In honour of Professor William Gear

Functional Calculus and Numerical Analysis Michel Crouzeix Université de Rennes 1 ICNAAM 2011 In honour of Professor William Gear Halkidiki, September 2011

The context Let us consider a closed linear operator A L(D(A), H) on a Hilbert space H. We assume D(A) H, with dense imbedding, and the spectrum satisfies σ(a) C. Example 1. A square matrix. H = D(A) = C d, A C d,d.

The context Let us consider a closed linear operator A L(D(A), H) on a Hilbert space H. We assume D(A) H, with dense imbedding, and the spectrum satisfies σ(a) C. Example 2. Dirichlet Laplace operator. H = L 2 (Ω), V = H0 1 (Ω), a(u, v) = Ω u. v dx, D(A) = {u V ; Au H, Au, v = a(u, v), v V } = H 2 (Ω) H0 1 (Ω), Au = u.

The context Let us consider a closed linear operator A L(D(A), H) on a Hilbert space H. We assume D(A) H, with dense imbedding, and the spectrum satisfies σ(a) C. Example 3. Robin Laplace operator. H = L 2 (Ω), V = H 1 (Ω), a C( Ω; R). a(u, v) = Ω u. v dx + a u v dσ, D(A) = {u V ; Au H, Au, v = a(u, v), v V } = {u H 2 (Ω); n u + au = 0, on Ω}, Au = u. Ω

The context Let us consider a closed linear operator A L(D(A), H) on a Hilbert space H. We assume D(A) H, with dense imbedding, and the spectrum satisfies σ(a) C. Example 4. Complex Robin-Laplace operator. H = L 2 (Ω), V = H 1 (Ω), a, b C( Ω; R). a(u, v) = Ω u. v dx + (a+ib) u v dσ, D(A) = {u V ; Au H, Au, v = a(u, v), v V } = {u H 2 (Ω); n u + (a+ib)u = 0, on Ω}, Au = u. Ω

The self-adjoint case Let us consider a closed linear operator A L(D(A), H) on a Hilbert space H. If A is a self-adjoint operator, spectral theory is very efficient. We know that the spectrum σ(a) is real and we have r(a) = sup r(x), x σ(a) if the rational function r is bounded on σ(a). By a density argument, this (in)equality allows to define f(a) L(H), for any function f that is continuous and bounded on σ(a). The previous inequality also holds with f instead of r.

The self-adjoint case Let us consider a closed linear operator A L(D(A), H) on a Hilbert space H. If A is a self-adjoint operator, spectral theory is very efficient. We know that the spectrum σ(a) is real and we have f(a) = sup f(x), x σ(a) for all continuous functions f that are bounded on σ(a). Furthermore the map f f(a), is a homomorphism from the algebra C b (σ(a)) into the algebra L(H).

Example of application Assume that we have a self-adjoint positive definite operator A L(D(A), H). We can define C(t) = cos(t A), and we are able to show that u(t) = C(t) u 0, with u 0 D(A), is the unique solution in C 2 (R, H) C 0 (R, D(A)) of u (t) + A u(t) = 0, t R, u(0) = u 0, u (0) = 0.

Continuity proof. Hyp : u 0 D(A), u(t) = C(t)u 0. With f 1 (z) = cos(t z), f 2 (z) = (cos((t+h) z) cos((t h) z))/z, we have, for z > 0, f 1 (z) 1 and f 2 (z) 2 ht. This allows to set C(t) = f 1 (A), and B(t, h) = f 2 (A), and we have C(t) 1, B(t, h) 2 ht. Furthermore B(t, h) = (C(t + h) C(t h))a 1, This yields u(t+h) u(t h) = B(t, h)au 0, and u(t+h) u(t h) 2 ht Au 0.

Differentiability proof. Hyp : u 0 D(A) and u(t) = C(t)u 0. We have cos(t t z)/z = 1/z (t s)cos(s z) ds. 0 Hence (homomorphism of algebra) C(t) A 1 = A 1 and u(t) = u 0 t 0 t 0 (t s)c(s) ds. (t s)au(s) ds.

Differentiability proof. Hyp : u 0 D(A) and u(t) = C(t)u 0. We have u(t) = u 0 t 0 (t s)au(s) ds. It follows u(0) = u 0, u (t) = t 0 Au(s) ds, u (0) = 0, u (t) = Au(t), and u C 2 (R; H).

Application 2. Heat equation Let A L(D(A), H) a self-adjoint, positive definite operator. We consider the heat equation u (t) + Au(t) = 0, u(0) = u 0 given in H, that we discretize in space by a finite elements method u h (t) + A hu h (t) = 0, u h (0) = u 0 h given in V h H, and then, in time, by the implicit scheme, t n = n t, u n+1 h u n h t + A h u n+1 h = 0.

Application 2. Heat equation Error estimate u(t n ) u n h u(t n) u h (t n ) + u h (t n ) u n h The first term u(t n ) u h (t n ) comes from the finite element method. We only consider the second u h (t n ) u n h = ( exp( n ta h ) r( ta h ) n) u 0 h. If A h is positive definite, (recall that r(x) = (1 + x) 1 ) exp( n ta h ) r( ta h ) n sup (e x ) n r(x) n C/n. x>0

Application 3. Convergence of GMRES. Let us consider a large linear system Ax = b, with b = 1. At the step n, GMRES algorithm builts x (n) Span{b, Ab,..., A n 1 b} element of the Krylov subspace spanned by b that minimizes the residual r (n) = b Ax (n). Therefore, we have r (n) min{ p(a) ;p(0) = 1, p P n }.

Application 3. Convergence of GMRES. Let us consider a large linear system Ax = b, with b = 1. At the step n, GMRES algorithm builts x (n) Span{b, Ab,..., A n 1 b} element of the Krylov subspace spanned by b that minimizes the residual r (n) = b Ax (n). If furthermore, A is self-adjoint with σ(a) [α, β], 0 < α < β, r (n) min{ p(a) ;p(0) = 1, p P n } min{ sup p(x) ;p(0) = 1, p P n }. x [α,β]

Application 3. Convergence of GMRES. Let us consider a large linear system Ax = b, with b = 1. At the step n, GMRES algorithm builts x (n) Span{b, Ab,..., A n 1 b} element of the Krylov subspace spanned by b that minimizes the residual r (n) = b Ax (n). If furthermore, A is self-adjoint with σ(a) [α, β], 0 < α < β, r (n) min{ p(a) ;p(0) = 1, p P n } min{ sup p(x) ;p(0) = 1, p P n } x [α,β] 2 β + α γ n + γ n, with γ =. β α

Spectral sets, J. von Neumann, 1951 A set X C, satisfying σ(a) X, is called a K-spectral set for the operator A if we have r(a) K sup z X r(z) for all rational functions that are bounded on X. If K = 1 we say that X is a spectral set.

Spectral sets A set X C, satisfying σ(a) X, is called a K-spectral set for the operator A if we have r(a) K sup z X r(z) for all rational functions that are bounded on X. Examples 1) If A is self-adjoint, or normal, then the spectrum σ(a) is a spectral set for A. 2) If A 1, then the unit disk { z 1} is a spectral set for A. 3) If Re Av, v 0, v H, then the right half-plane {Re z 0} is a spectral set for A.

Spectral sets A set X C, satisfying σ(a) X, is called a K-spectral set for the operator A if we have r(a) K sup z X r(z) Example 4. Let Ω be a domain that contains σ(a). If we may write the Cauchy formula r(a) = 2πi 1 Ω r(σ)(σ A) 1 dσ, then Ω is a K-spectral set for A with K = 1 2π Ω (σ A) 1 dσ.

There are many other ways to represent r(a) on an integral form : for instance by using Laplace transform (Hille-Phillips calculus), we can also use the Mellin transform, or the Poisson kernel, or the double layer potential... The following two lemmata may be useful in order to bound such integrals.

Lemma 1. Assume that dm(t) is a complex-valued measure, bounded on E, M(t) L(H), M(t) = M (t) 0, in E, r is a rational function bounded by 1 on E. Then E r(t)m(t) dm(t) E M(t) dm(t).

Lemma 1. Assume that dm(t) is a complex-valued measure, bounded on E, M(t) L(H), M(t) = M (t) 0, in E, r is a rational function bounded by 1 on E. Then E r(t)m(t) dm(t) Key of the proof. We have E M(t) dm(t) M(t)u, v M(t)u, u 1/2 M(t)v, v 1/2.

Lemma 2. Assume that dm(t) is a complex-valued measure, bounded on E, M(t), N(t) L(H), N(t) = N (t), in E, α > 0 Re M(t) = 1 2 (M(t) + M(t) ) N(t) α, in E, r is a rational function bounded by 1 on E. Then E r(t)(m(t)) 1 dm(t) E (N(t)) 1 dm(t).

Lemma 2. Assume that r is bounded by 1, α > 0, Re M(t) = 1 2 (M(t) + M(t) ) N(t) α, in E. Then E r(t)(m(t)) 1 dm(t) E (N(t)) 1 dm(t). Key of the proof. M(t) 1 u, v N(t) 1 u, u 1/2 N(t) 1 v, v 1/2.

The numerical range. Recall that the numerical range of a matrix A C d,d is defined by W(A) = { Av, v = v Av ; v C d, v = 1}. It is a closed convex subset of C, (Toeplitz & Hausdorff) that contains the spectrum σ(a). Let us consider a bounded convex domain Ω. We assume that W(A) Ω. If ν denotes the unit outward normal at the point σ Ω, then ν z Ω is equivalent to Re > 0, σ Ω. σ z

ν unit normal at σ Ω, s arclength on Ω, Π σ Ω ν σ dσ = i ν ds z Π σ is equivalent to Re ν σ z > 0.

The numerical range. Recall that the numerical range of a matrix A C d,d is defined by W(A) = { Av, v = v Av ; v C d, v = 1}. It is a closed convex subset of C, (Toeplitz & Hausdorff) that contains the spectrum σ(a). Let us consider a bounded convex domain Ω. We assume that W(A) Ω. If ν denotes the unit outward normal in the point σ Ω, then ν z Ω is equivalent to Re > 0, σ Ω. σ z Therefore W(A) Ω is equivalent to ν(σ A) 1 + ν( σ A ) 1 > 0, σ Ω.

The numerical range. The numerical range of an operator A L(D(A), H) is defined by W(A) = { Av, v = a(v, v);v D(A), v H = 1}. It is a convex subset of C, its closure contains the spectrum σ(a). If Ω is a convex domain, then W(A) Ω implies ν(σ A) 1 + ν( σ A ) 1 0, σ Ω.

Numerical Range and functional calculus An application of the two lemmata is the following theorem Theorem. If Ω is a closed convex domain of the complex plane with a conic boundary and if A is a closed operator with W(A) Ω, then Ω is a 3.2-spectral set for the operator A.

Numerical Range and functional calculus Theorem. If Ω is a closed convex domain of the complex plane with a conic boundary and if A is a closed operator with W(A) Ω, then Ω is a 3.2-spectral set for the operator A. Remark. Then f(a) is well defined for all functions that are holomorphic in the interior of Ω, bounded and continuous in Ω. Furthermore f(a) 3.2 sup z Ω f(z).

Numerical Range and functional calculus Application Corollary. If there exist a Hilbert space V H, V dense in H, and constants α > 0, λ, M, N R, such that then the problem α v 2 V Re Av, v + λ v 2 H M v 2 V Im Av, v N v H v V u (t) + A u(t) = 0, t R, u(0) = u 0 D(A), u (0) = v 0 D(A 1/2 ). has a unique solution in C 2 (R, H) C 0 (R, D(A)). Proof.

Numerical Range and functional calculus Application Proof. If v H = 1 and Av, v = x+iy, then we have x+λ α v 2 V α N 2y2, i.e. W(A) Ω := {x+iy ; x+λ α N 2y2 }. It can be seen that z Ω implies Im z 1/2 ω, with ω = max( λ +, N/2 α). Thus cos(t z) is a holomorphic function in Ω with bound cos(t z) e ω t... The same proof as in the self-adjoint context works.

Numerical Range and functional calculus Theorem. If Ω is a closed convex domain of the complex plane with a conic boundary and if A is a closed operator with W(A) Ω, then Ω is a 3.2-spectral set for the operator A. Proof in the strip case. We assume Ω = {x+iy; y 2 1 0}. We write A = B+iC, with B and C self-adjoint. The condition W(A) Ω reads C 1 and we have to show r(a) 3.2, if the rational function r is bounded by 1 in the strip Ω.

We assume r(z) 1 for z Ω = {x+iy; y 2 1 0}, A = B+iC, with B and C self-adjoint, C 1. We want to show r(a) 3.2.

We assume r(z) 1 for z Ω = {x+iy; y 2 1 0}, A = B+iC, with C 1. We want to show r(a) 3.2. Proof. We set R 1 = r(σ) 1 ( ) (σ A) 1 dσ ( σ A ) 1 d σ, Ω 2πi R 2 = 1 2πi Ω r(σ)( σ A ) 1 d σ,

We assume r(z) 1 for z Ω = {x+iy; y 2 1 0}, A = B+iC, with C 1. We want to show r(a) 3.2. Proof. We set R 1 = r(σ) 1 ( ) (σ A) 1 dσ ( σ A ) 1 d σ, Ω 2πi R 2 = 1 2πi Ω r(σ)( σ A ) 1 d σ, and note that, from the Cauchy formula, r(a) = 1 2πi Ω r(σ)(σ A) 1 dσ = R 1 + R 2.

We assume r(z) 1 for z Ω = {x+iy; y 2 1 0}, A = B+iC, with C 1. We want to show r(a) 3.2. Proof. We set R 1 = r(σ) 1 ( (σ A) 1 dσ ( σ A ) 1 d σ ), Ω 2πi R 2 = 1 2πi Ω r(σ)( σ A ) 1 d σ, and note that, from the Cauchy formula, r(a) = 1 2πi Ω r(σ)(σ A) 1 dσ = R 1 + R 2. Therefore, it suffices to show that R 1 2 and R 2 1.2.

Proof of R 1 2. We have R 1 = r(σ) 1 ( ) (σ A) 1 dσ ( σ A ) 1 d σ, Ω 2πi and it is easily seen that 1 ( ) (σ A) 1 dσ ( σ A ) 1 d σ 0. 2πi

Proof of R 1 2. We have R 1 = r(σ) 1 ( ) (σ A) 1 dσ ( σ A ) 1 d σ, Ω 2πi and it is easily seen that 1 ( (σ A) 1 dσ ( σ A ) 1 d σ ) 0. 2πi Thus it follows from Lemma 1 that 1 ( R 1 (σ A) 1 dσ ( σ A ) 1 d σ ) = 2. Ω 2πi

Proof of R 2 1.2 We have R 2 = 1 2πi Ω r(σ)( σ A ) 1 d σ. We note that Ω = Ω + Ω, with Ω + = R+i and Ω = R i. Ω + 0 Ω

Proof of R 2 1.2 We have R 2 = 1 2πi Ω r(σ)( σ A ) 1 d σ. We note that Ω = Ω + Ω, with Ω + = R+i and Ω = R i. Ω + 0 Ω On Ω +, we have σ = σ 2i and on Ω we have σ = σ+2i.

Proof of R 2 1.2 We obtain R 2 = 1 2πi = 1 2πi = 1 2πi ( ( Ω r(σ)( σ A ) 1 d σ Ω + r(σ)( σ A ) 1 d σ + Ω + r(σ)(σ 2i A ) 1 dσ + Ω r(σ)( σ A ) 1 d σ ) Ω r(σ)(σ+2i A ) 1 dσ We note that r(σ)(σ±2i A ) 1 is analytic for σ Ω. ).

Ω + Γ 0 Γ + We can replace and Ω + f(σ) dσ by Ω f(σ) dσ by + + f(x) dx, f(x) dx.

Proof of R 2 1.2 This yields R 2 = 1 2πi = 1 2πi = 2 π ( ( Ω + r(σ)(σ 2i A ) 1 dσ + r(x)(x 2i A ) 1 dx + r(x)((x A ) 2 + 4) 1 dx. Ω r(σ)(σ+2i A ) 1 dσ r(x)(x+2i A ) 1 dx ) )

Proof of R 2 1.2 This yields R 2 = 2 π r(x)((x A ) 2 + 4) 1 dx. Recall that A = B+iC and C 1, thus Re ( (x A ) 2 + 4 ) = (x B) 2 C 2 + 4 (x B) 2 + 3.

Proof of R 2 1.2 This yields R 2 = 2 π r(x)((x A ) 2 + 4) 1 dx. Recall that A = B+iC and C 1, thus Re ( (x A ) 2 + 4 ) = (x B) 2 C 2 + 4 (x B) 2 + 3. We may apply Lemma 2 and we get R 2 2 π 2 π sup λ R ((x B)2 + 3) 1 dx dy (y λ) 2 + 3 = 2 3 1.2

Numerical Range and functional calculus Theorem. If Ω is a convex domain with a conic boundary and if A is a closed operator with W(A) Ω, then Ω is a 3.2-spectral set for the operator A. We have only considered here the case of a strip, but the proof in the general case is very similar...

A more general result is : The numerical range of an operator A W(A) := { Av, v ;v D(A), v = 1} is a K-spectral set for A with K 11.08.

Application : Convergence of GMRES. We assume that W(A) Ω and 0 / Ω, Ω bounded convex domain. Let Φ be the Riemann mapping from C \ Ω onto the exterior of the unit disk, satisfying Φ(z) = γz + δ + O(z 1 ), as z, γ > 0. The Faber polynomial of degree n, F n, is defined by It is known that Similarly we have F n (A) = (Φ(z)) n = F n (z) + O(z 1 ), as z. F n (z) = Ω (Φ(σ))n 2πi 1 ( dσ σ z d σ σ z Ω Φ(σ)n 2πi 1 ( (σ A) 1 dσ ( σ A ) 1 d σ ). ).

Application : Convergence of GMRES. We have F n (A) = Ω Φ(σ)n 2πi 1 ( ) (σ A) 1 dσ ( σ A ) 1 d σ. Recall that ν = 1 i dσ ds, thus ( ) ( 1 2πi (σ A) 1 dσ ( σ A ) 1 d σ = 1 2π ν(σ A) 1 + ν( σ A ) 1 ) ds 0.

Application : Convergence of GMRES. We have F n (A) = Ω Φ(σ)n 2πi 1 ( ) (σ A) 1 dσ ( σ A ) 1 d σ. Recall that ν = 1 i dσ ds, thus ( ) ( 1 2πi (σ A) 1 dσ ( σ A ) 1 d σ = 1 2π ν(σ A) 1 + ν( σ A ) 1 ) ds 0. We can apply Lemma 1 F n (A) Ω ( ) 1 2πi (σ A) 1 dσ ( σ A ) 1 d σ = 1 + 1 = 2.

Application : Convergence of GMRES. We consider a large linear system Ax = b, with b = 1. GMRES algorithm built at step n, an element x (n) that minimizes the residual r (n) = b Ax (n). Therefore, r (n) min{ p(a) ;p(0) = 1, p P n }. With the choice p(z) = F n (z)/f n (0), we get, r (n) F n (A) / F n (0) 2/ F n (0).

Application : Convergence of GMRES. Setting γ = Φ(0) > 1, we have shown that, if the matrix satisfies W(A) Ω, then we have the estimate min{ p(a) ;p(0) = 1, p P n } 2 F n (0) 2 γ n γ 1. This generalizes to the matrices an estimate due to Kövari and Pommerenke in the scalar case min{sup p(z) ;p(0) = 1, p P n } z Ω 2 F n (0) 2 γ n γ 1.

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