb{T}$, are simple.

For $\alpha \in \mathbb{T}$, put

$$K_{\alpha }\left(z\right)=\sum _{\begin{array}{c}\zeta \in \partial \Omega \\ f\left(\zeta \right)=\alpha \end{array}}{\displaystyle \frac{n\left(\zeta \right)}{\kappa \left(\zeta \right)(\zeta -z)}}.$$

(6)

This kernel has positive real part at A. In fact, convexity gives

$$Re\left(\overline{n\left(\zeta \right)}(\zeta I-A)\right)\succ 0,$$

because $W\left(A\right)⋐\Omega$. If T is strictly accretive, then

$$Re\left(T^{-1}\right)=T^{-\ast }(ReT)T^{-1}⪰0.$$

(7)

Since

$$n\left(\zeta \right){(\zeta I-A)}^{-1}={\left(\overline{n\left(\zeta \right)}(\zeta I-A)\right)}^{-1},$$

we obtain

$$H_{\alpha }:=Re{K}_{\alpha }\left(A\right)⪰0.$$

(8)

Let $dm\left(\alpha \right)=dt/\left(2\pi \right)$ be normalized arclength measure on $\mathbb{T}$. The change of variables $\alpha =f\left(\zeta \right)$ and (5) give $dt=\kappa \left(\zeta \right)ds$ on every sheet of the boundary covering. Hence, for $m\ge 0$,

$$\begin{array}{cc}\hfill {\int }_{\mathbb{T}}{\alpha }^m{K}_{\alpha }\left(z\right)dm\left(\alpha \right)& ={\displaystyle \frac{1}{2\pi }}{\int }_{\partial \Omega }{\displaystyle \frac{f{\left(\zeta \right)}^{m}n\left(\zeta \right)}{\zeta -z}}ds\hfill \\ \hfill & ={\displaystyle \frac{1}{2\pi i}}{\int }_{\partial \Omega }{\displaystyle \frac{f{\left(\zeta \right)}^m}{\zeta -z}}d\zeta =f{\left(z\right)}^m,\hfill \end{array}$$

(9)

and, for $m\ge 1$,

$${\int }_{\mathbb{T}}{\overline{\alpha }}^m{K}_{\alpha }\left(z\right)dm\left(\alpha \right)={\displaystyle \frac{1}{2\pi i}}{\int }_{\partial \Omega }{\displaystyle \frac{\overline{f{\left(\zeta \right)}^m}}{\zeta -z}}d\zeta =:g_m\left(z\right),$$

(10)

where $g_m=\Gamma \left(f^m\right)$. These identities may be evaluated at A, since all functions involved are analytic on a neighborhood of $spec\left(A\right)$. If $F=f\left(A\right)$ and $G_m=g_m\left(A\right)$, they yield

$${\int }_{\mathbb{T}}{H}_{\alpha }dm\left(\alpha \right)=I,{\int }_{\mathbb{T}}{\alpha }^m{H}_{\alpha }dm\left(\alpha \right)={\displaystyle \frac{F^m+G_m^{\ast }}{2}}(m\ge 1).$$

(11)

The positivity in (8) and the first identity in (11) imply

$$\parallel F^m+G_m^{\ast }{\parallel }_2\le 2(m\ge 1).$$

(12)

To see this directly, for unit vectors $u,v$ use positivity first in the pointwise inequality $|v^{\ast }{H}_{\alpha }{u|}^2\le \left(v^{\ast }{H}_{\alpha }v\right)\left(u^{\ast }{H}_{\alpha }u\right)$, and then use Cauchy–Schwarz in $L^2\left(dm\right)$ to obtain

$$\left|v^{\ast }\left({\int }_{\mathbb{T}}{\alpha }^m{H}_{\alpha }dm\right)u\right|\le {\left({\int }_{\mathbb{T}}{v}^{\ast }{H}_{\alpha }vdm\right)}^{1/2}{\left({\int }_{\mathbb{T}}{u}^{\ast }{H}_{\alpha }udm\right)}^{1/2}=1.$$

The same scalar argument, with A replaced by any $z\in \Omega$, gives

$${|f\left(z\right)}^m+\overline{g_m\left(z\right)}|\le 2,\parallel g_m{\parallel }_{\infty ,\Omega }\le 3(m\ge 1).$$

(13)

We need one more consequence of the same kernel. Define

$$P_{\alpha }\left(z\right)={\displaystyle \frac{1}{1-\overline{\alpha }f\left(z\right)}},R_{\alpha }\left(z\right)=P_{\alpha }\left(z\right)-K_{\alpha }\left(z\right),C_{\alpha }=\alpha I-F.$$

(14)

The principal part of $P_{\alpha }$ at a boundary solution $f\left(\zeta \right)=\alpha$ is

$${\displaystyle \frac{n\left(\zeta \right)}{\kappa \left(\zeta \right)(\zeta -z)}},$$

by (5). Thus the apparent boundary poles cancel and $R_{\alpha }$ is analytic across $\partial \Omega$. Furthermore,

$$P_{\alpha }\left(z\right)=\sum _{m\ge 0}{\overline{\alpha }}^{m}f{\left(z\right)}^m(z\in \Omega ).$$

Comparing its Fourier coefficients with (9)–(10) gives

$$\begin{array}{cccc}\hfill {\int }_{\mathbb{T}}{\alpha }^m{R}_{\alpha }\left(z\right)dm\left(\alpha \right)& =0\hfill & \hfill & (m\ge 0),\hfill \end{array}$$

(15)

$$\begin{array}{cccc}\hfill {\int }_{\mathbb{T}}{\overline{\alpha }}^m{R}_{\alpha }\left(z\right)dm\left(\alpha \right)& =-g_m\left(z\right)\hfill & \hfill & (m\ge 1).\hfill \end{array}$$

(16)

Only these Fourier-coefficient identities will be used; therefore no pointwise convergence of a Fourier series is required.

Since $I-\overline{\alpha }F=\overline{\alpha }{C}_{\alpha }$, we have $P_{\alpha }\left(A\right)=\alpha C_{\alpha }^{-1}$. Here $C_{\alpha }$ is invertible: the spectral mapping theorem and the maximum principle give $spec\left(F\right)=f(spec(A\left)\right)\subset \mathbb{D}$. A direct multiplication shows

$$C_{\alpha }^{\ast }Re\left(P_{\alpha }\left(A\right)\right)C_{\alpha }=I-Re\left(\overline{\alpha }F\right).$$

(17)

Using $P_{\alpha }=K_{\alpha }+R_{\alpha }$, we therefore obtain the exact decomposition

$$E_{\alpha }:=I-Re\left(\overline{\alpha }F\right)=Q_{\alpha }+C_{\alpha }^{\ast }Re\left(R_{\alpha }\left(A\right)\right)C_{\alpha },Q_{\alpha }:=C_{\alpha }^{\ast }{H}_{\alpha }{C}_{\alpha }⪰0.$$

(18)

We now prove that every f of the form (4) satisfies ${\parallel f\left(A\right)\parallel }_2\le 2$. Suppose, to the contrary, that

$$\lambda :={\parallel F\parallel }_2>2.$$

(19)

Choose unit singular vectors $x,y$ such that

$$Fx=\lambda y,F^{\ast }y=\lambda x.$$

(20)

For every function h analytic on $\Omega$, put

$$\tau \left(h\right)=x^{\ast }h\left(A\right)x.$$

Because $h\left(A\right)$ commutes with F, (20) gives

$$y^{\ast }h\left(A\right)y=\tau \left(h\right),x^{\ast }h\left(A\right)y={\displaystyle \frac{1}{\lambda }}\tau \left(fh\right).$$

(21)

Indeed,

$$y^{\ast }h\left(A\right)y={\lambda }^{-2}{x}^{\ast }{F}^{\ast }h\left(A\right)Fx={\lambda }^{-2}{x}^{\ast }{F}^{\ast }Fh\left(A\right)x=x^{\ast }h\left(A\right)x,$$

and the second equality follows from $y=Fx/\lambda$.

For $m\ge 1$, define the real numbers

$${\beta }_m=Re\tau \left(f^m{g}_m\right).$$

(22)

They are bounded uniformly in m. To make this completely explicit, choose a positively oriented contour $\gamma ⋐\Omega$ of winding number one around $spec\left(A\right)$. The holomorphic functional calculus gives

$${\parallel h\left(A\right)\parallel }_2\le {M\parallel h\parallel }_{\infty ,\Omega },M:={\displaystyle \frac{length\left(\gamma \right)}{2\pi }}\underset{z\in \gamma }{max}{\parallel {(zI-A)}^{-1}\parallel }_2<\infty .$$

(23)

Consequently, by (13),

$$|{\beta }_m|\le \parallel \left(f^m{g}_m\right)\left(A\right){\parallel }_2\le 3M(m\ge 1).$$

(24)

The case $m=1$ of (12), together with (21), gives

$$\begin{array}{cc}\hfill 2& \ge Re\left(y^{\ast }(F+G_1^{\ast })x\right)\hfill \\ \hfill & =\lambda +{\displaystyle \frac{1}{\lambda }}Re\tau \left(f{g}_1\right)=\lambda +{\displaystyle \frac{{\beta }_1}{\lambda }}.\hfill \end{array}$$

(25)

Thus

$${\beta }_1\le -\lambda (\lambda -2)<1.$$

(26)

Fix an integer $N\ge 1$ and, for $\alpha \in \mathbb{T}$, set

$$v_N\left(\alpha \right)=x+{\displaystyle \sum _{j=1}^N}{\overline{\alpha }}^j{F}^{j-1}y.$$

(27)

The sum telescopes after multiplication by $C_{\alpha }$:

$$w_N\left(\alpha \right):=C_{\alpha }{v}_N\left(\alpha \right)=\alpha x-(\lambda -1)y-{\overline{\alpha }}^N{F}^{N}y.$$

(28)

Fourier orthogonality also gives the exact identity

$${\int }_{\mathbb{T}}{v}_N{\left(\alpha \right)}^{\ast }{E}_{\alpha }{v}_N\left(\alpha \right)dm\left(\alpha \right)=2-\lambda .$$

(29)

For completeness, the mean of $\parallel v_N{\parallel }_2^2$ is

$$1+{\displaystyle \sum _{k=0}^{N-1}}{\parallel F^{k}y\parallel }_2^2,$$

whereas the mean of $\overline{\alpha }{v}_N^{\ast }F{v}_N$ is

$$\lambda +{\displaystyle \sum _{k=1}^{N-1}}{\parallel F^{k}y\parallel }_2^2.$$

Their difference is precisely $2-\lambda$.

Applying the positive decomposition (18) to $v_N$ and using (29), we find

$$2-\lambda \ge {\int }_{\mathbb{T}}{w}_N{\left(\alpha \right)}^{\ast }Re\left(R_{\alpha }\left(A\right)\right)w_N\left(\alpha \right)dm\left(\alpha \right).$$

(30)

Put $d=\lambda -1$. The three Fourier exponents occurring in $w_N$ are $1,0,-N$. Expanding the right-hand side and using (15)–() shows that the only nonzero pairings are

$$\begin{array}{cc}\hfill {\int }_{\mathbb{T}}{w}_N^{\ast }{R}_{\alpha }\left(A\right)w_{N}dm& =d{x}^{\ast }{G}_{1}y+x^{\ast }{G}_{N+1}{F}^{N}y-d{y}^{\ast }{G}_N{F}^{N}y.\hfill \end{array}$$

(31)

By (21), the real part of (31) equals

$${\displaystyle \frac{\lambda -1}{\lambda }}{\beta }_1+{\displaystyle \frac{1}{\lambda }}{\beta }_{N+1}-(\lambda -1){\beta }_N.$$

(32)

It follows from (30) that

$$\begin{array}{|c|}\hline {\beta }_{N+1}\le -\lambda (\lambda -2)-(\lambda -1){\beta }_1+\lambda (\lambda -1){\beta }_N\\ \hline\end{array}(N\ge 1).$$

(33)

This recurrence contradicts the uniform bound (24). Indeed, set

$$q=\lambda (\lambda -1)>1,c=-\lambda (\lambda -2)-(\lambda -1){\beta }_1,$$

and define $U_1={\beta }_1$ and $U_{N+1}=c+q{U}_N$. Induction in (33) gives ${\beta }_N\le U_N$. If

$$s=-{\displaystyle \frac{c}{q-1}}$$

is the fixed point of this affine recurrence, then

$$(q-1)({\beta }_1-s)=\lambda (\lambda -2)({\beta }_1-1)<0$$

(34)

by (19) and (26). Therefore

$$U_N=s+q^{N-1}({\beta }_1-s)⟶-\infty .$$

For large N this gives ${\beta }_N<-3M$, contradicting (24). The contradiction proves

$${\parallel (B\circ \varphi )\left(A\right)\parallel }_2\le 2$$

(35)

for every finite Blaschke product B. The assertion is immediate for a constant unimodular B.

We now pass from finite Blaschke products to all bounded analytic functions. We record the elementary approximation argument. If $q:\mathbb{D}\to \overline{\mathbb{D}}$ is analytic, apply the Schur algorithm [9]

$$q_0=q,{\gamma }_j=q_j\left(0\right),q_{j+1}\left(z\right)={\displaystyle \frac{q_j\left(z\right)-{\gamma }_j}{z(1-\overline{{\gamma }_j}{q}_j\left(z\right))}}.$$

As long as $|{\gamma }_j|<1$, Schwarz’s lemma shows that $q_{j+1}$ is again a Schur function. At a chosen level N, replace $q_N$ by a unimodular constant and reverse the recursion by

$$b_j\left(z\right)={\displaystyle \frac{{\gamma }_j+z{b}_{j+1}\left(z\right)}{1+\overline{{\gamma }_j}z{b}_{j+1}\left(z\right)}}.$$

(36)

The resulting $b_0$ is a finite Blaschke product. Each reverse step in (36) raises by one the order of vanishing of the difference from the original Schur iterate. Hence $b_0-q=O\left(z^N\right)$. Since $|b_0-q|\le 2$, the generalized Schwarz lemma even gives

$$|b_0{\left(z\right)-q\left(z\right)|\le 2|z|}^N.$$

If a Schur parameter is unimodular, the algorithm has already terminated in a finite Blaschke product. We have thus proved that finite Blaschke products are locally uniformly dense in the Schur class.

Given $h\in H^{\infty }(\Omega )$ with ${\parallel h\parallel }_{\infty ,\Omega }\le 1$, apply this approximation to $q=h\circ {\varphi }^{-1}$. If $B_j\to q$ locally uniformly on $\mathbb{D}$, then $B_j\circ \varphi \to h$ locally uniformly on $\Omega$. A Cauchy integral over a fixed contour surrounding $spec\left(A\right)$ gives

$$(B_j\circ \varphi )\left(A\right)⟶h\left(A\right).$$

Passing to the limit in (35) proves (1).

It remains to let the regular convex domains approach $W\left(A\right)$. By the Toeplitz–Hausdorff theorem [10,11], $W\left(A\right)$ is convex. The standard compression proof recalls that, in dimension at most two, the numerical range is a (possibly degenerate) elliptical disk [12]; it therefore contains the segment joining any two prescribed points of $W\left(A\right)$. Compactness follows from continuity on the unit sphere. We briefly justify the needed geometric approximation. For the support-function and curvature-measure facts used below, see [13]. For any compact convex set $K\subset \mathbb{C}$, its support function

$$h_K\left(\theta \right)=\underset{z\in K}{max}Re\left(e^{-i\theta }z\right)$$

is continuous and the distribution $h_K+h_K^{\prime \prime }$ is a nonnegative measure. Convolve $h_K$ with a periodic Poisson kernel, add its uniform approximation error, and then add a positive constant tending to zero. The resulting real-analytic support functions $h_j$ satisfy

$$h_j>h_K,h_j+h_j^{\prime \prime }>0,h_j⟶h_K\mathrm{uniformly}.$$

The curves

$${\zeta }_j\left(\theta \right)=e^{i\theta }\left(h_j\left(\theta \right)+i{h}_j^{\prime }\left(\theta \right)\right)$$

bound real-analytic strictly convex domains ${\Omega }_j$ with $K⋐{\Omega }_j$, and ${\overline{\Omega }}_j\to K$ in Hausdorff distance.

Apply this construction to $K=W\left(A\right)$. From (1),

$${\parallel p\left(A\right)\parallel }_2\le 2\underset{z\in {\overline{\Omega }}_j}{max}\left|p\left(z\right)\right|.$$

Because p is continuous and ${\overline{\Omega }}_j\to W\left(A\right)$ in Hausdorff distance, the right-hand side tends to $2{max}_{z\in W\left(A\right)}\left|p\left(z\right)\right|$. This proves the theorem.

Finally, if r is rational and has no pole on $W\left(A\right)$, choose the domains ${\Omega }_j$ above sufficiently close to $W\left(A\right)$ that they contain no pole of r. Applying (1) directly to r on each ${\Omega }_j$ and then taking the same Hausdorff limit gives

$${\parallel r\left(A\right)\parallel }_2\le 2\underset{z\in W\left(A\right)}{max}\left|r\left(z\right)\right|.$$

This proves the stated spectral-set formulation as well.    □

Sharpness. 

The constant cannot be reduced. For

$$A=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),p\left(z\right)=z,$$

the numerical range is the closed disk of radius $1/2$ centered at the origin, whereas ${\parallel p\left(A\right)\parallel }_2={\parallel A\parallel }_2=1$. Thus the ratio in the theorem is exactly 2.