Journal of Mathematical Analysis and Applications 235, 180᎐191 Ž1999. Article ID jmaa.1999.6390, available online at http:rrwww.idealibrary.com on
Schrodinger Operators in L2 Ž ⺢. with ¨ Pointwise Localized Potential Ronan Pouliquen Departement de Mathematiques, Uni¨ ersite´ de Bretagne Occidentale, ´ ´ 29200 Brest, France Submitted by George A. Hagedorn Received March 5, 1998
1. INTRODUCTION Schrodinger operators with Dirac’s comb type potential have been ¨ largely studied by Albeverio et al. in their monograph w1x on the subject. In particular, they construct the self-adjoint realization of the formal operator y⌬ q ⌺k j ␦ a j in L2 Ž⺢., where a j 4j g ⺪ is a real sequence satisfying infŽ a jq1 y a j . G d ) 0. A study of self-adjointness under general conditions has also been done by Mikhailets in w5x Žsee also w6x.. In these two papers, the author connects this problem to the self-adjointness of an infinite Jacobian matrix as an operator in l 2 Ž⺪.. He also relates the semi-boundedness of the operator to the boundedness of k j 4j g ⺪ and Ž a jq1 y a j .y1 4j g ⺪ . Otherwise, a self-adjoint extension in L2 Ž⺢. of the formal operator y⌬ q W q ⌺k j ␦ a j with W g L1loc Ž⺢. is also presented by Gesztesy and Kirsch in w3x, with no conditions required on the increasing sequence a j 4j g ⺪ . However, the weights k j 4j g ⺪ are determined Žsee Eq. Ž3.8. in w3x. and the authors do not present any results of lower semiboundedness in their paper. Here, we give a rigorous rewriting of the operators y⌬ q ⌺k j ␦ a j and present a direct proof of self-adjointness and lower semi-boundedness under conditions sharper than those used in w1x, for the sequence a j 4j g ⺪ , which allow a jq1 y a j to tend to zero. Our condition is of the type
lim jª⬁
< kn < j
a n jq 1 y a n j 180
0022-247Xr99 $30.00 Copyright 䊚 1999 by Academic Press All rights of reproduction in any form reserved.
- q⬁,
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where k n j 4j g ⺪ is the negative subsequence of k j 4j g ⺪ and a n j 4j g ⺪ is the corresponding subsequence of a j 4j g ⺪ . Moreover, we give, explicitly, the domain of self-adjointness. We do not need any hypothesis on the positive part of k j 4j g ⺪ other than the boundedness of the whole sequence. The conditions required on this subsequence are, in fact, part of the domain itself. Explicitly, under the above conditions, we provide a self-adjoint realization of the operator y⌬ q ⌺k j ␦ a j on
½
D s H 1 Ž ⺢. l H 2 Ž ⍀ . l u
Ý k p < uŽ k p . < 2 - ⬁ j
j
jg⺪
5
y l ur Ž ᭙ j g ⺪ . u⬘ Ž aq j . y u⬘ Ž a j . s k j u Ž a j . 4 ,
where ⍀ s D j g ⺪ x a j , a jq1w.
¨ 2. SCHRODINGER OPERATORS WITH POINTWISE LOCALIZED POTENTIAL IN ⺢ Consider two real sequences a j 4j g ⺪ and k j 4j g ⺪ . Assume that a j 4j g ⺪ is strictly increasing and satisfies lim jªy⬁ a j s y⬁, lim jªq⬁ a j s q⬁, and that k j 4j g ⺪ is bounded. We fix the following notations:
Ž ᭙ j g ⺪.
⍀ j s x a j , a jq1 w , ⍀ s
D ⍀j. jg⺪
Define the Hermitian form B in H 1 Ž⺢. for compact supported functions in x. B Ž u, ¨ . s ² u⬘, ¨ ⬘:L2 Ž⺢ . q
Ý k j uŽ aj . ¨ Ž aj . . jg⺪
In order to specify the domain of definition of B, we need the following result whose proof is in w7x: LEMMA 1. Let the sequence a j 4j g ⺪ and k j 4j g ⺪ satisfy lim jªy⬁ a j s y⬁, lim jªq⬁ a j s q⬁, k j 4j g ⺪ g l⬁Ž⺪., and limjª⬁Ž k jrŽ a jq1 y a j .. - q⬁. Then, we can find C ) 0 such that, for any ) 0, the following holds in H 1 Ž⺢.:
Ý k j < uŽ aj . < 2 F jg⺪
1
5 u⬘ 5 2L2 Ž⺢ . q C Ž 1 q . 5 u 5 2L2 Ž⺢ . .
Ž 1.
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Construction of the Operator. We complete the hypotheses on the sequences a j 4j g ⺪ and k j 4j g ⺪ : Denote by k n j 4j g ⺪ the subsequence of negative terms of k j 4j g ⺪ and assume that < k n j <4j g ⺪ and a n j 4j g ⺪ satisfy the hypotheses of Lemma 1. Hence the sum Ý j g ⺪ k n j < uŽ a n j .< 2 converges. Note that we do not formulate any hypotheses other than boundedness on the positive subsequence k p j 4j g ⺪ of k j 4j g ⺪ . The Hermitian form B is defined on
½
DomŽ B . s H 1 Ž ⺢ . l u
Ý k p < uŽ ap . < 2 - ⬁ j
j
jg⺪
5
.
We have: THEOREM 2. Let the sequences a j 4j g ⺪ and k j 4j g ⺪ satisfy lim jªy⬁ a j s y⬁, lim jªq⬁ a j s q⬁, k j 4j g ⺪ g l⬁Ž⺪., and limjª⬁ Ž< k n j
½
Ý k p < uŽ ap . < 2 - ⬁
D s H 1 Ž ⺢. l H 2 Ž ⍀ . l u
j
j
jg⺪
5
y l ur Ž ᭙ j g ⺪ . u⬘ Ž aq j . y u⬘ Ž a j . s k j u Ž a j . 4
and the operator A on D by Au s y
Ý Ej Ž R j Ž u⬙ . . , jg⺪
where the operator R j , j g ⺪, restricts a function to the open set ⍀ j , while Ej , j g ⺪, extends a function of L2 Ž ⍀ j . by 0, outside ⍀ j . Then, the operator A is lower semi-bounded, is self-adjoint with domain D, and is the unique operator satisfying DomŽ A . ; DomŽ B .
Ž ᭙ u g DomŽ A . . Ž ᭙ ¨ g DomŽ B . .
² Au, ¨ :L2 s B Ž u, ¨ . .
Proof. Equality ² Au, ¨ :L2 s B Ž u, ¨ . on D = H 2 Ž⺢., and hence the symmetry of A on D, results from an integration by parts, using a family of truncation functions in the variable x, with a uniformly bounded derivative. Let u 0 g DomŽ A*.. We prove that u 0 g DomŽ A.. First, applying <² u 0 , A¨ :< F C1 5 ¨ 5 L2
Ž 2.
L2 Ž⺢.
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to test functions ¨ g D Ž ⍀ . ; D, shows that Au 0 g L2 Ž⺢.. Therefore, Au 0 s g L2 Ž ⺢ . , which is rewritten as
Ž ᭙ ) 0 .
Au 0 q 2 u 0 s g L2 Ž ⺢ . .
Ž 3.
The parameter will be given a fixed value later on. For any x in ⍀ j , the square integrable solutions of Eq. Ž3. are given by u0 Ž x . s
1
x
H 2 a
ey Ž syx . Ž s . ds y
jq1
1
x
y Ž xys.
Ž s . ds
He 2 a j
y Ž xya j .
q ␣ , j e
y Ž a jq 1 yx .
q  , j e
Ž 4.
.
Denote by u 0,c j the integral part Ž‘‘c’’ means ‘‘convolution’’. and by u 0,h j the rest of the right member of Ž4., which is the solution of the homogeneous equation associated with Ž3.. These are the restrictions on each ⍀ j of two functions denoted respectively by u 0c and u 0h. We are led to prove that u 0c and u 0h belong to H 2 Ž ⍀ .. Using Young’s inequality 5 f ) g 5 L2 F 5 f 5 L1 5 g 5 L2 we prove that for any j in ⺪, 5 u 0c , j 5 L2 Ž ⍀ j . F c du 0, j
dx
L ⍀ j. 2Ž
c d 2 u 0, j
dx 2
L2 Ž ⍀ j .
F
1
2 1
5 Rj Ž . 5 L2 Ž ⍀ j . ,
5 Rj Ž . 5 L2 Ž ⍀ j . ,
Ž 5.
F 2 5 Rj Ž . 5 L2 Ž ⍀ j . .
Ž 6.
On the other hand, we can also write h 5 2 5 u 0, j L Ž⍀ j. h du 0, j
dx
L2 Ž ⍀ j .
h d 2 u 0, j
dx 2
L2 Ž ⍀ j .
F
Ž ␣2, j q 2, j .
1r2
1r2
F Ž ␣2, j q 2, j . F 3 Ž ␣2, j q 2, j .
, 1r2
,
1r2
Ž 7. .
Ž 8.
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Therefore, we have u 0c Ž x . g H 2 Ž ⍀ . with 5 u 0c Ž x .5 H 2 Ž ⍀ . F C2 5 5 L2 Ž⺢ . and for any in D Ž⺢. Ž Ž x . ⭈ u 0h Ž x .. < ⍀ g H 2 Ž ⍀ .. However, as we do not know whether Ý j g ⺪ Ž a2, j q 2, j . converges or not, it is not possible to establish any global property for u 0h Ž x . Ži.e., a property on the whole ⺢.. The following boundary conditions hold for any j in ⺪:
⭸ u0
y u 0 Ž aq j . s u0 Ž a j . ;
⭸x
Ž aqj . y
⭸ u0 ⭸x
Ž ayj . s k j u 0 Ž a j . .
Let j g D be such that j Ž a j . s 0 and suppŽ j . ;x a jy1 , a jq1 w. Integrating ² u 0 , A j :L2 by parts in Ž2., we can write y < jX Ž a j . Ž u 0 Ž aq < 5 5 2 j . y u 0 Ž a j . . F C 3, j j L ,
Ž 9.
with C3 , j s 5 uY0 < xa jy 1 , a jq 1w 5 L2 Žx a jy 1 , a jq 1w. q C1 - q⬁. Let g D Ž⺢. be such that Ž0. s 0, ⬘Ž0. s 1. For any j g ⺪, define j, in ⺢ by
Ž ᭙ gx 0, ␦ j w . Ž ᭙ x g ⺢.
j, Ž x . s
ž
x y aj
/
.
Taking the constant ␦ j - 1 ensures that suppŽ j, . ;x a jy1 , a jq1 w, so that j, belongs to D. Using the inequality Ž9. with the family j, , we obtain y < 3r2 < u 0 Ž aq C3, j 5 5 L2 . j . y u0 Ž a j . F
Finally, letting tend to 0 proves that u 0 is continuous at the point a j . We shall now prove the boundary conditions X y uX0 Ž aq j . y u0 Ž a j . s k j u0 Ž a j . .
Ž ᭙ j g ⺪.
Let j g ⺪ and j g D be such that j Ž a j . / 0 and suppŽ j . ;x a jy1 , a jq1w. Proceeding as for the inequality Ž9. and using the continuity of u 0 , we obtain X y < j Ž a j . ⭈ Ž uX0 Ž aq < 5 5 2 j . y u 0 Ž a j . y k j u 0 Ž a j . . F C 3, j j L .
Ž 10 .
Let g D Ž⺢., Ž x . s 1, in a neighbourhood of the origin. Defining the family j, in ⺢ by
j, Ž x . s
ž
x y aj
/Ž
1 q k j ⭈ Ž x y aj . ⭈ H Ž x y aj . . ,
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the conditions on the parameter are the same as before to ensure that each j, belongs to D. Applying inequality Ž10. to this family of functions, we obtain X y 1r2 5 5 2 < uX0 Ž aq < L Ž⺢ . , j . y u 0 Ž a j . y k j u 0 Ž a j . F C 3, j ⭈ K j ⭈
where, for any j in ⺪, K j s 1 q < k j < Ž a jq1 y a j . . As tends to 0, we reach X y uX0 Ž aq j . y u0 Ž a j . s k j u0 Ž a j . .
We now show that ␣, j and , j satisfy
Ý ␣2, j q 2, j - ⬁. jg⺪
Replacing u 0 by its restrictions in the preceding boundary conditions, we obtain y u 0, j Ž aq j . s u 0, jy1 Ž a j .
du 0, j dx
du 0, jy1
Ž aqj . y
dx
Ž ayj . s k j Pj u 0, j
¦ ¥ Ž a . .§
Ž 11 .
j
Denote e , j s ey Ž a jq 1ya j . A , j s B , j s
aj
Ha
ey Ž a jys. jy1 Ž s . ds
jy1
a jq1
Ha
ey Ž sya j . j Ž s . ds.
j
The equations Ž11. can be rewritten as
M ⭈
⭈⭈⭈  , j
1
⭈⭈⭈ Ž k jr2 . B , j y A , j
2
Ž k jq1r2 . A , jq1 y B , jq1
0 ␣ , j ⭈⭈⭈
s
⭈⭈⭈
0
,
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where M is the following matrix: ⭈⭈⭈ M s
ž
1q
O kj
2
kj
/
2
ye , j
e , jy1
kj
ye , jy1
2
ž
e , j
1q
2 ⭈⭈⭈
O
We can write M s R q
1 2
kj
/
0
.
S , with R and S defined by
⭈⭈⭈ R s
O
0
ye , jy1
0
0
0 1
0 ye , jq1
1 0
0 0
0
O ⭈⭈⭈
⭈⭈⭈ S s
k j e , j
kj
k jq1
k jq1 e , j ⭈⭈⭈
0
.
The matrix R is explicitly invertible: ⭈⭈⭈ Ry1 s
0 e , jy1
O
0 0
0 1
1 0
0
e , jq1
0
0
O ⭈⭈⭈
0
.
Recall that if we denote by 5 5 L p the subordinate matrix norms to the vector norms 5 5 p , we have 5 A 5 L ⬁ s sup
Ý < ai j < ,
5 A 5 L 1 s sup
ig⺪ jg⺪
Ý < ai j < ,
and
jg⺪ ig⺪
5 A5 L 2 F 5 A5 L 1 5 A5 L ⬁ .
'
5 L 2 F 2, and 5 S 5 L 2 F 2 5 k j 5 l ⬁ . We can easily prove that 5 R 5 L 2 F 2, 5 Ry1 ⬁ 5 5 If ) 2 k j l , then M is invertible and we have 5 My1 5 L 2 F 5 Ry1 5 L2
1 1y
1 2
5 Ry1 5 L 2 5 S 5 L 2
F
2
y 25 kj5 l ⬁
.
L2 Ž⺢.
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Denoting the right member of this inequality by cŽ . and fixing a suitable value 0 ) 2 k 5 k j 5 l ⬁ for , we obtain ⭈⭈⭈  0 , j
0 ␣0 , j ⭈⭈⭈
⭈⭈⭈
c Ž 0 .
F
ž / 2 0
2
Ž k jr2 0 . B , j y A , j Ž k jq1r2 0 . A , jq1 y B , jq1 0
0
0
0
⭈⭈⭈
0
.
Ž 12 .
2
For any j in ⺪, simple computations yield 1
A0 , j F
ž
'2 Ha 0
1r2
aj
20 , jy1 Ž s . ds
/
jy1
,
and the corresponding inequality for B0 , j . From Ž12., we deduce that
Ý
␣20 , j
q
20 , j
F
jg⺪
F
c 2 Ž 0 .
Ž 2 0 .
ž ž /ž ž / Ý
3
Ž 2 0 .
jg⺪
5 k j 5 2l ⬁
c 2 Ž 0 .
Ž 2 0 .
< kj<2
3
Ž 2 0 .
2
q1
aj
Ha
jy1
q 1 5 0 5 2L2 Ž⺢ . ,
2
20 , jy1 Ž s . ds
// Ž 13 .
which gives the desired result. The function u 0 satisfies d 2 u 0rdx 2 < ⍀ g L2 Ž ⍀ ., du 0rdx < ⍀ g L2 Ž⺢.. The first property is a consequence of inequalities Ž5., Ž7., and Ž13.. For the second one, we first use Ž6., Ž8., and Ž13. in order to obtain du 0 dx
< ⍀ g L2 Ž ⍀ . .
Then for any in D Ž⺢ n ., we have du 0
¦ ; dx
,
D⬘ D
¦
s y u0 ,
d dx
;
.
L2
Integrating ² u 0 , ddx :L2 by parts on each I j and using the continuity of u 0 at the a j ’s, we obtain du 0
du 0
¦ ; ¦ dx
,
D⬘ D
s
dx
<⍀, <⍀
;
L2
F 5 u 0 < ⍀ 5 H 1 Ž ⍀ . 5 5 L2 . The conclusion follows. Hence, u 0 g H 1 Ž⺢..
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Let us prove that the sum Ý j g ⺪ k p j < u 0 Ž a p j .< 2 is finite. Let Ž᭙ l g ⺞. l g D Žx y l y 1, l q 1w. be such that 0 F l Ž x . F 1, in suppŽ l ., l < xy l , l ws 1 and Xl being uniformly bounded in l . We note u 0, l Ž x . s u 0 Ž x . l Ž x .. An integration by parts yields, for any l in ⺞, ² Au 0 , u 0, l :L2 Ž⺢ . s
du 0 du 0, , dx dx
¦
l
;
q
L2 Ž ⺢ .
Ý k j u 0 Ž a j . u 0, l Ž a j . . jg⺪
Lemma 1 gives
Ý < k n < < u 0 Ž an . < 2 F C4 5 u 0 5 2H Ž⺢ . . 1
j
j
jg⺪
Therefore,
Ý
k p j l Ž a p j . < u 0 Ž a p j . < 2 F C 5 u 0 5 2H 1 Ž⺢ . q ² Au 0 , u 0 , l :L2 Ž⺢ .
/
jg⺪
du 0 du 0, , dx dx
¦
y
l
;
.
L2 Ž ⺢ .
We conclude the proof using the monotone convergence theorem for the discrete sum and Lebesgue’s theorem for the terms in between brackets. Finally, we have shown that u 0 g D. Thus Ž A, D, L2 Ž⺢.. is a self-adjoint operator. Its uniqueness follows from a classical result about self-adjoint operators associated with quadratic forms Žsee w8, Theorem VIII.15x.. From inequality Ž1., we can find a constant C ) 0 such that, in DomŽ B ., the following holds:
Ý < k n < < u Ž an . < 2 F 5 ⵜu 5 2L
2
j
j
q 2C 5 u 5 2L2 .
Ž 14 .
jg⺪
Hence B is lower semi-bounded by y2C in DomŽ B .. There remains to prove that ŽDomŽ B ., 5 ⭈ 5 B . is closed Žwith 5 ⭈ 5 B s Ž B Ž⭈, ⭈ . q Ž2C q 1.5 ⭈ 5 2L2 .1r2 .. The norms 5 ⭈ 5 B and 5 ⭈ 5 H 1 Ž⺢ n . are not equivalent, as no hypothesis was made on the positive subsequence k p j 4j g ⺪ of k j 4j g ⺪ , with respect to 1 k jrŽ a jq1 y a j .. However, using inequality Ž1. with s 1 q 2C we prove that 5 ⭈ 5 2B G min
ž
1
,
4 2
2 2C q 1
/
5 ⭈ 5 2H 1 Ž⺢ . .
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Let u m 4m g ⺞ be a Cauchy sequence in DomŽ B . for 5 5 B . It converges to a function u in Ž H 1 Ž⺢., 5 5 H 1 . and we have, for any j in ⺪, lim < u m Ž a j . y u Ž a j . < s 0.
Ž 15 .
mªq⬁
Besides, using Ž14., we have
Ý < k n < < u Ž an . < 2 - q⬁. j
j
jg⺪
On the other hand, the sequence Žindexed by m. k p j < u mŽ a p j .< 2 4j g ⺪ is a Cauchy sequence in l 1 Ž⺪., since
Ý k p < um Ž ap . < 2 F 5 um 5 B . j
j
jg⺪
From Eq. Ž15. we deduce that the limit sequence is, in fact, equal to k p < uŽ a p .< 2 4j g ⺪ and that the latter belongs to l 1 Ž⺪.. Consequently, a j j Cauchy sequence in ŽDomŽ B ., 5 5 B . is convergent in this set, which concludes the proof. Remarks. This method applies in the same way to a finite sequence of points a j 4j g ⺪ . In this case the only requirement is that the sequence of weights k j 4j g ⺪ should be bounded. A similar result can be constructed in ⺢ q with a potential localized on parallel hyperplanes. Its proof follows the same principle modulo a few technical arrangements: The adequate operator is A q u s yÝ j g ⺪ EjŽ RjŽ ⌬ u.., where Rj and Ej relatively are the restriction and extension operators to the open set ⍀ q, j s ⺢ qy 1 = x a j , a jq1w. Denote ⍀ q s Dj g ⺪ ⍀ q, j . The operator A q is to be studied in the set
½
D s H 1Ž⺢q . l H 2 Ž ⍀q . l u
½
l u Ž ᭙ j g ⺪.
⭸u ⭸ Njq
q
Ý k p 5 p u 5 2L Ž⺢ 2
j
j
qy 1
.
-⬁
jg⺪
⭸u ⭸ Njy
s yk j j u
5
5
with j the trace operator on the hyperplane x q s a j 4 and the normal derivatives
⭸u ⭸ Njq
s y j
ž
⭸u ⭸ xq
<⍀j ,
/
⭸u ⭸ Njy
s j
ž
⭸u ⭸ xq
< ⍀ jy1 .
/
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A partial Fourier transformation in the first q y 1 variables, denoted by ‘‘n’’, is applied to the equation A q u 0 s g L2 Ž⺢ q . in order to obtain the equivalent of Eq. Ž3. in ⺢ q : 2
2
Au ˆ0 q Ž 2 . ⬘ 4 uˆ0 s ˆ g L2 Ž ⺢ q . , with
Ž᭙u g D.
Asy
Ý Ej jg⺪
⭸ 2u
ž ž // Rj
⭸
⬘ 4 s Ž 2 q < < 2 .
and
x q2
1r2
.
The solution of this equation is as follows: u ˆ0 Ž ⬘, x q . s
1
xq
H 4 ² ⬘: a y
1
ey2 ² ⬘:Ž syx q. ˆ Ž ⬘, s . ds
jq1
xq
H 4 ² ⬘: a
ey2 ² ⬘:Ž x qys. ˆ Ž ⬘, s . ds
j
q ␣ j Ž ⬘ . ey2 ² ⬘:Ž x qya j. q  j Ž ⬘ . ey2 ² ⬘:Ž a jq1yx q. . Fix O , a bounded open set in ⺢ qy 1, and , a function in D Ž O .. The function of one variable ² u ˆ0 , :L2 Ž⺢ qy 1 . has the same regularity and satisfies the same trace equalities as in the monodimensional case. Hence, we prove local regularity for u 0 . Global regularity is obtained by showing that Ý j g ⺪ ␣ j Ž ⬘. 2 q  j Ž ⬘. 2 satisfies some integration properties. The method is the same as in the monodimensional case Žinversion of an infinite system in l 2 Ž⺪.. and leads to the inequality
Ý ␣ j Ž ⬘. jg⺪
2
2
q j Ž ⬘. F
c 2 Ž .
Ž 4 ⬘ 4 .
3
ž
5 k j 5 2l ⬁
Ž 4 .
2
q 1 5 Ž ⬘, ⭈ . 5 2L2 Ž⺢ . ,
/
with c Ž . s
2
y 5 k j 5 l ⬁
.
There remains to prove the uniqueness of A q with Theorem VIII.15 of w8x, using a q-dimensional version of inequality Ž1. in order to conclude.
REFERENCES 1. S. Albeverio, F. Gesztesy, R. Hoegh-Krohn, and H. Holden, ‘‘Solvable Models in Quantum Mechanics,’’ Springer-Verlag, Berlin, 1988.
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L2 Ž⺢.
191
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