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 : 47A11 ; 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 when H = Cd and A is a d d matrix. × The main aim of this article is to prove that the inequality

p(A) 11.08 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) 11.08 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 11.08, and the inequality (2) that it is completely bounded with constant 11.08. 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 11.08. k k∞ It will be clear from the proof that the value 11.08 is not optimal. Denoting by and Q Qcb the 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 polynomials p : C C, P : C Cm,n and any values of ∈ → → d, m, n, we have

2 11.08. ≤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 demonstrate that the inequalities (3) are still valid for any bounded linear operator A (H) on any Hilbert space ∈ L H. For a convex subset E of C we introduce the algebra (E) of continuous and bounded Hb functions in E which are holomorphic in the interior of E. The following result shows the existence of a functional calculus [10] 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 result 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 in Section 3 by application to the theory of Cosine functions. Section 4 is devoted to an integral representation formula for P (A), which is the key to proving (2). In Section 5 we prove some preliminary bounds which allow us to treat a subclass of matrices. The remaining case, which corresponds to thin numerical ranges, is more complicated and is considered in Section 6. 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 . It then follows from (3), used on , that n Kn 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 on H. 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). Writing r in simple partial fraction form it is clear that r(A) is well deﬁned and belongs to (H). In the case W (A) = C, the space L (W (A)) is reduced to constant functions and Theorem 3 follows immediately (but has no Hb interest). In the other cases, replacing, if needed, A by α + eiθA, α C, θ R, we only have ∈ ∈ 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 r ε ε ⊂ be a rational function which is bounded in W (A). We then clearly have r (W (A )). From ∈Hb ε Theorem 2 we conclude

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 find that the map r r(A) is bounded with constant . → 7→ Q The space C (z) is dense in the subset (W (A)) of functions in (W (A)) which have a limit b Hb,0 Hb at . This therefore allows a first extension of the map to this subspace. ∞ Let us now consider 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 ≥ } 1 z f(z) λ A(1+εA)− gλ(A) sup | | f ,A, with ε> 0, k k ≤ Q z W (A) 1+ εz λ z ≤ Q d(λ, W (A)) k k∞ ∈ − we deduce, by taking the limit as ε 0, → λ A gλ(A) | | f ,A and gλ(A) (H, D(A)). k k ≤ Q d(λ, W (A)) k k∞ ∈L We can therefore define f(A) (H) by f(A) = (λ A)g (A). Passing to the limit in the ∈ L − λ 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.

3 An application to second order problems

Let us consider now two complex Hilbert spaces V H, and a continuous sesquilinear form ⊂ a(., .) on V V . We assume that V is continuously and densely embedded in H and that there × exist α> 0, λ, M, N R such that ∈ α v 2 Re a(v, v)+ λ v 2 M v 2 , k kV ≤ k kH ≤ k kV v V. (5) and Im a(v, v) N v v , ∀ ∈ ≤ k kV k kH To this form we associate the linear operator A (D(A),H) defined by ∈L D(A) := u V ; there exists K such that a(u, v) K v , v V , { ∈ u | |≤ uk kH ∀ ∈ } and Au, v = a(u, v). h iH It is then an easy matter to see that D(A) is dense in H and σ(A) W (A). ⊂ We now introduce the parabola α := x+iy ; x + λ y2 . P { ≥ N 2 } The assumptions (5) clearly implies W (A) . Note also that z implies Im z1/2 ω, ⊂ P ∈ P | | ≤ with ω := max(λ, N/(2√α)). Thus the function f (z) := cos(t√z) satisfies f (z) eω t , for all t | t |≤ | | z and all t in R. Also f is holomorphic in C, since the cosine function is even. Theorem 3 ∈ P t allows us to define C(t) := cos(t√A) (H) and we have the following properties : ∈L

4 Theorem 4. The function deﬁned by C(t) := cos(t√A) is a cosine function on the Hilbert space H, i.e.

C(t) (H), t R, C(.) is strongly continuous on H, ∈L ∀ ∈ C(0) = I and C(t+h)+ C(t h) = 2 C(t) C(h). − Furthermore A is the inﬁnitesimal generator of the cosine function C(.). − Proof. Since the map f f(A) is an algebraic homomorphism, the two equations are direct 7→ consequences of the well known formulas cos 0 = 1 and cos(a+b) + cos(a b) = 2 cos a cos b. − Theorem 3 also provides the bound ω t C(t) e | |. k k ≤ Q Let us now consider u H and set u(t) := C(t) u . The strong continuity of C(.) on H means 0 ∈ 0 that u(.) C0(R ; H), for all u H. Using the previous bound for C(.) and the density of ∈ 0 ∈ k k D(A) in H, it suﬃces to show that u(.) C0(R ; H) if u D(A). For that we introduce the ∈ 0 ∈ function cos(t+h)√z cos(t h)√z g (z) := − − , t,h = 0. t,h t h z 6 Assuming z and t + h T we have ∈ P | | | |≤ sin t√z sin h√z 1 1 gt,h(z) = 2 = 2 cos(xt√z) dx cos(xh√z) dx . t√z h√z 0 0 Z Z Thus g (z) 2 eωT and g (A) 2 eωT . | t,h |≤ k t,h k≤ Q From the relation of cos(t+h)√z cos(t h)√z = thzg we then deduce that − − t,h C(t+h) C(t h)= thAg (A), − − t,h which implies, if u D(A), 0 ∈ u(t+h) u(t h)= t h g (A) Au . − − t,h 0 Thus u(t+h) u(t h) 2 t h Au eωT , k − − k≤ Q | | | | k 0k which completes the proof of the strong continuity. 0 In a similar way, we can see that u C (R ; D(A)) as soon as u0 D(A). From the relation t ∈ ∈ t cos t√z = 1 z (t s) cos s√z ds we then deduce that C(t)= I A (t s)C(s) ds, and thus − 0 − − 0 − R t R u(t)= u (t s)Au(s) ds. 0 − − Z0 Since Au C0(R ; H) this clearly implies u C2(R ; H), u(0) = u , u (0) = 0, ∈ ∈ 0 ′ t u′(t)= Au(s) ds and u′′(t)+ Au(t) = 0, t R. − ∀ ∈ Z0 In particular this implies that A is the generator of the cosine function C(.). −

5 Remark. This theorem admits an interesting converse. Markus Haase [7] has proved the following result. If A is the generator of a cosine function on the Hilbert space H, then, with − respect to an equivalent scalar product ., . ◦ , A has its numerical range in a horizontal parabola h i := x+iy ; x + µ ω2y2 for some ω = 0,µ R. P { ≥ } 6 ∈ Then using this new scalar product, setting a(u, v)= Au, v ◦ , and denoting by V the closure h1/2 i of D(A) for the norm v = Re a(v, v) + (µ+1) v, v ◦ . Assumptions (5) are satisﬁed with k kV h i α = M = 1, λ = µ+1, and N = ω 1. This shows that the framework used in this section is − well suited for a general study of cosine functions in Hilbert spaces. We refer to [8] and [1] for a classical and a modern presentation of the theory of cosine functions. The cosine functions allows the treatment of linear diﬀerential equations of second order.

Theorem 5. Assume that u D(A), v V , and set S(t) := t C(s) ds. Then the function 0 ∈ 0 ∈ 0 u(t) := C(t)u + S(t)v is the unique solution of 0 0 R u C2(R; H) C0(R; D(A)), ∈ ∩ (u′′(t)+ A u(t) = 0, t R, u(0) = u0, u′(0) = v0. ∀ ∈ Proof. a) Existence. We have seen in the proof of the previous theorem that, if v0 = 0, then u(t) = C(t)u0 satisfies the requirements. Thus it suffices to consider the case u0 = 0. We set v(t) = S(t)v0. The initial conditions v(0) = 0 and v′(0) = v0 clearly hold. If v0 D(A) then, 2 0 ∈ with w(t) = C(t)v0, we have w′′ + Aw = 0 and w C (R; H) C (R; D(A)). This yields that t ∈ ∩ v(t)= w(s) ds satisfies v + Av = 0 and v C3(R; H) C1(R; D(A)). 0 ′′ ∈ ∩ We refer to [9] for the proof that the accretive operator (λ+1+A) 1/2 is an isomorphism R − from H into V . We then can write v = (λ+1+A) 1/2w with w H and w C v . 0 − 0 0 ∈ k 0kH ≤ k 0kV Thus we have 1/2 v′′(t) H = Av(t) H = AS(t) (λ+1+A)− w H k k k k k 0k z t sup cos s√z ds w 1/2 0 H ≤ Q z W (A) (1+λ+z) 0 k k ∈ Z 1 z 1/2 t ω sup cos st√z ds w0 H C e| | v0 V . ≤ Q z (1+λ+z) 0 k k ≤ Q k k ∈P Z Together with the density of D(A) in V , this shows that v C0(R; D(A)) C2(R; H) for all ∈ ∩ v V . 0 ∈ b) Uniqueness. It is sufficient to prove that, if u C2(R; H) C0(R; D(A)) satisfies ∈ ∩ u′′(t)+ A u(t) = 0, u(0) = 0, u′(0) = 0, then u = 0. Replacing if needed u by (λ+A) 1u we can assume that u C2(R; D(A)) C0(R; D(A2)). We − ∈ ∩ then set

v(t,s)= C(s)u(t) S(s)u′(t), w(t,s)= C(s)u′(t)+ S(s)A u(t). − 2 2 We deduce from the equality C(t) + S(t) A = Id D A (which is valid on D(A)) that | ( ) u(t)= C(t) v(t,t) S(t) w(t,t). − We have ∂v ∂v (t,s)= C(s)u′(t)+ S(s)Au(t), (t,s)= C(s)u′(t) S(s)A u(t). ∂t ∂s − −

6 Thus d ∂v ∂v v(t,t)= (t,t)+ (t,t) = 0. dt ∂t ∂s Since v(0, 0) = 0 it follows that v(t,t) = 0 for all t. Similarly we have w(t,t) = 0 for all t and thus u(t)= C(t) v(t,t) S(t) w(t,t) = 0. − 4 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 . (6) 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−

(ii) Thick domains : h(1) α = 0.15. 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 are deﬁned by σ (x) ∂Ω, Im σ+(x) > 0 and x = Re σ (x), ± → ± ∈ ± ± dσ θ = arg( ± ). ± ds

7 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 − }

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 ∂t (ξ(θ,t) σ(θ)+ e2iθ(¯σ(θ) z¯))2 Z0 − − (1+e 2iθ+ )(1+e 2iθ− ) dτ ν(x) = − − , 2π dx (¯z τ¯+ i(1+e 2iθ+ )h)(¯z τ¯ i(1+e 2iθ− )h) R(x, z¯) = − − − − − . (¯z τ¯)h +τ ¯ h − ′ ′ The proof of our main result, Theorem 1, is based on the following lemma.

Lemma 6. For any polynomial p and any z Ω, we have the representation formula ∈ 2π b p(τ) p(z)= p(σ) µ(σ, z) ds + Jp(θ, z¯) dθ + ν(x) dx. (7) R(x, z¯) Z∂Ω Z0 Za Proof. From the Cauchy formula we have 1 dσ dσ¯ 1 dσ¯ p(σ) µ(σ, z) ds = p(σ) = p(z) p(σ) . ∂Ω 2πi ∂Ω σ z − σ¯ z¯ − 2πi ∂Ω σ¯ z¯ Z Z − − Z − The relation (7) is therefore equivalent to 2π 1 dσ¯ b p(τ) Jp(θ, z¯) dθ = p(σ) ν(x) dx. (8) 2πi σ¯ z¯ − R(x, z¯) Z0 Z∂Ω − Za 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 = v (θ, 1) dθ v (θ, 0) dθ. 2 − 2 Z0 Z0 Z0 Z0 Zπ/2 We clearly have 2π 1 p(σ) 1 dσ¯ v (θ, 1) dθ = dσ = p(σ) , 2 2πi e2iθ(¯σ z¯) 2πi σ¯ z¯ Z0 Z∂Ω − Z∂Ω −

8 3π/2 1 θ+(a) p(τ) dτ v (θ, 0) dθ = dθ, 2 2πi τ σ + e2iθ(¯σ z¯) dθ Zπ/2 Zθ+(b) − − and thus, since σ τ¯ = τ σ = i h, + − − + − 3π/2 a 2iθ+ 1 p(τ) e− dτ v2(θ, 0) dθ = dx. 2πi τ¯ z¯ i (1+e 2iθ+ ) h dx Zπ/2 Zb − − − Similarly, 5π/2 1 b p(τ) e 2iθ− dτ v (θ, 0) dθ = − dx, 2 2πi τ¯ z¯ + i (1+e 2iθ− ) h dx Z3π/2 Za − − and hence

5π/2 b 2iθ− 2iθ+ 1 e− e− v2(θ, 0) dθ = p(τ) dτ 2iθ− 2iθ+ π/2 2πi a τ¯ z¯ + i (1+e− ) h − τ¯ z¯ i (1+e− ) h Z Z − − − 1 b i(¯τ z¯)(e 2iθ+ e 2iθ− )+ h(e 2iθ+ +e 2iθ− +2e 2i(θ++θ−)) = p(τ) − − − − − − − dτ 2π (¯τ z¯ i (1+e 2iθ+ ) h)(¯τ z¯ + i (1+e 2iθ− ) h) Za − − − − − 1 b (1+e 2iθ+ )(1+e 2iθ− )((¯τ z¯)h +τ ¯ h) dτ = p(τ) − − − ′ ′ dx 2π (¯τ z¯ i (1+e 2iθ+ ) h)(¯τ z¯ + i (1+e 2iθ− ) h) dx Za − − − − − b 1 = p(τ) ν(x) dx. R(x, z¯) Za Formula (7) now easily follows. In the previous calculations we have used the relations

2iθ+ 2iθ− 2iθ+ 2iθ− tan θ+ tan θ 1 1 e− 1 e− i(e− e− ) h′(x)= − − = − − = − , 2iθ+ 2iθ− 2iθ+ 2iθ− 2 2i 1+e− − 1+e− (1+e− )(1+e− ) and

2iθ+ 2iθ− tan θ+ + tan θ 1 1 e− 1 e− τ¯′(x) = 1 i − = 2 − − 2iθ+ 2iθ− − 2 2 − 1+ e− − 1+ e− e 2iθ+ + e 2iθ− + 2e 2iθ+ e 2iθ− = − − − − . 2iθ+ 2iθ− (1 + e− )(1 + e− )

We now consider a square matrix A satisfying W (A) Ω. Then the matrix ⊂ 1 iθ 1 iθ 1 µ(σ, A)= e (σ A)− e− (¯σ A∗)− 2πi − − − is well defined and, as is easily verified, µ(σ, A) v, v 0 for v Cd. The matrix µ(σ, A) is thus h i ≥ ∈ positive semi-definite.

Theorem 7. Assume that W (A) Ω. Let P be a polynomial with values in Cm,n. We then ⊂ have 2π b 1 P (A)= P (σ) µ(σ, A) ds + JP (θ, A∗) dθ + P (τ) (R(x, A∗))− ν(x) dx. (9) ⊗ ⊗ Z∂Ω Z0 Za

9 Proof. The relation (9) is a consequence of

1 1 P (A)= P (σ) (σ A)− dσ 2πi ∂Ω ⊗ − and Z 2π b 1 1 1 P (σ) (¯σ A∗)− dσ¯ = JP (θ, A∗) dθ + P (τ) (R(x, A∗))− ν(x) dx. 2πi ⊗ − ⊗ Z∂Ω Z0 Za The ﬁrst equality is the Cauchy formula and the second follows from the relation (8), i.e., 1 dσ¯ 2π b 1 P (σ) = JP (σ, z¯) dθ + P (τ) ν(x) dx, 2iπ σ¯ z¯ R(x, z¯) Z∂Ω − Z0 Za 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 the last term is missing since a = b.

5 Some preliminary bounds

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

Lemma 8. Assume that the polynomial P satisﬁes P (z) 1, z Ω. Then we have k k≤ ∀ ∈ P (σ) µ(σ, A) ds 2. (10) ∂Ω ⊗ ≤ Z

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

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

µ(σ, A) ds sup P (ζ) . ≤ ∂Ω ζ Ω k k Z ∈

We conclude the proof of (10) by noticing that, by the Cauchy formula, µ(σ, A) ds = 2I. Z∂Ω

In order to treat the second term on the right of (9), we need some new notations. With the angle θ we associate the angle ψ and the real number ρ> 0 deﬁned by σ τ(a) if Re σ(θ) a, − ≤ θ ψ (0,π) and ρeiψ = σ Re σ if a< Re σ(θ) < b, − ∈  − σ τ(b) if Re σ(θ) b.  − ≥ π We note that ρ is then a smooth function of ψ, except possibly for the values ψ = 2 (mod. π), and we have ξ(t,θ)= σ(θ)+(t 1)ρeiψ. The following lemma gives a preliminary bound for the − second term of (9).

10 Lemma 9. Assume that the polynomial P satisﬁes P (z) 1, z Ω. Then we have k k≤ ∀ ∈ 2π 2π ρ′(ψ) JP (θ, A∗) dθ g ρ(ψ) dψ, (11) 0 ≤ 0 Z Z 1 arctan t √ 2 where g(t) := ( 2 + | π | ) 1+t .

Proof. From the von Neumann inequality [12] for the half-plane Πθ we have

JP (θ, A∗) sup JP (θ, z¯) . k k≤ z Πθ | | ∈ It follows from the maximum principle that 1 iθ ρ y JP (θ, A∗) sup JP (θ,σ + ye ) sup P (ξ(θ,t)) |iψ| iθ 2 dt. k k ≤ y R k k≤ y R 0 | k π (t 1)ρe + e y ∈ ∈ Z | − | Using the change of variables (t 1)ρ = s y , the hypothesis P 1, and the relation ρ /ρ = − | | k k ≤ ′ cot(ψ θ), we get − ∞ ds max(θ ψ,π θ+ψ) ρ′(ψ) JP (θ, A∗) = − − = g . k k≤ π seiψ + eiθ 2 π sin(θ ψ) ρ(ψ) Z0 | | − We now remark that ′ g ρ (ψ) d(θ ψ) = 0, ρ(ψ) − Z∂Ω and thus 2π 2π 2π ρ′(ψ) ρ′(ψ) JP (θ, A∗) dθ g dθ = g dψ. k k ≤ ρ(ψ) ρ(ψ) Z0 Z0 Z0

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

Lemma 10. Assume that h(1) α = 0.15. Then we have ≥ 2π ′ g ρ (ψ) dψ 9.01. ρ(ψ) ≤ Z0 Proof. Recall that in the ‘thick’ case a = b = 1. We remark that the function g is even, positive and convex. We introduce the notations R = sup σ τ(a) ; σ ∂Ω , r = inf σ τ(a) ; σ ∂Ω t {| − | ∈ } {| − | ∈ } G(t) = g(tan ϕ) dϕ. Z0 Let us consider a real number χ [ π , π ). Our estimation is based on the following inequality ∈ 3 2 proved in [5]

2π ′ g ρ (ψ) dψ 4 G(χ) + 2 G(π 2χ), if r/R cos χ. ρ(ψ) ≤ − ≥ Z0 We now need to get a lower bound for r/R. We set d = h(a)/a = h(1). We clearly have α d 1, and R 2 a. From the inequalities ≤ ≤ ≤ Re(σ τ(a)) a and Im(σ τ(a)) 3 d a, | − |≤ | − |≤ we deduce R a√1+9d2. ≤

11 σ+(a) • It also follows from the convexity that Ω contains the quadrilateral with vertices da rq 0, σ+(a), 2 a, σ (a). • − τ(a) Therefore the r corresponding to Ω is greater • a • 0 2 that the rq which would correspond to this η−(a) quadrilateral. The worst case corresponds to • η (a) = 0, which gives r a d/√4 d2 + 1. σ−(a) − ≥ Therefore r/R d/(√1 + 4 d2√1 + 9 d2), and we deduce, with χ = arccos α , ≥ √1+4α2√1+9α2 2π ′ g ρ (ψ) dψ 4 G(χ) + 2 G(π 2χ) 9.01. ρ(ψ) ≤ − ≤ Z0

Recall that the last term on the right of (9) vanishes for thick domains. The three previous lemmas immediatly imply the following.

Corollary 11. In the ‘thick’ case, i.e. when h(1) 0.15, we have ≥ P (A) 11.01, for all polynomials P with P (z) 1 in Ω. (12) k k≤ k k≤ 6 Bounds for a thin domain

We turn now to the ‘thin’ case. The next lemma will bound the right hand side of (11) in this situation.

Lemma 12. We assume that h(1) < α = 0.15 and that the polynomial P satisﬁes P (z) 1, k k≤ z Ω. Then we have ∀ ∈ 2π 2π ρ′(ψ) JP (θ, A∗) dθ g ρ(ψ) dψ 7.25. 0 ≤ 0 ≤ Z Z Proof. We deﬁne

ra := inf σ τ(a) , Ra := sup σ τ(a) , π/2<ψ<3π/2 | − | π/2<ψ<3π/2 | − |

rb := inf σ τ(b) , Rb := sup σ τ(b) . π/2<ψ<π/2 | − | π/2<ψ<π/2 | − | − − Similarly as in the previous lemma, we have, but now with d = h(a)/a = h(b)/(2 b) = 2α, −

12 ad bd ra , rb . ≥ √1+4d2 ≥ √1+4d2 da σ+(a) Using that • tan θ+(a) tan θ+(1) 2 h(1) d, ≥ ≥− ≥− da we deduce that the points σ ∂Ω such that ∈ ψ (π/2, 3π/2) satisfy • τ(a) ∈ a 0 Re(τ(a) σ) a • ≤ − ≤ 0 and Im(τ(a) σ) 2 a d, • | − |≤ σ−(a) which shows da R a 1+4d2. a ≤ In the same wayp we get R b√1+4d2. b ≤ 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 ρ′(ψ) ρ′(ψ) ρ′(ψ) ρ′(ψ) g ρ(ψ) dψ = g ρ(ψ) dψ + g ρ(ψ) dψ = g ρ(ψ) dψ. Z0 Z∂Ωa Z∂Ωb Z∂Ω1 Furthermore, we have ad r := inf σ τ(a) and R := sup σ τ(a) a 1+4d2. 1 2 1 σ ∂Ω1| − |≥ √1+4d σ ∂Ω1| − |≤ ∈ ∈ p d Hence, with χ = arccos and d = 2α = 0.3, 1+4d2

2π ′ g ρ (ψ) dψ 4 G(χ) + 2 G(π 2χ) 7.25. ρ(ψ) ≤ − ≤ Z0

We now use the notation B := Re A = 1 (A + A ) and C := Im A = 1 (A A ), thus 2 ∗ 2i − ∗ A = B + iC. The following lemma provides a preliminary bound for the third term on the right in (9) which we denote P3. Lemma 13. Assume that the polynomial P satisfies P (z) 1, z Ω. Then we have k k≤ ∀ ∈ b b 2 1 1 P3 Re R(x, A∗) − dx , with P3 := P (τ) R(x, A∗) − ν(x) dx. (13) k k≤ π a a ⊗ Z Z Proof. We assume preliminary that the selfadjoint matrix S(x) := Re R(x, A∗) is positive definite. This will be proved in the subsequent lemma which yields that each term in (13) is well defined. We first note that γ′ = (tan θ+ + tan θ )/2, this yields − 2 2 2 2 2 π ν(x) = cos θ+ cos θ 4 + (tan θ+ + tan θ ) − − | | 2 2 2 2 2 2 = 4 cos θ+ cos θ + sin θ+ cos θ +2sin θ+ sin θ cos θ+ cos θ + cos θ+ sin θ − − − − − 2 2 = cos θ+ + cos θ + 2 cos θ+ cos θ cos(θ+ θ ) 4. − − − − ≤

13 This shows that ν 2/π. We now set | |≤ 1/2 1/2 D(x)= S(x)− Im R(x, A∗) S(x)− , so that 1 1/2 1 1/2 1 R(x, A∗) − = S(x)− (I +iD(x))− S(x)− , and (I +iD(x))− 1. k k≤ There exist two vectors u (Cd)n and v (Cd)m such that ∈ ∈ P = P u, v and u = v = 1. k 3k h 3 i k k k k We have b 1/2 1 1/2 P = P (τ) S− u, (I iD)− S− v ν dx. k 3k h ⊗ − i Za Therefore, using that P (τ) 1, (I +iD(x)) 1 1 and ν(x) 2/π, k k≤ k − k≤ | |≤ b 2 1/2 1/2 P S(x)− u S(x)− v dx. k 3k≤ π k k k k Za We now remark that b b b 1/2 2 1 1 S(x)− u dx = S(x)− u, u dx = S(x)− dx u, u a k k a h i a Z Z b DZ E 1 S(x)− dx . ≤ a Z We conclude from the Cauchy-Schwarz inequality

b 1 P3 S(x)− dx . k k≤ a Z

We have to show that S(x) = Re R(x, A∗) is positive deﬁnite. This is the object of the following lemma which provides also a lower bound for S(x).

Lemma 14. Let us consider the function ϕ deﬁned by

2 4 3 β2 2 (λ x) + 2 h′(x) h(x) (λ x)+ − 2 h(x) ϕ(x, λ) := − 1+β − 1+β , (λ x)h (x)+ h(x) − ′ where β = 4α = 0.6. Then we have

Re R(x, A∗) ϕ(x,B) > 0, x [a, b]. (14) ≥ ∀ ∈ Proof. It follows from the strict convexity of Ω that the function h is a strictly concave function on [0, 2], and furthermore h(0) = h(2) = 0. This implies, for x [a, b], ∈ (λ x)h′(x)+ h(x) h(λ) > 0, λ (0, 2), − ≥ ∀ ∈ and 2 α h′(b) h′(x) h′(a) 2 α. − ≤ ≤ ≤ ≤ It follows that ϕ(x, λ) > 0 for all λ (0, 2), and hence the matrix ϕ(x,B) is well deﬁned and ∈ positive deﬁnite.

14 We have (¯z τ¯+ i(1+e 2iθ+ )h)(¯z τ¯ i(1+e 2iθ− )h) R(x, z¯)= − − − − − . (¯z τ¯)h +τ ¯ h − ′ ′ Simple calculations give 2 2 2 2 2 (¯z τ¯) (h′) (¯τ ′) h 2 i(θ++θ−) (h′(x)) R(x, z¯) = − − + 4 h(h′) cos θ+ cos θ e− (¯z τ¯) h +τ ¯ h − − ′ ′ 2 2 i(θ++θ−) 2 (¯τ ′) + 4 i (h′) γ′ cos θ+ cos θ e− +h − . (¯z τ¯) h +τ ¯ h − ′ ′ Noting that 4iγ = τ 2 τ¯ 2 and ′ ′ − ′ i(θ++θ−) 2 2 e = cos θ+ cos θ (1 tan θ+tan θ + 2 i γ′) = (τ ′ + h′ ) cos θ+ cos θ − − − − we obtain 2 2 2 2 2 2 2 i(θ++θ−) 2i(θ++θ−) 2 τ ′ + h′ (τ ′ τ¯′ )h′ (¯τ ′) + 4 i (h′) γ′ cos θ+ cos θ e− = e τ¯′ 2 2 + 2− 2 − τ¯′ + h′ τ¯′ + h′ 2i(θ++θ−) 2 = e τ ′ . Therefore we have

2 2 2i(θ++θ−) 2 2 i(θ++θ−) h τ ′ e− h′ R(x, z¯)=(¯z τ¯)h′ τ¯′h + 4 hh′ cos θ+ cos θ e− + , − − − (¯z τ¯)h +τ ¯ h − ′ ′ and thus 2 2 2 h′ Re R(x, A∗)= h′(B x) h + 4 c h h′ + h Re E, (15) − − 1 with

2 2i(θ++θ−) 1 c1(x) := cos θ+ cos θ cos(θ+ + θ ), et E(x) := τ ′ e− (A∗ τ¯)h′ +τ ¯′h − . − − − We assume temporarily the following lemma Lemma 15. We have the estimates

1 2 1 c (x)) and Re E(x) (1 h′ ) h′(B x)+ h − , x [a, b]. (16) 1 ≥ 1+ β2 ≥ − − ∀ ∈ Going back to (15) we have demonstrated

2 4 2 2 2 1 h′ Re R(x, A∗) h′(B x) h + hh′ + h (1 h′ ) h′(B x)+ h − ≥ − − 1+ β2 − − 2 = h′ ϕ(x,B). This shows (14) if h (x) = 0. At a point where h (x) = 0 (such a point is unique since h is ′ 6 ′ strictly concave), (14) holds by continuity.

Proof of Lemma 15. a) We remark that, for x (a, b), max( h (x) , γ (x) ) 2α = 0.3 1/√5. Thus ∈ | ′ | | ′ | ≤ ≤ 1 tan θ+ tan θ cos θ+ cos θ cos(θ+ + θ ) = − 2 −2 − − (1 + tan θ+)(1 + tan θ ) − 1+ h 2 γ 2 = ′ − ′ (1 + (h +γ )2)(1 + (h γ )2) ′ ′ ′ − ′ 1 1 2 2 2 . ≥ 1+4max(h′ , γ′ ) ≥ 1+ β

15 b) We set 1/2 1/2 ϕ = arg τ ′, B = h′(B x)+ h, C = h′(C γ)+ γ′h, D = B− C B− , 1 1 − 1 − 1 1 1 1 and note that iϕ e 1 1/2 1/2 τ ′ = , h′(A∗ τ¯) +τ ¯′h = B1 + iC1 = B1 (I iD1)B1 . cos ϕ1 − − This allows us to write

2i(ϕ1 θ+ θ−) e − − 1/2 1/2 1/2 − − E = 2 B1 (I iD1)− B1 . cos ϕ1 −

Since D1 is selfadjoint, there exists λ in the spectrum of D1 such that

2i(ϕ1 θ+ θ−) 2i(ϕ1 θ+ θ−) e − − 1 e − − Re (I iD )− c (x) := Re . cos2ϕ − 1 ≥ 2 cos2ϕ (1 iλ) 1 1 − From this we deduce 1 Re E c (x) B− , ≥ 2 1 and it suﬃces to show that c (x) 1 h (x)2. 2 ≥ − ′ With λ = tan ζ we have

i(2(ϕ1 θ+ θ−)+ζ) cos ζ e − − cos ζ cos(2(ϕ1 θ+ θ )+ ζ) − c2(x) = Re 2 = −2 − cos ϕ1 cos ϕ1 2 2 cos(ϕ1 θ+ θ ) sin (ϕ1 + ζ θ+ θ ) − − = −2 − 2− − . cos ϕ1 − cos ϕ1

Using that tan θ+ = γ′ +h′, tan θ = γ′ h′, tan ϕ1 = γ′, we obtain − − 2 2 2γ′ 1+γ′ h′ tan(θ+ +θ )= , tan(ϕ1 θ+ θ )= γ′ − − 1 γ 2 +h 2 − − − − 1+γ 2 +h 2 − ′ ′ ′ ′ 2 2 2 2 2 cos (ϕ1 θ+ θ ) (1+γ′ )(1 + h′ + γ′ ) − − − = cos2ϕ (1+γ 2 +h 2)2 + γ 2(1+γ 2 h 2)2 1 ′ ′ ′ ′ − ′ 4 h 2 γ 2(1+γ 2) = 1+ ′ ′ ′ 1. (1+γ 2 +h 2)2 + γ 2(1+γ 2 h 2)2 ≥ ′ ′ ′ ′ − ′ Therefore in order to complete the proof of this lemma it suﬃces to show that 2 2 2 sin (ϕ1 + ζ θ+ θ ) h′ cos ϕ1. − − − ≤ For this we remark that, if X + iY Ω and x (a, b), we have ∈ ∈ (γ′(x) h′(x))(X x)+ γ(x) h(x) Y (γ′(x)+h′(x))(X x)+ γ(x)+h(x), − − − ≤ ≤ − i.e. , with respect to its boundary, the convex set Ω lies under the tangent at σ+ and over the tangent at σ . Therefore the condition W (A) Ω implies − ⊂ (γ′(x) h′(x))(B x)+ γ(x) h(x) C (γ′(x)+h′(x))(B x)+ γ(x)+h(x), − − − ≤ ≤ − which yields

(γ′(x) h′(x) ) B C (γ′(x)+ h′(x) ) B , −| | 1 ≤ 1 ≤ | | 1 or equivalently

(γ′(x) h′(x) ) D (γ′(x)+ h′(x) ). −| | ≤ 1 ≤ | |

16 Recall that tan θ = γ h and that λ = tan ζ is an eigenvalue of D . Thus ζ lies between θ ± ′ ± ′ 1 + and θ . Consequently − 2 2 2 2 h′ sin (ϕ1 + ζ θ+ θ ) max sin (ϕ1 θ+), sin (ϕ1 θ ) = − − − ≤ − − − (1+γ 2)(1+( γ h )2) ′ | ′|−| ′| 2 2 h′ cos ϕ1 2 2 h′ cos ϕ . ≤ 1+( γ h )2 ≤ 1 | ′|−| ′|

We are now able to bound the last term, P3, in (9). Lemma 16. We have the estimate P 1.83. k 3k≤ Proof. We remark that the spectrum of B satisﬁes σ(B) (0, 2). We deduce from (13) and (14) ⊂ b b 2 1 2 1 P3 Re R(x, A∗) − dx ϕ(x,B) − dx k k ≤ π a ≤ π a Z Z 2 b 1 sup dx. ≤ π λ (0,2) a ϕ(x, λ) ∈ Z x λ Setting u(x)= − and using that h (x) 2 α = β/2, we obtain h(x) | ′ |≤ 1 u (x) u (x) = ′ ′ . 4 3 β2 2β 3 β2 ϕ(x, λ) u(x)2 + h (x) u(x)+ − ≤ u(x)2 u(x) + − 1+β2 ′ 1+β2 − 1+β2 | | 1+β2 We ﬁnally get 2 du 4 du P ∞ = ∞ 3 2β 3 β2 2β 3 β2 k k ≤ π u2 u + − π 0 u2 u + − Z−∞ − 1+β2 | | 1+β2 Z − 1+β2 1+β2 1+ β2 2 β = 2 (1 + arctan ). 3+ β2 β4 π 3+ β2 β4 − − p p This inequality, P 1.83, combined with Lemmas 8, 12 and Corollary 11, completes the k 3k≤ proof of Theorem 1.

7 Some concluding remarks

Remark 1. The estimate 11.08 obtained in this paper is not optimal. There is no doubt Qcb ≤ that refinements are possible which would decrease this bound. We are convinced that our estimate is very pessimistic, but to improve it drastically (recall that our conjecture is = 2), Qcb it is clear that we have to find a completely different method. In the simple case when Ω is a disk, the constant 2 in the inequality (6) has been obtained using a sophisticated representation formula due to Ando (cf. [3]).

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 [11], gives : k Akcb ≤ Qcb

17 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 [2], [11]) 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. − Remark 4. Only two items in the reference list below are related to the core of this work. I have always been fascinated by the von Neumann inequality for the half-plane [12] and inequality (1) may be considered as a generalization of it. I have become interested to work with the operator form µ(σ, A) of the double layer potential by the article [6] of Bernard and Fran¸cois Delyon ; this paper is clearly at the origin of the present work. The proof proposed here for inequality (2) is still intricate and only uses old-fashioned mathematics, but I have found no help in the modern literature on operator theory for this speciﬁc problem. It curiously seems to belong to an uncharted area of research. Acknowledgements. I am endebted to B. Beckermann for a valuable remark which has allowed me to obtain an improved estimate together with a more transparent proof.

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