Salim Noreddine
email: salim.noreddine@polytechnique.org

Fluctuations of the traces of complex-valued iid random matrices

Keywords: Central limit theorems, Nualart-Peccati criterion of asymptotic normality, Invariance principles, i.i.d. random matrices, Normal approximation

In this paper, we extend to the complex-valued case a recent result by Nourdin and Peccati on the gaussian uctuations of traces of i.i.d. matrices.

Introduction

Let X be a centered random variable with unit variance, taking its values in C and admitting moments of all orders. Let {X i,j } i,j 1 be a family of independent and identically distributed copies of X. We denote by X n the random matrix dened as

X n = X i,j √ n 1 i,j n .
In this paper, we aim to nd the limit (in law) of

trace(X d n ) -E trace(X d n ) , ( 1.1)
where X d n denotes the d-th power of X n . To achieve this goal, we will make use of the following identity:

trace(X d n ) = n -d 2 n i 1 ,...,i d =1 X i 1 ,i 2 X i 2 ,i 3 . . . X i d ,i 1 .
In the case where X is real-valued, the problem was solved by Nourdin and Peccati [START_REF] Nourdin | Universal Gaussian uctuations of non-Hermitian matrix ensembles[END_REF]. The present study extends [START_REF] Nourdin | Universal Gaussian uctuations of non-Hermitian matrix ensembles[END_REF] to the more general case of a complex-valued random variable X.

When d = 1, the expression (1.1) is very simple; we have indeed

trace(X n ) -E trace(X n ) = 1 √ n n i=1 X i,i .
As a consequence, if X is real-valued then a straightforward application of the standard cental limit theorem (CLT) yields the convergence in law to N (0, 1). The case where X 1 is complex-valued is not much more dicult, as one only needs to use the bidimensional CLT to get the convergence in law to Z = Z 1 + iZ 2 , where (Z 1 , Z 2 ) is a gaussian vector with the same covariance matrix as that of (Re(X), Im(X)).

When d 2, we have:

trace(X d n ) -E trace(X d n ) = n -d 2 n i 1 ,...,i d =1 X i 1 ,i 2 ....X i d ,i 1 -E[X i 1 ,i 2 ....X i d ,i 1 ] . (1.2)
If X is real-valued, it is shown in [START_REF] Nourdin | Universal Gaussian uctuations of non-Hermitian matrix ensembles[END_REF] that there is convergence in law of (1.2) to the centered normal law with variance d. The idea behind the proof is to separate the sum in the right-hand side of (1.2) into two parts: a rst part consisting of the sum over the diagonal terms, i.e. the terms with indices i 1 , . . . , i d such that there is at least two distinct integers p and q satisfying (i p , i p+1 ) = (i q , i q+1 ); and a second part consisting of the sum over non-diagonal terms, i.e. the sum over the remaining indices. Using combinatorial arguments, it is possible to show that the sum over diagonal terms converges to 0 in L 2 . Thus, the contribution to the limit comes from the non-diagonal terms only. In order to tackle the corresponding sum, the idea [START_REF] Nourdin | Universal Gaussian uctuations of non-Hermitian matrix ensembles[END_REF] is to focus rst on the particular case where the entries X i,j are gaussian. Indeed, in this context calculations are much simpler because we then deal with a quantity belonging to the d-th Wiener chaos, so that the Nualart-Peccati [START_REF] Nualart | Central limit theorems for sequences of multiple stochastic integrals[END_REF] criterion of asymptotic normality may be applied. Then, we conclude in the general case (that is, when the entries are no longer supposed to be gaussian) by extending the invariance principle of Nourdin, Peccati and Reinert [START_REF] Nourdin | Invariance principles for homogeneous sums: universality of Gaussian Wiener chaos[END_REF], so to deduce that it was actually not a loss of generality to have assumed that the entries were gaussian.

In this paper we study the more general case of complex-valued entries. As we will see, the obtained limit law is now that of a random variable Z = Z 1 + iZ 2 , where (Z 1 , Z 2 ) is a gaussian vector whose covariance matrix is expressed by means of the limits of the expectations of the square of (1.1), as well as the modulus of the square of (1.1). To show our result, our strategy consists to adapt, to the complex case, the same method used in the real case. Specically, we show the following theorem. Theorem 1.1 Let {X ij } i,j 1 be a family of centered, complex-valued, independent and identically distributed random variables, with unit variance and admitting moments of all orders. Set

X n = X i,j √ n 1 i,j n .
Then, for any integer k 1,

trace(X d n ) -E trace(X d n ) 1 d k law -→ {Z d } 1 d k .
The limit vector {Z d }

with a + b = 1 and a -b + i2c = E(X 2 1,1 ) d .
The closest result to ours in the existing literature, other than the previously quoted reference by Nourdin and Peccati [START_REF] Nourdin | Universal Gaussian uctuations of non-Hermitian matrix ensembles[END_REF], is due to Rider and Silverstein [START_REF] Rider | Gaussian Fluctuations for non-hemitian random matrix ensembles[END_REF]. At this stage of the exposition, we would like to stress that Theorem 1.1 already appears in the paper [START_REF] Rider | Gaussian Fluctuations for non-hemitian random matrix ensembles[END_REF], but under the following additional assumption on the law of X 1,1 : Re(X 1,1 ) and Im(X 1,1 ) must have a joint density with respect to Lebesgue measure, this density must be bounded, and there exists a positive α such that E((X 1,1 ) k ) k αk for every k > 2. These assumptions can sometimes be too restrictive, typically when one wants to deal with discrete laws. Nevertheless, it is fair to mention that Rider and Silverstein focus more generally on gaussian uctuations of trace f (X n ) -E[trace f (X n ) ], when f : C → C is holomorphic and satises some additional technical assumptions (whereas, in our paper, we `only' discuss the polynomial case f ∈ C[X]).

The rest of the paper is devoted to the proof of Theorem 1.1. To be in position to do so in Section 4, we need to establish some preliminary combinatorial results in Section 2, as well as some results related to the gaussian approximation in Section 3.

Some preliminary combinatorial results

As we said, before giving the proof of theorem 1.1 we need to present some combinatorial results. In what follows, we assume that d 2 is xed. Note that the pairs (i 1 , i 2 ), . . . , (i d , i 1 ) appearing in formula 1.2 are completely determined by the d-tuple (i 1 , . . . , i d ). Indeed, it is straightforward that the set C n of elements

((i 1 , i 2 ), .., (i d , i 1 )) in [1, n] 2 d is in bijection with [1, n] d via the application ((i 1 , i 2 ), .., (i d , i 1 )) → (i 1 , . . . , i d ).
The cardinality of C n is therefore equal to n d . We denote by D n the set of diagonal terms of C n , i.e. the set of elements of C n such that there exist (at least) two distinct integers j and k such that

(i j , i j+1 ) = (i k , i k+1 ), with the convention that i d+1 = i 1 . We denote by N D n = C n \ D n the set of non-diagonal terms. If i 1 , . . . , i d are pairwise distinct, then ((i 1 , i 2 ), .., (i d , i 1 )) be- longs to N D n . Thus, the cardinality of N D n is greater or equal to n(n -1) . . . (n -d + 1), or, equivalently, the cardinality of D n is less or equal to n d -n(n -1) . . . (n -d + 1). In particular, the cardinality of D n is O(n d-1 ).
For any integer p 1, any integers α, β ∈ [1, n] and any element I p having the following form

I p = (x 1 , y 1 ), . . . , (x p , y p ) ∈ [1, n] 2 p , ( 2.3)
we denote by mc I p (α, β) the number of times that the pair (α, β) appears in (2.3). Furthermore, we denote by mp I p (α) the number of occurrences of α in {x i , y j } 1 i,j p . For example, if I 4 = ((1, 3), [START_REF] Nualart | Central limit theorems for sequences of multiple stochastic integrals[END_REF][START_REF] Rider | Gaussian Fluctuations for non-hemitian random matrix ensembles[END_REF], [START_REF] Nourdin | Universal Gaussian uctuations of non-Hermitian matrix ensembles[END_REF][START_REF] Nualart | Central limit theorems for sequences of multiple stochastic integrals[END_REF], (5, 7)) then mp

I 4 (3) = 3, mp I 4 (1) = 2, mc I 4 (1, 3) = 2
and mc I 4 (3, 4) = 1. For r elements

J k = (i (k) 1 , i (k)
2 ), . . . , (i

(k) d , i (k) 1 ) ∈ C n , k = 1, . . . , r,
we dene the concatenation J 1 . . . J r as being

(i (1)
1 , i

2 ), . . . , (i

d , i (1)
1 ), (i

(2) 1 , i (2)
2 ), . . . , (i

(2) d , i (2)
1 ), . . . , (i

(r) 1 , i (r) 2 ), . . . , (i (r) d , i (r) 1
) .

As such, J 1 . . . J r is an element of (C n ) r .

From now on, we denote by #A the cardinality of a nite set A. The following technical lemma will allow us to estimate the moments of (1.2). More precisely, (i), (ii), (iii), (iv) will imply that the variance of the sum of the diagonal terms converges in L 2 to 0, (v) and (vi) will allow us to show that the variance of the sum of the non-diagonal terms converges to d, and (vii) and (viii) will be used in the computation of the fourth moment of that sum. Lemma 2.1 Let the notations previously introduced prevail, and consider the following sets:

A n = I 2d = J 1 J 2 ∈ (D n ) 2 : mc I 2d (α, β) = 1 for every α, β ∈ [1, n] B n = I 2d ∈ A n : mp I 2d (α) ∈ {0, 4} for every α ∈ [1, n] = I 2d ∈ A n : mc I 2d (α, β) ∈ {0, 2} for every α, β ∈ [1, n] E n = I d ∈ D n : mc I d (α, β) = 1 for every α, β ∈ [1, n] F n = {I d ∈ E n : mp I d (α) ∈ {0, 4} for every α ∈ [1, n] = {I d ∈ E n : mc I d (α, β) ∈ {0, 2} for every α, β ∈ [1, n] G n = I 2d = J 1 J 2 ∈ (N D n ) 2 : mc I 2d (α, β) ∈ {0, 2} for every α, β ∈ [1, n] H n = I 2d ∈ G n : mp I 2d (α) ∈ {0, 4} for every α ∈ [1, n] = I 2d ∈ G n : mc I 2d (α, β) ∈ {0, 2} for every α, β ∈ [1, n] K n = I 4d = J 1 J 2 J 3 J 4 ∈ (N D n ) 4 : mc I 4d (α, β) ∈ {0, 2, 4} for every α, β ∈ [1, n] L n = {I 4d ∈ H n : mp I 4d (α) ∈ {0, 4} for every α ∈ [1, n] = {I 4d ∈ H n : mc I 4d (α, β) ∈ {0, 2} for every α, β ∈ [1, n] .
As n → ∞, we have:

(i) # (A n \ B n ) = O(n d-1 ). (ii) If d is even, #B n = n . . . (n -d + 1); if d is odd, #B n = 0. (iii) #(E n \ F n ) = O(n d-1 2 ). (iv) If d is even, #F n = n . . . (n -d 2 + 1); if d is odd, #F n = 0. (v) #G n \ H n = O(n d-1 ). (vi) #H n = d × n . . . (n -d + 1). (vii) #(K n \ L n ) = O(n 2d-1 ). (viii) #L n = 3d 2 × n . . . (n -2d + 1).
Proof.

(i) Let I 2d = (i 1 , i 2 ), . . . , (i d , i 1 ), (i d+1 , i d+2 ), . . . , (i 2d , i d+1 ) ∈ A n \ B n . By denition of A n , we have mp I 2d (i j ) 4 for any j = 1, . . . , 2d. Furthermore, the fact that I 2d / ∈ B n ensures the existence of at least one integer j 0 between 1 and 2d such that

mp I 2d (i j 0 ) > 4. Let σ : [1, 2d] → [1, 2d] be dened by j → σ(j) = min{k : i k = i j }. It is readily checked that 4d = α∈Im(σ) mp I 2d (i α ). We conclude that #Im(σ) < d. Therefore, # (A n \ B n ) = O(n d-1 ).
(ii) Assume that B n is non-empty. Let

I 2d = (i 1 , i 2 ), . . . , (i d , i 1 ), (i d+1 , i d+2 ), . . . , (i 2d , i d+1 ) ∈ B n .
For every integer j ∈ [1, 2d], we have mp I 2d (i j ) = 4. Dening σ and proceeding as in point (i) above, we obtain that #Im(σ) = d. We set m = min{l ∈ Im(σ) | l + 1 / ∈ Im(σ)}. Since #Im(σ) = d, it follows that m d. In fact m d -1, otherwise the elements of the d-tuple (i 1 , . . . , i d ) would be all distinct, and ((i 1 , i 2 ), . . . , (i d , i 1 )) could not be in D n , which would yield a contradiction. In the case d = 2, I 2d = (i 1 , i 2 ), (i 2 , i 1 ), (i 3 , i 4 ), (i 4 , i 3 ) ∈ B n if and only if i 1 = i 2 , i 3 = i 4 and i 1 = i 3 . Thus, the cardinality of B n is equal to n(n -1).

In what follows we suppose that d 3.

Let us show that d is even (which will prove that B n is empty if d is odd) and that I 2d can be written as

(l 1 , l 2 ), . . . , (l d 2 -1 , l d 2
), (ld are pairwise distinct integers in [1, n], which will prove that the formula for #B n given in (ii) holds true. The proof is divided in several parts.

(a) Using a proof by contradiction, let us assume that there exists an integer q in [m + 1, d] such that i q does not belong to {i 1 , . . . , i m }. We denote by γ the smallest element verifying this. Note that γ m + 2 necessarily, and that there exists an integer p m such that i γ-1 = i p . Therefore, i γ-1 appears in the four pairs

(i p-1 , i p ), (i p , i p+1 ), (i γ-2 , i γ-1 ), (i γ-1 , i γ ).
Note that for the four pairs above, it is possible that the two pairs in the middle are the same. By denition of B n , we have mp I 2d (i γ-1 ) = 4 so these pairs are the only pairs of I 2d containing the integer i γ-1 . Moreover, by denition of A n , we have mc

I 2d (i γ-1 , i γ )
2. Thus, we necessarily have either

(i γ-1 , i γ ) = (i p , i p+1 ); or (i γ-1 , i γ ) = (i γ-2 , i γ-1 ); or (i γ-1 , i γ ) = (i p-1 , i p ). If we had (i γ-1 , i γ ) = (i γ-2 , i γ-1
), then we would have i γ-2 = i γ-1 = i γ and i γ would appear at least six times in the writing of I 2d , which is not possible . Similarly,

(i γ-1 , i γ ) = (i p-1 , i p ) is impossible. Thus, it must hold that (i γ-1 , i γ ) = (i p , i p+1
). We can therefore state that i γ = i p+1 . Since we also have that p + 1 m + 1 and i m+1 ∈ {i 1 , . . . , i m }, we can conclude that i γ ∈ {i 1 , . . . , i m }, which yields the desired contradiction. Hence,

{i m+1 , . . . , i d } ⊂ {i 1 , . . . , i m }.
(2.4)

(b) Let us show that if l, k d -1 are two distinct integers satisfying i k = i l , then (i k , i k+1 ) = (i l , i l+1 ). Let l, k d -1 be two integers such that l = k et i k = i l .
The integer i l appears in the four pairs {(i l-1 , i l ), (i l , i l+1 ), (i k-1 , i k ), (i k , i k+1 )} (or only in three pairs, if both pairs in the middle are the same, which happens whether l = k -1). As mp I 2d (i k ) = 4, these pairs are the only pairs of I 2d containing the integer i k . By denition of A n , all pairs of I 2d must have at least two occurrences in

I 2d . If we have (i k , i k+1 ) = (i k-1 , i k ) then we have i k = i k+1 = i k-1 and i k appears at least six times in I 2d , which cannot be true. Similarly, (i k , i k+1 ) = (i l-1 , i l ) is impossible. Therefore, it must hold that (i k , i k+1 ) = (i l , i l+1 ). (c) It follows from the denition of m that there exists an integer r ∈ [1, m] satisfying i m+1 = i r . Let us show that (i 1 , . . . , i d ) = (i 1 , . . . , i m , i r , . . . , i r+d-m-1 ).
(2.5)

If m = d-1, then (i 1 , . . . , i d ) = (i 1 , . . . , i m , i r ) and (2.5) is veried. If m d-2 then, being given that i m+1 = i r and that we already showed in (b) that if l, k d -1 are two distinct integers satisfying i k = i l then i k+1 = i l+1 , we can state that i m+2 = i r+1 . Thus, if m = d -2 then (i 1 , . . . , i d ) = (i 1 , . . . , i m , i r , i r+1 )
, and (2.5) is once again veried. Finally, if m d -3, we iterate this process as many times as necessary until we get (2.5).

(d)

Let us now prove that the elements of (i r , . . . , i r+d-m-1 ) are all distinct. Once again, we use a proof by contradiction. Thus, let us assume that there exists an integer p in [1, n] which appears at least twice in the uplet (i r , . . . , i r+d-m-1 ). We then have {i r , . . . , i r+d-m-1 } = {i m+1 , . . . , i d } ⊂ {i 1 , . . . , i m }, see (2.4) and (2.5). Thus, p appears at least three times overall in the uplet (i 1 , . . . , i d ). This latter fact implies mp I 2d (p) 6, which contradicts the assumption mp I 2d (p) = 4. (e) Finally, let us establish that 2m = d and r = 1. The elements of (i r , . . . , i r+d-m-1 ) being all distinct, the couple (i r+d-m-1 , i 1 ) = (i d , i 1 ) cannot belong to the set of pairs

{(i r , i r+1 ), . . . , (i r+d-m-2 , i r+d-m-1 )} = {(i m+1 , i m+2 ), . . . , (i d-1 , i d )}.
(because, by (d), no pair of this set can have i r+d-m-1 as a rst coordinate.) Moreover, since i 1 does not belong to {i 2 , . . . , i m } then the pair (i r+d-m-1 , i 1 ) cannot belong to the set of pairs {(i 1 , i 2 ), . . . , (i m-1 , i m )} (because no pair of this set can have i 1 as a second coordinate). Also, the integer i d appearing at least twice in the uplet (i 1 , . . . , i d ), it cannot belong to the uplet (i d+1 , . . . , i 2d ) (otherwise, i d would appear at least six times in the pairs of I 2d ). Thus, the only way for the occurrence of the pair (i d , i 1 ) in I 2d to be greater or equal than 2 is that

(i r+d-m-1 , i 1 ) = (i m , i m+1 ). Therefore, i 1 = i m+1 = i r . As i 1 , . . . , i m are all distinct and r m, it must hold that r = 1. Hence, r + d -m -1 = d -m and i d-m = i m . Since i 1 , . . . , i d-m are all distinct, see indeed (d), it must be true that m d -m. Since i d-m = i m , we conclude that d-m = m, that is, d = 2m. As such, we establish that (i d+1 , . . . , i 2d ) = (i d+1 , . . . , i 3d 2 , i d+1 , . . . , i 3d 2 ). Let us nally note that i 1 , . . . , i d 2 , i d+1 , . . . , i 3d 2 are neces- sarily distinct because mp I 2d (i j ) = 4. This completes the proof of part (ii). (iii) Consider I = (i 1 , i 2 ), . . . , (i d , i 1 ) ∈ E n \ F n . Let ξ : [1, d] → [1, d] be dened by j → ξ(j) = min{k | i k = i j }. From the equation 2d = α∈Im(ξ)
mp I (i α ), and using the fact that all mp I (i α ) are greater or equal than 4, as well as there exists an α in Im(ξ)

satisfying mp I (i α ) > 4, we conclude that #Im(ξ) < d 2 .
Therefore, the cardinality of

E n \ F n is equal to O(n d-1 2 ).
(iv) Let us assume that F n is not empty. Consider

I d = (i 1 , i 2 ), . . . , (i d , i 1 ) ∈ F n .
Proceeding as in point (ii) above, we conclude that F n is empty in the case where d is odd, and that the elements of F n have the following form when d is even:

(l 1 , l 2 ), . . . , (ld 2 -1 , l d 2 )(ld 2 , l 1 ), (l 1 , l 2 ), . . . , (l d 2 -1 , l d 2 )(l d 2 , l 1 ) .
Here, l 1 , . . . , l d

2 are pairwise distinct integers in [1, n]. The formula of #F n given in (iv) follows directly from that. (v) Consider I = (i 1 , i 2 ), . . . , (i d , i 1 ), (i d+1 , i d+2 ), . . . , (i 2d , i d+1 ) ∈ G n \ H n . Let ζ : [1, 2d] → [1, 2d] be dened by j → ζ(j) = min{k | i k = i j }. From the identity 4d = α∈Im(ζ)
mp I (i α ), and using the fact that mp I (i α ) are greater or equal than 4, as well as that there exists an α in Im(ζ) satisfying mp

I (i α ) > 4, we conclude as in (i) that #Im(ζ) < d. Therefore, the cardinality of G n \ H n is O(n d-1
[d + 1, 2d] satisfying (i p , i p+1 ) = (i k , i k+1 ),
which would yield i p = i q = i k and, consequently, mp I (i p ) 6. This would contradict the fact that mp I (i p ) = 4. (c) Let us establish that, for every p ∈ [1, d] and q ∈ [d + 1, 2d] such that i p = i q , we have i p+1 = i q+1 . Using a proof by contradiction, let us assume that there exists an integer q ∈ [d + 1, 2d] dierent from q such that (i p , i p+1 ) = (i q , i q +1 ). Then it must hold that i p = i q = i q and mp I (i p ) 6, which contradicts the fact that mp

I (i p ) = 4.
The

= d × n . . . (n -d + 1). (vii) Consider I = (i 1 , i 2 ), . . . , (i d , i 1 ), . . . , (i 3d+1 , i 3d+2 ), . . . , (i 4d , i 3d+1 ) ∈ K n \ L n . Let η : [1, 4d] -→ [1, 4d
] be the application dened by η(j) = min{k| i k = i j }. From the identity 8d = α∈Im(η) mp I (i α ), and using the fact that mp I (i α ) are all greater or equal than 4, as well as for at least one α ∈ Im(η) it must hold that mp I (i α ) > 4, we conclude that #Im(η) < 2d. Therefore, the cardinality of K n \ L n is O(n 2d-1 ).

(viii) Consider I = (i 1 , i 2 ), . . . , (i d , i 1 ), . . . , (i 3d+1 , i 3d+2 ), . . . , (i 4d , i 3d+1 ) ∈ L n . For every j 4d, we have mp I (i j ) = 4. Then 2d = #Im(η), with η as in point (vii).

(a) Using a proof by contradiction, let us show that, for every k ∈ [0, 3], the integers i kd+1 , . . . , i (k+1)d are all distinct. Assume that there exist two distinct integers l and h in [1, d], as well as an integer k ∈ [0, 3], satisfying i kd+l = i kd+h . By denition of the set L n , we have (i kd+1 , i kd+2 ), . . . , (i (k+1)d , i kd+1 ) ∈ N D n . Then, the pairs

(i kd+1 , i kd+2 ), . . . , (i (k+1)d , i kd+1 )
are all distinct, and we have mc I (i kd+h , i kd+h+1 ) = 2, which implies that there exists k ∈ [0, 3], dierent from k, and h ∈ [1, d] satisfying i kd+h = i k d+h . It follows that i kd+h appears at least six times in I, which contradicts the fact that mp I (i kd+h ) = 4. (b) For any p = 0, . . . , 3, let us introduce M p = {i pd+1 , . . . , i (p+1)d }. For any integers p, q in [0, 3], we have either M p M q = ∅ or M p = M q . Otherwise there would exist an integer j such that i qd+j ∈ M p and i qd+j+1 / ∈ M p and, since mc I (i qd+j , i qd+j+1 ) = 2, there would exist q ∈ [0, 3], dierent from p and q, and j ∈ [1, d] such that (i qd+j , i qd+j+1 ) = (i q d+j , i q d+j +1 ); therefore i qd+j would appear at least six times in I, which would yield a contradiction. (c) If M p = M q , then proceeding as in point (vi), we show that there exists j ∈ [1, d] such that (i pd+1 , . . . , i (p+1)d ) = (i qd+j , . . . , i (q+1)d , i qd+1 , . . . , i dq+j-1 ).

The results (a), (b) et (c) allow us to conclude that a generic element of L n is characterized by: -the choice of one case among the following three cases: either

M 0 = M 1 and M 2 = M 3 ; or M 0 = M 2 and M 1 = M 3 ; or M 0 = M 3 and M 1 = M 2 .
In what follows, we consider the case M 0 = M 1 and M 2 = M 3 (we can proceed similarly in the other two cases); -the choice of 2d integers i 1 , . . . , i d , i 2d+1 , . . . , i 3d that are pairwise distinct in [1, n]; -the choice of an integer k ∈ [1, d] such that (i d+1 , . . . , i 2d ) = (i k , . . . , i d , i 1 , . . . , i k-1 ); -the choice of an integer k ∈ [1, d] such that (i 3d+1 , . . . , i 4d ) = (i 2d+k , . . . , i 3d , i 2d+1 , . . . , i 2d+k -1 ).

It is now easy to deduce that #L n = 3d 2 n . . . (n -2d + 1).

Gaussian approximations

Let X = {X i } i 1 be a family of centered independent random variables taking values in R r and having pairwise uncorrelated components with unit variance. Let G = {G i } i 1 be a family of independent standard gaussian random variables taking values in R r and having independent components. Suppose also that X and G are independent, and set

X = (X 1 1 , . . . , X 1 r , X 2 1 , . . . , X 2 r , . . .) = (X 1 , . . . , X r , X r+1 , . . . , X 2r , . . .). i.e., X j+(i-1)r = X i j . Consider integers m 1, d m . . . d 1 2, N 1 , .
. . , N m , as well as real symmetric functions f 1 , . . . , f m such that each function f i is dened on [1, rN i ] d i and vanishes at the points (i 1 , ..., i d i ) such that ∃j = k for which i j /r = i k /r (we remind that x means the unique integer k such that k < x k + 1). Let us dene

Q i (X) = Q d i (f i , X) = rN i i 1 ,...,i d =1 f i (i 1 , . . . , i d i ) X i 1 . . . X i d i .
In the case of complex-valued matrices, the real and imaginary parts of the entries X i,j are not necessarily independent. Therefore, we will need to modify the results used by Nourdin and Peccati in the paper [START_REF] Nualart | Central limit theorems for sequences of multiple stochastic integrals[END_REF]. The following lemma is a variant, weaker in terms of assumptions, of the hypercontractivity property. Lemma 3.1 Let the notations previously introduced prevail. Assume that α = sup

i E(|X i | 4 ) < ∞ and set K = 36 × 25 r × (1 + 2α 3 4 ) 2 . Then E(Q d (X) 4 ) K d E(Q d (X) 2 ) 2 . (3.6) Proof. Set      U = ∀k: i k / ∈{(N -1)r+1,...,N r} f (i 1 , . . . , i d )X i 1 . . . X i d V j = ∃!k:i k =(N -1)r+j f (i 1 , . . . , i d )X i 1 . . . X (N -1)r+j . . . X i d
The notation X (N -1)r+j means that this term is removed from the product. Observe that X (N -1)r+j = X N j according to the notation that we adopted previously, and that the quantity Q d (X) is given by:

Q d (X) = U + r j=1 X N j V j
(as f vanishes at the previously specied points). Note that, for every p N and every i, j ∈ [1, r], X p j is independent from U and V i . Thus, by choosing p = N , we have

E(Q d (X) 4 ) = s 0 +...+sr=4 24 s 0 ! . . . s r ! E(U s 0 r j=1 (V j X N j ) s j ) = E(U 4 ) + s 1 +...+s r =2 12 s 1 ! . . . s r ! E(U 2 r j=1 V s j j )E( r ( j=1 X N j ) s j ) + s 1 +...+s r =3 24 s 1 ! . . . s r ! E(U r j=1 V s j j )E( r j=1 (X N j ) s j ) + s 1 +...+sr=4 24 s 1 ! . . . s r ! E( r j=1 V s j j )E( r j=1 (X N j ) s j ).
In the equation above, we used that

s 1 +...+sr=1 4
s 1 !...sr! E(U 3 r j=1
(V j X N j ) s j ) = 0 since X N j are centered. By using the generalized Hölder inequality, we obtain:

E(U s 0 r j=1 V s j j ) E U 4 s 0 4 r j=1 E(V 4 j ) s j 4 .
Since the terms E(V 4 j ) s j 4 are upper bounded by

r j=1 E(V 4 j ) 1 2
s j 2 , we obtain:

s 1 +...+s r =4-s 0 E(U s 0 r j=1 V s j j ) 5 r E U 4 s 0 4 r j=1 E(V 4 j ) 1 2
4-s 0 2

.

10

Using the generalized Hölder inequality again, we have E(

r j=1 (X N j ) s j ) r j=1 E (X N j ) 4 s j 4 α s j
4 . Therefore:

E(Q d (X) 4 ) E(U 4 ) + 12 × 5 r E(U 4 ) 1 2 r j=1 E(V 4 j ) 1 2
(3.7)

+24 × 5 r α 3 4 E(U 4 ) 1 4 r j=1 E(V 4 j ) 1 2 3 2 +24 × 5 r α r j=1 E(V 4 j ) 1 2 2 .
Note that α does not appear in the second term of the right-hand side of the inequality above because X N j are random variable with unit variance and zero covariance. Using the inequality x

1 4 y 3 2
x 1 2 y + y 2 , obtained by separating the cases x y 2 and x y 2 , we get:

E(U 4 ) 1 4 r j=1 E(V 4 j ) 1 2 3 2 E(U 4 ) 1 2 r j=1 E(V 4 j ) 1 2 + r j=1 E(V 4 j ) 1 2

2

.

Then

E(Q d (X) 4 ) E(U 4 ) + 12 × 5 r (1 + 2α 3 4 )E(U 4 ) 1 2 r j=1 E(V 4 j ) 1 2
(3.8)

+24 × 5 r (α 3 4 + α) r j=1 E(V 4 j ) 1 2 2 .
To prove the hypercontractivity property (3.6), we will use an induction on N . When N = 1, because f vanishes at the previously specied points, then the only case where the value taken by Q d (X) is not zero is when d = 1, that is, when Q d (X) has the form r j=1 a j X 1 j . In this case, U = 0 and V j = a j . Thus, by (3.7), we have

E(Q d (X) 4 ) 24 × 5 r α r j=1 a 2 j 2 . It follows that E(Q d (X) 4 ) KE(Q d (X) 2 ) 2 .
Let us now assume that the result holds for N -1. Then, because U and V j are functions of X 1 , . . . , X N -1 , we can apply the recursive hypothesis to E(U 4 ) and E(V 4 j ), and obtain that:

E(Q d (X) 4 ) K d E(U 2 ) 2 + 12 × 5 r (1 + 2α 3 4 ) K 1 2 E(U 2 ) r j=1 E(V 2 j ) +K d 24 × 5 r (α + α 3 4 ) K r j=1 E(V 2 j ) 2 K d   E(U 2 ) 2 + 2E(U 2 ) r j=1 E(V 2 j ) + r j=1 E(V 2 j ) 2   = K d E(U 2 ) + r j=1 E(V 2 j ) 2 .
Furthermore, since the X N j are centered, unit-variance and independent of U and of V j , we have

E(Q d (X) 2 ) = E((U + r j=1 X N j V j ) 2 ) = E(U 2 ) + 2 r j=1 E(U V j )E(X N j ) + i,j=1,...,r E(V i V j )E(X N i X N j ) = E(U 2 ) + r j=1 E(V 2 j ),
which completes the proof.

The following two lemmas will be used to prove the convergence in law of the sum of the non-diagonal terms in (1.2), and to show that the limit does not depend on the common law of X i,j . Lemma 3.2 Let {X i } i 1 , {G i } i 1 and Q i (X) be as in the beginning of section 3. Let us

assume that β = sup i E(|X i | 3 ) < ∞, E(Q i (X) 2 ) = 1, and that V is the symmetric matrix dened as V (i, j) = E(Q i (X)Q j (X)). Consider Z V = (Z 1 V , . . . , Z m V ) ∼ N m (0, V ) (i.
e. Z V is a gaussian vector with a covariance matrix equal to V ).

1. If ϕ : R m → R is a function of class C 3 such that ϕ ∞ < ∞ then E(ϕ(Q 1 (X), . . . , Q m (X))) -E(ϕ(Q 1 (G), . . . , Q m (G))) ϕ ∞ β + 8 π K 3 4 (dm-1) r 3 m 4 d m ! 3 d 1 !(d 1 -1)! max 1 k m max 1 j N k inf j f k ,
where

inf j f k = rN k i 1 ,...,i d k -1 =1 f k (j, i 1 , . . . , i d k -1 ) 2 . 2. If ϕ : R m → R is a function of class C 3 such that ϕ ∞ < ∞ then E(ϕ(Q 1 (X), . . . , Q m (X))) -E(ϕ(Z V ) ϕ ∞ m i=1 ∆ i,i + 2 1 i<j m ∆ i,j + ϕ ∞ m 4 d m ! 3 d 1 !(d 1 -1)! β + 8 π K 3 4 (d m -1) r 3 + 32 π 64 π dm-1 max 1 k m max 1 j N k inf j f k
where inf j f k as above and ∆ i,j given by

d j d i -1 s=1 (s-1)! d i -1 s -1 d j -1 s -1 (d i + d j -2s)! f i d i -s f i 2 + f j d j -s f j 2 +1 d i <d j d j ! d j d i f j d j -d i f j 2 , with f j r f j (i 1 , . . . , i 2d j -2r ) = rN j k 1 ,...,k r =1 f j (k 1 , . . . , k r , i 1 , . . . , i d j -r )f j (k 1 , . . . , k r , i d j -r+1 , . . . , i 2d j -2r ).
Proof. Set Q(X) = (Q 1 (X), . . . , Q m (X)) and, for any

1 p N + 1, consider            Z (p) = G 1 , . . . , G (p-1)r , X (p-1)r+1 , . . . , X rN U (i) p = ∀k: i k / ∈{(p-1)r+1,...,pr} f i (i 1 , . . . , i d )Z (p) i 1 . . . Z (p) i d V (i) p,j = ∃!k:i k =(p-1)r+j f i (i 1 , . . . , i d )Z (p) i 1 . . . Z (p) (p-1)r+j . . . Z (p) i d
The notation Z (p) (p-1)r+j means that this term is removed from the product. Let us set

U p = (U (1) p , . . . , U (m) p ) and V p,j = (V (1) p,j , . . . , V (m) p,j ). Note that Q(Z (p)
) can be written as

Q(Z (p) ) = U p + r j=1 X p j V p,j .
Similarly, we have:

Q(Z (p+1) ) = U p + r j=1 G p j V p,j . For a vector Y = (Y 1 , . . . , Y m ) in R m and a vector s = (s 1 , . . . , s m ) in N m , we set Y s = m i=1 Y s i i .
1. Let ϕ be a function of class C 3 . The Taylor formula gives:

E ϕ(Q(Z (p) )) -E   |s| 2 1 s! ∂ s ϕ(U p ) r j=1 X p j V p,j s   ϕ ∞ E   |s|=3 r j=1 X p j V p,j s   .
Note that, for every p, X p j is independent from U p and from V p,i . Thus, we have:

E |s|=3 r j=1 X p j V p,j s ) = E m k,l,q=1 r j 1 =1 X p j 1 V (k) p,j 1 r j 2 =1 X p j 2 V (l) p,j 2 r j 3 =1 X p j 3 V (q) p,j 3 = m k,l,q=1 r j 1 =1 r j 2 =1 r j 3 =1 E X p j 1 X p j 2 X p j 3 E V (k) p,j 1 V (l) p,j 2 V (q) p,j 3 .
The Hölder inequality ensures that:

E X p j 1 X p j 2 X p j 3 E X p j 1 3 1 3 E (X p j 2 ) 3 1 3 E (X p j 3 ) 3 1 3 β.
Using the Hölder inequality, as well as the lemma 3.1 and the relation E (V

(k) p,n ) 2 = d k ! 2 inf pr+n f k , we obtain E V (k) p,j 1 V (l) p,j 2 V (q) p,j 3 E V (k) p,j 1 4 1 4 E V (l) p,j 2 4 1 4 E V (q) p,j 3 4 1 4 K 3 4 (d m -1) E V (k) p,j 1 2 1 2 E V (l) p,j 2 2 1 2 E V (q) p,j 3 2 1 2 K 3 4 (dm-1) d m ! 2 max 1 j r max 1 k m inf pr+j f k 3 2 . Then, E ϕ(Q(Z (p) )) -E   |s| 2 1 s! ∂ s ϕ(U p ) r j=1 X p j V p,j s   ϕ ∞ βK 3 4 (dm-1) (r m) 3 d m ! 2 max 1 j r max 1 k m inf pr+j f k 3 2
.

By writing the same formula for Q(Z (p+1) ) we obtain this time

E ϕ(Q(Z (p+1) )) -E   |s| 2 1 s! ∂ s ϕ(U p ) r j=1 G p j V p,j s   ϕ ∞ 8 π K 3 4 (dm-1) (r m) 3 d m ! 2 max 1 j r max 1 k m inf pr+j f k 3 2
.

In the last inequality, the term 8 π comes from the fact that G p j are standard gaussian which implies that E G p j 3 = 8 π . Since the vectors X p and G p are centered, have the same covariance matrix and are independent from U p and from V p j , then by putting the two inequalities together, we obtain:

E ϕ Q(Z (p+1) ) -E ϕ Q(Z (p) ) ϕ ∞ β + 8 π K 3 4 (d m -1) (r m) 3 d m ! 2 max 1 j r max 1 k m inf pr+j f k 3 2 . (3.9) Since r max i N i j=1 inf j f k = E((Q k (X)) 2 ) d k !(d k -1)! then max i N i p=1 max 1 j r max 1 k m inf pr+j f k max i N i p=1 r j=1 m k=1 inf pr+j f k m k=1 r max i N i j=1 inf j f k m k=1 E (Q k (X)) 2 d k !(d k -1)! m d 1 !(d 1 -1)! .
By summing over p in (3.9), we nally obtain that:

|E (ϕ (Q(X))) -E (ϕ (Q(G)))| ϕ ∞ β + 8 π K 3 4 (d m -1) r 3 m 4 d m ! 3 d 1 !(d 1 -1)! max 1 k m max 1 j N k inf j f k .
2. Let ϕ be a function of class C 3 . We have

|E (ϕ(Q(X))) -E (ϕ(Z V )))| |E (ϕ(Q(X))) -E (ϕ(Q(G))))|+|E (ϕ(Q(G))) -E (ϕ(Z V )))| .
For the rst term we use the point 1 of lemma 3.2 to nd an upper bound. For the second term we observe that the vector G have independent components, which allows us to use Theorem 7.2 in [START_REF] Nourdin | Invariance principles for homogeneous sums: universality of Gaussian Wiener chaos[END_REF] to get the following inequality :

E(ϕ(Q 1 (X), . . . , Q m (X))) -E(ϕ(Z V ) ϕ ∞ m i=1 ∆ i,i + 2 1 i<j m ∆ i,j +C ϕ ∞ 32 π m j=1 ( 64 π ) d j -1 3 d j ! 3 max 1 k m max 1 j N k inf j f k .
The constant C is such that

max k N k i=1 max 1 j m inf i f j C and since max k N k i=1 max 1 j m inf i f j m j=1 max k N k i=1 inf i f j m j=1 E ((Q j (X)) 2 ) d j !(d j -1)! m d 1 !(d 1 -1)! then we can choose the constant C equal to m d 1 !(d 1 -1)! . Thus, we obtain E(ϕ(Q 1 (X), . . . , Q m (X))) -E(ϕ(Z V ) ϕ ∞ m i=1 ∆ i,i + 2 1 i<j m ∆ i,j + ϕ ∞ 32 π 64 π dm-1 m 4 (d m !) 3 d 1 !(d 1 -1)! max 1 k m max 1 j N k inf j f k .
Lemma 3.3 Let the notations used in lemma 3.2 prevail. Consider the class H of indicator functions on measurable convex sets in R m . Let us dene

B 1 = m i=1 ∆ i,i + 2 1 i<j m ∆ i,j B 2 = m 4 d m ! 3 d 1 !(d 1 -1)! β + 8 π K 3 4 (d m -1) r 3 + 32 π 64 π dm-1 max 1 k m max 1 j N k inf j f k 1.
Let us assume that the covariance matrix V is the m-dimensional identity matrix. Then

sup h∈H(R m ) E h(Q 1 (X), . . . , Q m (X)) -E [h(Z V )] 8 3 6 7 + 4 3 13 7 (5B 1 + 5B 2 ) 1 7 m 3 7 .
2. Let us assume that the covariance matrix V is invertible and let Λ = diag(λ 1 , . . . , λ k ) be the diagonal matrix of the eigenvalues of V . Let B be an orthogonal matrix (i.e.

B T B = I m and BB T = I m ) such that V = BΛB T , and let b = max i,j (Λ -1 2 B T ). Then sup h∈H(R m ) E h(Q 1 (X), . . . , Q m (X)) -E [h(Z V )] 8 3 6 7 + 4 3 13 7 (5b 2 B 1 +5b 3 B 2 ) 1 7 m 3 7 .
Proof. 1. Let us assume that the covariance matrix V is the m-dimensional identity matrix. Denote by Φ the standard normal distribution in R m , and by φ the corresponding density function. Consider h ∈ H(R m ) and dene the following function:

h t (x) = R m h( √ ty + √ 1 -tx)Φ(dy), 0 < t < 1.
The key result is Lemma 2.11 in [START_REF] Goetze | On the rate of convergence in the multivariate CLT[END_REF] which states that, for every probability measure Q on R m , every random variables W ∼ Q and Z ∼ Φ, and any 0 < t < 1, we have

sup h∈H(R m ) |E [h(W )] -E [h(Z V )]| 4 3 sup h∈H(R m ) |E [h t (W )] -E [h t (Z V )]| + 2 √ m √ t . (3.10) Let us dene u(x, t, z) = (2πt) -m 2 exp - m i=1 (z i - √ 1-tx i ) 2 2t
. Using the change of variable

z = √ ty + √ 1 -tx in h t (x) leads to h t (x) = R m h(z)u(x, t, z)dz.
By the dominated convergence theorem, we may dierentiate under the integral sign and obtain

∂ 2 h t ∂x 2 i (x) = - 1 -t t R m h(z)u(x, t, z)dz + 1 -t t 2 R m h(z)(z i - √ 1 -tx i ) 2 u(x, t, z)dz.
Since h ∞ 1 then we have

∂ 2 h t ∂x 2 i (x) 1 -t t + 1 -t t 2 R m (z i - √ 1 -tx i ) 2 u(x, t, z)dz. If (Y 1 , . . . , Y m ) is a gaussian vector with covariance matrix tI m then R m (z i - √ 1 -tx i )u(x, t, z)dz = E(Y 2 i ) = t.
Therefore, we have

∂ 2 h t ∂x 2 i (x) 2 1 -t t .
Furthermore, for i = j we have

∂ 2 h t ∂x i ∂x j (x) = 1 -t t 2 R m h(z)(z i - √ 1 -tx i )(z j - √ 1 -tx j )u(x, t, z)dz, so that ∂ 2 h t ∂x i ∂x j (x) 1-t t 2 E(|Y i |)E(|Y i |) = 2(1-t) πt . We conclude that h ∞ 2 t 5 t 3 .
Similarly, for i, j, k in [1, m] it holds that:

∂ 3 h t ∂x i ∂x j ∂x k (x) (1 -t) 3 2 t 3 max 3E(|Y i |)t + E(|Y i | 3 ); E(|Y j |)t + E(|Y i | 2 )E(|Y j |); E(|Y i |)E(|Y j |)E(|Y k |).
Therefore h ∞ 5 t 3 . Combining the latter inequality with the result (3.10) and point 2 of Lemma 3.2, we obtain

sup h∈H(R m ) |E [h(Q(X))] -E [h(Z V )]| 4 3 sup h∈H(R m ) |E [h t (Q(X))] -E [h t (Z V )]| + 2 √ m √ t 8 3 √ m √ t + 4 3 (5B 1 + 5B 2 )t -3 .
The function in the right-hand side of the inequality reaches its minimum at

t = 15(B 1 +B 2 ) √ m 2 7 , hence sup h∈H(R m ) |E [h(Q(X))] -E [h(Z V )]| 8 3 6 7 + 4 3 13 7 (5B 1 + 5B 2 ) 1 7 m 3 7 . 2. Set Q(X) = (Q 1 (X), . . . , Q m (X)). For any h ∈ H(R m ), we have E(h(Q(X))) -E(h(Z v )) = E(h(BΛ 1 2 Λ -1 2 B T Q(X))) -E(h(BΛ 1 2 Λ -1 2 B T Z v )). Dene g(x) = h(BΛ 1 2 x), x ∈ R m . Since g ∈ H(R m
) then, using inequality (3.10), we get

sup h∈H(R m ) |E [h(Q(X))] -E [h(Z V )]| sup g∈H(R m ) E g(Λ -1 2 B T Q(X)) -E g(Λ -1 2 B T Z V ) 4 3 sup g∈H(R m ) E g t (Λ -1 2 B T Q(X)) -E g t (Λ -1 2 B T Z V ) + 2 √ m √ t .
We can nd an upper bound for the second and third derivatives of f t (x) = g t (Λ -1 2 B T x). Indeed, f t ∞ 5b 2 t -3 and f t ∞ 5b 3 t -3 . By using the same reasoning as in point 1 and replacing B 1 by b 2 B 1 and B 2 by b 3 B 2 in (3.10), we obtain the result. [START_REF] Rider | Gaussian Fluctuations for non-hemitian random matrix ensembles[END_REF] Proof of Theorem 1.1

We use hereafter the notation adopted in the beginning of Section 2. If we separate the diagonal terms from the non-diagonal terms in (1.2), we obtain

trace(X d n ) -E(trace(X d n )) = 1 n d 2 ((i 1 ,i 2 ),...,(i d ,i 1 ))∈D n (X i 1 ,i 2 . . . X i d ,i 1 -E(X i 1 ,i 2 . . . X i d ,i 1 )) + 1 n d 2 ((i 1 ,i 2 ),...,(i d ,i 1 ))∈N D n X i 1 ,i 2 . . . X i d ,i 1 .
The expectation in the second sum is equal to zero because the X i,j are independent and centered. The variance of the term containing the diagonal terms is upper bounded by O 1 √ n and, therefore, goes to 0 as n goes to innity. Indeed, if we set M = sup i,j

E(|X i,j | 2d ), then Var   1 n d 2 ((i 1 ,i 2 ),...,(i d ,i 1 ))∈Dn X i 1 ,i 2 . . . X i d ,i 1   = 1 n d   E     ((i 1 ,i 2 ),...,(i d ,i 1 ))∈D n X i 1 ,i 2 . . . X i d ,i 1   2   -   E   (i 1 ,i 2 ),...,(i d ,i 1 )∈D n X i 1 ,i 2 . . . X i d ,i 1     2   .
Keeping the notation introduced in lemma 2.1, we have:

E   ((i 1 ,i 2 ),...,(i d ,i 1 ))∈Dn X i 1 ,i 2 . . . X i d ,i 1 2   = ((i 1 ,i 2 ),...,(i d ,i 1 ),(i d+1 ,i d+2 ),...,(i 2d ,i d+1 ))∈A n E X i 1 ,i 2 . . . X i d ,i 1 X i d+1 ,i d+2 . . . X i 2d ,i d+1 .
Since E X i 1 ,i 2 . . . X i d ,i 1 X i d+1 ,i d+2 . . . X i 2d ,i d+1 is equal to 1 over the subset B n of A n , and is upper bounded by M over the subset A n \ B n , then we can state that:

E     ((i 1 ,i 2 ),...,(i d ,i 1 ))∈D n X i 1 ,i 2 . . . X i d ,i 1   2   -#B n M #(A n \ B n ). (4.11)
Furthermore, since the X i,j are centered and independent, then

E(X i 1 ,i 2 . . . X i d ,i 1 ) = 0 if ((i 1 , i 2 ), . . . , (i d , i 1 )) ∈ D n \ E n . Thus, E   ((i 1 ,i 2 ),...,(i d ,i 1 ))∈D n X i 1 ,i 2 . . . X i d ,i 1   = ((i 1 ,i 2 ),...,(i d ,i 1 ))∈E n E(X i 1 ,i 2 . . . X i d ,i 1 ).
On the other hand, E(X i 1 ,i 2 . . . X i d ,i 1 ) is equal to 1 over the subset F n of E n , and bounded by

√ M over E n \ F n . Then, E   ((i 1 ,i 2 ),...,(i d ,i 1 ))∈Dn X i 1 ,i 2 . . . X i d ,i 1   -#F n √ M #(E n \ F n ). (4.12) whose covariance matrix V is given by V (k, k ) = lim ∞ E(Q d (g k n , Y )Q d (g k n , Y )).
To do so, it is sucient to check the assumptions of Lemma 3.3, that is, (i) max

i=1,...,2N inf (a,b),p gk N → 0, (ii) for every 1 s d -1, gk N s gk N 2 → 0, (iii) E(Q d (g k n , Y )Q d (g k n , Y ) → δ i,j
(with δ i,j the Kronecker symbol), and (iv)

E(Q d (g k n , Y ) 2 ) → σ 2 . We can rewrite Q d (g k n , Y ) as Q d (g k n , Y ) = 1 n d 2 (j 1 ,...,j d )∈{0,1} n j 1 +...+j d =k (i 1 ,...,i d )∈N D n Y j 1 i 1 ,i 2 . . . Y j d i d ,i 1 .
The second-order moment of

Q d (g k n , Y ) is equal to 1 n d (j 1 ,...,j 2d )∈{0,1} n j 1 +...+j d =j d+1 +...+j 2d =k (i 1 ,i 2 ),...,(i d ,i 1 ) ∈N D n (i d+1 ,i d+2 ),...,(i 2d ,i d+1 ) ∈N D n E(Y j 1 i 1 ,i 2 . . . Y j d i d ,i 1 Y j d+1 i d+1 ,i d+2 . . . Y j 2d i 2d ,i d+1 ). (4.13)
For the expectation corresponding to the indices i 1 , . . . , i 2d , j 1 , . . . , j 2d in (4.13) to be dierent from zero, it must hold that (i 1 , . . . , i 2d ) belongs to G n , where G n has been dened in Lemma 2.1. Furthermore, since the subset

G n \ H n is of cardinality O(n d-1 ), its contribution to the moment of order 2 of Q d (g k n , Y ) is O( 1 n ).
It remains then to see what happens when (i 1 , . . . , i 2d ) belongs to H n . In this case, let us recall from the proof of point (vi) of Lemma 2.1 that the elements of the set H n are completely characterized by d given pairwise distinct integers i 1 , . . . , i d ∈ [1, n] and a given integer k ∈ [1, d] such that (i d+1 , . . . , i 2d ) = (i k , . . . , i d , i 1 , . . . , i k-1 ). Moreover, if the expectation corresponding to the indices i 1 , . . . , i 2d , j 1 , . . . , j 2d in (4.13) is dierent from zero, then it must hold that (j d+1 , . . . , j 2d ) = (j k , . . . , j d , j 1 , . . . , j k-1 ) and this expectation is equal to 1. Thus,

E Q d (g k n , Y ) 2 = 1 n d (j 1 ,...,j d )∈{0,1} n j 1 +...+j d =k d × n . . . × (n -d + 1) + O 1 n = dC k d × n . . . × (n -d + 1) n d + O( 1 n ), which yields E Q d (g k n , Y ) 2 -→ n→∞ dC k d . Moreover, E(Q d (g k N , Y )Q d (g j N , Y )) is equal to 1 n d (j 1 ,...,j 2d )∈{0,1} n j 1 +...+j d =k,j d+1 +...+j 2d =j (i 1 ,i 2 ),...,(i d ,i 1 ) ∈N D n (i d+1 ,i d+2 ),...,(i 2d ,i d+1 ) ∈N D n E(Y j 1 i 1 ,i 2 . . . Y j d i d ,i 1 Y j d+1 i d+1 ,i d+2 . . . Y j 2d i 2d ,i d+1 ).
Similarly to the computation of the second-order moment of Q d (g k n , Y ), the set of elements for which the expectation in (4.14) is dierent from zero is the set

G n of Lemma 2.1. The subset G n \ H n is of cardinality O(n d-1 ), which implies that its contribution to E(Q d (g k N , Y )Q d (g j N , Y )) is O( 1 n ).
Furthermore, the elements of the set H n are characterized by d given pairwise distinct integers i 1 , . . . , i d ∈ [1, n] and a given integer k ∈ [1, d] such that (i d+1 , . . . , i 2d ) = (i k , . . . , i d , i 1 , . . . , i k-1 ). Moreover, for E(Y

j 1 i 1 ,i 2 . . . Y j d i d ,i 1 Y j d+1
i d+1 ,i d+2 . . . Y j 2d i 2d ,i d+1 ) to be dierent from zero, it must hold that (j d+1 , . . . , j 2d ) = (j k , . . . , j d , j 1 , . . . , j k-1 ), which is impossible in the case j = k. We conclude that E

(Q d (g k N , Y )Q d (g j N , Y )) → 0 for every j = k. From the denition of gk n , it is clear that gk n ∞ g k n ∞ 1 n d 2 . Then, inf (a,b),p gk N = (x 1 ,...,x d-1 )∈[1,n] d-1 (y 1 ,...,y d-1 )∈[1,n] d-1 (p 1 ,...,p d-1 )∈{0,1} d-1 gk n (((a, b), p), ((x 1 , y 1 ), p 1 ), . . . , ((x d-1 , y d-1 ), p d-1 )) 2 (i 1 ,...,i d )∈[1,n] d (p 1 ,...,p d )∈{0,1} d σ∈S d g k n ((i σ(1) , i σ(1)+1 ), p σ(1) ), . . . , ((i σ(d) , i σ(d)+1 ), p σ(d) ) 2 ×1 {a=i σ(1) } × 1 {b=i σ(1)+1 } × 1 {p=p σ(1) } 2 d-1 (d)!n d-2 gk n 2 ∞ 2 d-1 (d)! n 2 .
Therefore max For the sake of notational simplicity and because this case is representative of the difculty, in the rest of the proof we assume that σ 1 = σ 2 = σ 3 = σ 4 = I d , where I d stands for the identity permutation over [1, d] Then, all the assumptions of Lemma 3.3 are fullled by Q d (g k N , Y ). Therefore,

Q d (f N , Y ) law -→ d k=1 i + ρ R 1 -R k R k 0 R d-k 1 dC k d G k ,

1 n s g k,σ 2 n((x 1 4 σ 1 (

 12141 1 s d -1 and σ 1 , σ 2 ∈ S d . Then g k,σ , y 1 ), p 1 ), . . . , ((x d-s , y d-s ), p d-s ), ((x 1 , y 1 ), p 1 ), . . . , ((x d-s , y d-s ), p d-s ) = (x d-s+1 ,...,x d )∈[1,n] s (y d-s+1 ,...,y d )∈[1,n] s (p d-s+1 ,...,p d )∈{0,1} s g k,σ 1 n [((x 1 , y 1 ), p 1 ), . . . , ((x d-s , y d-s ), p d-s ), ((x d-s+1 , y d-s+1 ), p d-s+1 ), . . . , ((x d , y d ), p d )] ×g k,σ 2 n ((x 1 , y 1 ), p 1 ), . . . , ((x d-s , y d-s ), p d-s ), ((x d-s+1 , y d-s+1 ), p d-s+1 ), . . . , ((x d , y d ), p d ) ,σ 2 ,σ 3 ,σ 4 ∈S d (x 1 ,...,x d-s )∈[1,n] d-s (y 1 ,...,y d-s )∈[1,n] d-s (p 1 ,...,p d-s )∈{0,1} d-s (x 1 ,...,x d-s )∈[1,n] d-s (y 1 ,...,y d-s )∈[1,n] d-s (p1 ,...,p d-s )∈{0,1} d-s (x d-s+1 ,...,x d )∈[1,n] s (y d-s+1 ,...,y d )∈[1,n] s (p d-s+1 ,...,p d )∈{0,1} s (x d-s+1 ,...,x d )∈[1,n] s x 1 , y 1 ), p 1 ), . . . , ((x d-s , y d-s ), p d-s ), ((x d-s+1 , y d-s+1 ), p d-s+1 ), . . . , ((x d , y d ), p d )] ×g k,σ 2n ((x 1 , y 1 ), p 1 ), . . . , ((x d-s , y d-s ), p d-s ), ((x d-s+1 , y d-s+1 ), p d-s+1 ), . . . , ((x d , y d ), p d ) ×g k,σ 3 n ((x 1 , y 1 ), p 1 ), . . . , ((x d-s , y d-s ), p d-s ), ((x d-s+1 , y d-s+1 ), p d-s+1 ), . . . , ((x d , y d ), p d ) ×g k,σ 4 n ((x 1 , y 1 ), p 1 ), . . . , ((x d-s , y d-s ), p d-s ), ((x d-s+1 , y d-s+1 ), p d-s+1 ), . . . , ((x d , y d ), p d ) .

.

  Since g k n is equal to zero at point [((x 1 , y 1 ), p 1 ), . . . , ((x d , y d ),p d )] if y i = x i+1 or y d = x 1 , then (x 1 ,...,x d-s )∈[1,n] d-s (y 1 ,...,y d-s )∈[1,n] d-s (p 1 ,...,p d-s )∈{0,1} d-s (x 1 ,...,x d-s )∈[1,n] d-s (y 1 ,...,y d-s )∈[1,n] d-s (p 1 ,...,p d-s )∈{0,1} d-s (x d-s+1 ,...,x d )∈[1,n] s (y d-s+1 ,...,y d )∈[1,n] s (p d-s+1 ,...,p d )∈{0,1} s (x d-s+1 ,...,x d )∈[1,n] s x 1 , y 1 ),p1 ), . . . , ((x d-s , y d-s ), p d-s ), ((x d-s+1 , y d-s+1 ), p d-s+1 ), . . . , ((x d , y d ), p d )] ×g k,I d n ((x 1 , y 1 ), p 1 ), . . . , ((x d-s , y d-s ), p d-s ), ((x d-s+1 , y d-s+1 ), p d-s+1 ), . . . , ((x d , y d ), p d ) ×g k,I d n ((x 1 , y 1 ), p 1 ), . . . , ((x d-s , y d-s ), p d-s ), ((x d-s+1 , y d-s+1 ), p d-s+1 ), . . . , ((x d , y d ), p d ) ×g k,I d n ((x 1 , y 1 ), p 1 ), . . . , ((x d-s , y d-s ), p d-s ), ((x d-s+1 , y d-s+1 ), p d-s+1 ), . . . , ((x d , y d ), p d ) = (α 1 ,...,α d-s+1 )∈[1,n] d-s+1 (p 1 ,...,p d-s )∈{0,1} d-s(α 1 ,...,α d-s+1 )∈[1,n]

  (vi) Consider I = (i 1 , i 2 ), . . . , (i d , i 1 ), (i d+1 , i d+2 ), . . . , (i 2d , i d+1 ) ∈ H n .(a) By denition of N D n , there is no redundancy neither among the pairs (i 1 , i 2 ), . . . , (i d , i 1 ) nor among the pairs (i d+1 , i d+2 ), . . . , (i 2d , i d+1 ). Therefore, to satisfy the constraint dening H n , it is necessary and sucient that each couple of (i 1 , i 2 ), . . . , (i d , i 1 ) matches one and only one couple among (i d+1 , i d+2 ), (i d+2 , i d+3 ), . . . , (i 2d , i d+1 ). (b) Using a proof by contradiction, let us show that the elements of {i 1 , . . . , i d } are pairwise distinct. If p and q were two distinct integers in[1, d] such that i p = i q then, according to (a), there would exist k ∈

).

  ×1 {α 1 =α 1 } × 1 {α d-s+1 =α d-s+1 } ×g k n [((α 1 , α 2 ), p 1 ), . . . , ((α d-s , α d-s+1 ), p d-s ), ((α d-s+1 , i 1 ), p d-s+1 ), . . . , ((i s-1 , x 1 ), p d )] ×g k n ((α 1 , α 2 ), p 1 ), . . . , ((α d-s , α d-s+1 ), p d-s ), ((α d-s+1 , i 1 ), p d-s+1 ), . . . , ((i s-1 , x 1 ), p d ) ×g k n ((α 1 , α 2 ), p 1 ), . . . , ((α d-s , α d-s+1 ), p d-s ), ((α d-s+1 , i 1 ), p d-s+1 ), . . . , ((i s-1 , x 1 ), p d ) ×g k n ((α 1 , α 2 ), p 1 ), . . . , ((α d-s , α d-s+1 ), p d-s ), ((α d-s+1 , i 1 ), p d-s+1), . . . , ((i s-1 , x 1 ), p d ) 2 2d n 2d-2 g k

						d-s+1
				(p 1 ,...,p d-s	)∈{0,1} d-s	(p d-s+1	,...,p d	)∈{0,1} s
	n	4 ∞	2 2d n -2 .
	We conclude that gk n s gk n	2 2	→ 0.

(i 1 ,...,i s-1 )∈[1,n] s-1 (p d-s+1 ,...,p d )∈{0,1} s (i 1 ,...,i s-1 )∈[1,n] s-1

Acknowledgments. This work is part of my forthcoming PhD dissertation. I am extremely grateful to my advisor Ivan Nourdin for suggesting this topic, as well as for many advices and encouragements.

Finally, by combining the estimations (4.11) and (4.12), and using points (i) to (iv) of Lemma 2.1 and the fact that #D n = O(n d-1 ), we get the following result, with Z n dened by Z n = ((i 1 ,i 2 ),...,(i d ,i 1 ))∈D n X i 1 ,i 2 . . . X i d ,i 1 :

From points (ii) and (iv)

). Using point (i) and the relation (4.11), we obtain the estimation

). Finally, using points (iii) -(iv) and the relation (4.12), we get the following estimations:

. From these estimations, we conclude that:

). When R = 1, the X i are real-valued, which corresponds exactly to the result of Nourdin et Peccati [START_REF] Nourdin | Universal Gaussian uctuations of non-Hermitian matrix ensembles[END_REF] (there is then nothing more to prove). By contrast, when R = 0, the X i are purely imaginary-valued; factoring out by i d in the trace formula shows that the result in this case can be derived from the case R = 1. In what follows, we can then freely assume that R ∈ (0, 1).

, and dene:

and

We have:

which yields

We dene for any two elements (i

a family of random variables dened by

We can then conclude that

If gk n stands for the symmetrization of


Vahagn Nersesyan

Hayk Nersisyan

Global exact controllability in infinite time of Schrödinger equation: multidimensional case

We prove that the multidimensional Schrödinger equation is exactly controllable in infinite time near any point which is a finite linear combination of eigenfunctions of the Schrödinger operator. We prove that, generically with respect to the potential, the linearized system is controllable in infinite time. Applying the inverse mapping theorem, we prove the controllability of the nonlinear system.

Introduction

This paper is concerned with the problem of controllability for the following Schrödinger equation

i ż = -∆z + V (x)z + u(t)Q(x)z, x ∈ D, (1.1)
z| ∂D = 0, (1.2) z(0, x) = z 0 (x), (1.3) where D ⊂ R d , d ≥ 1 is a rectangle, V, Q ∈ C ∞ (D, R) are given functions, u is the control, and z is the state. We prove that (1.1)-(1.3) is exactly controllable in infinite time near any point which is a finite linear combination of eigenfunctions 1 of the Schrödinger operator, extending the results of [START_REF] Nersesyan | Global exact controllability in infinite time of Schrödinger equation[END_REF] to the multidimensional case.

Recall that in the papers [START_REF] Beauchard | Local controllability of a 1D Schrödinger equation[END_REF][START_REF] Beauchard | Controllability of a quantum particle in a moving potential well[END_REF][START_REF] Beauchard | Local controllability of linear and nonlinear Schrödinger equations with bilinear control[END_REF] it is proved that the 1D Schrödinger equation is exactly controllable in finite time in a neighborhood of any finite linear combination of eigenfunctions of Laplacian. In [START_REF] Chambrion | Controllability of the discrete-spectrum Schrödinger equation driven by an external field[END_REF][START_REF] Privat | The squares of Laplacian-Dirichlet eigenfunctions are generically linearly independent[END_REF][START_REF] Mason | Generic controllability properties for the bilinear Schrödinger equation[END_REF], approximate controllability in L 2 is proved for multidimensional Schrödinger equation, generically with respect to functions V, Q and domain D. In [START_REF] Mirrahimi | Lyapunov control of a particle in a finite quantum potential well[END_REF][START_REF] Beauchard | Approximate stabilization of a quantum particle in a 1D infinite square potential well[END_REF][START_REF] Nersesyan | Global approximate controllability for Schrödinger equation in higher Sobolev norms and applications[END_REF][START_REF] Nersesyan | Growth of Sobolev norms and controllability of the Schrödinger equation[END_REF][START_REF] Morancey | Explicit approximate controllability of the Schrödinger equation with a polarizability term[END_REF], stabilization results and approximate controllability properties are proved. In particular, combination of the results of [START_REF] Nersesyan | Global approximate controllability for Schrödinger equation in higher Sobolev norms and applications[END_REF] with the above mentioned local exact controllability properties gives global exact controllability in finite time for 1D case in the spaces H 3+ε , ε > 0. See also papers [START_REF] Ramakrishna | Controllability of molecular systems[END_REF][START_REF] Turinici | Quantum wavefunction controllability[END_REF][START_REF] Altafini | Controllability of quantum mechanical systems by root space decomposition of su(n)[END_REF][START_REF] Albertini | Notions of controllability for bilinear multilevel quantum systems[END_REF][START_REF] Agrachev | An estimation of the controllability time for single-input systems on compact Lie groups[END_REF][START_REF] Beauchard | Implicit Lyapunov control of finite dimensional Schrödinger equations[END_REF] for controllability of finite-dimensional systems and papers [START_REF] Lebeau | Contrôle de l'équation de Schrödinger[END_REF][START_REF] Machtyngier | Stabilization of the Schrödinger equation[END_REF][START_REF] Baudouin | Uniqueness and stability in an inverse problem for the Schrödinger equation[END_REF][START_REF] Zuazua | Remarks on the controllability of the Schrödinger equation[END_REF][START_REF] Dehman | Stabilization and control for the nonlinear Schrödinger equation on a compact surface[END_REF][START_REF] Ervedoza | Approximate controllability for a system of Schrödinger equations modeling a single trapped ion[END_REF] for controllability properties of various Schrödinger systems.

The linearization of (1.1)-(1.3) around the trajectory e -iλ k,V t e k,V with u = 0 and z 0 = e k,V (e k,V is an eigenfunction of the Schrödinger operator -∆ + V corresponding to some eigenvalue λ k,V ) is of the form

i ż = -∆z + V (x)z + u(t)Q(x)e -iλ k,V t e k,V , x ∈ D, (1.4)
z| ∂D = 0, (1.5) z(0, x) = 0.

(

Writing this in the Duhamel form

z(T ) = -i T 0 S(T -s)[u(s)Qe -iλ k,V s e k,V ]ds, (1.7)
where S(t) = e it(∆-V ) is the free evolution, we see that (1.4)- (1.6) is equivalent to the following moment problem for d mk := ie iλ m,V T Qem,V ,e k,V z(T ), e m,V d mk =

T 0 e iω mk s u(s)ds, m ≥ 1, ω mk = λ m,V -λ k,V .

(1.8)

It is well known that a gap condition for the frequencies ω mk is necessary for the solvability of this moment problem when T < +∞ (e.g., see [START_REF] Young | An Introduction to Nonharmonic Fourier Series[END_REF]). The asymptotic formula for the eigenvalues λ m,V ∼ C d m 2 d implies that there is no gap in the case d ≥ 3 (when d = 2, existence of a domain for which there is a gap between the eigenvalues is an open problem). Moreover, it follows from [START_REF] Avdonin | On the question of Riesz bases of exponential functions in L 2[END_REF] that there is a linear dependence between the exponentials: there is a non-zero

{c m } ∈ ℓ 2 such that +∞ m=1 c m e iω mk s = 0 for t ∈ [0, T ]. Hence (1.4)-(1.6) is non- controllable in finite time T < +∞.
The situation is different when T = +∞. Indeed, by Lemma 3.10 in [START_REF] Nersesyan | Growth of Sobolev norms and controllability of the Schrödinger equation[END_REF], the exponentials are independent on [0, +∞), and moreover, (1.4)-(1.6) is controllable, by Theorem 2.6 in [START_REF] Nersesyan | Global exact controllability in infinite time of Schrödinger equation[END_REF]. In [START_REF] Nersesyan | Global exact controllability in infinite time of Schrödinger equation[END_REF], we used the controllability of linearized system (1.4)-(1.6) to prove the controllability of nonlinear system only in the case d = 1. In the multidimensional case, we were able to prove the controllability of (1.4)-(1.6) in a more regular Sobolev space than the one where nonlinear system (1.1)-(1.3) is well posed. We do not know if this difficulty of loss of regularity can be treated using the Nash-Moser inverse function theorem in the spirit of [START_REF] Beauchard | Local controllability of a 1D Schrödinger equation[END_REF]. More precisely, in the multidimensional case, it is very difficult to prove that the inverse of the linearization satisfies the estimates in the Nash-Moser theorem. In this paper, we find a space H (see (1.11) for the definition), where the nonlinear problem is well posed and the linearized problem is controllable. Applying the inverse inverse function theorem in the space H, we get controllability for (1.1)- (1.3). Let us notice that H is a sufficiently large space of functions, it contains the Sobolev space H 3d . Thus, in particular, we prove controllability in H 3d . The result of this paper is optimal in the sense that it seems that the multidimensional Schrödinger equation (1.1)-(1.3) is not exactly controllable in finite time.

Acknowledgments. The authors would like to thank J.-P Puel for providing them in privat communication [27] some results about regularity questions for the Schrödinger equation.

Notation

In this paper, we use the following notation. Let us define the Banach spaces

ℓ 2 := {{a j } ∈ C ∞ : {a j } 2 ℓ 2 = +∞ j=1 |a j | 2 < +∞}, ℓ 2 0 := {{a j } ∈ ℓ 2 : a 1 ∈ R}, ℓ ∞ := {{a j } ∈ C ∞ : {a j } ℓ ∞ = sup j≥1 |a j | < +∞}, ℓ ∞ 0 := {{a j } ∈ ℓ ∞ : lim j→+∞ a j = 0}, ℓ ∞ 01 := {{a j } ∈ ℓ ∞ 0 : a 1 ∈ R}.
We denote by H s := H s (D) the Sobolev space of order s ≥ 0. Consider the Schrödinger operator

-∆ + V , V ∈ C ∞ (D, R) with D(-∆ + V ) := H 1 0 ∩ H 2 .
Let {λ j,V } and {e j,V } be the sets of eigenvalues and normalized eigenfunctions of this operator. Let •, • and • be the scalar product and the norm in the space L 2 . Define the space H s (V ) := D((-∆ + V ) s 2 ) endowed with the norm

• s,V = (λ j,V ) s 2 •, e j,V ℓ 2 . When D is the rectangle (0, 1) d and V (x 1 , . . . , x d ) = V 1 (x 1 ) + . . . + V d (x d ), V k ∈ C ∞ ([0, 1], R), the eigenvalues and eigenfunctions of -∆ + V on D are of the form λ j1,...,j d ,V = λ j1,V1 + . . . + λ j d ,V d , (1.9) e j1,...,j d ,V (x 1 , . . . , x d ) = e j1,V1 (x 1 ) • . . . • e j d ,V d (x d ), (x 1 , . . . , x d ) ∈ D, (1.10)
where {λ j,V k } and {e j,V k } are the eigenvalues and eigenfunctions of operator

-d 2
dx 2 + V k on (0, 1). Define the spaces

H = {z ∈ L 2 : (j 3 1 • . . . • j 3 d ) z, e j1,...,j d ,V ∈ ℓ ∞ 0 , z H := (j 3 1 • . . . • j 3 d ) z, e j1,...,j d ,V ℓ ∞ < +∞}, (1.11) V = {z ∈ L 2 : z V := +∞ j1,...,j d =1 (j 3 1 • . . . • j 3 d )| z, e j1,...,j d ,V | < +∞}.
(1.12)

The eigenvalues and eigenfunctions of Dirichlet Laplacian on the interval (0, 1) are λ k,0 = k 2 π 2 and e k,0 (x) = √ 2 sin(kπx), x ∈ (0, 1). It is well known that for any

V ∈ L 2 ([0, 1], R) λ k,V = k 2 π 2 + 1 0 V (x)dx + r k , (1.13)
e k,V -e k,0 L ∞ ≤ C k , (1.14) de k,V dx - de k,0 dx L ∞ ≤ C, (1.15)
where +∞ k=1 r 2 k < +∞ (e.g., see [START_REF] Pöschel | Inverse Spectral Theory[END_REF]). For a Banach space X, we shall denote by B X (a, r) the open ball of radius r > 0 centered at a ∈ X. The integer part of x ∈ R is denoted by [x]. We denote by C a constant whose value may change from line to line.

Main results

Well-posedness of Schrödinger equation

We assume that V (x 1 , . . . ,

x d ) = V 1 (x 1 ) + . . . + V d (x d ), x k ∈ [0, 1] and V k ∈ C ∞ ([0, 1], R), k = 1, . . . , d.
Let us consider the following Schrödinger equation

i ż = -∆z + V (x)z + u(t)Q(x)z + v(t)Q(x)y, (2.1) z| ∂D = 0, (2.2)
z(0, x) = z 0 (x). (2.
3)

The following lemma shows the well-posedness of this system in H 2 (V ) .

Lemma 2.1. For any

z 0 ∈ H 2 (V ) , u, v ∈ L 1 loc ([0, ∞), R) and y ∈ L 1 ([0, ∞), H 2 (V ) ) problem (2.1)-(2.3) has a unique solution z ∈ C([0, ∞), H 2 (V ) ). Furthermore, if v = 0, then for all t ≥ 0 we have z(t) = z 0 .
(2.4)

See [START_REF] Cazenave | Semilinear Schrödinger equations[END_REF] for the proof. In [START_REF] Beauchard | Local controllability of linear and nonlinear Schrödinger equations with bilinear control[END_REF] it is proved that this problem is well posed in H 3 (V ) for d = 1, and in [27] the well-posedness in

H 3 (V ) is proved for d ≥ 1.
For any integer l ≥ 3, let m = m(l) := [ l-1 2 ] and define the space

C m 0 := {u ∈ C m ([0, ∞), R) : d k u dt k (0) = 0, k ∈ [0, m]} endowed with the norm of C m ([0, ∞), R).
The following lemma shows that problem (2.1)-(2.3) is well posed in higher Sobolev spaces when u, v and y are more regular.

Lemma 2.2. For any integer l ≥ 3, any

z 0 ∈ H l (V ) , any y ∈ W m,1 loc ([0, ∞), H 2 (V ) ) and any u, v ∈ C m 0 the solution z in Lemma 2.1 belongs to the space C([0, ∞), H l )∩ C 1 ([0, ∞), H l-2 ). Moreover, there is a constant C > 0 such that z(t) H l + z W m,1 ([0,t],H 2 (V ) ) ≤C( z 0 l,V + v C m 0 y W m,1 ([0,t],H 2 (V ) ) ) × e C( u C m 0 +1)t .
(2.5) See Appendix of [START_REF] Beauchard | Local controllability of a 1D Schrödinger equation[END_REF] for the proof.

Lemma 2.3. Denote by U t (•, •) : H 2 (V ) × L 1 loc (R + , R)→H 2 (V ) the resolving oper- ator of (1.1), (1.2). Then U t (•, •) is locally Lipschitz continuous: there is C > 0 such that U t (z 0 , u) -U t (z ′ 0 , u ′ ) H l ≤ C( z 0 -z ′ 0 l,V + u -u ′ C m 0 z ′ 0 l,V )e C( u C m 0 +1)t .
(2.6)

Proof. Notice that z(t) := U t (z 0 , u) -U t (z ′ 0 , u ′ ) is a solution of problem i ż = -∆z + V (x)z + u(t)Q(x)z + (u(t) -u ′ (t))Q(x)U t (z ′ 0 , u ′ ), z| ∂D = 0, z(0, x) = z 0 (x) -z ′ 0 (x).
Applying Lemma 2.2, we get

z(t) H l ≤ C( z 0 -z ′ 0 l,V + u -u ′ C m 0 U • (z ′ 0 , u ′ ) W m,1 ([0,t],H 2 (V ) ) )e C( u C m 0 +1)t , (2.7)
U • (z ′ 0 , u ′ ) W m,1 ([0,t],H 2 (V ) ) ≤ C z ′ 0 l,V e C( u C m 0 +1)t . (2.8)
Replacing (2.8) into (2.7), we get (2.6).

Let us rewrite (1.1)-(1.3) in the Duhamel form

z(t) = S(t)z 0 -i t 0 S(t -s)[u(s)Qz(s)]ds, (2.9)
where S(t) = e it(∆-V ) is the free evolution. Let us take any w ∈ L 1 (R + , R) and estimate the following integral

G t (z) := t 0 S(-s)[w(s)Qz(s)]ds.
We take controls from the weighted space space

G := {u ∈ L 1 (R + , R) : u(•)e B• ∈ L 1 (R + , R)} endowed with the norm u G = u(•)e B• L 1
, where the constant B > 0 will be chosen later. For B > C + 1, where C is the constant in Lemma 2.2, we have the following result.

Proposition 2.4. Let us take any l ≥ 4d, z 0 ∈ H l (V ) , w ∈ G and u ∈ C m 0 , and let z(t) := U t (z 0 , u). Then there are constants δ, C > 0 such that for any u ∈ B C m 0 (0, δ) and for any t > s ≥ 0

G t (z) -G s (z) H ≤ C t s z(τ ) H l |w(τ )|dτ, (2.10
)

and the following integral converges in H G ∞ (z) := +∞ 0 S(-τ )[w(τ )Qz(τ )]dτ. (2.11)
Proof. Using (2.5) with v = 0, the definition of G, and choosing δ > 0 sufficiently small, we see that

+∞ 0 z(τ ) H l |w(τ )|dτ < +∞.
Combining this with (2.10), we prove the convergence of the integral in (2.11).

Let us prove (2.10). To simplify the notation, let us suppose that d = 2; the proof of the general case is similar. Let (-

V (x 1 , x 2 ) = V 1 (x 1 )+V 2 (x 2
∂ 2 ∂x 2 1 + V 1 )(Qz), e j1,V1 e j2,V2 = 1 λ 2 j1,V1 (- ∂ 2 ∂x 2 1 + V 1 )(Qz), (- ∂ 2 ∂x 2 1 + V 1 )e j1,V1 e j2,V2 = 1 λ 2 j1,V1 1 0 ∂ 2 ∂x 2 1 (Qz)e j2,V2 dx 2 ∂ ∂x 1 e j1,V1 x1=1 x1=0 + 1 λ 2 j1,V1 V 1 (- ∂ 2 ∂x 2 1 + V 1 )(Qz), e j1,V1 e j2,V2 + ∂ ∂x 1 (- ∂ 2 ∂x 2 1 + V 1 )(Qz), ∂ ∂x 1 e j1,V1 e j2,V2
= :

I j + J j .
Let us estimate I j . Since ∂ 2 ∂x 2 1 (Qz(s)) = 0 for all x 1 ∈ [0, 1] and for x 2 = 0 and

x 2 = 1, integration by parts in x 2 implies

I j = 1 λ 2 j1,V1 λ j2,V2 1 0 (- ∂ 2 ∂x 2 2 + V 2 ) ∂ 2 ∂x 2 1 (Qz) e j2,V2 dx 2 ∂ ∂x 1 e j1,V1 x1=1 x1=0 = 1 λ 2 j1,V1 λ 2 j2,V2 1 0 (- ∂ 2 ∂x 2 2 + V 2 ) ∂ 2 ∂x 2 1 (Qz) (- ∂ 2 ∂x 2 2 + V 2 )e j2,V2 dx 2 ∂ ∂x 1 e j1,V1 x1=1 x1=0 = 1 λ 2 j1,V1 λ 2 j2,V2
(-

∂ 2 ∂x 2 2 + V 2 ) ∂ 2 ∂x 2 1 (Qz) ∂ ∂x 2 e j2,V2 ∂ ∂x 1 e j1,V1 x2=1 x2=0 x1=1 x1=0 + 1 λ 2 j1,V1 λ 2 j2,V2 1 0 V 2 (- ∂ 2 ∂x 2 2 + V 2 ) ∂ 2 ∂x 2 1 (Qz) e j2,V2 dx 2 ∂ ∂x 1 e j1,V1 x1=1 x1=0 + 1 λ 2 j1,V1 λ 2 j2,V2 1 0 ∂ ∂x 2 (- ∂ 2 ∂x 2 2 + V 2 ) ∂ 2 ∂x 2 1 (Qz) ∂ ∂x 2 e j2,V2 dx 2 ∂ ∂x 1 e j1,V1 x1=1 x1=0
=: I j,1 + I j,2 + I j,3 .

(2.12)

Let us consider the term I j,1 :

I j,1 = 2j 1 j 2 π 2 λ 2 j1,V1 λ 2 j2,V2
(-

∂ 2 ∂x 2 2 + V 2 ) ∂ 2 ∂x 2 1 (Qz) cos(j 1 πx 1 ) cos(j 2 πx 2 ) + 1 λ 2 j1,V1 λ 2 j2,V2 (- ∂ 2 ∂x 2 2 +V 2 ) ∂ 2 ∂x 2 1 (Qz) ∂ 2 ∂x 1 ∂x 2 (e j1,V1 e j2,V2 -e j1,0 e j2,0 ) x2=1 x2=0 x1=1 x1=0 .
Using (1.13), (1.15) and the Sobolev embedding

H s ֒→ L ∞ , s > d 2 , we get sup j1,j2≥1 j 3 1 j 3 2 t s e i(λj 1 ,V 1 +λj 2 ,V 2 )τ w(τ )I j,1 dτ ≤ C t s z(τ ) H l |w(τ )|dτ.
The Riemann-Lebesgue theorem and (1.15) imply that

j 3 1 j 3 2 t s e i(λj 1 ,V 1 +λj 2 ,V 2 )τ w(τ )I j,1 dτ →0 as j 1 + j 2 → + ∞.
Thus

j 3 1 j 3 2 t s e i(λj 1 ,V 1 +λj 2 ,V 2 )τ w(τ )I j,1 dτ ∈ ℓ ∞ 0 .
The terms I j,2 , I j,3 and J j are treated exactly in the same way. We omit the details. Thus we get that

G t (z) -G s (z) H = t s S(-τ )[w(τ )Qz(τ )]dτ H ≤ C t s z(τ ) H l |w(τ )|dτ.
Let T n → + ∞ be a sequence such that e -iλV,j Tn →1 as n→∞ for any j ≥ 1 (e.g., see Lemma 2.1 in [START_REF] Nersesyan | Global exact controllability in infinite time of Schrödinger equation[END_REF]). Then S(T n )z→z as n→ + ∞ in H for any z ∈ H and t ≥ 0.

(2.13) Indeed, since

S(t)z = +∞ j=1 e -iλj,V t z, e j,V e j,V , (2.14)
we have

S(T n )z -z H ≤ sup λj 1 ,...,j d ,V ≤N (j 3 1 • . . . • j 3 d )|e -iλj 1 ,...,j d ,V Tn -1|| z, e j1,...,j d ,V | + 2 sup λj 1 ,...,j d ,V >N (j 3 1 • . . . • j 3 d )| z, e j1,...,j d ,V | ≤ ε 2 + ε 2 = ε
for sufficiently large integers N, n ≥ 1.

Let us take t = T n in (2.9) and pass to the limit n→∞. Using Proposition 2.4, the embedding H l (V ) ֒→ H and (2.13), we obtain the following result.

Lemma 2.5. Let us take any l ≥ 4d and z 0 ∈ H l (V ) . There is a constant δ > 0 such that for any u ∈ B C m 0 (0, δ) ∩ G the following limit exists in H

lim n→+∞ U Tn (z 0 , u) =: U ∞ (z 0 , u).
(2.15)

Exact controllability in infinite time

Let l ≥ 4d be the integer in Proposition 2.4. Take any integer s ≥ l and let

H s 0 (R + , R) := {u ∈ H s (R + , R) : u (k) (0) = 0, k = 0, . . . , s -1}.
The set of admissible controls is the Banach space

F := G ∩ H s 0 (R + , R) (2.16)
endowed with the norm u F := u G + u H s . Equality (2.4) implies that it suffices to consider the controllability properties of (1.1), (1.2) on the unit sphere S in L 2 . We prove the controllability of (1.1), (1.2) under below condition.

Condition 2.6. Suppose that the functions

V, Q ∈ C ∞ (D, R) are such that (i) inf p1,j1,...,p d ,j d ≥1 |(p 1 j 1 • . . . • p d j d ) 3 Q pj | > 0,Q pj := Qe p1,...,p d ,V , e j1,...,j d ,V , (ii) λ i,V -λ j,V = λ p,V -λ q,V
for all i, j, p, q ≥ 1 such that {i, j} = {p, q} and i = j.

See [START_REF] Nersesyan | Global exact controllability in infinite time of Schrödinger equation[END_REF] and [START_REF] Privat | The squares of Laplacian-Dirichlet eigenfunctions are generically linearly independent[END_REF][START_REF] Nersesyan | Global approximate controllability for Schrödinger equation in higher Sobolev norms and applications[END_REF][START_REF] Mason | Generic controllability properties for the bilinear Schrödinger equation[END_REF] for the proof of genericity of (i) and (ii), respectively. Let us set E := span{e j,V }.

(2.17)

Below theorem is the main result of this paper. (z, σ) there is a control u ∈ F such that limit (2.15) exists in H and z 1 = U ∞ (z, u).

Remark 2.9. As in the case d = 1 (see Theorems 3.7 and 3.8 in [START_REF] Nersesyan | Global exact controllability in infinite time of Schrödinger equation[END_REF]) here also one can prove controllability in higher Sobolev spaces with more regular controls, and a global controllability property using a compactness argument.

3 Proof of Theorem 2.8

Controllability of linearized system

In this section, we study the controllability of the linearization of (1.1), (1.2) around the trajectory U t (z, 0), z ∈ S ∩ E:

i ż = -∆z + V (x)z + u(t)Q(x)U t (z, 0), (3.1)
z| ∂D = 0, (3.2) z(0, x) = z 0 . (3.3)
The controllability in infinite time of this system is proved in [START_REF] Nersesyan | Global exact controllability in infinite time of Schrödinger equation[END_REF], Section 2.

For the proof of Theorem 2.8 we need to show controllability of (3.1)-(3.3) in H which is larger than the space considered in [START_REF] Nersesyan | Global exact controllability in infinite time of Schrödinger equation[END_REF]. Hence a generalization of the arguments of [START_REF] Nersesyan | Global exact controllability in infinite time of Schrödinger equation[END_REF] is needed.

Let S be the unit sphere in L 2 . For y ∈ S, let T y be the tangent space to S at y ∈ S:

T y = {z ∈ L 2 : Re z, y = 0}.

By Lemma 2.1, for any

z 0 ∈ H 2 (V ) and u ∈ L 1 loc (R + , R), problem (3.1)-(3.3) has a unique solution z ∈ C(R + , H 2 (V ) ). Let R t (•, •) : H 2 (V ) × L 1 ([0, t], R) → H 2 (V ) , (z 0 , u) → z(t) be the resolving operator. Then R t (z 0 , u) ∈ T Ut(z,0) for any z 0 ∈ T z ∩ H 2 (V ) and t ≥ 0. Indeed, d dt Re R t , U t = Re Ṙt , U t + Re R t , Ut = Re i(∆ -V )R t -iu(t)Q(x)U t , U t + Re R t , i(∆ -V )U t = Re i(∆ -V )R t , U t + Re R t , i(∆ -V )U t = 0.
Since Re R 0 , U 0 = Re z 0 , z = 0, we get R t (z 0 , u) ∈ T Ut(z,0) . As (3.1)-(3.3) is a linear control problem, the controllability of system with z 0 = 0 is equivalent to that with any z 0 ∈ T z . Henceforth, we take z 0 = 0 in (3.3). Let us rewrite this problem in the Duhamel form

z(t) = -i t 0 S(t -s)u(s)Q(x)U s (z, 0)ds. (3.4)
Let T n →∞ be the sequence defined in Section 2.1. For any u ∈ F the following limit exists in H

R ∞ (0, u) := lim n→+∞ z(T n ) = lim n→+∞ R Tn (0, u). (3.5)
Using (2.14) and (3.4), we obtain

z(t), e m,V = -i +∞ k=1 e -iλm,V t z, e k,V Q mk t 0 e iω mk s u(s)ds, m ≥ 1, (3.6)
where ω mk = λ m,V -λ k,V and Q mk := Qe m,V , e k,V . Let us take t = T n in (3.6) and pass to the limit as n → +∞. The choice of the sequence T n implies that

R ∞ (0, u), e m,V = -i +∞ k=1 z, e k,V Q mk +∞ 0 e iω mk s u(s)ds. (3.7)
Moreover, R ∞ (0, u) ∈ T z . Indeed, using (3.5) and the convergence U Tn (z, 0)→z in H, we get Re R ∞ (0, u), z = lim n→∞ Re R Tn (0, u), U Tn (z, 0) = 0.

Lemma 3.1. The mapping R ∞ (0, •) is linear continuous from F to T z ∩ H.

Proof. By (2.24) in [START_REF] Nersesyan | Global exact controllability in infinite time of Schrödinger equation[END_REF], there is a constant C > 0 such that for any m j , k j ≥ 1, j = 1, . . . , d we have

(m 1 • . . . • m d ) 3 (k 1 • . . . • k d ) 3 Qe k1,...,k d ,V , e m1,...,m d ,V ≤ C. (3.8)
Then (3.7), (3.8) and the Schwarz inequality imply that

R ∞ (0, u) H = sup m1,...,m d ≥1 |(m 3 1 • . . . • m 3 d ) R ∞ (0, u), e m1,...,m d ,V | ≤ C sup m1,...,m d ≥1 (m 3 1 • . . . • m 3 d ) z, e m1,...,m d ,V Qe m,V , e m,V +∞ 0 u(s)ds + C z V sup m,k≥1,m =k (m 1 • . . . • m d ) 3 (k 1 • . . . • k d ) 3 Qe k1,...,k d ,V , e m1,...,m d ,V +∞ 0 e iω mk s u(s)ds ≤ C z 2 V u 2 F < +∞, where V is defined by (1.12).
Let us introduce the set E 0 := {z ∈ S : ∃p, q ≥ 1, p = q,z = c p e p,V + c q e q,V , |c p | 2 Qe p,V , e p,V -|c q | 2 Qe q,V , e q,V = 0}. Theorem 3.2. Under Condition 2.6, for any z ∈ S ∩ E \ E 0 , the mapping R ∞ (0, •) : F → T z ∩ H admits a continuous right inverse, where the space

T z ∩ H is endowed with the norm of H. If z ∈ S ∩ E 0 , then R ∞ (0, •) is not invertible.
Remark 3.3. The invertibility of the mapping R T (0, •) with finite T > 0 and z = e 1 is studied by Beauchard et al. [START_REF] Beauchard | Spectral controllability of 2D and 3D linear Schrödinger equations[END_REF]. They prove that for space dimension d ≥ 3 the mapping is not invertible. By Beauchard

Proof of Theorem 3.2. Let us take any z ∈ S ∩ E \ E 0 and y ∈ T z ∩ H. There is an integer N ≥ 1 such that z, e k,V = 0 for any k ≥ N + 1. Let us define

d mk := i y, e m,V e k,V , z -i e k,V , y z, e m,V Q mk + C mk , for k ≤ N , where C mk ∈ C. Notice that sup m,k≥1
y, e m,V e k,V , z

Q mk ≤ C y H z H < +∞.
Repeating the arguments of the proof of Theorem 2.6 in [START_REF] Nersesyan | Global exact controllability in infinite time of Schrödinger equation[END_REF], one can show that the constants C mk can be chosen such that sup m,k≥1

|d mk | < +∞, d mm = d 11 , d mk = d km for all 1 ≤ m, k ≤ N, d mk →0 as m→∞ for any fixed k ≥ 1,
and y = R ∞ (0, u) holds for any solution u ∈ F of system

d mk = ǔ(ω mk ) for all m ≥ 1 and k ∈ [1, N ].
It remains to use the following proposition, which is proved in next subsection.

Proposition 3.4. If the strictly increasing sequence ω m ∈ R, m ≥ 1 is such that ω 1 = 0 and ω m → + ∞ as m→ + ∞, then there is a linear continuous operator

A from ℓ ∞ 01 to F such that { Ǎ (d)(ω m )} = d for any d ∈ ℓ ∞ 01 .
The proof of the non-invertibility of R ∞ (0, •) is a remark by Beauchard and Coron [START_REF] Beauchard | Controllability of a quantum particle in a moving potential well[END_REF] (cf. Step 2 of the proof of Theorem 2.6 in [START_REF] Nersesyan | Global exact controllability in infinite time of Schrödinger equation[END_REF]).

Remark 3.5. The proof of Theorem 3.2 does not work in the multidimensional case for a general z / ∈ E. Indeed, assume that z, e kn,V = 0 for some sequence k n → + ∞. Then the well-known asymptotic formula for eigenvalues λ k,V ∼ C d k 2 d implies that the frequencies ω mnkn →0 for some integers m n ≥ 1 for space dimension d ≥ 3. Thus the moment problem ǔ(ω mk ) = d mk cannot be solved in the space L 1 (R + , R) for a general d mk . Clearly, this does not imply the non-controllability in infinite time of linearized system.

Proof of Proposition 3.4

The proof of Proposition 3.4 is close to that of Proposition 2.9 in [START_REF] Nersesyan | Global exact controllability in infinite time of Schrödinger equation[END_REF]. Let (2.16). Moreover, F is a Hilbert space. The construction of the operator A is based on the following lemma. Step 1. First, let us show that for any M > 0 there is

G := {u ∈ L 1 (R + , R) : u 2 (•)e B• ∈ L 1 (R + , R)} endowed with the norm u G = u 2 (•)e B• L 1 , where the constant B > 2B. Then F := G ∩ H s 0 (R + , R) is a subspace of F defined by
u 0 ∈ B F (0, M ) such that F (u 0 ) = inf u∈B F (0,M) F (u). (3.10)
To this end, let u n ∈ B F (0, M ) be an arbitrary minimizing sequence. Since F is reflexive, without loss of generality, we can assume that there is u 0 ∈ B F (0, M ) such that u n ⇀ u 0 in F . Using the compactness of the injection H 1 ([0, N ])→C([0, N ]) for any N > 0 and a diagonal extraction, we can assume that u n (t)→u 0 (t) uniformly for t ∈ [0, N ]. Again extracting a subsequence, if it is necessary, one gets {ǔ n (ω m )}→{ǔ 0 (ω m )} in ℓ ∞ as n→ + ∞. Indeed, the tails on [T, +∞), T ≫ 1 of the integrals (3.9) are small uniformly in n (this comes from the boundedness of u n in G), and on the finite interval [0, T ] the convergence is uniform. This implies that

F (u 0 ) ≤ inf u∈B F (0,M) F (u).
Since u 0 ∈ B F (0, M ), we have (3.10).

Step 2. To complete the proof, we need to show that F (u 0 ) = 0.

Lemma 3.7. Under the conditions of Proposition 3.4, the set

U := {{ǔ(ω m )} : u ∈ F } is dense in ℓ ∞ 10 .
Combining this with the Baire lemma, we get that for sufficiently large Since ω i = ω j for i = j, by Lemma 3.10 in [START_REF] Nersesyan | Growth of Sobolev norms and controllability of the Schrödinger equation[END_REF], we have h m = 0 for any m ≥ 1. This proves that U is dense.

M > 0 Ũ := {{ǔ(ω m )} : u ∈ B F (0, M )} is dense in B ℓ ∞ 10 (0, 1). Thus F (u 0 ) = 0. Proof of Lemma 3.7. It is well known that the dual of ℓ ∞ 0 is ℓ 1 . Let us suppose that h = {h m } ∈ ℓ 1 is such that h, {ǔ(ω m )} ℓ 1 ,ℓ ∞ 0 =

Application of the inverse mapping theorem

The proof is based on the inverse mapping theorem. We project the system onto the tangent space T z and apply the inverse mapping theorem to the following mapping

Ũ∞ (•) : F →T z ∩ H, u→P U ∞ (z, u),
where P is the orthogonal projection in L 2 onto T z , i.e., P z = z -Re z, z z, z ∈ L 2 . Notice that P -1 : B Tz (0, δ)→S is well defined for sufficiently small δ > 0. The following result proves that Ũ∞ is C 1 . Proposition 3.8. For a sufficiently small δ > 0 the mapping

U ∞ (z, •) : B F (0, δ)→H, u→U ∞ (z, u), is C 1 . Moreover, dU ∞ (z, u)v = R ∞ (u, v), where R ∞ (u, v) := lim n→+∞ R Tn (u, v) in H, (3.11)
and R t is the resolving operator of

i ż = -∆z + V (x)z + u(t)Q(x)z + v(t)Q(x)U t (z, u), (3.12)
z| ∂D = 0, (3.13) z(0, x) = z 0 .

(3.14)

This proposition implies that Ũ∞ ∈ C 1 (B F (0, δ)). By the definition of T n , we have lim n→+∞ U Tn (z, 0) = z. Hence U ∞ (z, 0) = z and Ũ∞ (0) = 0. We have d Ũ∞ (0)v = R ∞ (0, v), which is invertible for z / ∈ E 0 in view of Theorem 3.2. Thus applying the inverse mapping theorem, we complete the proof of Theorem 2.8 for z / ∈ E 0 . In the case z ∈ E 0 the linearized system is not controllable, and R ∞ is not invertible. Controllability near z in finite time and for d = 1 is proved by Beauchard and Coron [START_REF] Beauchard | Controllability of a quantum particle in a moving potential well[END_REF]. They show that the linearized system is controllable up to codimension one. This implies that the nonlinear system is also controllable up to codimnsion one. The controllability in the missed directions is proved using the intermediate values theorem. In the case d ≥ 1 and T = +∞, the proof repeats literally the arguments of [START_REF] Beauchard | Controllability of a quantum particle in a moving potential well[END_REF]. We omit the details.

Proof of Proposition 3.8. See [START_REF] Beauchard | Local controllability of linear and nonlinear Schrödinger equations with bilinear control[END_REF] for the proof the fact that U T (z, •) is C 1 when T is finite, d = 1 and phase space is H 3 . Let us show that U ∞ (z, •) is differentiable at any u ∈ B F (0, δ) for sufficiently small δ > 0. We need to prove that

U ∞ (z, u + v) -U ∞ (z, u) -R ∞ (u, v) H = o( v F ).
(3.15)

Notice that h = U t (z, u + v) -U t (z, u) -R t (u, v) is a solution of It remains to prove that R ∞ (u, •) is continuous in B F (0, δ). For g := R t (u 1 , v) -R t (u 2 , v) we have i ġ = -∆g + V (x)g + u 1 (t)Q(x)g + (u 1 (t) -u 2 (t))Q(x)R t (u 2 , v)

i ḣ = -∆h + V (x)h + (u(t) + v(t))Q(x)h + v(t)Q(x)R t (u,
+ v(t)Q(x)(U t (z,

  [6], R T is invertible in the case d = 1 and z = e 1 . The case d = 2 is open to our knowledge. For any u ∈ L 1 (R + , R), denote by ǔ the inverse Fourier transform of the function obtained by extending u as zero to R * -: ǔ(ω) := +∞ 0 e iωs u(s)ds.

Lemma 3 . 6 .

 36 Under the conditions of Proposition 3.4, for any d ∈ ℓ ∞ 01 there is u ∈ F such that {ǔ(ω m )} = d. Proof of Proposition 3.4. By Lemma 3.6, the mapping u→{ǔ(ω m )} is surjective linear bounded form Hilbert space F onto Banach space ℓ ∞ 01 . Hence it admits a linear bounded right inverse A : ℓ ∞ 01 → F . Proof of Lemma 3.6. Let us show that there is a constant M > 0 such that for any d ∈ ℓ ∞ 01 , d ℓ ∞ 01 ≤ 1 there is u ∈ B F (0, M ) satisfying {ǔ(ω m )} = d. Let us introduce the functional F (u) := {ǔ(ω m )} -d ℓ ∞ defined on the space F .

e

  0 for all u ∈ F . Then replacing in this equality ǔ(ω m ) by its integral representationiωms h m ds.

( v C m 0 RU+ v C m 0 U

 00 v), h| ∂D = 0, h(0, x) = 0. Using Proposition 2.4 and Lemma 2.2, we geth(∞) H ≤ C +∞ 0 ( h(τ ) H l |u(τ ) + v(τ )| + R τ (u, v) H l |v(τ )|)dτ ≤ C +∞ 0 • (u, v) W m,1 ([0,τ ],H 2 (V ) ) |u(τ ) + v(τ )|e C( u+v C m 0 +1)τ + R τ (u, v) H l |v(τ )|)dτ • (z, u) W m,1 ([0,τ ],H 2 (V ) ) |u(τ ) + v(τ )|e • (z, u) W m,1 ([0,τ ],H 2 (V ) ) |v(τ )|e C( v C m 0 +1)τ )dτ ≤ C v 2F , for any v ∈ B F (0, ε), sufficiently small ε > 0, and for sufficiently large B > 0 in the definition of G.

u 1 )

 1 -U t (z, u 2 )), g| ∂D = 0, g(0, x) = 0. By Proposition 2.4, g(∞) H ≤ C +∞ 0 ( g(τ ) H l |u 1 (τ )| + R τ (u 2 , v) H l |u 1 (τ ) -u 2 (τ )| + U τ (z, u 1 ) -U τ (z, u 2 )) H l |v(τ )|)dτ =: I 1 + I 2 + I 3 .Lemmas 2.2 and 2.3 implyI 1 ≤ C +∞ 0 ( R • (u 2 , v) W m,1 ([0,τ ],H 2 (V ) ) u 1 (τ ) -u 2 (τ ) C m 0 + U • (z, u 1 ) -U • (z, u 2 ))) W m,1 ([0,τ ],H 2 (V ) ) ||v(t) C m 0 )|u 1 (τ )|e C( u1 C m 0 +1)τ dτ ≤ C u 1 -u 2 F .The terms I 2 , I 3 are treated in a similar way. Thus we get the continuity of R ∞ (u, •).

  Theorem 2.7. Under Condition 2.6, for any z ∈ S ∩ E there is σ > 0 such that problem (1.1), (1.2) is exactly controllable in infinite time in S ∩ B H (z, σ), i.e., for any z 1 ∈ S ∩ B H (z, σ) there is a control u ∈ F such that limit (2.15) exists in H and z 1 = U ∞ (z, u).

	See Section 3.3 for the proof. Since the space H 3d (V ) is continuously embedded into H, we obtain
	Theorem 2.8. Under Condition 2.6, for any z ∈ S ∩ E there is σ > 0 such that for any z 1 ∈ S ∩ B H 3d (V )