Numerical Range and Functional Calculus in Hilbert Space

Numerical range and functional calculus in Hilbert space

Michel Crouzeix

Abstract We prove an inequality related to polynomial functions of a square matrix, involving the numerical range of the matrix. We also show extensions valid for bounded and also unbounded operators in Hilbert spaces, which allow the development of a functional calculus.

Mathematical subject classiﬁcation : 47A12 ; 47A25

Keywords : functional calculus, numerical range, spectral sets.

1 Introduction

In this paper H denotes a complex Hilbert space equipped with the inner product . , . and h i corresponding norm v = v, v 1/2. We denote its unit sphere by Σ := v H; v = 1 . For k k h i H { ∈ k k } a bounded linear operator A (H) we use the operator norm A and denote by W (A) its ∈ L k k numerical range :

A := sup A v , W (A) := Av,v ; v ΣH . k k v ΣH k k {h i ∈ } ∈ Recall that the numerical range is a convex subset of C and that its closure contains the spectrum of A. We keep the same notation in the particular case where H = Cd and A is a d d matrix. × The main aim of this article is to prove that the inequality

p(A) 56 sup p(z) (1) k k≤ z W (A) | | ∈ holds for any matrix A Cd,d and any polynomial p : C C (independently of d). ∈ → It is remarkable that the completely bounded version of this inequality is also valid. More precisely, we consider now matrix-valued polynomials P : C Cm,n, i.e., for z C, P (z) = → ∈ (p (z)) is a matrix, with each entry p (.) a polynomial C C. Then the inequality ij ij → P (A) 56 sup P (z) , (2) k k≤ z W (A) k k ∈ holds for any matrix A Cd,d, any polynomial P : C Cm,n, and any values of d, m, n. ∈ → We now ﬁx a matrix A Cd,d and consider the algebra C[z] of polynomials p from W (A) C ∈ into , provided with the norm p ,A = supz W (A) p(z) . The map p p(A) is then a k k∞ ∈ | | 7→ homomorphism from the algebra C[z] into the algebra Cd,d. The inequality (1) means that this homomorphism is bounded with constant 56, and the inequality (2) that it is complely bounded with constant 56. These inequalities may be summarized in the following statement.

1 Theorem 1. For any matrix A Cd,d the homomorphism p p(A) from the algebra C[z], with ∈d,d 7→ norm . ,A, into the algebra C is completely bounded, with constant 56. k k∞ It will be clear from the proof that the value 56 is not optimal. Denoting by and the Q Qcb smallest constants such that the inequalities

p(A) sup p(z) , P (A) cb sup P (z) , (3) k k ≤ Q z W (A) k k k k ≤ Q z W (A) k k ∈ ∈ hold for any matrix A Cd,d, any polynomial p : C C, P : C Cm,n and any values of ∈ → → d, m, n, we have

2 56. ≤Q≤Qcb ≤ 0 2 The lower bound is attained by the polynomial p(z)= z and the matrix A = , in which 0 0 case W (A) is the unit disk. Using the fact that these constants are independent of d, we will deduce that the inequalities (3) are still valid for any bounded linear operator A (H) on any Hilbert space H. For a ∈ L convex subset E of C we introduce the algebra (E) of continuous and bounded functions in Hb E which are holomorphic in the interior of E. The following statement shows the existence of a functional calculus based on the numerical range.

Theorem 2. Let H be a Hilbert space. For any bounded linear operator A (H) the ho- ∈ L momorphism p p(A) from the algebra C[z], with norm . ,A, into the algebra (H), is 7→ k k∞ L bounded with constant . It admits a unique bounded extension from (W (A)) into (H). Q Hb L This extension is bounded with constant and completely bounded with constant . Q Qcb For many applications it is useful to consider unbounded operators. Assume now that A ∈ (D(A),H) is a closed linear operator with domain D(A) densely and continuously embedded in L H. Its numerical range is then deﬁned by W (A) := Av,v ; v Σ D(A) . We assume that {h i ∈ H ∩ } the spectrum σ(A) is included in W (A). The numerical range is still convex but unbounded and we hence cannot apply polynomials. We therefore instead consider the algebra Cb(z) of rational functions which are bounded on W (A), and provide it with the norm r ,A = supz W (A) r(z) . k k∞ ∈ k k We then have the following statement similar to Theorem 2

Theorem 3. For any closed linear unbounded operator A such that σ(A) W (A), the ho- ⊂ momorphism r r(A) from the algebra Cb(z), with norm . ,A, into the algebra (H), is 7→ k k∞ L bounded with constant . It admits a unique bounded extension from the algebra (W (A)) into Q Hb (H). This extension is bounded with constant and completely bounded with constant . L Q Qcb An outline of this paper is the following : in Section 2 we deduce Theorems 2 and 3 from the inequality (3), and illustrate them by application to the theory of Cosine functions. Section 3 is devoted to an integral representation formula for P (A), which is the key to proving (2). In Section 4 we prove some preliminary bounds which allow us to treat most matrices. The remaining case, which corresponds to very thin numerical ranges, is more complicated and is considered in Section 5. Section 6 is devoted to the proof of a geometrical inequality for convex domains in the plane, which is needed in Section 4. Some concluding remarks are given in Section 7. The author conjectures that, in fact, = = 2. Q Qcb

2 2 The extension to inﬁnite dimension

Proof of Theorem 2. We assume that (3) is valid if H = Cd, for all integers d. This implies that (3) holds for any ﬁnite dimensional Hilbert space. Let now H be inﬁnite-dimensional and consider an operator A (H) and a polynomial p of degree n. For u H given, we introduce ∈L ∈ the Krylov space = Span u, Au, . . . , Anu H, denote by Π the orthogonal projection Kn { } ⊂ n onto n, and set An := ΠnA n . We clearly have p(A) u = p(An) u and W (An) W (A), and K |K ⊂ A acts on the n+1 dimensional space . We then deduce that n Kn p(A) u = p(An) u sup p(z) u sup p(z) u , k k k k ≤ Q z W (An) | | k k ≤ Q z W (A) | | k k ∈ ∈ which implies p(A) sup p(z) . The same proof works also for polynomials with matrix k k ≤ Q z W (A) | | ∈ values, thus the inequalities (3) are still valid. The Mergelyan theorem states that C[z] is dense in (W (A)), which completes the proof. Hb

Proof of Theorem 3. We consider now a closed linear unbounded operator A such that σ(A) ⊂ W (A) and a rational function r bounded on W (A). It is clear that r(A) is well deﬁned and belongs to (H) (write r in simple partial fraction form). In the case W (A) = C, the space L (W (A)) is reduced to constant functions and Theorem 2 follows immediately (but has no Hb interest). In the other cases, changing, if needed, A to α + eiθA, α C, θ R, we have only ∈ ∈ two possibilities to consider : a) The numerical range is a strip : W (A)= z C ; 0 Re z a , a> 0. { ∈ ≤ ≤ } 1 1 We then set Aε = (1+εA∗)− A(1+εA)− , for ε> 0. b) x ; x> 0 W (A) z C ; Re z 0 . { }⊂ ⊂ { ∈ ≥ } 1 We then set Aε = A(1+εA)− , for ε> 0. In both cases it is easy to verify that A is a bounded operator and that W (A ) W (A). Let ε ε ⊂ r be a rational function bounded in W (A), clearly r (W (A )). From Theorem 2 we have ∈Hb ε

r(Aε) sup r(z) sup r(z) . (4) k k ≤ Q z W (Aε) | | ≤ Q z W (A) | | ∈ ∈

We now note that limε 0 r(Aε)= r(A), strongly on H. Indeed, it suﬃces to verify that → 1 1 lim(λ Aε)− u = (λ A)− u for all λ / W (A) and all u D(A). ε 0 − − ∈ ∈ → We consider the case of a strip, the other situation being simpler. We have by a simple calculation 1 1 1 1 1 1 (λ Aε)− u (λ A)− u = ε(λ Aε)− (1+εA∗)− (A+A∗)(1+εA)− (λ A)− Au − − − − − 1 1 1 − 1 ε(λ A )− (1+εA∗)− (εA∗)(1+εA)− A(λ A)− Au. − − ε − The convergence to 0 of the second member follows from the bounds 1 1 A + A∗ 2 a, (1 + εA∗)− 1, (1 + εA∗)− (εA∗) 1, k k≤ k k≤ k k≤ 1 1 1 1 (λ A )− , (λ A)− , k − ε k≤ d(λ, W (A)) k − k≤ d(λ, W (A))

1 λ A (λ A)− | | . k − k≤ d(λ, W (A))

3 Taking the limit as ε 0 in (4), we obtain that the map r r(A) is bounded with constant → 7→ . The space C (z) is dense in the subset (W (A)) of functions of (W (A)) which have a Q b Hb,0 Hb limit at . This therefore allows a first extension of the map to this subspace. ∞ Let us consider now f (W (A)) and λ / W (A). With g (z) = (λ z) 1f(z) we have ∈ Hb ∈ λ − − g (W (A)), and from the previous extension, g (A) (H) is well defined. If, in addition, λ ∈Hb,0 λ ∈L f (W (A)), then f(A) (H) and g (A) = (λ A) 1f(A). This shows that g (A) ∈ Hb,0 ∈ L λ − − λ ∈ (H, D(A)) and f(A) = (λ A)g (A). L − λ We return to the general case f (W (A)). Recall that we have assumed W (A) z ∈ Hb ⊂ { ∈ C ; Re z 0 . From the inequality, with ε> 0, ≥ } 1 z f(z) λ A(1+εA)− gλ(A) sup | | f ,A, k k ≤ Q z W (A) 1+ εz λ z ≤ Q d(λ, W (A)) k k∞ ∈ − λ we deduce, by taking the limit as ε 0, that A gλ(A) | | f ,A and gλ(A) → k k ≤ Q d(λ,W (A)) k k∞ ∈ (H, D(A)). We can therefore define f(A) (H) by f(A) = (λ A)g (A). Passing to the limit L ∈L − λ in the inequality

1 λ z f(z) (1+εA)− f(A) sup − f ,A, k k ≤ Q z W (A) 1+ εz λ z ≤ Q k k∞ ∈ − we deduce f(A) f ,A. It is easily seen that our deﬁnition does not depend on the k k ≤ Qk k∞ particular choice of λ. The same arguments work for the complete bound.

In order to illustrate the previous results, we describe the following application, which has been mentioned previously in [3]. Assume that the unbounded operator A satisﬁes the hypothe- ses of Theorem 3, and that there exists α> 0 such that

W (A) := x + iy C ; y2 α2x . ⊂ P { ∈ ≤ } As a result of Theorem 3 the operators exp(it√A) and cos(t√A) are well deﬁned for all t R ∈ and strongly continuous in H. The problem

u′′(t)+ A u(t) = 0, for t R, u(0) = u , u′(0) = v , ∈ 0 0 with u D(A), v D(A1/2) given, therefore admits a unique solution u C2(R; H) 0 ∈ 0 ∈ ∈ ∩ C0(R; D(A)), which may be written

t u(t) = cos(t√A) u0 + cos(s√A) v0 ds. Z0 Examples of such operators A can be found among the ones associated with convection diﬀusion equations

Av := div (a ~ v)+~b.~ v + cv, − ∇ ∇ with suitable assumptions on the coeﬃcients a, ~b and c, and appropriate choices of the spaces H and D(A). To our knowledge, until now no other methods have provided such an existence result.

4 3 The representation formula

From now on, A will be a d d matrix . Let Ω be a bounded convex domain of C such that × W (A) Ω; we assume that its boundary ∂Ω is smooth and has a strictly positive curvature at ⊂ each point. We then deduce from (2) that

P (A) cb sup P (z) , for any matrix-valued polynomial P . (5) k k ≤ Q z Ω k k ∈ Conversely, if this estimate holds for all such convex domains, we easily obtain (2). After the use of a similarity transformation, we can assume that there exist two functions 0 η C ([0, 2]) C∞((0, 2)) such that ± ∈ ∩ 0, 2 ∂Ω, diam (Ω) = 2, Ω= x+iy ; 0

Πθ θ • ∂Ω σ • + • ψ σ σ+ (b) h • τ(a) • Γ • τ •τ(b) • • • • • 0 a x b 2 • • • σ−

In order to make the figure more clear, we have enlarged the scale in the vertical direction. (ii) Thick domains : h(1) 10 5. In this case we will use a = b = 1. ≥ − We denote by σ the generic point of arclength s on the boundary ∂Ω, oriented counterclock- dσ dσ iθ wise, and set θ = arg( ds ) so that ds = e . Due to the strict convexity, the point σ can be considered as a C∞ function of θ. We define the following functions of the real variable x : σ : (0, c) C is defined by σ (x) ∂Ω, Im σ (x) > 0 and x = Re σ (x), + → + ∈ + + σ : (0, c) C is defined by σ (x) ∂Ω, Im σ (x) < 0 and x = Re σ (x), − → − ∈ − − dσ+ dσ θ+ = arg( ), θ = arg( − ). ds − ds We also use the following notations 1 d 1 eiθ e iθ µ(σ, z)= arg(σ z) = ( − ), π ds − 2πi σ z − σ¯ z¯ − − eiθ Π = z ; µ(σ, z) > 0 = z ; Re (¯σ z¯) > 0 . θ { } { i − }

5 ξ(θ,t) = tσ(θ) + (1 t) τ(a) if Re σ(θ) a, − ≤ = tσ(θ) + (1 t) τ(Re σ(θ)) if Re σ(θ) [a, b], − ∈ = tσ(θ) + (1 t) τ(b) if Re σ(θ) b, − ≥

1 1 ∂ξ e2iθ(¯σ(θ) z¯) Jp(θ, z¯) = p(ξ(θ,t)) − dt, π Z ∂t (ξ(θ,t) σ(θ)+ e2iθ(¯σ(θ) z¯))2 0 − − b 2iθ+ 1 p(τ) e− dτ G+p(¯z) = dx, 2πi Z τ¯ z¯ i (1+e 2iθ+ ) h dx a − − − 1 b p(τ) e 2iθ− dτ G p(¯z) = − dx. − 2πi Z τ¯ z¯ + i (1+e 2iθ− ) h dx a − − The proof of our main result, Theorem 1, is based on the following lemma.

Lemma 4. For any polynomial p and any z Ω, we have the representation formula ∈ 2π p(z)= p(σ) µ(σ, z) ds + Jp(θ, z¯) dθ G+p(¯z)+ G p(¯z). (6) Z∂Ω Z0 − − Furthermore,

1 sup Jp(θ, z¯) ( 2 + cot(θ ψ) ) sup p(ζ) , (7) z Πθ | |≤ | − | ζ Ω | | ∈ ∈ where arg(σ τ(a)) if Re σ a, − ≤ ψ =  arg(σ σ¯)) if a< Re σ < b, − arg(σ τ(b)) if Re σ b, − ≥  defined in such a way that θ ψ (0,π). − ∈ Proof. From the Cauchy formula we have 1 dσ p(z)= p(σ) . 2πi Z σ z ∂Ω − The relation (6) is therefore equivalent to 1 dσ¯ 2π p(σ) = Jp(θ, z¯) dθ G+p(¯z)+ G p(¯z). (8) 2πi Z σ¯ z¯ Z − − ∂Ω − 0 We first note that Jp(θ, z¯) is well defined, continuous and analytic inz ¯, for all θ R and all ∈ z Ω. We set ∈ 1 p(ξ(θ,t)) ∂ξ 1 p(ξ(θ,t)) ∂ξ v (θ,t)= , v (θ,t)= . 1 −2πi ξ(θ,t) σ + e2iθ(¯σ z¯) ∂t 2 2πi ξ(θ,t) σ + e2iθ(¯σ z¯) ∂θ − − − − Using that ∂ σ + e2iθ(¯σ z¯) = 2 i e2iθ(¯σ z¯), we get ∂θ − − − ∂v ∂v 1 ∂ξ e2iθ(¯σ z¯) div ~v = 1 + 2 = p(ξ(θ,t)) − , ∂θ ∂t π ∂t (ξ(θ,t) σ + e2iθ(¯σ z¯))2 − − and hence by Green’s formula, 2π 2π 1 2π 5π/2 Jp(θ, z) dθ = div ~vdt dθ = v2(θ, 1) dθ v2(θ, 0) dθ. Z0 Z0 Z0 Z0 − Zπ/2

6 We clearly have 2π 1 p(σ) 1 dσ¯ v2(θ, 1) dθ = dσ = p(σ) , Z 2πi Z e2iθ(¯σ z¯) 2πi Z σ¯ z¯ 0 ∂Ω − ∂Ω − 3π/2 1 θ+(a) p(τ) dτ v2(θ, 0) dθ = dθ, Z 2πi Z τ σ + e2iθ(¯σ z¯) dθ π/2 θ+(b) − − and thus, since σ τ¯ = τ σ = i h, + − − + − 3π/2 a 2iθ+ 1 p(τ) e− v2(θ, 0) dθ = dτ = G+p(¯z). Z 2πi Z τ¯ z¯ i (1+e 2iθ+ ) h − π/2 b − − − Similarly, 5π/2 v2(θ, 0) dθ = G p(¯z), Z3π/2 − which shows (8). It remains to prove (7). From the maximum principle we have, if, for instance, a Re(σ) b, ≤ ≤ that, for z Π , ∈ θ Jp(θ, z¯) sup Jp(θ,σ + yeiθ) | | ≤ y R | | ∈ 1 σ σ¯ − y sup p(ξ(θ,t)) | 2 || | dt σ σ¯ iθ 2 ≤ y R Z0 | | π (t 1) − + e y ∈ | − 2 | σ σ¯ − y sup ∞ | 2 || | ds sup p(ζ) σ σ¯ iθ 2 ≤ y R Z0 π s − + e y ζ Ω | | ∈ | 2 | ∈ ∞ dτ max(θ ψ,π θ + ψ) max iψ iθ 2 sup p(ζ) = − − sup p(ζ) . ≤ ε= 1 Z0 π τe + ε e ζ Ω | | π sin(θ ψ) ζ Ω | | ± | | ∈ − ∈ The estimate (7) then follows from the inequality max (u, π u) π sin u+π cos u , which holds − ≤ 2 | | for u = θ ψ (0,π). The other possible cases for σ may be treated in the same way. − ∈ We consider now a square matrix A satisfying W (A) Ω. Then the operator ⊂ 1 iθ 1 iθ 1 µ(σ, A)= e (σ A)− e− (¯σ A∗)− 2πi − − − is well defined and, it is easily verified that, for v Cd, µ(σ, A) v, v 0. The operator µ(σ, A) ∈ h i≥ is thus positive semi-definite.

Theorem 5. Assume that W (A) Ω. Let P be a polynomial with values in Cm,n. We then ⊂ have 2π P (A)= P (σ) µ(σ, A) ds + JP (θ, A∗) dθ G+P (A∗)+ G P (A∗). (9) Z∂Ω ⊗ Z0 − − Proof. The relation (9) is a consequence of

1 1 P (A)= P (σ) (σ A)− dσ, 2πi Z∂Ω ⊗ − and 2π 1 1 P (σ) (¯σ A∗)− dσ¯ = JP (θ, A∗) dθ G+P (A∗)+ G P (A∗). 2πi Z∂Ω ⊗ − Z0 − −

7 The ﬁrst equality is the Cauchy formula and the second follows from the relation (8), i.e., 1 dσ¯ 2π P (σ) = JP (σ, z¯) dθ G+P (¯z)+ G P (¯z), 2iπ Z σ¯ z¯ Z − − ∂Ω − 0 since each term in this relation is an analytic function ofz ¯, for z Ω. ∈ Remark. Note that the representation formula (9) is valid for thin as well as for thick domains Ω, but in the latter case we have the simpliﬁcation G+P (A∗)= G P (A∗) = 0. − 4 Some preliminary bounds

For the proof of Theorem 1, we have to bound each term on the right of (9). The following lemma gives preliminary estimates for the ﬁrst two terms.

Lemma 6. Assume that the polynomial P satisﬁes P (z) 1, z Ω. Then we have k k≤ ∀ ∈ 1 P (σ) µ(σ, A) ds 2, (10) 2iπ Z∂Ω ⊗ ≤ and 2π 2π JP (θ, A∗) dθ π + cot(θ ψ) dψ. (11) Z0 ≤ Z0 | − |

(The relation between θ and ψ was deﬁned in Lemma 4.)

Proof. a) Since the operator µ(σ, A) is positive, we have the inequality

P (σ) µ(σ, A) ds P (σ) µ(σ, A) ds Z∂Ω ⊗ ≤ Z∂Ω k k

µ(σ, A) ds sup P (ζ) . ≤ Z∂Ω ζ Ω k k ∈ We conclude the proof of (10) by noticing that, by the Cauchy formula,

µ(σ, A) ds = 2I. Z∂Ω

b) From the von Neumann inequality [5] for the half-plane Πθ and the estimate (7), we have 1 JP (θ, A∗) sup JP (θ, z¯) 2 + cot(θ ψ) sup P (ζ) . k k≤ z Πθ | |≤ | − | ζ Ω k k ∈ ∈ Since

cot(θ ψ) d(θ ψ) = 0, Z∂Ω | − | − we infer that 2π 2π 1 2π JP (θ, A∗) dθ ( + cot(θ ψ) ) dθ π + cot(θ ψ) dψ. k Z0 k ≤ Z0 2 | − | ≤ Z0 | − |

The next lemma will complete the bound for the right hand side of (11) in the ‘thick’ case.

8 Lemma 7. Assume that h(1) 10 5. Then we have ≥ − 2π cot(θ ψ) dψ 48.83. Z0 | − | ≤ Proof. a) Recall that in the ‘thick’ case a = b = 1. With ρ(ψ) = σ τ(a) , we have the polar | − | representation σ = τ(a)+ ρeiψ, and note that cot(θ ψ)= ρ /ρ. Therefore, − − ′ 2π cot(θ ψ) dψ = d log σ τ(a) = TV(log σ τ(a) ). Z | − | Z | − | | − | 0 ∂Ω We introduce the notations R = sup σ τ(a) ; σ ∂Ω and r = inf σ τ(a) ; σ ∂Ω . {| − | ∈ } {| − | ∈ } We conclude from Lemma 14, which will be stated and proved in Section 6, the estimate R4 R d log σ τ(a) 2 log , if 2, ( 6 log 2 otherwise). Z | − | ≤ r2(R2 2r2) r ≥ ≤ ∂Ω − b) We set d = h(a)/a = h(1). Since the diameter is 2 we have 10 5 d 1, and R 2 a. − ≤ ≤ ≤ We also deduce from the convexity that Ω σ+(a) contains the quadrilateral with vertices •

0, σ+(a), 2 a, σ (a). − da Therefore the r corresponding to Ω is greater rq • that the rq which would correspond to this τ(a) • a • quadrilateral. The worst case corresponds to 0 2 η (a) = 0, which gives r a d/√4 d2 + 1, η−(a) − ≥ • and therefore R/r 2 √4 d2 + 1/d. ≤ σ−(a) Hence, by a simple calculation, 8(4d2 + 1)2 d log σ τ(a) 2 max log 48.83. Z | − |≤ 10−5 d 1 d2(7d2 + 2) ≤ ∂Ω ≤ ≤

Recalling that the last two terms on the right of (9) vanish for thick domains. The two previous lemmas immediatly imply the following. Corollary 8. In the ‘thick’ case, i.e. when h(1) 10 5, we have ≥ − P (A) 54, for all polynomials P with P (z) 1 in Ω. (12) k k≤ k k≤ 5 Bounds for a thin domain

We turn now to the ‘thin’ case. From the convexity of Ω we have, for σ ∂Ω, ∈ Im σ Im σ tan θ max | |, | | and Im σ 2 h(Re σ). | |≤ Re σ 2 Re σ | |≤ n − o Recall that the numbers a, b with a < b have been deﬁned by h(a)/a = 1/6= h(b)/(2 b). This − implies the following remark which will be frequently used in this section. Remark. We have the estimate 1 tan θ , as soon as a Re σ b. (13) | |≤ 3 ≤ ≤

9 The next lemma will bound the right hand side of (11) in the ‘thin’ case.

5 Lemma 9. We assume that h(1) < 10− . We then have

2π cot(θ ψ) dψ 7.551. Z0 | − | ≤ Proof. Similarly as in the previous lemma, we have, with d = h(a)/a = 1/6, ad a ra := inf σ τ(a) ; π/2 <ψ< 3π/2 = . {| − | }≥ √4d2 + 1 2√10 Using that tan θ (a) tan θ (1) 2 h(1), + ≥ + ≥ − we deduce that the points σ ∂Ω such that ∈ ψ (π/2, 3π/2) satisfy ∈ 2a h(1) σ+(a) 0 Re(τ(a) σ) a • ≤ − ≤ a a 6 and Im(τ(a) σ) + 2 a h(1), • τ(a) | − |≤ 6 • a which shows 0 • σ a π 3π √38 −( ) R := sup σ τ(a) ; <ψ< a . 2a h(1) a {| − | 2 2 }≤ 6 In the same way we get for τ(b), b √38 rb := inf σ τ(b) and Rb := sup σ τ(b) b . π π | − |≥ √ π π | − |≤ 6 ψ ( 2 , 2 ) 2 10 ψ ( , ) ∈ − ∈ − 2 2 We now introduce the sets a Ω = z Ω;Re z a τ(a)+ (z τ(b)) ; z Ω and Re z b , 1 { ∈ ≤ } ∪ { b − ∈ ≥ } ∂Ω = σ ∂Ω;Re σ < a , and ∂Ω = σ ∂Ω;Re σ > b . a { ∈ } b { ∈ } It is easily seen that Ω1 is convex and that 2π cot(θ ψ) dψ = d log σ τ(a) + d log σ τ(b) , Z | − | Z | − | Z | − | 0 ∂Ωa ∂Ωb

= d log σ τ(a) . Z | − | ∂Ω1 Furthermore, we have a √38 r1 := inf σ τ(a) and R1 := sup σ τ(a) a . σ ∂Ω1 | − |≥ 2√10 σ ∂Ω1 | − |≤ 6 ∈ ∈ Applying Lemma 14 this yields 2π 3802 cot(θ ψ) dψ 2 log < 7.551. Z | − | ≤ 368 9 0 ×

We now write A = B + iC, with B = 1 (A + A ) and C = 1 (A A ) self-adjoint. Note 2 ∗ 2i − ∗ that, from the assumption h(1) 10 5, it follows that C 2 10 5 and B 2, thus A is ≤ − k k ≤ − k k ≃ very close to B. We easily obtain a bound for P (B) by spectral theory. The following lemma k k uses this for estimating P (A) . k k

10 Lemma 10. Assume that the polynomial P satisﬁes P (z) 1, z Ω. Then we have the k k ≤ ∀ ∈ estimate

P (A) 26.4+ G+P (A∗) G+P (B) + G P (A∗) G P (B) . k k≤ k − k k − − − k Proof. We deduce from (9), applied to both A and B,

P (A) P (B)= p(σ) µ(σ, A) ds P (σ) µ(σ, B) ds − Z∂Ω ⊗ − Z∂Ω ⊗ 2π 2π + JP (θ, A∗) dθ JP (θ,B) dθ Z0 − Z0 G+P (A∗)+ G+P (B)+ G P (A∗) G P (B). − − − − From spectral theory we have P (B) 1, and, from Lemma 6 and Lemma 9, k k≤ P (σ) µ(σ, A) ds 2, P (σ) µ(σ, B) ds 2, Z∂Ω ⊗ ≤ Z∂Ω ⊗ ≤

2π 2π JP (θ, A∗) dθ 10.7, JP (θ,B) dθ 10.7. Z0 ≤ Z0 ≤

The result stated now follows from the triangle inequality.

The following is the key tool for explaining how the numerical range enters into the proof. Lemma 11. Let

Γ(λ, x) := h′(x) (λ x)+ h(x) and γ(x) := Im τ(x), for x (0, 2). − ∈ Then the condition W (A) Ω is equivalent to : 0 B 2 and ⊂ ≤ ≤ 1/2 1/2 (C γ′(x) (B x) γ(x))u, v (Γ(B,x)) u Γ(B,x)) v , u, v H, x (0, 2). |h − − − i| ≤ k k k k ∀ ∈ ∀ ∈ This condition implies C u, v 2 (Γ(B,x))1/2u Γ(B,x))1/2v , u, v H. |h i| ≤ k k k k ∀ ∈ Proof. a) We ﬁrst remark that X +iY Ω is equivalent to ∈ 0 X 2 and Y γ′(x) (X x) γ(x) Γ(X,x), x (0, 2). ≤ ≤ | − − − |≤ ∀ ∈ Indeed, Y γ (X x) γ(x) = Γ(X,x) are the equations of the tangents to ∂Ω at the points − ′ − − ± σ+ and σ . Therefore the condition W (A) Ω means 0 B 2 and − ⊂ ≤ ≤ (C γ′(x)(B x) γ(x))v, v Γ(B,x) v, v , v H, x (0, 2). |h − − − i| ≤ h i ∀ ∈ ∀ ∈ Thus Γ(B,x) 0 and the lemma follows from the generalized Cauchy-Schwarz inequality. ≥ b) We note that, if X (0, 2), then ∈ Γ(X,x) γ′(x) (X x) γ(x)= η′ (x) (X x) η (x) > 0, − − − − − − − − Γ(X,x)+ γ′(x) (X x)+γ(x)=+η′ (x) (X x)+η (x) > 0. − + − + Therefore X +iY Ω implies Y 2 Γ(X,x). The proof is completed as in part a). ∈ | |≤ We now introduce the operator, for t [0, 1], x [a, b], ∈ ∈ 1/2 2iθ+ 1 Ψ(t,x) := (Γ(B,x)) (¯τ i (1+e− ) h B + itC)− (14) − − 1/2 2iθ+ 1 = (Γ(B,x)) (x iη i h e− B + itC)− . − + − − The following technical lemma will be used to bound the third term on the right in (9).

11 Lemma 12. We have, for all u H, ∈ Ψ(0, .) u 1.33√π u , Ψ(1, .) u 10.93√π u . k kL2 (a,b) ≤ k k k kL2 (a,b) ≤ k k Proof. a) For brevity we will omit the subscript + in the proof. Since C does not appear in Ψ for t = 0, we deduce from spectral theory b Γ(λ, x) 1/2 Ψ(0, .) u K u , with K = sup dx . L2 (a,b) 1 1 2iθ 2 k k ≤ k k λ (0,2) Za x λ i η i h e− ∈ | − − − | We have 2iθ 2 2 2 x λ i η i h e− = (x λ h sin 2θ) + (η + h cos 2θ) | − − − | − − (x λ h sin 2θ)2 + h2 cos2 2θ ≥ − − = (x λ)2 + h2 2 h sin 2θ(x λ) − − − 6 (x λ)2 + h2 x λ h. ≥ − − 5 | − | x λ Here we have used sin 2θ 3 , which follows from tan θ 1 . Setting u(x)= − we deduce | |≤ 5 | |≤ 3 h(x) b (λ x)h + h b u K2 − ′ dx = ′ dx 1 ≤ Z (x λ)2 + h2 6 x λ h Z u2 + 1 6 u a − − 5 | − | a − 5 | | du 5π 5 3 2 ∞ = + arctan 1.763 π. ≤ Z u2 + 1 6 u 4 2 4 ≤ 0 − 5 | |

b) We will now bound ∂tΨ(t,x) u and note that 1/2 2iθ 1 2iθ 1 ∂tΨ(t,x) = i Γ(B,x) (x iη i h e− B+itC)− C(x iη i h e− B+itC)− − − −1/2 − − − − = i Ψ(t,x) C Γ(B,x)− Ψ(t,x). − Using Lemma 11 we get

∂tΨ(t,x) u = sup ∂tΨ(t,x) u, v k k v 1 h i k k≤ 1/2 sup C Γ(B,x)− Ψ(t,x)u, Ψ(t,x)∗v ≤ v 1 h i k k≤ 2 Ψ(t,x ) u K (t), (15) ≤ k k 2 where 1/2 K2(t) = sup Γ(B,x) Ψ(t,x)∗v v 1 k k k k≤ 1/2 2iθ 1 1/2 = sup Γ(B,x) (x+iη+i h e B itC)− Γ(B,x) v . v 1 k − − k k k≤ Let us consider 2iθ 1 1/2 w := Ψ(t,x)∗v = (x+iη+i h e B itC)− Γ(B,x) v − − and ∆ := (x+iη+i h e2iθ B itC) w, w = v, Γ(B,x)1/2w . h − − i h i We deduce from the Cauchy-Schwarz inequality ∆ Γ(B,x)1/2w v . (16) | | ≤ k k k k

12 We also have ∆ 2 = (x B sin 2θ h) w, w 2 + (η+cos 2θ h tC) w, w 2 = ∆ 2 + ∆ 2. | | h − − i h − i | 1| | 2| Using that Cw,w (η+ η (B x))w, w and η = tan θ, we get h i ≤ h ′ − i ′ ∆ = (η+cos 2θ h tC) w, w (cos 2θ h tan θ (B x)) w, w . 2 h − i ≥ h − − i We introduce X := (B x) w, w and X := h w, w . 1 h − i 2 h i We then have X 0 and 2 ≥ 1/2 2 ∆ = X +sin 2θ X , ∆ cos 2θ X tan θ X , Γ(B,x) w = h′ X + X . | 1| | 1 2| 2 ≥ 2 − 1 k k 1 2

Assume ﬁrst cos2θ X tan θ X 0. We then have 2 − 1 ≥ ∆ 2 X +sin 2θ X 2 + (cos 2θ X tan θ X )2 | | ≥ | 1 2| 2 − 1 (1 + tan2 θ)X2 + 2 tan θ X X + X2. ≥ 1 1 2 2 Therefore 2 1/2 4 2 2 2 2 10 ∆ 9 Γ(B,x) w 10(1+tan θ) 9 h′ X1 + 2(10 tan θ 9h′) X1X2 + X2 | | − k k ≥ − 2 − 2 2 (X + (10 tan θ 9h′)X ) + 10 (1 9(tan θ h′) )X . ≥ 2 − 1 − − 1 Using that tan θ h 1/3 we ﬁnd | − ′|≤ 10 10 Γ(B,x)1/2w 4 ∆ 2 Γ(B,x)1/2w 2 v 2, k k ≤ 9 | | ≤ 9 k k k k which implies Γ(B,x)1/2w √10 v . k k≤ 3 k k Consider now the remaining case, tan θ X1 > cos 2θ X2. We then have 1/2 2 Γ(B,x) w h′X1 + X2 h′X1 tan θ + X2 tan θ k k = | | = | 2 | ∆1 X1 + sin 2θ X2 X tan θ +2 sin θ X | | | 1 2| h′ cos 2θ + tan θ max( h′ , | |) = max( h′ , h′ cos 2θ + tan θ ) 1. ≤ | | cos2θ + 2sin2 θ | | | | ≤ | | This shows Γ(B,x)1/2w ∆ and consequently k k ≤ | | √10 Γ(B,x)1/2w v v . k k ≤ k k≤ 3 k k

Returning to K2 we hence deduce

1/2 1/2 √10 K2(t)= sup Γ(B,x) Ψ(t,x)∗v = sup Γ(B,x) w , v 1 k k v 1 k k≤ 3 k k≤ k k≤ and from (15) √10 ∂ Ψ(t,x) u 2 Ψ(t,x) u . k t k≤ 3 k k By integration of this diﬀerential inequality we conclude √10 Ψ(1,x) u exp(2 ) Ψ(0,x) u 10.93√π u . k k≤ 3 k k≤ k k

13 We are now able to bound the last two terms in (9).

Lemma 13. We have the estimate

G P (A∗) G P (B) 14.74. k + − + k≤ Proof. As in the previous lemma we omit the subscript +. With u Hn, v Hm, we start ∈ ∈ from

2π GP (A∗) GP (B) = k − k b 2iθ 2iθ 1 sup e− P (τ) (¯τ ih(1+e− ) A∗)− u 1, v 1 Za h − − k k≤ k k≤ 2iθ 1 (¯τ ih(1+e− ) B)− u, v dτ − − − i b 2iθ 1 2iθ 1 sup C P (τ) (¯τ ih(1+e− ) A∗)− u, (τ +ih(1+e ) B)− v dτ . ≤ u 1, v 1 Za h − − − i | | k k≤ k k≤

We remark that dτ /dx = τ (x) √37/6. Recalling the deﬁnition (14) we obtain, using | | | ′ | ≤ P (τ) 1 and Lemma 11, k k≤ √37 b 2π GP (A∗) GP (B) 2 sup Ψ(1,x) u Ψ(0,x)∗ v dx. k − k≤ 6 u 1, v 1 Za k k k k k k≤ k k≤ In view of Lemma 12 this yields √37 2π GP (A∗) GP (B) 2 π 1.33 10.93 < 2 π 14.74. k − k≤ 6 × ×

With a similar proof we ﬁnd

G P (A∗) G P (B) 14.74. k − − − k≤ This inequality combined with Lemma 10 and Corollary 8 completes the proof of Theorem 1.

6 Appendix : an inequality for plane convex domains

We set B(ω, r)= z C ; z ω < r . { ∈ | − | } Lemma 14. Let Ω be a convex domain in the complex plane satisfying B(ω, r) Ω B(ω, R), ⊂ ⊂ and let R 2r. Then we have ≥ R4 d log σ ω 2 log . (17) Z | − | ≤ r2(R2 2r2) σ ∂Ω ∈ − This estimate is best possible.

14 Proof. It is suﬃcient to consider a polygonal convex domain Ω. Using a counterclockwise ori- entation of the boundary, we denote by σ0,σ 1 ,σ1,...,σ 1 ,σn+1 = σ0, the points on ∂Ω cor- 2 n+ 2 responding to the successive local maxima “ σj ω ” and local minima “ σj+ 1 ω ” of σ ω . | − | | 2 − | | − | Note that n 2. We then have ≥ n σ ω σ ω Φ(Ω) := d log σ ω = log | j − | + log | j+1 − | Z∂Ω | − | σj+ 1 ω σj+ 1 ω Xj=0 2 2 | − | | − | = log J(α) + log J(β) , | | | | where we have set n σj+ 1 ω σj+ 1 ω α = arccos | 2 − |, β = arccos | 2 − |, J(α)= cos α . j σ ω j σ ω j | j − | | j+1 − | jY=0 r We remark that, with γ = arccos , R π π J(α)= J(β), 0 < α γ, 0 < β γ, γ ( , ), j ≤ j ≤ ∈ 3 2 n σ ω 0 < α + β arg j+1 − , (α + β ) 2π. j j ≤ σ ω j j ≤ j − Xj=0 Since Φ(Ω) = 2 log J(α), the inequality (17) is equivalent to − r2(R2 2r2) J(α) cos2γ cos(π 2γ)= − . (18) ≥ − R4 Note that, with α = α = β = β = γ, α = β = π 2γ, n = 2, equality is attained in (18), 0 1 0 1 2 2 − and thus also in (17), which shows the “best possible part” of the lemma. We can assume that n α π (otherwise we exchange α and β). We then set j=0 j ≤ n P J = min J(α) ; α π, 0 α γ, for j = 0, 1,...,n . 0 { j ≤ ≤ j ≤ } Xj=0 It is easily seen that the minimum is attained and, since n 2, ≥ n J = min J(α) ; α = π, 0 α γ, for j = 0, 1,...,n . 0 { j ≤ j ≤ } Xj=0 From the optimality conditions we have, at a minimizing point α, J tan α + λ = 0, for all j such that 0 < α < γ, − 0 j j where λ is the Lagrange multiplier related to the constraint j αj = π. We deduce that λ P either αj = 0, or αj = arctan (ℓ times), or αj = γ (k times). J0 We consider the only three diﬀerent possibilities for k : π π 1 k = 0, then l 3 and J(α)= cos ℓ cos 3 = . • ≥ ℓ ≥ 3 8 π γ ℓ π γ 2 k = 1, then l 2 and J(α) = cos γ cos − cos γ cos − . • ≥ ℓ ≥ 2

15 π 2γ ℓ k = 2, then l 1 and J(α) = cos2γ cos − cos2γ cos(π 2γ). • ≥ ℓ ≥ − Hence 1 π γ 2 J(α) min , cos γ cos − , cos2γ cos(π 2γ) = cos2γ cos(π 2γ). ≥ 8 2 − − This shows (18) and thus completes the proof.

7 Some concluding remarks

Remark 1. The estimate 56 obtained in this paper is not optimal. There is no doubt Qcb ≤ that refinements are possible in Lemmas 9, 12, 13, and especially in Lemma 10, which would decrease this bound. We know how to replace 56 by 34 but we have favoured simplicity in the proofs. We are convinced that our estimate is very pessimistic, but to improve it drastically (recall that our conjecture is = 2), it is clear that we have to find a completely different Qcb method. In the simple case when Ω is a disk, the constant 2 in the inequality (5) has been obtained using a sophisticated representation formula due to Ando (cf. [2]).

Remark 2. We have shown that the map uA from the algebra of rational functions bounded on W (A) into the C -algebra (H), deﬁned by u (r) = r(A), is completely bounded with a ∗ L A complete norm u . A direct application of Paulsen’s theorem, see [4], gives : k Akcb ≤ Qcb There exists an invertible operator S (H) such that S S 1 and ∈L k k k − k ≤ Qcb 1 r(S− AS) sup r(z) , for any rational function r bounded in W (A), k k≤ z W (A) | | ∈ 1 or, equivalently, W (A) is a spectral set for the operator S− AS. Note that we have a reduction of the numerical range : W (S 1AS) W (A). − ⊂ Remark 3. We also deduce from a result due to Arveson (see [1], [4]) that there exist a larger Hilbert space , containing H as a subspace, and a normal operator N acting on , with K K spectrum σ(N) ∂W (A), such that ⊂ 1 r(A)= S P r(N) S− , for any rational function r bounded in W (A), H |H where P denotes the orthogonal projection from onto H. In other words A is similar to an H K operator having a normal ∂W (A) dilation. − References

[1] W. Arveson : Subalgebras of C∗-algebras. II, Acta Math. 128 (1972), 271-308. [2] M. Crouzeix, Bounds for analytic functions of matrices, Int. Equ. Op. Th. 48 (2004), 461–477.

[3] M. Crouzeix, Operators with numerical range in a parabola, Arch. Math., 82 (2004), 517-527.

[4] V. Paulsen, Completely bounded maps and operator algebras, Cambridge Univ. Press, 2002.

[5] J. von Neumann, Eine Spektraltheorie f¨ur allgemeine Operatoren eines unit¨aren Raumes, Math. Nachrichten 4 (1951) 258–281.