Parameter regularity of dynamical determinants of expanding maps of the circle and an application to linear response

In order to adapt to the differentiable setting a formula for linear response proved by Pollicott and Vytnova in the analytic setting, we show a result of regularity of dynamical determinants of expanding maps of the circle. The main tool is the decomposition of a transfer operator as a sum of a nuclear part and a"small"bounded part.


Introduction
In this work we adapt to the differentiable setting the formula for linear response obtained by Pollicott and Vytnova [8] in the analytic setting (based on an idea of Cvitanovic [3]). We recall the main argument: if τ → T τ is an analytic curve of analytic expanding maps of the circle, defined on a neighbourhood ]−ǫ, ǫ[ of 0, and g : S 1 → R is an analytic function then for all τ ∈ ]−ǫ, ǫ[ and u ∈ R the map (1) z → exp Investigating this formula, it is easy to write it in terms of the value of the derivative at τ = 0 of τ → T τ on the periodic points of T 0 (see Remark 6.3, in particular formula (35)).
Reading the proof of Lemma 3.1 of [8], it appears that if there is some R > 1 such that, for all u, τ sufficiently close to 0, the map defined by (1) extends on the disc of center 0 and radius R to a holomorphic function z → d (z, u, τ )), then, if d is C 1 , the formula (2) holds and, if d is C 2 , the formula (3) holds.
Consequently, our main result is a result of regularity for dynamical determinants of expanding maps of the circle: Theorem 6.1 which implies in particular (taking t = (τ, u), T t = T τ,u = T τ and g t,u = g τ,u = −ug − log |T ′ τ |) that (2) and (3) hold under some assumptions of regularity of τ → T τ and g (see Corollary 6.2). The hypotheses of Theorem 6.1 are not the weakest possible, they have been chosen to make the exposition as simple and self-contained as possible (see remark 2.5).
To define the dynamical determinant and prove its main properties in the analytic case, one may use the work of Ruelle (see [9]). We will use here the approach exposed in the first part of [1] (see also [2]). In particular, the main ingredient of our proof will be the decomposition of a transfer operator (Proposition 3.1), which is an adaptation of Proposition 3.15 of [1].
Date: July 15, 2021. 1 In §1, we set the notations involved in the Paley-Littlewood decomposition. In §2, we use the Paley-Littlewood decomposition to state Lemma 2.1, which is a local version of the decomposition mentioned above. Section 3 is dedicated to the definition of the transfer operator and its decomposition as described above.
In §4, we define a "flat trace" for some operators. It may be proved that it coincides with the flat trace defined in Section 3.2.2 of [1] in most cases.
In §5, we prove some results of regularities of the eigenelements of the transfer operator, using a method due to Gouëzel, Keller and Liverani (see [7] and [5]).
In §6, we state and prove our main result, Theorem 6.1 and show how to deduce formulae (2) and (3) from it.
In the analytic setting, one can prove a similar statement in the case of Anosov diffeomorphisms of the torus using the work of Rugh (see [12] and [13]). It is likely that the method presented here adapt to the case of differentiable expanding maps in higher dimensions (see Remark 2.5) and to the case of Anosov diffeomorphisms (using the approach exposed in the second part of [1]).
This work is part of a master degree internship under the supervision of Viviane Baladi.

The Paley-Littlewood decomposition
We shall denote by S the Schwartz class on R and by S ′ the space of tempered distributions on Recall that if s, t ∈ R, θ ∈ ]0, 1[ and ϕ ∈ S ′ , Hölder's inequality implies that In particular, if B is some Banach space and A : B → S ′ is a linear operator then we have where the space (B, B ′ ) of bounded linear operator from a Banach B to another B ′ is equipped with the operator norm . (B,B ′ ) . If ψ : R → R is some C ∞ function of at most polynomial growth, denote by Op (ψ) the pseudodifferential operator defined by for x ∈ R and ϕ ∈ S. Thus Op (ψ) is the operator of multiplication by ψ in Fourier transform. We shall extend Op (ψ) to an operator between Sobolev spaces as often as we may.
For all m ∈ N * there is a constant C m such that for all n ∈ N Finally, using Plancherel's formula, for all s ∈ R there is a constant C > 0 such that for all ϕ ∈ H s we have In particular for all s, s ′ ∈ R there is a constant C such that for all n ∈ N we have

Local decomposition of the transfer operator
This section aims at proving the following lemma which is a variation on Lemma 2.22 of [1]. In this statement, F t must be thought as an inverse branch of an expanding map of the circle T t .
be a C N function on U whose values are diffeomorphisms with derivative bounded by Λ on supp f t . For all ϕ ∈ S and t ∈ U set Then for all t ∈ U the operator M t can be written as a sum such that the following properties hold : i. for all s < r there is a constant c > 0 that only depends of s and r such that for all t ∈ U : ii. for all integers 0 k N , all s > k + 3 2 and all s ′ < r − 1 the map iii. for all integers 0 k N , all k + 3 2 < s < r − 1 and all ǫ > 0 the map In this statement, nuc H s , H s ′ denotes the space of nuclear operators from H s to H s ′ equipped with the nuclear operator norm (see for instance [4] for definitions and main results).
In all this section, fix a real number Λ > 0 and if l, n are natural integers write l ֒→ n if 2 n Λ2 l+6 , l ֒→n otherwise .

Then set
First, we investigate the "bounded" term (M t ) b .
Lemma 2.2. Let r ∈ N * and s ∈ R * + . There exists a constant c > 0 that does not depend of Λ such that for any C r+1 -diffeomorphism F : R → R and any compactly supported C r function f : R → R, setting for all ϕ ∈ H s then for any finite subset E of A we have Proof. Let ϕ ∈ S. For all l ∈ N set ϕ l = Op (ψ l ) ϕ. Then Since for all ξ ∈ R there are at most three values of n for which ψ n (ξ) = 0, Cauchy-Schwartz implies that Applying Cauchy-Schwartz again we get We used (6) in the last line.
Proof. The net of the partial sums of this series is bounded according to Lemma 2.2 and has a unique accumulation point since it converges for the operator norm topology on H 1 , H −1 for instance. Now, we study the "nuclear" term (M t ) c of the decomposition given by lemma 2.1. One of its most important property is being regularizing.
Lemma 2.4. Let r ∈ N * and m ∈ N. Let s, s ′ > 0 with s > m + 3 2 and s ′ < r − 1. Let f : R → R be a C r function compactly supported in a bounded open interval ]a, b[ and let F : R → R be a C r+1diffeomorphism with derivative bounded by Λ on supp f . Then there is a summable sequence (b n,l ) l ֒→n that only depends on a, b, s, s ′ , m, Λ, the C r norm of f , and the C r+1 norm of F on [a, b], such that, setting for all ϕ ∈ S Mϕ = f ϕ (m) • F , for all l ֒→n the operator Op (ψ n ) MOp (ψ l ) extends to a nuclear operator from H s to H s ′ with nuclear operator norm bounded by b n,l .
Proof. Since ϕ → ϕ (m) is bounded from H s to H s−m and commutes with Op (ψ l ), one only has to deal with the case m = 0.
In the following, all the C i are constants depending only on a, b, s, s ′ , m, Λ, the C r norm of f on [a, b] and the C r+1 norm of F on F −1 [a, b].
Fix ǫ > 0 such that s > 3 2 + ǫ and set a : where V n,l is defined by , integrating by parts r times, we get Using this trick r times we write with Φ bounded by C 2 2 −rn when ξ ∈ supp ψ n and η ∈ supp ψ l , and thus, From (10) and (12), we find for all y ∈ R : , we may improve (13) by integrating by parts twice on η in (9) and (11). This way we get where f is a positive integrable function that only depends on a, b, the C r norm of f on [a, b], and the Using the same kind of arguments, one can easily show that the function y ∈ R → V n,l (·, y) ∈ L 2 is (Lipschitz-)continuous. Furthermore the function y ∈ R → δ y • Op (a) ∈ (H s ) ′ is (Hölder-)continuous. Thus the integral 1 is well-defined and extends Op ψ n MOp (ψ l ) according to (8). Moreover, its nuclear operator norm is bounded by 1 Thus, the nuclear operator from H s to H s ′ defined by (7)).
Remark 2.5. The constant 3 2 that appears in Lemma 2.4 must be understood as 1 + 1 2 and may be replaced by 1 + 1 p by working with spaces H t p (R) as in [1] (here, 1 is thought as the dimension and thus may be replaced by some integer D to get a more general result). This is one way of weakening the hypotheses of Theorem 6.1.
We immediately deduce the following result from Lemmas 2.3 and 2.4.
Moreover its operator norm between these spaces is bounded by a constant that only depends of a, b, s, m, λ, the C r norm of f , and the C r+1 norm of F on Finally, we investigate the regularity of the dependence in t of the operator M t . Lemma 2.7. Let r ∈ N * and m ∈ N. Let s ∈ R with m + 3 2 < s < m + r − 1. Let U be an open set of R D for some integers D. Let t → f t ∈ C r (R) be a continuous function on U whose values are supported in a bounded open interval ]a, b[. Let t → F t ∈ C r+1 (R) be a continuous function on U whose values are diffeomorphisms of R into itself. For all ϕ ∈ S and t ∈ U set Then for all ǫ > 0 the function t → M t ∈ (H s , H s−m−ǫ ) is continuous on U .
Proof. As previously, we may and will suppose that m = 0. If V is an open set of R, we denote by H s 0 (V ) the closure in H s of the set of C ∞ functions compactly supported in V . Then, we can write where in the left-hand side M t is seen as an operator from H s ro H s−ǫ and in the right-hand side M t is seen as an operator from H s [ is continuous. Thanks to Corollary 2.6, we only need to prove this when ϕ is C ∞ function supported in ]a, b[. But for such a function, the continuity is obvious (it even holds in C r Then for all ǫ > 0 the function Proof. The proof is an induction on N . The case N = 0 has been dealt with in Lemma 2.7. Let N 1. As usual, we may suppose m = 0. Furthermore, we only need to deal with the case ǫ < s − N − 1 2 . Let i ∈ {1, . . . , D} and for all t = (t 1 , . . . , t D ) ∈ U and ϕ ∈ H s , write Consequently, we only need to check that A t is the partial derivative of t → M t with respect to t i . Let T = (T 1 , . . . , T d ) ∈ U and set for t i sufficiently close to T i : Thus Now, we can prove our "local decomposition" lemma.
Proof of Lemma 2.1. Set Then the first point is a consequence of Lemma 2.3. The second point is deduced from Lemma 2.8 by a standard argument of dominated convergence (the domination being a consequence of Lemma 2.4). Finally, the third point is an immediate consequence of the second point and Lemma 2.8.

Decomposition of the transfer operator
Let U be an open set of R D . Let r ∈ N * , N ∈ N. Let 0 < λ < 1. Let t → T t ∈ C r+1 S 1 , S 1 be a C N function on U whose values are expanding maps of the circle with expansion constant λ −1 that is Our main object of study is the transfer operator defined for all t ∈ U and all ϕ ∈ C S 1 by We shall associate to L t an operator K t with similar properties. Then we shall apply Lemma 2.8 to get a similar decomposition for the operator K t . The properties of this decomposition are stated in Proposition 3.1.
We need further notation to do so. Let K be a compact subset of U andK be a compact neighbourhood of K in U . We may choose a finite cover α = (V ω ) ω∈Ω of S 1 by open intervals with the following properties : 1. for all ω ∈ Ω, the canonical projection π : R → S 1 has a C ∞ local inverse κ ω defined on V ω ; 2. for all t ∈K and all ω ∈ Ω, the map T t induces a diffeomorphism from a neighbourhood of V ω to a neighbourhood of T t V ω ; 3. for all t ∈K 4. for all t ∈K and m ∈ N * the elements of m−1 i=0 T −i t α are open intervals; 5. denoting for all ω ∈ Ω by W ω the open interval of S 1 with same center than V ω but three times as long, the coverα = (W ω ) ω∈Ω also satisfies the four properties above.
Indeed, these properties hold as soon as the diameter of the elements of α is small enough, "small enough" being uniform in t thanks to the compacity ofK. For all m ∈ N * , t ∈ U and − → ω = (ω 0 , . . . , Replacing, V ω by W ω , we define in the same way W− → ω ,t and α m,t . For all m ∈ N * , t ∈ U and x ∈ S 1 write By a standard bounded distortion argument, there is a constant M > 0 such that for all t ∈K, m ∈ N * and V ∈α m,t , if x, y ∈ V then For all t ∈ U define K t on ω∈Ω S by Proposition 3.1. For all t ∈ K and all 3 2 < s < r − 1 the operator K t extends to a bounded operator from B s to itself. Moreover, for all m ∈ N * the operator K m t can be written as a sum 1 (18) K m t = (K m t ) b + (K m t ) c such that the following properties hold: i. for all 1 2 < s < r, there exists a constant c > 0 such that for all m ∈ N * and all t ∈ K: ii. for all integers 0 k N , all 3 2 + k < s < r − 1, all m ∈ N * and all ǫ > 0 the map extends to a C k function on a neighbourhood of K.
Proof. The boundedness of K t on B s will be a consequence of the decomposition (18) for m = 1. Fix an integer m ∈ N * . If t ∈ K choose a subset I = I t,m of Ω m that reaches the infimum Then find a neighbourhood U t,m ⊆K of t in U with the following properties: 1. for all t ′ ∈ U t,m (21) 3. for all t ′ ∈ U t,m and − → ω ∈ I We explain briefly how to find such a neighbourhood. The first point is easy, one only needs to notice that (21) is equivalent to The third point is an argument of continuity, using the fact that for all t ′ ∈K the set W− → ω ,t ′ is an interval. The second point is more complicated.
But the infimum on the left-hand side of this inequality is taken on a set that does not depend of t ′ , so we can use the same kind of argument as for the third point, recalling (20). Now, (U t,m ) t∈K is an open cover of K and consequently, one only needs to get the decomposition (18) on each of its elements separately (then glue the different decompositions using a partition of unity). So fix t 0 ∈ K and write I = I t0,m the subset of Ω m that appears in the definition of U t0,m . For all ω ∈ Ω choose a C ∞ functionχ ω : S 1 → R such that 0 χ ω 1 andχ ω (x) > 0 if and only if x ∈ W ω . Then for all m ∈ N * , t ∈ U t0,m and − → ω = (ω 0 , . . . , ω m−1 ) ∈ I, set: which is well-defined thanks to (21). Thus we have for all t ∈ U t0,m : Then for all − → ω ∈ I, t ∈ U t0,m and ϕ ∈ C ∞ S 1 , define and then Now, fix − → ω ∈ I and write K m − → ω ,t as a matrix of operators (A ω,ω ′ ,t ) ω,ω ′ ∈Ω that is, for all ϕ = (ϕ ω ) ω∈Ω ∈ ω∈Ω S, we have If ω, ω ′ ∈ Ω and ϕ ∈ S, we have • π may be extended in a C r+1 -diffeomorphism F t of R in a consistent way: , recalling (23). Thus, using Lemma 2.1, we get a decomposition of A ω,ω ′ ,t in with the expected regularity and, for all 1 2 < s < r, a constant c s , that only depends of s, such that where M has been introduced in (17) (that we used on second and third line). We also used (23) on second line and (15) on the last line. From this, we deduce a decomposition of K m − → ω ,t with the expected regularity and the same estimate of the operator norm up to a multiplicative factor (#Ω) 2 . Summing over − → ω ∈ I, we get the decomposition (18) with the expected regularity and thanks to (22).

Remark 3.2.
As a consequence of Theorem 3.1 for all integers 0 k N , for all 3 2 < s < r − 1 , all m ∈ N * and all ǫ the map t → K t ∈ B s , B s−k−ǫ is C k on a neighborhood of K.

Flat trace of the transfer operator
This section is dedicated to the definitions and basic properties of the "flat trace" and "flat determinant", which are key tools in the study of dynamical determinants in [1] or [2]. Although we will not define these objects in the same way, it could be shown that both definitions agree in most cases.
For all ǫ > 0 we set for all x ∈ R: where F −1 is the inverse of the Fourier transform and ρ : R → R is a C ∞ function, taking values in [0, 1], compactly suported, of integral 1 and identically equals to 1 one a neighbourhood of 0.
Lemma 4.1. For all s, s ′ > 1 2 the following properties hold: i. for all ǫ > 0 , J ǫ (resp. I ǫ ) defines a nuclear operator of order 0 from H s to H s ′ (resp. from B s to B s ′ ); ii. there is a constant C such that for all ǫ ∈ ]0, 1] we have J ǫ H s →H s C (resp. I ǫ B s →B s C); iii. for all ϕ ∈ H s (resp. B s ) , J ǫ ϕ (resp. I ǫ ϕ) tends to ϕ in H s (resp. B s ) as ǫ tends to 0.
Proof. The first point is immediate since J ǫ factorizes through S which is a nuclear space (see for instance the second part of [6]).
The norm in (H s , H s ) of ϕ ∈ H s → ρ ǫ * ϕ ∈ H s is bounded by ρ ǫ L ∞ ρ ǫ L 1 = 1. We get a uniform bound on the norm in (H s , H s ) of ϕ ∈ H s → χ ǫ ϕ by a classical Leibniz inequality (for instance corollary 4.2.2 of [14]).
The third point is a consequence of the second one and the fact that the convergence holds for function in S by an argument of density.
Thus if s ∈ R and A is a bounded operator from H s to itself, A • J ǫ is a nuclear operator and since the Hilbert H s as the approximation property, we can set and then the "flat trace" of A is defined as that is n 0 a n z n is the formal power series recursively defined by (26) a 0 = 1 and a n = − 1 n n−1 k=0 a k tr ♭ A n−k for n 1.
Proof. As in the proof of Proposition 3.1, write Choose − → ω ∈ I and write K m − → ω ,t as a matrix of operators (A ω,ω ′ ,t ) ω,ω ′ ∈Ω . For all ω ∈ Ω, ϕ ∈ H s and x ∈ R we can write thanks to (24) which expresses A ω,ω,t as an integral of nuclear operators. Thus Since F t has its derivative bounded by λ m < 1, the map x → F t (x) − x is a diffeomorphism from R to itself, let G be its inverse. Denote by x * the unique fixed point of F t and perform the change of variables "u = x * − F t (x) + x" to get it has at most one fixed point in W− → ω ,t . If y is such a fixed point and y ∈ V ω then κ ω (y) is a fixed point of F t , thus y = π (x * ) and Summing over ω ∈ Ω and then − → ω ∈ I, we get Formula (27) of Lemma 4.2 implies that the flat determinant of K t is the dynamical determinant of Theorem 6.1 (and is given by (1) in the application).
We want now to show that the product of "bounded" terms (K m t ) b of the decompositions (18) of large enough powers of K t have almost no trace. That's the point of Lemma 4.4. To do that, we need first to state an abstract property of the flat trace. Notice that the convergence in weak operator topology is the convergence that appears in Lemma 2.3. , it comes that, for all ω ∈ Ω, the operator B ω,ω,t can be written as a sum (in weak operator topology) of terms of the form with the M j as in the first part and (29) 2 nj λ mj 2 lj+6 λ L 2 lj +6 for all j ∈ {1, . . . , J}, which implies n j L log 2 λ + l j + 6. We shall show that, provided ǫ is small enough and L large enough, the "epsilon trace" of all these operators is zero, which ends the proof with Lemma 4.3.
Let u be the operator defined by (28) composed by J ǫ . If there is j ∈ {1, . . . , J − 1} such that ψ lj ψ nj = 0 then u is zero and so is its trace. Otherwise, for all j ∈ {1, . . . , J − 1} we have l j n j+1 + 1, which leads with (29) to n 1 J (L log 2 λ + 1) + l J + 6. Thus, provided L is large enough Suppose that ϕ is an eigenvector of u corresponding to a non-zero eigenvalue. Then ϕ is supported in supp ψ n1 . But, provided ǫ is small enough, ρ ǫ is supported in [−1, 1] and thus J ǫ ϕ = ρ ǫ (ρ ǫ * ϕ) is supported in [−2 n1 − 1; 2 n1 + 1] that doesn't intersect supp ψ lJ according to (30). Consequently, ϕ must be zero, which is absurd, and thus the spectrum of u is {0}. Since u is nuclear of order 0 (J ǫ is), its trace is zero (see for example corollary 4 page 18 of the second part of [6]).

Linear response a priori
In this section, we prove some properties of the spectrum of the transfer operator K t , using methods introduced by Gouëzel, Keller and Liverani in [7] and [5] (see also Paragraph A.3 in [1] for a sum-up). Notice that the spaces B s are not the same as those used, for instance, in [1]. In particular, there is no compact injection of B s in B s ′ when s ′ < s, which implies for instance that we cannot use Hennion's theorem. However, the results of [7] and [5] do not require compact injection and thus we can prove that the spectrum of the transfer operator has the expected behaviour on the Banach spaces B s .
First, we recall that the transfer operator has a spectral gap. We shall denote by P t the topological pressure associated with the dynamics T t (see §3.4 of [10] for a definition).
Lemma 5.1. For all t ∈ K and all 3 2 < s < r − 1, the real number e Pt(gt) is an eigenvalue of K t : H s → H s . Moreover, this eigenvalue is simple and there is η < e Pt(gt) that does not depend of s (but may depend of t) such that e Pt(gt) is the only eigenvalue of modulus larger than η of K t acting on B s .
Proof. Just notice that S maps B s in C 1 2 , and so if ϕ is an eigenvector of K t for a non-zero eigenvalue then Sϕ is an eigenvector of L t acting on C 1 2 for the same eigenvalue. Next, note that the lemma is true while replacing K t by L t and B s by C 1 2 or C r (see for instance Theorem 3.6 of [11]). Reciprocally, we know that e Pt(gt) is an eigenvalue of L t acting on C r . Let ϕ be a corresponding eigenvector. Then the coordinates of P ϕ are C r and compactly supported, thus P ϕ ∈ B s , and P ϕ is an eigenvector of K t for the eigenvalue e Pt(gt) .
We need some technical estimates to apply the results from [7] and [5].
Lemma 5.2. Let e P0(g0) < R. There is a constant C > 0 such that for all t ∈ K sufficiently close to 0 and all m ∈ N * we have Proof. We know that e P0(g0) = inf m∈N * Z (0, m) 1 m . Thus there exists m 0 ∈ N * such that Z (0, m 0 ) < R m0 . Since Z (., m 0 ) is upper semi-continuous, this inequality still holds replacing 0 by t sufficiently close. For such a t, we can write for m = qm 0 + r: Lemma 5.3. For each ρ > λe P0(g0) there exists a neighbourhood W of 0 in K such that for all s, s ′ ∈ 3 2 , r − 1 , there are some constants C 1 , C 2 , M such that for all m ∈ N * , t ∈ W and ϕ ∈ B s we have Choose ρ ′ ∈ λe P0(g0) , ρ . Applying Lemma 5.2 and recalling Theorem 3.1, we have for t sufficiently close to 0 and σ ∈ {s, for some constant C. Choose L large enough so that C 1 (ρ ′ ) L ρ L and then write for m = qL + r ∈ N * (with 0 r < L) Thus if ϕ ∈ B s and t sufficiently close to 0, we have From which we get From now on, we suppose that K is a rectangle. Fix η > λe P0(g0) as in Lemma 5.1 for t = 0 and 0 < δ < e P0(g0) − η. Set V δ,η = z ∈ C : |z| η or z − e P0(g0) δ .
We state a result of continuity and then a result of differentiability.
Lemma 5.4. If N 1 then for all 5 2 < s < r − 1 and all ǫ > 0, there is a neighbourhood W of 0 in K such that for all t ∈ W , the spectrum of K t acting on B s is contained in V δ,η . Moreover there exists a constant C such that for all t ∈ W and z ∈ C \ V δ,η we have Proof. We want to appply Theorem 1 of [7] twice with . = . B s and |.| = . B s−ǫ and δ replaced by δ 2 . This theorem is stated in a one-dimensional setting but in view of the dependences in the data of the constants appearing in the results (that are explicitly given), it may be applied here. Thanks to Theorem 3.1 and Lemma 5.3, there is a neighbourhood W ′ of 0 in K such that the conditions (2) and (3) of [7] are fulfilled by the family (K t ) t∈W ′ (with α = ρ < η). Since t → K t ∈ B s , B s−1−ǫ is C 1 on a neighbourhood of K, we get by interpolating between B s and B s−1−ǫ (that is applying the inequality (4)) for all t 0 , t 1 ∈ K and some constants C and β > 0. Thus the condition (5) of [7] is fulfilled on W ′ . As pointed out in Remark 6 of [7], the condition (4) is unnecessary here since C \ V δ 2 ,η is connected. Thus applying this theorem and the remark, we find a neighbourhood W of 0 in t such that for all t ∈ W the spectrum of K t acting on B s is contained in V δ 2 ,η . Now, we may apply the theorem from [7] again, taking each point of W as the origin, to end the proof of the lemma.
Lemma 5.5. For all integers 0 k N − 1, all real 5 2 + k < s < r − 1, and all ǫ > 0, there is a neighbourhood W of 0 in K such that for all z ∈ V δ,η the map Moreover, for all multi-indices α with |α| k we have Proof. The case k = 0 has been dealt with in Lemma 5. is C 1 (as pointed out in Remark 3.2), (A.6) and (A.7) are a consequence of the fact that t → K t ∈ B s , B s−2−ǫ is C 2 . Applying this theorem in each direction and interpolating between B s−1− ǫ 2 and B s−2−ǫ , it comes that the map defined by (32) admits partial derivatives given by (33) on a neighborhood of 0. These partial derivatives are continuous on a neighborhood of 0 as a consequence of Lemma 5.4.
The end of the proof is an induction. We show show how to get k = 2 from k = 1. Notice that we cannot calculate the second partial derivatives by differentiating the partial derivatives as products. However, the expected formula will stand. Indeed, fix i, j ∈ {1, . . . , d} and, setting A t = (z − K t ) −1 , write the growth rate where e j is the j-th vector of the canonical basis of R d , as The norm of this expression as an operator from B s to B s−2−ǫ is smaller than A t+hej (B s 2 ,B s 2 ) A t+hej (B s 2 ,B s 2 ) where s 1 = s − 1 − ǫ 2 , s 2 = s − 2 − ǫ and s ′ 2 = s − 2 − ǫ 2 . From the previous cases, we get that, provided t is in some neighbourhood of 0, this expression tends to 0 as h tends to 0. Consequently, A t has the second partial derivatives announced. Since they are continuous, the result is proved for k = 2. The idea of the proof in the general case is the same, up to more notational issues.
We also need some information about the greatest eigenvalue of K t and the associated spectral projection.
Lemma 5.6. For all 5 2 < s < r − 1, there is a neighbourhood W of 0 in K such that for all t ∈ W , e Pt(gt) ∈ D e P0(g0) , δ and e Pt(gt) is the only element of the spectrum of K t that has modulus greater than η.
Lemma 5.7. For all 0 k N − 1, all 5 2 + k < s < r − 1 and all 5 2 < s ′ < r − 1 − k, there is a neighbourhood W of 0 in K such that for all t ∈ W the spectrum of K t is contained in V δ,η and setting for all t ∈ W where Γ is a circle of center e P0(g0) and of radius slightly larger than δ, the map Proof. The first point is only a reminder of Lemma 5.4. Recall that Π t is a spectral projection. Choose ǫ > 0 sufficiently small and ϕ 0 ∈ B s ′ +k+ǫ ∩ B s such that Π 0 ϕ 0 = 0 (for instance take an eigenvector of K 0 for the eigenvalue e P0(g0) ). Then choose a linear form l on B 0 such that l (Π 0 ϕ 0 ) = 0. Set ρ t = Π t ϕ 0 and notice that lemma 5.5 and dominate convergence imply that t → ρ t ∈ B s ′ is C k on a neighbourhood of 0. In particular ρ t = 0 for t close enough of 0. Applying Lemma 5.6, the spectral projector Π t has rank one when acting on B s , providing t is small enough, and can thus be written Finally using Cauchy's formula and differentiation under the integral, the partial derivatives of d are holomorphic on the disc of center and radius R. Thus the power series that appear in formula (35) converge exponentially fast (we get a much faster convergence in the analytic setting, see [8]).