Pick Functions Related to the Multiple Gamma Functions of order $n$

0.19 MB·

by  Sourav Das

PICK FUNCTIONS RELATED TO THE MULTIPLE GAMMA FUNCTIONS
OF ORDER n
SOURAVDASAND A.SWAMINATHAN
Abstract. Let Gn betheBarnesmultipleGammafunction ofordernandthefunctionfn(z)
bedefined as
fn(z)= logzGnnL(ozg+z1), z∈C\(−∞,0].
6
1 In this work, a conjecture to find the Stieltjes representation is proposed such that fn(z) is a
0 Pick function. The conjecture is established for the particular case n = 3 by examining the
2
properties of f3(z).
n
a
J

3
1 1. Introduction

] The class of functions g(z) = u(z) + iv(z) analytic in the upper-half plane with positive
A
C imaginary part are called Pick class [14, p. 18] and denoted by P. A Pick function f(z) P
∈
. can be extended by reflection to holomorphic functions in C R and they have the integral
h \
t representation as following [10, p. 508]
a
m ∞
1 t
[ f(z) = az+b+ dµ(t), (1.1)
−∞ t z − t2+1
Z (cid:18) − (cid:19)
1
v where a 0, b R and µ is a nonnegative Borel measure in R satisfying
≥ ∈
7
∞
6 dµ(t)
< ,
1 −∞ t2+1 ∞
3 Z
0 with
.
1
f(iy)
0 a = lim , b = f(i), (1.2)
6 y→∞ (iy) ℜ
1
f(t+iy)dt
: µ = lim ℑ . (1.3)
v
y→0+ π
i
X
It is well known that the limit for µ is in vague topology [14, p. 32] (see also [10, p. 508], [20, p.
r
a 96]). The reciprocal of a Stieltjes function belongs to P is observed in [10, p. 507]. For more
information on Pick functions one can see [8–11,14,18,19].
One of the recent interest, among researchers, is to find the members of the class of Pick
functions. For example, the second author posed the problems of finding Pick functions related
to generalized polylogarithm and generalized hypergeometric functions in [6, p.91]. Functions
that are related to the logarithm of Gamma functions are of special interest.
In 1899, E.W. Barnes [4] introduced multiple Gamma functions, denoted by Γ (z) or G (z) ,
n n
using multiple Hurwitz zeta functions. The Gamma function of order n, denoted by Γ can be
n
2010 Mathematics Subject Classification. 33B15, 41A60.
Key words and phrases. Multiple Gamma function, Pick function, Stieltjes integral representation, logarithm
function.
1
2 SouravDasandA.Swaminathan
defined by the recurrence relation
Γ (z)
n+1
Γ (z+1) = ; Γ (z) = Γ(z); Γ (1) = 1.
n+1 1 n
Γ (z)
n
Here Γ(z) denotes the usual Gamma function, studied by Euler, which is defined by
∞
Γ(z) := e−ttz−1dt, Rez > 0.
Z0
We remark that equivalent forms of Γ(z) exist in the literature, (see [12] for details).
By defining
Γ (z) = (G (z))(−1)n−1, n N.
n n
∈
it is easy to see that
1
G (z) = Γ (z) = Γ(z), G (z) = , G (z) = Γ (z).
1 1 2 3 3
Γ (z)
2
These G (z)’s are known as multiple Gamma functions of order n and studied first in detail by
n
1
Vign´eras [22] in 1978 (see also [4,5]). Note that, in [5] Barnes had shown that G (z) = =
2
Γ (z)
2
G(z), where G(z) is the Barnes G- function.
Clearly, these G (z)’s satisfy the recurrence relation
n
G (z+1) = G (z)G (z), G (1) = 1, z C, n N. (1.4)
n+1 n+1 n n
∈ ∈
Further information about G (z), required for the results given in this paper, are provided in
n
Section 2. For various applications of G (z) we refer to [12] and references therein.
n
Consider the function
logG (z+1)
f (z) = n , z C ( ,0], n N.
n znLogz ∈ \ −∞ ∈
Berg and Pedersen already proved that f (z) P and found its Stieltjes representation in
1
∈
[9,10]. In [18, p. 364] Pedersen found the Stieltjes representation of f (z) and proved that it
2
belongs to P.
Motivated by the earlier results on f (z) and f (z), we are interested in finding the Stieltjes
1 2
representation for f (z) so that it belongs to the class of Pick functions. The problem we are
n
interested in is given below as a conjecture as it is challenging to establish this as a result.
Conjecture 1. The function
logG (z+1)
f (z) = n , z C ( ,0], n N,
n znLogz ∈ \ −∞ ∈
is a Pick function with Stieltjes representation
logG (z+1) 1 ∞ d˜ ( t)
n n
= − dt.
znLogz n! − t+z
Z0
where
( 1)n−1N (t )log t +log G (t+1)
n n
− | | | | | |, t < 0;
d˜ (t) = − tn((log t )2+π2) ,
n
 | |
0, t 0.

≥
[t]([t]+1)([t]+2) ([t]+n 1)
N (t) =  ··· − for t > 0. (1.5)
n
n!
PickfunctionsrelatedtoMultipleGammafunctionsofordern 3
As mentioned earlier, this result is established for n = 1 in [10](see also [9] and for n = 2
in [18]. In Section 3, we establish theresult for n = 3. Even though it is difficult to establish the
above conjecture for general n, numerical and graphical evidences suggest that the conjecture
is true. We provide below some results to support our claim. The first one we have in this line
is due to Theorem 3 of [21, p. 436] which we state as a lemma.
Lemma 1.1.
G (x+1) 1
n
lim =
n→∞ xnlogx n!
Thevalue of logG (z) at a regular point, in the upperhalf plane, is provided by the following
n
result.
Proposition 1. We have for any k,n N,
∈
(k)
lim logG (z) = log G (t) +i( 1)nπ n
n n
| | − n!
z t
→
z > 0
ℑ
for t ( k, k+1), where (k) = k(k+1)(k+2) (k+n 1) is the Pochhammer symbol.
n
∈ − − ··· −
The above result, for the case n = 1 is given in [9]. We state this case explicitly as it is used
in further discussion.
Lemma 1.2. [9, Lemma 2.1] We have for any k 1,
≥
lim logΓ(z) = log Γ(t) iπk
| |−
z t
→
z > 0
ℑ
for t ( k, k+1) and
∈ − −
lim logΓ(z) =
| | ∞
z t
→
z > 0
ℑ
for t = 0, 1, 2,....
− −
The following result would be useful in establishing the Stieltjes representation for f (z).
n
Proposition 2. For any n,k N we have
∈
(−1)n
(k 1) log t log G (t+1)
lim f (z) = π n! − n | |− | n |
ℑ n tn((log t )2 +π2)
z t | |
→
z > 0
ℑ
for t ( k, k+1).
∈ − −
The paper is organized as follows. Proofs of Proposition 1 and Proposition 2 and further
details on G (z) are given in Section 2. In Section 3 results for the case n = 3 related to
n
Conjecture 1 are provided. Concluding remarks with related problems for further research are
provided in Section 4.
4 SouravDasandA.Swaminathan
2. Properties of Multiple Gamma function
A result on obtaining the unique meromorphic function G (z), by the analogy of Bohr-
n
Mollerup theorem, given in [12, Theorem 1.2, p.100]), is as follows.
Lemma 2.1. [12] For all n N, there exists a unique meromorphic function G (z) satisfying
n
∈
each of the following properties
1. Gn(z+1) = Gn−1(z)Gn(z), z C.
∈
2. G (1) = 1 and G (x) = x.
n 0
3. For x 1, G (x) are infinitely differentiable and
n
≥
dn+1
logG (x) 0.
dxn+1 n ≥
(cid:16) (cid:17)
By analyzing Lemma 2.1, it is easy to see that (Γ (z))−1 is an entire function with zeros at
n
z = k, k N 0 with multiplicities given by
− ∈ ∪{ }
n+k 1
− n N, k N 0 . (2.1)
n 1 ! ∈ ∈ ∪{ }
−
Using (2.1), the following explicit form of the multiple Gamma function of order n, in terms
of Weierstrass canonical product of Γ , is given in [12].
n
 n+k 2 
∞ z  n −1  n+k 2 n ( 1)j−1 zj
Γn(1+z)=exp[Qn(z)]kY=1(cid:16)1+ k(cid:17) − exp n−−1 !Xj=1 − j kj,
 
   
 (2.2)
where Q (z) is a polynomial of degree n given by
n
n−1
p (z)
Q (z) :=( 1)n−1 zA (1)+ k f(k) (1) ,
n − − n k! n−1
" #
Xk=1 (cid:16) (cid:17)
n−1
p (z)
k (k)
f (z) := zA (1)+ f (1) +A (z),
n − n k! n−1 n
Xk=1 (cid:16) (cid:17)
∞ n+k 2 z n ( 1)j−1 zj
A (z) := ( 1)n−1 − log 1+ + −
n − n 1 !− k j kj
Xk=1 − (cid:16) (cid:17) Xj=1
 
and
n+1
1 n+1
pn(z) = Bn+1−kzk, n N
n+1 k ! ∈
k=1
X
with B being the Bernoulli numbers.
k
To prove Proposition 1, we need a behaviour of logG which is given in the following result.
2
Lemma 2.2. We have for any k 1,
≥
k(k+1)
lim logG (z) = log G (t) +iπ
2 2
| | 2
z t
→
z > 0
ℑ
PickfunctionsrelatedtoMultipleGammafunctionsofordern 5
for t ( k, k+1) and
∈ − −
lim logG (z) =
2
| | ∞
z t
→
z > 0
ℑ
for t = 0, 1, 2,... .
− −
Proof. Using recurrence relation (1.4) and Lemma 1.2 we have
k−1
logG2(z)=logG2(z+k) logG1(z+l)
−
l=0
X
k−1
= lim logG2(z)=logG2(t+k) (log G1(t+l) i(k l)π)
⇒ − | |− −
z t l=0
→ X
z >0
ℑ
G2(t+k)
=log +iπ(k+(k 1)+ +2+1)
G1(t)G1(t+1) G1(t+k 1) − ···
(cid:12) ··· − (cid:12)
(cid:12) (cid:12)
(cid:12) k(k+1) (cid:12)
=log(cid:12)G2(t) +iπ (cid:12)
| | 2
The other part follows from the fact that logG (z) log G (z). (cid:3)
2 2
| | ≥ | |
Proof of Proposition 1. We have already shown that the statement is true for the case n = 1
as given in Lemma 1.2. The case n = 2 is established in Lemma 2.2. Lemma 3.3 in Section 3
reflects the case n = 3.
Let the statement be true for n = m N. Then
∈
(k)
lim logG (z) = log G (t) +i( 1)mπ m. (2.3)
m m
| | − m!
z t
→
z >0
ℑ
Now with the help of (2.3) we have
k−1
logG (z) = logG (z)+ logG (z+l)
m+1 m+1 m
l=0
X
= lim logG (z)
m+1
⇒
z t
→
z > 0
ℑ
k−1
(k l)
= logG (t+k) log G (t+l) +i( 1)mπ − m
m+1 m
− | | − m!
l=0(cid:18) (cid:19)
X
k−1
G (t+k) (k l)
= log m+1 +i( 1)m+1π − m
G (t)G (t+1) G (t+k 1) − m!
(cid:12) m m ··· m − (cid:12) l=0
(cid:12) (cid:12) X
(cid:12) (cid:12)
Again with the help(cid:12)of southeast diagonal sum proper(cid:12)ty (see [15, eq. (4.1.6a)] for details) and
a simple computation we obtain
k−1 k
(k l) l+m 1 m+k (k)
m m+1
− = − = = .
m! m ! m+1 ! (m+1)!
l=0 l=1
X X
6 SouravDasandA.Swaminathan
Therefore,
(k)
lim logG (z) = log G (t) +i( 1)m+1π m+1
m+1 m+1
| | − (m+1)!
z t
→
z > 0
ℑ
which shows that the statement is true for n =m+1 when it is true for n = m.
Hence by the principle of mathematical induction the statement of the proposition is true for
all n N. (cid:3)
∈
Note. From Proposition 1, we recall the remark given in [9, p.223] that the holomorphic branch
of logH is real on the positive half line, whenever the meromorphic function H in C with zeros
and poles on ( ,0] is real and positive on the positive half line. Further, the limit of logH
−∞
at a regular point is iπ multiplied by the number of zeros minus the number of poles in (t,0]
counted according to the multiplicity.
logG (z+1)
n
Proof of Proposition 2. Given that f (z) = Hence, for t ( k, k +1), k 1,
n znLogz ∈ − − ≥
we have
( 1)n
log G (t+1) +iπ − (k 1)
n n
| | n! −
lim f (z) = ,
n tn(log t +iπ)
z t | |
→
z > 0
ℑ
(−1)n
(k 1) log t log G (t+1)
= π n! − n | |− | n |,
tn((log t )2+π2)
| |
which completes the proof. (cid:3)
Note that for establishing Conjecture 1 for f (z), we need a result similar to Lemma 3.4 and
n
a result for growth at infinity. We are unable to establish the same at this stage. Hence in the
next section, we provide the results for the case n = 3.
3. The triple Gamma function related to Pick functions
In this section we provide the Stieltjes representation for the function
logΓ (z+1)
f (z) = 3 , z C ( ,0]
3 z3Logz ∈ \ −∞
so that f (z) belongs to the class of Pick functions.
3
Note that, the triple Gamma function Γ (z) can be defined by the Weierstrass canonical
3
product form which will be a particular case of (2.2) for n = 3. Taking logarithm on both sides
of that particular case, for z C ( ,0], we have
∈ \ −∞
logΓ (1+z) = Dz3+Ez2+Fz+logP (z)
3 Γ3
where
∞
k(k+1) z k+1 1 1 1 1
logP (z) = Log 1+ + z 1+ z2+ 1+ z3 ,
Γ3 − 2 k 2 − 4 k 6k k
Xk=1(cid:18) (cid:16) (cid:17) (cid:18) (cid:19) (cid:18) (cid:19) (cid:19)
(3.1)
PickfunctionsrelatedtoMultipleGammafunctionsofordern 7
with
1 π2 3 1 1 3 log(2π)
D = γ+ + , E = γ+log(2π)+ and F = logA (3.2)
−6 6 2 4 2 8 − 4 −
(cid:18) (cid:19) (cid:18) (cid:19)
With the help of maximum principle we will prove that f has nonnegative imaginary part
3
in the upper half plane. For this we require to analyze the behaviour on the real line as well as
the growth at infinity. To prove our required results we need the following lemmas.
Lemma 3.1. There exist a constant and a sequence r tending to infinity such that
n
{ }
logP (z) Const z 3log z
| Γ3 | ≤ | | | |
holds for all z C ( ,0] of absolute value r .
n
∈ \ −∞
Proof. Expression (3.1) can also be represented in the following canonical product form
k(k+1)
∞
z − k(k+1)z k(k+1) z2 k(k+1) z3
P (z) = 1+ 2 exp +
Γ3 k 2 k − 2 2k2 2 3k3
kY=1(cid:16) (cid:17) (cid:18) (cid:19)
k(k+1)
∞
1 z k(k+1) z z2 z3
= 1+ 2 exp + (3.3)
⇒ P (z) k 2 −k 2k − 3k
Γ3 kY=1(cid:16) (cid:17) (cid:26) (cid:18) (cid:19)(cid:27)
1
which implies is of genus 3.
P (z)
Γ3
1 k(k+1)
Since has zeros at k of multiplicity for k 1 we obtain the zero counting
P (z) − 2 ≥
Γ3
1
function n N (t) associated with the zeros of as
3
≡ P (z)
Γ3
[t]([t]+1)([t]+2)
N (t) = , t > 0,
3
6
where [t] denotes the integer part of t.
Hence using Appendix A and Proposition A.2 in [18, p. 368] we have
1
log Constant z 3log z
P (z) ≤ | | | |
(cid:12) Γ3 (cid:12)
=⇒(cid:12)(cid:12)(cid:12)|logPΓ3(z)(cid:12)(cid:12)(cid:12)| ≤ Constant|z|3log|z| (cid:3)
Lemma 3.2.
logΓ (x+1) 1
3
lim = .
x→∞ x3logx 6
Proof. The Stirling formula for triple Gamma function [21, p. 437] is given by
x3 x2 1 11 5 x 13
logΓ (x+1) + log(x+1) x3+ x2+
3
∼ 6 − 4 24 − 36 24 3 − 72
(cid:18) (cid:19)
x2 x 2x 1 1
′ ′ ′
− ζ (0)+ − ζ ( 1) ζ ( 2)
− 2 2 − − 2 −
∞
1 1
+ + x2 (6r 11)x+(4r2 16r+16)
12x+1 { − − − }
r=2
X
Now dividing both sides by x3logx and taking x we have the required result. (cid:3)
→ ∞
8 SouravDasandA.Swaminathan
Lemma 3.3. We have for any k 1,
≥
k(k+1)(k+2)
lim logG (z) = log G (t) iπ
3 3
| |− 6
z t
→
z > 0
ℑ
for t ( k, k+1) and
∈ − −
lim logG (z) =
3
| | ∞
z t
→
z > 0
ℑ
for t = 0, 1, 2,... .
− −
Proof. Using recurrence relation (1.4) and Lemma 2.2 we have
k−1
logG (z) = logG (z+k) logG (z+l)
3 3 2
−
l=0
X
k−1
(k l)(k l+1)
= lim logG (z) = logG (t+k) log G (t+l) +iπ − −
3 3 2
⇒ − | | 2
z t l=0(cid:18) (cid:19)
→ X
z >0
ℑ
k
G (t+k) iπ
3
= log n(n+1)
G (t)G (t+1) G (t+k 1) − 2
(cid:12) 2 2 ··· 2 − (cid:12) n=1
(cid:12) (cid:12) X
(cid:12) k(k+1)(k+2) (cid:12)
= log(cid:12)G (t) iπ (cid:12)
3
| |− 6
The other part follows from the fact that logG (z) log G (z). (cid:3)
3 3
| | ≥ | |
Lemma 3.4. Let k 1 and d˜: R R be such that
≥ → ∪{∞}
k(k 1)(k+1)
( 1)3 − log t log G (t+1)
3
− 3! | |− | |
, for t ( k, k+1);
d˜ (t) = t3((log t )2+π2) ∈ − −
3  | |
 0, for t 0;
 ≥
, for t = k.
 ∞ −


Then d˜ (t) 0.
3
≥
k(k 1)(k+1)
Proof. Itisenoughtoshowthat, fort ( k, k+1), log G (t+1) + − log t 0,
3
∈ − − | | 6 | | ≥
which we obtain by induction hypothesis.
First we show that the statement is true for k = 1. i.e;
g(t) = log G (t+1) 0 for t ( 1,0).
3
| | ≥ ∈ −
Clearly, g(t) is decreasing for 1< t < 0, as
′ −
g (t)
t2 π2 3 t 1 3 log2π t3 ∞ k+2
= γ+ + + γ+log2π+ + logA +
−2 6 2 2 2 8 − 4 − 2 (k+1)2(k+t+1)
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) k=0
X
is negative for 1 < t < 0. Hence g(t) g(0) = 0, as G (1) = 1.
3
− ≥
PickfunctionsrelatedtoMultipleGammafunctionsofordern 9
Now we will show that the statement is true for k = 2. It is easy to see that
1 1
log G (t+1) = D+ t3+ E t2+(F +1)t Log 1+t
3
| | 3 − 2 − | |
(cid:18) (cid:19) (cid:18) (cid:19)
∞
k(k+1) t t 1 t2 1 t3
+ Log 1+ + ,
− 2 k − k 2k2 − 3k3
k=2(cid:18) (cid:18) (cid:12) (cid:12) (cid:19)(cid:19)
X (cid:12) (cid:12)
(cid:12) (cid:12)
where D, E and F are given, as in (3.2). Clearly
(cid:12) (cid:12)
∞
k(k+1) t t 1 t2 1 t3
Log 1+ +
− 2 k − k 2k2 − 3k3
k=2(cid:18) (cid:18) (cid:12) (cid:12) (cid:19)(cid:19)
X (cid:12) (cid:12)
is positive, because (cid:12) (cid:12)
(cid:12) (cid:12)
t t t2 t3
Log 1+ < + for t < k.
k k − 2k2 3k3 | |
(cid:12) (cid:12)
Now, considering the remainin(cid:12)g part(cid:12), it is enough to prove
(cid:12) (cid:12)
(cid:12) (cid:12)
1 1
g(t) := D+ t3+ E t2+(F +1)t Log 1+t > 0
3 − 2 − | |
(cid:18) (cid:19) (cid:18) (cid:19)
as log t > 0 for t ( 2, 1).
| | ∈ − −
′
Since in 2< t < 1, g (t) = 0 is equivalent to
− −
α t3+α t2+α t+α = 0
3 2 1 0
where, α = F, α = 2E +F, α = 3D+2E and α = 3D+1.
0 1 2 3
Approximating α ,α ,α ,α to numerical values and using Descartes’ rule of signs, it can be
3 2 1 0
′
observed that g (t) has two positive real roots and a negative real root.
′ ′
Clearlyg ( 2)g ( 1.2) < 0impliesthenegative realrootliesintherequiredinterval( 2, 1),
− − − −
which we denote as t 1.50615....
0
≃ −
′′
Now, g (t )> 0 justifies that t is the minimum value.
0 0
By means of simple computation, we conclude that g(t )> 0 which implies
0
log G (t+1) > 0 for k = 2.
3
| |
Again log t > 0 for t ( 2, 1). which proves that the statement is true for k = 2. Let the
| | ∈ − −
statement be true for k = m 2. Then
≥
m(m 1)(m+1)
log G (t+1) − log t
3
| | ≥ − 6 | |
m(m+1)(m+2)
log t ( as m 1< m+2)
≥ − 6 | | −
which shows that the statement is true for k = m+1.
Hence by induction hypothesis we have
k(k 1)(k+1)
log G (t+1) + − log t 0 for t ( k, k+1). (3.4)
3
| | 6 | | ≥ ∈ − −
Therefore, d˜ (t) 0 for t < 0. (cid:3)
3
≥
Remark 3.1. d˜(t) can be expressed as
3
log G (t+1) +n(t )log t
3
| | | | | |, for t < 0;
d˜ (t)= − t3((log t )2+π2) (3.5)
3
 | |
0, for t 0;

≥

10 SouravDasandA.Swaminathan
where
[t]([t]+1)([t]+2)
n(t)= , for t > 0. (3.6)
6
Theorem 3.1. We have
liminf f (z) 0 and f (z) πd˜ (t)
3 3 3
ℑ ≥ ℑ →
as z t R within the upper half plane.
→ ∈
Proof. First we will calculate the imaginary part of f (z) as z t for t ( k, k+1), k 1
3
→ ∈ − − ≥
within the upper half plane with the help of Lemma 3.3.
Let t ( k, k+1) for k 1. Then
∈ − − ≥
lim f (z)
3
z t
→
z > 0
ℑ
iπ
log G (t+1) k(k 1)(k+1)
3
| |− 6 −
=
t3(log t +iπ)
| |
π2 k(k 1)(k+1)
log G (t+1) log t + k(k 1)(k+1) log G (t+1) + − log t
3 3
| | | | 6 − | | 6 | |
= iπ .
t3((log t )2+π2) − t3((log t )2+π2)
| | | |
Hence,
k(k 1)(k+1)
log G (t+1) + − log t
3
| | 6 | |
lim f (t+iy) = π
y→0+ℑ 3 − t3((log t )2+π2)
| |
For t = k, k = 1,2,3,... , we have
−
log G ( k+1+iy)
3
f ( k+iy) | | − ||
| 3 − | ≥ k+iy 3 Log( k+iy) −→ ∞
|− | | − |
for y 0+ because G (z) has poles at z = 0, 1, 2,....
3
→ − −
Since f (z) is real on the positive real axis, f (z) 0 as z t > 0. Clearly, with the help
3 3
ℑ → ∈
of (3.4) we have
liminf f (z) 0 as z t R 0 .
3
ℑ ≥ → ∈ \{ }
logG (z+1) 1
3
Now we will analyze the behaviour of f (z) = at z = 0.
3 z · z2Logz
Since,
logG (z+1)
3 ′
lim = (logG )(1) = F < 0,
3
z→0 z
the behaviour of f (z) at origin is determined by the behaviour of F/(z2Logz).
3
Let z = reiθ. Then
1 θcos2θ+logrsin2θ
=
ℑ z2Logz − r2((logr)2+θ2)
(cid:18) (cid:19)
π
= for z within upper half plane.
−r2((logr)2+π2)
Hence, liminf f (z) 0, as z t R within the upper half plane.
3
ℑ ≥ → ∈
Now with the help of Lemma 3.3 and Lemma 3.4, we can conclude that f (z) πd˜ (t) as
3 3
ℑ →
z t R within the upper half plane. (cid:3)
→ ∈