Runge–Kutta approximation for \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C_0$$\end{document}C0-semigroups in the graph norm with applications to time domain boundary integral equations

We consider the approximation of an abstract evolution problem with inhomogeneous side constraint using A-stable Runge–Kutta methods. We derive a priori estimates in norms other than the underlying Banach space. Most notably, we derive estimates in the graph norm of the generator. These results are used to study convolution quadrature based discretizations of a wave scattering and a heat conduction problem.


Introduction
Many time dependent partial differential equations can be conveniently described in the language of strongly continuous semigroups.In this language, these initial boundary value problems resemble systems of ordinary differential equations, which suggests that they are amendable to the standard discretization schemes of multistep or Runge-Kutta type.Unlike the ODE case, one needs to pay special attention to the boundary conditions imposed by the generator of the semigroup.This, in most cases, leads to a reduction of order phenomenon, meaning that the convergence rates are (mainly) determined by the stage order of the Runge-Kutta method instead of the classical order.The a priori convergence of Runge-Kutta methods for semigroups has been extensively studied in the literature.Starting with the early works [BT79,Cro76], it has been established that conditions of the form u(t) ∈ dom(A µ ), where A is the generator of the semigroup, determine the convergence rates.In [OR92], this has been generalized to the case of non-integer µ ≥ 1 using the theory of interpolation spaces.Finally, in [AMP03], the case of µ ∈ [0, 1] was adressed, which is the case needed for PDEs with inhomogeneous boundary conditions.All of these works focus on establishing convergence rates with respect to the norm of the underlying Banach space.In many applications one needs to establish convergence with respect to other norms, for example, in order to be able to bound boundary traces of the solution.Most notably, one might be interested in convergence of A u, where A is an extension of the generator that disregards boundary conditions.If u is assumed to be in dom(A), we get A u = Au and the convergence result can be easily established by using the fact that the time-evolution commutes with the generator of the underlying semigroup (both in the continuous and discrete settings).If the boundary conditions are inhomogeneous, such a strategy cannot be pursued.It is the goal of this paper to establish convergence results for A u also for the case u(t) ∈ dom(A µ ) for µ ∈ [0, 1], again using the theory of interpolation spaces.
Similarly it is sometimes useful to compute discrete integrals of the time evolution, by reusing the same Runge-Kutta method.Also in this case, we establish rigorous convergence rates.
Our interest in such estimates originally arose from the study of time domain boundary integral equations and their discretization using convolution quadrature(CQ).It has already been noticed in the early works (see e.g.[LO93]) that such discretizations have a strong relation to the Runge-Kutta approximation of the underlying semigroup.This approach of studying TDBIEs in a strictly time-domain way has recently garnered a lot of interest, see [HS16b,BLS15,HQSVS17] and the monograph [Say16], as it potentially allows sharper bounds than the more standard Laplace domain based approach.Similar techniques have even been extended to the case of certain nonlinear problems in [BR17].This paper can be seen as our latest addition to this effort.While the convergence rates provided by the Laplacedomain approach in [BLM11] and the results in this current paper are essentially the same, the present new approach provides better insight into the dependence on the end-time of the computation.It also fits more naturally with the time-domain analysis of the continuous problem and space discretization, as for example presented in [HQSVS17].
The paper is structured as follows.Section 2 introduces the abstract setting and fixes notation, most notably for working with Runge-Kutta methods.Section 3 then contains the main estimates.Starting by summarizing known results from [AMP03] in Section 3.1, we then formulate the main new results of this article in Section 3.2.After proving some preparatory Lemmas related to Runge-Kutta methods in Section 4 and 5, we then provide the proofs of the main estimates in Section 6.In Section 8, to showcase how the theory developed in this paper is useful for this class of problems, we consider a simple exterior scattering problem in Section 8.3 and a heat transmission problem in Section 8.5.We note that Section 8.3 showcases the need for the bound on the discrete integral of the result, whereas Section 8.5 was chosen because, in order to bound the main quantity of interest on the boundary, we need to apply a trace theorem.This necessitates the use of the graph norm estimate.

Problem setting
We start by fixing the general setting used for the rest of the paper, first with respect to the equation to be solved and then with respect to its discretization.

Operator equation, functional calculus, and Sobolev towers
Assumption 2.I.We are given: (a) a closed linear operator A : dom A ⊂ X → X in a Banach space X , (b) and a bounded linear operator B : dom A → M.
We assume that A := A | ker B generates a C 0 -semigroup and that B admits a bounded right inverse, denoted by E , such that range E ⊂ ker(I − A ), where I : X → X is the identity operator.
Proposition 2.1.If A is the generator of a C 0 -semigroup on a Banach space X , then there exist constants ω ≥ 0 and M ≥ 1 such that the spectrum σ(A) of A satisfies σ(A) ⊆ {z ∈ C : Re z ≤ ω} and the resolvent satisfies the estimates (2.2) Proof.The case ω = 0 is shown in [Paz83,Corollary 3.6].The more general case follows as usual by considering the shifted semigroup e ωt e At or directly from the integral representation of the resolvent.
When working with Runge-Kutta methods, it is useful to use a calculus that allows us to apply rational functions to (unbounded) operators, as long as the poles of the function are compatible with the spectrum of the operator.Definition 2.2 (Rational functions of operators).Let q be a rational function that is bounded at infinity, Λ be the set of poles of q, which we can write in the form (note that we allow for some of the factors in the numerator to be constant) It is easy to see that different reorderings of the factors in the numerator and denominator of q produce the same result and that each factor in the definition of q(A) is a bounded linear operator in X since λ i ∈ σ(A).The bounded linear operator q(A) : X → X satisfies (2.4) The error estimates of this paper use the theory of interpolation spaces.For Banach spaces X 1 ⊂ X 0 with continuous embedding and µ ∈ (0, 1), we define the space [X 0 , X 1 ] µ,∞ using real interpolation with the following norm: (2.5) We will not go into details of the definitions, they can be found in [Tri95,Tar07] or [McL00,Appendix B].For simplicity of notation we often drop the second parameter ∞ and just write [X 0 , X 1 ] µ .
The most important property is the following: a bounded linear operator T : We also note that for µ 1 ≤ µ 2 , the spaces are nested, i.e., [X 0 , X 1 ] µ 2 ⊆ [X 0 , X 1 ] µ 1 with a continuous embedding.For notational convenience we write [X 0 , X 1 ] 0 := X 0 and [X 0 , X 1 ] 1 := X 1 .We will be interested in a collection of spaces defined by interpolating the domains of the powers of the operator A. The details of this construction can be found, for example in [EN06].
Definition 2.3 (Sobolev towers).Let A be a closed operator on a Banach space X .For µ ∈ N 0 , we define the following spaces X 0 := dom A 0 := X and X µ := dom A µ , equipped with the following norm

Runge-Kutta approximation and discrete stage derivative
An m-stage Runge-Kutta method is given by its Butcher tableau, characterized by Q ∈ R m×m and b, c ∈ R m .The Runge-Kutta approximation of the Problem (2.1) starts at u k 0 := u 0 and then computes for n ≥ 0 the stage vector U k n ∈ X m and the step approximation u k n+1 ∈ X by solving (2.7c) We have used the following notation (the spaces Y and Z are generic): The following lemma involving inversion of matrices of operators associated to an operator can be proved by taking the Jordan canonical form of the matrix S.
Lemma 2.4.If A : dom A ⊂ X → X is a linear operator on a Banach space X and S ∈ C m×m satisfies σ(A) ∩ σ(S) = ∅, then is invertible.Furthermore, there exists a constant C S , depending only on S, such that Under Assumption 2.I, the internal stage computation in the RK method can be decomposed in the following form: (2.8c) In (2.8b) we look for Z k n ∈ (dom A) m .The stability function of the Runge-Kutta method is the rational function r(z) := 1 + zb (I − zQ) −1 1.We will not consider the full class of Runge-Kutta methods, but will restrict our considerations to those satisfying the following Assumptions: Assumption 2.II.(i) The matrix Q is invertible (ii) The stability function r does not have poles in {z : Re z < 0}, and |r(it)| ≤ 1 for all t ∈ R (i.e., the method is A-stable).Equivalently, |r(z)| < 1 for all z with negative real part.
We note that Assumption 2.II (i) implies that the following limit exists and that r is a rational function with poles only in C + and bounded at infinity.The computation of the internal stages in the numerical approximation (2.7) requires the inversion of as can be seen from the equivalent form (2.8).
If A is the infinitesimal generator of a C 0 -semigroup, ω and M are given by Proposition 2.1, and we choose (recall that σ(Q) ⊂ C + ) then the RK method can be applied for any k ≤ k 0 .By Proposition 2.1 and Lemma 2.4, it follows that (2.10) Using Definition 2.2, we can define r(k A) for an RK method satisfying Assumption 2.II and k ≤ k 0 satisfying (2.9).We then define This quantity is relevant for the study of the error propagation in the Runge-Kutta method.
Given an RK method, we consider the following matrix-valued rational function (2.12) (The verification that these two formulas correspond to the same matrix is simple by using the Sherman-Morrison-Woodbury formula.)This matrix is related to the discrete differentiation process associated to an RK method satisfying Assumption 2.II: on the one hand k −1 δ(z) is the discrete symbol associated to the Discrete Operational Calculus built with the RK method [LO93]; on the other hand, a direct interpretation of this symbol is possible using Z-transforms (see [HS16a,Section 6]).Given a sequence U := {U n } (tagged from n ≥ 0) on a space, its Z-transform is the formal series For a detailed treatment on formal power series, see [Hen88].
Definition 2.5.Let U := {U n } and V := {V n } be two sequences in X m and let U and V be their respective Z-transforms.If we write The above definition is consistent with the RK discrete operational calculus of Lubich and Ostermann, see Section 8.1 and [LO93].We now show an explicit form of the computation of ∂ k and its inverse.
Lemma 2.6.If U = {U n } is a sequence in X m , then X := (∂ k ) −1 U can be computed with the recurrence and V := ∂ k U can be computed with the inverse recurrence (2.14) Proof.The proof of (2.13) is a simple exercise in Z-transforms, while (2.14) follows from (2.13) by writing U n in terms of X n (and changing names to the sequences).
Note that the first result of Lemma 2.6 says that if we apply the RK method to the equation and X := {X n } is the sequence of vectors of internal stages, then X = (∂ k ) −1 U , where U n := u(t n + kc).
Finally we note that we call a Runge-Kutta method stiffly accurate, if it satisfies b Q −1 = e m := (0, . . ., 0, 1).Stiffly accurate methods satisfy (we use that and r(∞) = 0.For stiffly accurate methods, the computation of the discrete stage derivative of a vector of stage of samples of a continuous function is particularly simple: Lemma 2.7.For stiffly accurate RK methods, the sequence G := ∂ k F with F n = F (t n + kc), satisfies Proof.For stiffly accurate methods we have r(∞) = 0 and therefore However, since c m = 1, we have e m F (t n−1 + kc) = F (t n−1 + kc m ) = F (t n ), which proves the result.
We also make the following optional assumption, which allows to increase the convergence order in some cases.

Error estimates
We are now in a position to formulate the main results of this article and put them into context with previous results, most notably from [AMP03].

The estimates of Alonso-Mallo and Palencia
The following two results summarize the results of Alonso-Mallo and Palencia [AMP03], rewritten with the notation of the present paper.The 'proofs' which we provide clarify how notation needs to be adapted and how the hypotheses of the main results of [AMP03] are satisfied in our context.Proposition 3.1 ([AMP03, Theorem 1]).Let Assumption 2.I hold and assume that the exact solution u satisfies u ∈ C p+1 ([0, T ] , X µ ) for some µ ≥ 0. Let {u k n } denote the Runge-Kutta approximation from (2.7).There exist constants k 0 > 0, C > 0, such that for 0 < k ≤ k 0 and 0 < nk ≤ T the following estimate holds: The constant C depends on the Runge-Kutta method, µ, and the constants M and ω from (2.2).The constant k 0 depends only on ω and the Runge-Kutta method.
Proof.We only make remark on the differences in notation.A different definition of interpolation spaces is given in [AMP03], but the proof only relies on estimates of the form (2.6).The choice of k 0 follows from the fact that it is only needed to ensure that (I − k Q ⊗ A) is invertible, see (2.10).The assumption µ ≤ p − q in [AMP03, Theorem 1] can be replaced by using the rate min{p, q + µ} in (3.1) as the spaces X µ ⊆ X p−q are nested for µ ≥ p − q.We also lowered the regularity requirements on the highest derivative compared to their stated result.The fact that this holds true follows from inspection of the proof.Compare also to Lemma 5.9 for the key ingredient.
For a subset of Runge-Kutta methods, these estimates can be improved: Let the assumptions of Proposition 3.1 hold and assume that, in addition, the RK-method satisfies Assumption 2.III.There exist constants k 0 > 0, C > 0 such that for 0 < k ≤ k 0 and 0 < nk ≤ T the following improved estimate holds: The constant C depends on the Runge-Kutta method, µ, and the constants M and ω from (2.2); k 0 depends only on the constant ω and the Runge-Kutta method.
Proof.Again, this is just a reformulation of [AMP03, Theorem 2].We first note that, due to our assumption on r(∞), we are always in the case m = 0 of [AMP03].
Since we assumed that on the imaginary axis |r(it)| < 1 for 0 = t ∈ R , we directly note that for sufficiently small k ≤ k 0 , all the zeros of r(z) − 1 except z = 0 satisfy Re z > kω.By the resolvent bound (2.2) we can therefore estimate i.e., we have a uniform resolvent bound in the set Z α,δ in [AMP03].We also note that we reformulated the convergence rate such that we do not have the restriction µ ≤ p − q − 1, since the exceptional cases are already covered by Proposition 3.1.

New results in this article
In this section, we present some a priori estimates for the convergence of Runge-Kutta methods when applied to the abstract problem (2.1).These can be seen as a continuation of [AMP03], to the case where the boundary conditions are not given exactly but stem from computing discrete integrals and differentials using the same Runge-Kutta method.
Theorem 3.4 (Integrated estimate).Let u solve (2.1) with u 0 = 0, assume that for some µ ≥ 0 we have and let x := ∂ −1 u.Let U k = {U k n } and let u k = {u k n } be the discrete approximation given by (2.7) for a method satisfying Assumption 2.II.If X k := (∂ k ) −1 U k and we define x k = {x k n } with the recurrence then there exists a constant k 0 > 0 such that for all k < k 0 and n ∈ N with nk ≤ T the following estimate holds: If Assumption 2.III holds and if we assume the stronger regularities The constant k 0 depends only on ω from (2.2) and the Runge-Kutta method.If ω = 0 then k 0 can be chosen arbitrarily large.C depends on ω, M from (2.2), the Runge-Kutta method and µ.Theorem 3.5 (Differentiated estimate).Let u solve (2.1) with u 0 = 0 and assume u(0) = 0. Assume that for some µ ≥ 0 we have be the discrete approximation given by (2.7) for a stiffly accurate method satisfying Assumption 2.II.
, then there exists a constant k 0 > 0 such that for all k < k 0 and n ≥ 1 such that nk ≤ T the following estimate holds: If, in addition, the method satisfies Assumption 2.III and T,0 The constant k 0 depends only ω from (2.2) and the Runge-Kutta method.If ω = 0, then k 0 can be chosen arbitrarily large.C depends on ω, M from (2.2), the Runge-Kutta method and µ.
Theorem 3.6 (Strong norm estimate).If u, Ξ, and F satisfy the hypotheses of Theorem 3.5, and {u k n } is the approximation provided by (2.7) for a stiffly accurate method satisfying Assumption 2.II (same restrictions with respect to k 0 ), then for n such that nk ≤ T , there holds If, in addition, the method satisfies Assumption 2.III, then T,0 C depends on ω, M from (2.2), the Runge-Kutta method and µ.
Remark 3.7.Most of the effort in proving the above theorem is in order to obtain a convergence rate higher than q, even though the constraint in the stages only approximate with order q.This is possible by exploiting the additional structure of the discretization error of the side constraint.
Remark 3.8.We formulated all our results for homogeneous initial conditions, since it is sufficient for our purposes in time domain BEM and convolution quadrature.It should be possible to generalize these results to the case of u 0 ∈ dom(A s ) for sufficiently large s ≥ 1 by considering the evolution of the semigroup with inhomogeneous side-constraint but homogeneous initial condition and the semigroup of homogeneous constraint but inhomogeneous u 0 separately.
Remark 3.9.The loss of order by 1 in Theorem 3.6 compared to Propositions 3.1 and 3.2 is to be expected.Indeed, if we look at the case u ∈ dom(A µ ) for µ ≥ 1, this means A u ∈ dom(A µ−1 ).Applying Proposition 3.2 to this semigroup then also gives a reduced order of k min(q+µ,p) .

Some computations related to the main theorems
We will collect the sampled data and the stage and step parts of the solutions in four formal series If the data functions are polynomially bounded in time, the series in (4.1a) are convergent (in X m and M m respectively) with at least unit radius of convergence.
Because of the equivalent formulation of the numerical method in the form (2.8), and using (2.10), it follows that for k ≤ k 0 (with k 0 chosen using (2.9)), the numerical solution is at least bounded in the form U k n X C n .Thus, the two series in (4.1b) also converge on a sufficiently small disk.
Proof.Let us start by proving a simple result: the discrete equations (2.7a) and (2.7c) hold if and only if (2.7a) and hold.To see this, note that (2.7a) is equivalent to and therefore (2.7a) and (2.7c) imply or equivalently (4.3).The reciprocal statement is proved similarly.The recurrence (4.3) is equivalent to (4.2c).At the same time, the recurrence (2.7a) is equivalent to Plugging (4.2c) into (4.4), the formula (4.2a) follows.
Proposition 4.1 is a rephrasing of [Rie17, Lemma 3.19], where the computation is also laid out in more detail.Note how equations (4.2a)-(4.2b)relate strongly to (2.1), with the discrete symbol k −1 δ(z) playing the role of the time derivative and playing the role of a discrete Dirac delta at time t = 0.
We need a corollary to the previous result: Corollary 4.3.Let Assumption 2.II hold.Then, for all r 0 < 1, there exists a constant d > 0 such that for all |z| < r 0 it holds that Proof.In view of of Lemma 4.2, since σ(Q) is finite, independent of z, and contained in C + , we are mainly concerned with the set {w ∈ C : r(w)z = 1}.We first note that Second, we observe that by taking d 0 small enough, we can ensure that w → r(w) is continuous for Re(w) ≤ d 0 and thus is a closed set.Third, by considering the limit along the imaginary axis, we get Thus, for |w| sufficiently large, it holds that |r(w)| ≤ 1/r 0 .Overall, we get that is a compact set with empty intersection with the imaginary axis.Thus, it must have a positive distance from it.Applying these observations and Lemma 4.2 concludes the proof.
Lemma 4.4.Let Assumptions 2.I and 2.II hold.For r 0 < 1, there exists k 0 = k 0 (ω, r 0 ) > 0 such that for all k ≤ k 0 and |z| ≤ r 0 the problem has a unique solution for arbitrary F ∈ X m and Ξ ∈ M m .If ω = 0 in Proposition 2.1, then there are no restrictions on k, and the results holds for all |z| < 1.
Proof.Assume first that S ∈ C m×m is such that σ(S) ⊂ {z : Re z > ω} and consider the problem This problem is uniquely solvable by Lemma 2.4, since σ(A) ⊂ {z : Re z ≤ ω} and therefore σ(A) ∩ σ(S) = ∅.We then define U := V + W , which solves (4.6).To see uniqueness, one observes that the difference of two solutions solves the homogeneous problem ( Ξ = 0 and F = 0) for which uniqueness was established in Lemma 2.4.By Corollary 4.3, the union of the spectra of δ(z) for |z| ≤ r 0 has a positive distance d(r 0 ) > 0 from the imaginary axis.If we take k 0 < d(r 0 )/ω, then σ(k −1 δ(z)) ⊂ {s : Re s > ω} for all |z| ≤ r 0 and k ≤ k 0 .When ω = 0, we can take any k 0 .By the previous considerations this implies unique solvability.
Proof.Recall that equations (2.7a) and (2.7c) are equivalent to (2.7a) and (4.3), as shown in the proof of Proposition 4.1.Moreover, the latter equations are equivalent to (4.2) in the Z-domain.In the present case we have u 0 = 0.For a given square matrix P ∈ C m×m and an operator C, we have which proves that By Proposition 4.1, equations (4.9) are equivalent to (4.7).Finally (4.8a) follows from (4.2a), while (4.8b) follows from (4.4) and (4.8a).
for data Proof.Follow the proof of Proposition 4.5.

Some Lemmas regarding Runge-Kutta methods
In order to shorten the statements of the results of this section, in all of them we will understand that: (1) We have an RK method with coefficients Q, b, c satisfying Assumption 2.II (invertibility of Q and A-stability).The method has classical order p and stage order q.
(2) We have an operator A in X that is the generator of a C 0 -semigroup, characterized by the quantities M and ω of Proposition 2.1.The associated Sobolev tower {X µ }, obtained by interpolation of dom A µ for positive integer values of µ, will also be used.
The following lemma will be used at a key point in the arguments below.
Lemma 5.1.Let A be a linear operator in X and q be a rational function bounded at infinity with poles outside σ(A).The following properties hold: (a) The operator q(A) maps dom A to dom A for all .
Proof.To prove (a), show first by induction on that (A − λI) −1 maps dom A into dom A +1 .Using this result for each of the factors in the definition (2.3) the result follows.To prove (b) note first that p is rational, bounded at infinity, and that σ(A) does not intersect the set of poles of p.Using Definition 2.2, we have p(A) = q(A)A − = A − q(A), and the result follows.
We start by recalling some simple facts about RK methods that we will need in the sequel.Using the notation c := (c 1 , . . ., c m ) , the following equalities (order conditions) hold (see e.g.[AMP03,OR92])): For a stiffly accurate method we have (2.15) and therefore (5.3) Lemma 5.2 (Discrete antiderivative and RK quadrature).
} and {g k n } be given by the recursion For the errors d k n := g(t n ) − g k n , and for n such that nk ≤ T , we have the estimates (5.4b) Additionally, at the stage level we have (5.4c) Proof.Since and the RK method defines a quadrature formula with degree of precision p − 1 (as follows from the order conditions (5.2b)), the bound (5.4b) follows.Using a telescopic sum argument (and d k 0 = 0), (5.4a) is shown to be a consequence of (5.4b).Since G k n = 1g k n + kQf (t n + kc) (see Lemma 2.6), we can write The order conditions (5.2a) with j = 0 prove that if π is a polynomial of degree p − 1, then b (π(kc) − 1π(0) − kQ π(kc)) = 0.
A Taylor expansion of degree p − 1 about t n can then be used to show that and (5.4c) follows.

Estimates on rational functions of the operator
We will use the rational functions ) (5.6) Note that these rational functions are bounded at infinity and that r ,β (0) = 0. We will also use the vector-valued rational function and note that g(0) = 0 and r(z) = 1 + g(z) 1.
Proof.Using the Neumann series expansion for the inverse in the definition or r ,−1 , valid for |z| ≤ 1/ Q , we can write after using (5.2) and (5.3) (for which we needed stiff accuracy).With the same technique we write which finishes the proof.
Lemma 5.4.If the RK method satisfies Assumption 2.III, then there exists a constant k 0 > 0 depending on the RK method and on ω such that for β ∈ {0, 1}, ≤ p − β, and all 0 < k ≤ k 0 with 0 ≤ n k ≤ T , we have the estimate with ρ k (T ) defined in (2.11).If = p and β = 1, the left-hand side of (5.9) is bounded by Cρ k (T ).The constant C > 0 in (5.9) depends only on the Runge-Kutta method, M and ω, k 0 , , and µ, but is independent of n and k.If the Runge-Kutta method is stiffly accurate, the estimate (5.9) also holds for β = −1.If ω = 0, then k 0 can be chosen arbitrarily.
We then write From its definition, we observe that the rational function r ,β is bounded at infinity, has a zero of order p − − β + 1 at z = 0 (Lemma 5.3), and its poles are contained in σ(Q −1 ) ⊂ C + (by A-stability).Note that r(0) = 1 with nonzero derivative, since r(z) is an approximation of exp(z) with order p + 1.Therefore, the rational function (r(z) − 1)z µ has a zero of order µ + 1 ≤ p − − β + 1 at z = 0.All its other zeros are in C + , since by Assumption 2.III, |r(z)| < 1 for all z = 0 with Re z ≤ 0. This implies that the rational function q ,β,µ , which is bounded at infinity (we are using here that r(∞) = 1, which is part of Assumption 2.III), has its poles in (5.10) Take now k 0 > 0 such that k 0 ω < λ min := min{Re λ : λ ∈ Λ} and note that by Proposition 2.1 Therefore, using (2.4) where C depends on M , ω, k 0 , and the RK method.By Lemma 5.1 we have This and (5.11) proves (5.9) for integer µ ≤ p − − β.For larger integer values of µ, the result does not need to be proved as the maximum rate is already attained.We just have to estimate the X p− −β norm by the stronger X µ norm.For real values of µ, we use interpolation.
We still need to prove the result when p − − β = −1, which can only happen when = p and β = 1.We note that r p,1 (0) = 0 and we can therefore argue as in the previous case for µ = 0. Lemma 5.5.If the RK method satisfies Assumption 2.III and k 0 is the value given in Lemma 5.4, then (5.12) for all k ≤ k 0 and n such that nk ≤ T .
Proof.Since g(0) = 0, we can adapt the proof of Lemma 5.4 to each of the components of the vector-valued function g.The key step is to show that h(z) := (r(z) − 1) −1 g(z) is bounded at infinity and has all its poles in the set defined in (5.10) and therefore Since the operator s n (k A)g(k A) on the left-hand side of (5.12) can be rewritten as (r(kA) n+1 − I)h(k A) , the bound (5.12) follows readily.
We can then use the same argument to prove (5.11) for this redefined new function q ,β,µ .(Note that we do not use Assumption 2.III in this argument.)Using that r ,β (k A) = k µ q ,β,µ (kA)A µ in dom A µ , the result follows.Stiff accuracy of the method is used in the case β = −1 when we apply Lemma 5.3, dealing with the zeros of r ,−1 .The proof of (5.14) is a similar adaptation of the proof of Lemma 5.5.

Estimates on discrete convolutions
The RK error will naturally induce several types of discrete convolutions that we will need to estimate separately.In all of them we will have the structure (5.15) We first deal with the simplest cases.
Lemma 5.7.For nk ≤ T , the sequence defined by (5.15) can be bounded by If η n := g(kA) ξ n for ξ n ∈ X m , then Proof.The sequence defined by the recurrence (5.15) can be written as the discrete convolution r(kA) n−j η j . (5.16) The estimates then follow from the definition of ρ k (T ) and from (5.14).
The next estimate is related to the consistency error of the RK method in the sense of how the RK method approximates derivatives at the stage level.We introduce the operator D k (y; t) := y(t + kc) − y(t)1 − kQ ẏ(t + kc). (5.17) Proof.We will prove the result for scalar-valued functions, since the extension for Banach-space valued functions is straightforward.We use a Taylor expansion Plugging this decomposition in the definition of D k (y; t), we obtain Defining R k (t) := r(kc; t) − kQ ṙ(kc; t) and using (5.1), we have (5.18).Since We are almost ready for the two main lemmas of this section, the first one without Assumption 2.III and the second one with it.These results and their proofs follow [AMP03, Theorem 1 and 2], where only the case β = 0 is covered.Lemma 5.9.Let y ∈ C p+1 ([0, T ]; X ) ∩ C p ([0, T ]; X µ ), and we consider the sequence ω n defined by the recurrence (5.15) with η n := k −1 g(kA) (kQ) β D(y; t n ).Then there exists a constant k 0 > 0 depending only on ω from (2.2) and the RK-method such that for nk ≤ T for β ∈ {0, 1}, µ ≥ 0, and k ≤ k 0 .The estimate also holds for β = −1 if the method is stiffly accurate.If ω = 0, then k 0 can be chosen arbitrarily.
Proof.Considering the functions we have that η n = k −1 e k β (t n ).Using Lemmas 5.6 and 5.8 (recall the definition of r j,β in (5.5)), we can bound By (5.19), we then have (5.21) and the result then follows from Lemma 5.7.
Proof.We will use the function e k β defined in (5.20) and an expression involving the rational functions s n defined in (5.6) (recall that s 0 = 1).We first apply Lemmas 5.4, 5.5 and 5.8 to estimate and using that y j (t) − y (j) (t − k) Xµ ≤ k max t−k≤τ ≤t+k y (j+1) Xµ , the analogous computation to the above bound, but using y − y(• − k) as data implies and therefore Note that if β ∈ {0, 1} we can make a simpler estimate for the term originating from R k , (i.e., the one containing the highest derivative) using less regularity for y by not taking advantage of the difference between y (p+1) (t j ) and y (p+1) (t j−1 ) and thus end up requiring less regularity.Using the estimate (5.21) for the last term in (5.22), we have thereby already derived estimates for all three terms in (5.22).

Proofs
The two different cases (with or without Assumption 2.III) will be collected by using the parameter α := 1, if Assumption 2.III holds, 0, otherwise.(6.1) 6.1.Proof of Theorem 3.4 Recall that u solves (2.1) with u(0) = 0.The functions Ξ and F are the given boundary and volume data.If Γ := ∂ −1 Ξ and On the other hand, {X k n } = (∂ k ) −1 {U k n } solves by Proposition 4.6: Before we can estimate the difference between the functions x and x k n , we need one final lemma.Lemma 6.1.Let x solve Assume that for some µ ≥ 0 we have Proof.We set y := x − E Γ.By assumption we have y ∈ C p ([0, T ]; X µ ) and B (x − E Γ) = 0. Since x ∈ dom(A ) and range E ⊂ dom(A ) this implies y(t) ∈ dom(A) for all t ∈ [0, T ].We further calculate using (6.4) and range E ⊆ ker I − A : Each of the contributions is assumed in C p ([0, T ]; X µ ), thus y ∈ C p ([0, T ]; X µ+1 ).
We will need the sequences {γ k n } and {h k n } with the scalar parts of the computations of {Γ k n } and {H k n } respectively, namely (see Lemma 2.6) ) We then consider Using (6.5a), the definition Γ k = (∂ k ) −1 Ξ, and (2.14), we can write Lemma 5.2 (take f = E Ξ for the first three inequalities and f = F for the last one) proves that T,0 , (6.7a) The error analysis is derived by tracking the evolution of the following differences (compare (6.2) and (6.3b) to see the vanishing boundary condition for E k n ) and note that by (6.7a) T,0 , which shows that we only need to estimate e k n to prove Theorem 3.4.We start with the observation that x solves the following equation, as can be easily derived from the equation (6.2): Recalling that Assumption 2.I included the hypothesis range E ⊂ ker(I − A ), we have (Q ⊗ A )∆ k n = Q∆ k n .Combining (6.8) and (6.3a), we get Naive estimation of the terms D k (x, t n ) and ∆ k n − 1δ k n would yield convergence rates similar to Propositions 3.1 and 3.2.In order to get an increased rate, as stated in Theorem 3.4, we combine these two terms using the function Y (t) := x(t) − E Γ(t).By Lemma 6.1 and the assumptions of Theorem 3.4, this function satisfies We can thus further simplify (6.10) This then immediately gives (recall (5.7) for the definition of g) It is easy to see from (6.2) that x satisfies Subtracting (6.3c) from this, plugging in (6.11), using that (b ⊗ A )∆ k n = b ∆ k n , and setting we have What is left is the careful combination of terms so that we can bound everything using Lemmas 5.7, 5.9, and 5.10 by writing Since the above recurrence defining {e k n } is linear as a function of the right-hand side, we can estimate its norm by adding the effects of each of the terms.In the order in which they appear in the last expression, we use: Lemmas 5.9-5.10 with β = 0, but noting that Y (t) ∈ dom(A µ+1 ); Lemmas 5.9-5.10 with β = 1; Lemma 5.7 combined with (6.7a) and (6.7d); Lemma 5.7 combined with (6.7c) and (6.7b); for the first term of ϕ k h we use Lemma 5.7 combined with Lemma 5.2 with f := ẋ.Finally, for the second contribution to ϕ k h , we use (5.4c).Combined, these results give T,0 If we apply Lemma 6.1 to bound the X µ+1 -norm, we arrive at the stated estimate.

Proof of Theorem 3.5
This proof is very similar to the one for Theorem 3.4, while slightly simpler.We will point out the main steps of the proof.Note that we use the simple form of ∂ k for stiffly accurate RK methods given in Lemma 2.7.We define G := Ḟ and Θ := Ξ so that v = u satisfies Its RK approximation and ) and (note (6.12b) and (6.12c)) By (6.12a) and (6.13a), using that (Q ⊗ A )∆ k n = Q∆ k n (assumption on the lifting) and Lemma 2.7 to represent G k n , we have and therefore, from (6.12c) and (6.13c) The final term can be shown using to be of order O(k p+1 ) by combining (5.18) with (5.2b) and (5.3).Use then Lemmas 5.9 and 5.10 with β = −1 and β = 0 as well as Lemma 5.7 to bound Finally Propositions 3.1 and 3.2 are used to bound T,0 .(6.15)

Proof of Theorem 3.6
Thanks to Proposition 4.5, Theorem 3.6 can be proved as an easy corollary of Theorem 3.5.Since the last stage of a stiffly accurate method is the step, we have that (4.8a) implies that

Maximal dissipative operators in Hilbert space
In this short section we summarize some results that show that the hypotheses on the abstract equation and its discretization are simpler for maximal dissipative operators on Hilbert spaces.These results are well-known and will be needed when applying the theory developed in the previous sections to some model problems in Section 8.If A is maximal dissipative in the Hilbert space X , i.e., Re Ax, x X ≤ 0 ∀x ∈ dom A, and if A − I : dom A → X is invertible with bounded inverse, then the constants in Proposition 2.1 can be chosen as M = 1 and ω = 0.In this case A generates a contraction semigroup in H. See [Paz83, Section 1.4].
In particular, if the RK method satisfies Assumption 2.II and then the equations (2.7) (or equivalently (2.8)), defining the RK approximation of (2.1) are uniquely solvable for any k > 0 (apply Lemma 2.4 with S = k −1 Q −1 ).The following lemma gives a bound for ρ k (T ) in this specific setting.
Lemma 7.1 (Discrete Stability).Let A be a linear, maximally dissipative operator on a Hilbert space H.For A-stable Runge-Kutta methods and arbitrary k > 0, we can bound r(k A) H→H ≤ 1, ( and therefore ρ k (T ) ≤ 1 for all k and T > 0.
Finally, using that c(c(z)) = z for all z, it follows that r = q • c.It is then an easy computation to prove that r(A) = q(c(A)).(Note that this equality can also be proved using functional calculus.) In Propositions 3.1 and 3.2, if A is maximally dissipative, k 0 can be chosen arbitrarily.In Lemma 5.4, if A is maximally dissipative, k 0 can be chosen arbitrarily.

Applications
In this section, Ω is a bounded Lipschitz open set in R d (d = 2 or 3) with boundary Γ.
We will consider the two-sided bounded surjective trace operator γ ± : H 1 (R d \ Γ) → H 1/2 (Γ) and we will denote H −1/2 (Γ) for the dual of the trace space.The angled bracket • , • Γ will be used for the H −1/2 (Γ) × H 1/2 (Γ) duality pairing and (•, •) R d will be used for the inner product in L 2 (R d ) and L 2 (R d ) d .We will also use the normal trace γ ± ν : Here we make the convention that the normal derivative points out of Ω for both interior and exterior trace.
We note that the applications in this section are chosen for their simplicity.More complicated applications, also involving full discretizations by convolution quadrature and boundary elements of systems of time domain boundary integral equations can be found in [RSM20] and [QRSZ19].

Boundary Integral Equations and Convolution Quadrature
In this section, we give a very brief introduction to boundary integral equations and their discretization using convolution quadrature.In that way, we can later easily state our methods for both the heat and wave equations in a concise and unified language.We present the result mostly formally, but note that they can be made rigorous under mild assumptions on the appearing functions.This theory can be found in most monographs on boundary element methods, see e.g., [SS11,McL00,Ste08] or [Say16].
For s ∈ C + , we consider solutions u ∈ H 1 (R d \ Γ) to the Helmholtz equation Using the representation formula, u can be rewritten using only its boundary data: where the single layer and double layer potentials are given by (S(s)ϕ) (x) := Γ Φ(x − y; s)ϕ(y) dy, We note that both S(s)λ and D(s)ψ solve the Helmholtz equation for any given densities λ ∈ H −1/2 (Γ) and ψ ∈ H 1/2 (Γ).
We will need the following four boundary integral operators: When solving problems in the time domain, we can leverage our knowledge of the Helmholtz equation using the Laplace transform L .For an operator valued analytic function F with dom(F ) ⊃ C + , we can then define the convolution operator where L is the Laplace transform in the sense of causal distributions.(Precise definitions can be found in [Say16, Chapter 3] and [Lub94]).
Given a Runge-Kutta method, it is then easy to define the convolution quadrature approximation to such operators, as was introduced in [LO93].We just replace the Laplace transform by the Z-transform and s with the function δ/k, i.e., we define: where g denotes a sequence in the shared domain of F (s) and k > 0 denotes the stepsize.The matrix-valued function F ( δ(z) k ) is defined using the Riesz-Dunford calculus, but can be computed in practice by diagonalizing the argument.
Remark 8.1.We note that our use of the notation ∂ k and (∂ k ) −1 is consistent with this definition by using the functions F (s) := s and F (s) := s −1 .

An exotic transmission problem
In this section we show how to apply Theorems 3.4-3.6 to a transmission problem in free space associated to the infinitesimal generator of a group of isometries (both ±A are maximal dissipative) with some exotic transmission conditions which impose partial observation of a trace.In Section 8.3 we will explain how this problem is related to a boundary integral representation of a scattering problem and how the current results yield the analysis of a fully discrete method for that integral representation.We keep the presentation brief.For more details and exemplary applications we refer to [HQSVS17].
Let Y h be finite dimensional subspace of H 1/2 (Γ) and consider the spaces The expression γv := γ − v − γ + v denotes the jump of the trace of v across Γ.The condition γv ∈ Y h is equivalent to We then set In X we use the natural inner product, in V we use the norm of , and in M we use the usual norm.We will define A : dom A = V → X and B : V → M by understanding that A can also be extended to As we did in Assumption 2.I, we consider dom A = ker B = V h × W 0 h and define A as the restriction of A to this subset.
Proof.The identity (8.5) shows that A(v, w), (v, w) X = 0 for all (v, w) and defining w = ∇v + f , we have a pair (v, w) ∈ V h × W 0 h such that (v, w) − A(v, w) = (f, f ) and thus A is maximal dissipative.The proof of the maximal dissipativity of −A is similar.(Note that this is a particular case of what appears in [HQSVS17].) We consider the standard problem (2.1) with vanishing initial conditions and data ∀t > 0, (8.6a) Proof.Given ξ ∈ M = H −1/2 (Γ), we solve the coercive problem and then define w := ∇v.This problem is equivalent to (note (8.5)) it follows that the norm of the solution operator for (8.7) is independent of the space Y h .
Proposition 8.4.The lifting E from Proposition 8.3 is a bounded linear map C depends only of Ω.
Proof.Thanks to Propositions 8.2 and 8.3, this problem fits in the abstract framework described in [HQSVS17], which proves existence and uniqueness of solution to (8.6).Propositions 8.2 -8.3 have some consequences.First of all, Assumption 2.I holds.Secondly, assuming g(t) ∈ L 2 (Γ), any solution to (2.1) with the above data (F = 0, Ξ = g) is in X 1/2 , and therefore, solutions to (8.6) take values in X 1/2 as well.Finally, We also need a regularity result that allows us to bound time derivatives of the solution in terms of the data.The continuity condition for the (s + 2)-th derivative of g in Proposition 8.6 can be relaxed to local integrability, but then the norms on the right-hand side of (8.9) have to be modified.Proposition 8.6.If g ∈ C s+2 ([0, ∞); L 2 (Γ)) satisfies g ( ) (0) = 0 for ≤ s + 1, then the unique solution to (8.6) satisfies ) for all ≤ s, there exists C, independent of the choice of Y h such that for all t ≥ 0 (v ( ) (8.9) Proof.This result follows from [HQSVS17, Theorem 3.1].To see 8.6, we note that (v h , w h ) is constructed by writing (v h (t), w h (t)) = (v 0 h (t), w 0 h (t)) + E g(t), with (v 0 h (t), w 0 h (t)) ∈ dom(A).The statement then follows from Proposition 8.4.We now consider the RK approximation of (8.6) in a finite time interval [0, T ], providing pairs of stage-values (V k h,n , W k h,n ) ∈ X m and step approximations (v k h,n , w k h,n ) ∈ X .We then define with u k h,0 = 0 (see Lemma 2.6) and ψ k h,n := γu k h,n .Proposition 8.7.For sufficiently smooth g, with RK approximations using a method satisfying Assumption 2.II, and with α given by (6.1), for nk ≤ T we have the estimates and ≤ CT 2 k min{q+1/2+α,p} p+3 =q max t≤T g ( ) (t) L 2 (Γ) .(8.12) The constants depend on Γ and the Runge-Kutta method, but do not depend on T or on the choice of Y h .
Proof.We will use Theorems 3.4 and 3.6 as well as Propositions 3.1 and 3.2.We note that ρ k (T ) ≤ 1 by Lemma 7.1 and Proposition 8.2.Also, with the E operator of Proposition 8.3 , we have with C independent of Y h .The bound (8.11) follows from Theorem 3.4, using (8.9) and (8.13) to estimate the right-hand side.The bound follows from Propositions 3.1 and 3.2, using (8.9) for the estimate in terms of the data.The H 1 (R d \ Γ) estimate (8.12) is then a direct consequence of (8.11) and (8.14), the estimate for ψ h − ψ k h follows from the standard trace theorem.

Scattering
We stay in the geometric setting of the previous section.Assume that d ∈ R d is a unit vector (direction of propagation) and that c ∈ R is such that Ω ⊂ {x ∈ R d : x•d > c}.Let φ : R → R be a function such that φ(r) = 0 for all r ≥ c.The incident wave u inc (x, t) := φ(x • d − t), propagates in the direction d at unit speed and has not reached the scatterer given by Ω at time t = 0.The data for our problem will be the function g : [0, T ] → L 2 (Γ) given by g(t) := −∂ ν u inc (•, t).
The scattering problem by a sound-hard obstacle occupying the domain Ω looks for the scattered field u : (Note that we can take the trace of the normal derivative of the incident wave, since it is locally smooth.) A direct formulation for solving this problem is equivalent to an extension of u to the interior domain by zero.This means we solve With some additional hypotheses on the growth of g (which is needed to have a well-defined distributional Laplace transform), we can represent the solution to (8.15) as u = S(∂)g − D(∂)ψ, where ψ := γu .Note that, to be precise with the use of weak distributional definitions, all functions have to be extended by zero to t < 0 (we say that they are causal) and the time interval is extended to infinity.
Taking the trace in this representation formula, the solution of (8.15) can be found by solving an equation for ψ and then postprocessing with the potential operators: and we still have that ψ = γu .We can equivalently write (8.15) and the equivalent (8.16) by using the variables v := u and w := ∇u.We note that u = ∂ −1 v and ψ = ∂ −1 γv .Here, (v, w) solve (we restrict t to the interval [0, T ] again) that is, (8.6) with Y h = H 1/2 (Γ).For the discretization, we consider a finite dimensional space Y h and the Galerkin approximation to (8.16), so that we look for ψ The functions v h := uh and w h := ∇u h satisfy (8.6).The difference between the solutions of (8.15) and (8.17) can be studied by comparing the solutions to (8.6) when Y h = H 1/2 (Γ) and when Y h is a finite dimensional space, see [HQSVS17] for details.For our purposes, it is sufficient to note that we get quasi-optimal estimates for the discretization in space.Discretization in time is performed by applying convolution quadrature to (8.17).The fully discrete solution reads The approximations ψ k h and u k h are then computed by the usual post-processing, i.e.
Lemma 8.8.The sequences u k h and ψ k h computed via (8.18) coincide with the Runge-Kutta approximations to (8.6) and their traces respectively.
Proof.The details of the computation can be found in the appendix of [MR17].The basic idea is to take the Z-transform and show that both approaches solve the matrix-valued Helmholtz problem (4.2).This gives the following immediate corollary, representing an apriori bound for the fully discrete method: Corollary 8.9.Let the assumptions of Proposition 8.7 hold.Then for u h and ψ h , approximated using convolution quadrature, we can estimate: The constants depend on Γ and the Runge-Kutta method, but do not depend on T or on the choice of Y h .Remark 8.10.There is another approach for analyzing convolution quadrature methods, which is based on estimates in the Laplace domain.It can be shown that the Neumann-to-Dirichlet map, realized by the boundary integral equations (8.18), satisfies a bound of the form Applying the abstract theory of [BLM11] then implies convergence rate min(q + 1, p) for the boundary data ψ h .Modifying their proof, one can also get for g(t) ∈ L 2 (Γ) that which would yield the same convergence rate as Corollary (8.19), but without insight into the dependence on the end-time T .

Numerical example
We solve (8.18) on a "hollow square", as depicted in Figure 8.1.The geometry was chosen to be non-convex and not simply connected, in order to test if the rate observed is a general result, or if our estimates might prove sharp in some situation.We prescribe the exact solution as a traveling wave, given by u(x, t) := φ(x • d − t), φ(s) := cos(π s/2) exp(−4(s 0 − s) 2 ).
s 0 := 4 is chosen so that φ(0) is sufficiently small in the domain.We set 2 ] and solve up to an end time of T = 12.An approximation of the H 1/2error is computed via i.e., we compare to the L 2 -projection of the exact solution.Since we are interested in the convergence rate with respect to the timestep size k, we consider a fixed, but sufficiently fine mesh.We used 3 and 5 stage Radau IIA methods, with orders (q, p) of (3, 5) and (5, 9), respectively (see [HW10] for their definition).Our theory predicts convergence rates of 4.5 and 6.5.In Figure 8.2, we observe a rate that is closer to 5 and 8.This means that (just like the standard Laplace-domain estimates) our estimates do not appear to be sharp in this case.Further investigations into the cause of this phenomenon are required.Results trying to explain this phenomenon, initially prompted by the work on this article, can be found in [MR19] but with a different model problem.temperature distribution and radiates heat into a medium Ω + := R d \ Ω.We make the simplifying assumption that at t = 0 the temperature is 0. Since the problem is posed on an unbounded domain, it is a good candidate for boundary integral equations, while being simple enough to showcase our more general results.We only briefly give the mathematical setting.More details and a more involved physical example can be found in [QRSZ19] In order to derive the boundary integral formulation, we take the Laplace transform of (8.20a), giving for κ := √ s: −∆ u(s) + κ 2 u(s) = 0, which is Helmholtz's equation for a complex wave number κ.We make an ansatz of the form u = S(κ) λ for some unknown density λ, which can be determined by applying the trace operator, giving the equation V (κ) λ = L (g).
The solution u can then be recovered by computing u = S κ (∂), where S κ (s) := S( √ s).
The discrete version of this is then given by solving It can be shown that plugging the discrete into the representation formula U k := S κ (∂ k )Λ k gives back the Runge-Kutta approximation of (8.20).The approximations at the endpoints t n = n k, denoted by λ k and u k respectively can be computed by the usual post-processing.We refer to the appendix of [MR17] for an analogous computation in the context of the Schrödinger equation, which easily transfers to our situation.For simplicity, we do not consider any discretization in space.A Galerkin approach could easily be included into the analysis, analogously to Section 8.2.
Proof.It is easy to see that H 2 0 (R d \ Ω) ⊆ dom(A).Using the Besov spaces introduced in Appendix A, we can write, if µ ≤ 1/4: where in the last step, we used [McL00, Theorem B.9].
Proof.We first note that we can control the derivatives u ( ) by the data.This can be done completely analogous to Proposition 8.6 by the techniques of [HQSVS17].
For simplicity of notation, we only consider the case q < p − 1.All the other cases follow analogously but giving different rates when applying the abstract theory.By Proposition 8.11, we can apply Propositions 3.1 or 3.2 with µ = 1/4, depending on whether we are in the setting α = 0 or α = 1.This gives estimate (8.22).
Applying Theorem 3.6, we get the following convergence in the graph norm of A : Since γ + u k (t n ) = γ + u(t n ), integration by parts and the Cauchy-Schwarz inequality give: Estimate (8.23) then follows from (8.22) and (8.25).For the estimate (8.24) of the density, we fix ξ ∈ H 1/2 (Γ), and let v denote a lifting to H 1 (R d ).We calculate We are still free to pick the precise lifting v. Doing so as in [Say16, Proposition 2.5.1],we get The result then follows from the previous estimates.

Proposition 4. 1 .
The sequences {U k n } and {u k n } satisfy equations (2.7) if and only if

(
v h (0), w h (0)) = (0, 0).(8.6c) Uniqueness of the solution to (8.6) follows from Proposition 8.2.We will handle existence of a solution below.The quantities of interest are u h := ∂ −1 v h and its Dirichlet traceψ h := γu h : [0, ∞) → Y h .Proposition 8.3.There exists a linear bounded right inverse of B, E : M → dom A such that range E ⊂ ker(I − A ).The norm of E is independent of the space Y h .