General Mathematics Vol. 15, No. 1 (2007), 67–74
The Hlawka–Djoković Inequality and Points
in Unitary Spaces 1
Walther Janous
Dedicated to Professor Ph.D. Alexandru Lupaş on the occasion
of his 65th birthday
Abstract
In this note several inequalities are proved. They are all in connection with the Hlawka–Djoković inequality. At the end a problem
for further research is posed.
2000 Mathematics Subject Classification: 26D10.
1
Introduction
Starting points of this note was the following problem for planar triangles
recently posed by O. Furdui in [5].
Let G be the centroid of ∆ABC, and let A1 , B1 , C1 be the mid-points
of BC, CA, AB, respectively. If P is an arbitrary point in the plane of
∆ABC, show that
P A + P B + P C + 3P G ≥ 2(P A1 + P B1 + P C1 ) .
1
Received 15 January, 2007
Accepted for publication (in revised form) 28 February, 2007
67
68
Walther Janous
(It should be noted that this result dates back at least to the paper [3] by
M. Chiriţă and R. Constantinescu. See also the referential source [6], p.
410.)
The proof of this inequality is as follows.
−→ −−→
−→
Let a, b and c denote the three vectors P A, P B and P C, respectively.
Then the original Hlawka inequality says
|a| + |b| + |c| + |a + b + c| ≥ |a + b| + |b + c| + |c + a|
(See for instance [7], p. 521.)
Because of |a + b + c| = 3P G and |a + b| = 2P A1 etc, the result is
immediate.
In this note we extend this result to a finite number of points in arbitrary
unitary spaces X.
We then go on to exemplify the general inequality for some special
spaces.
At the end we pose a problem concerning the asymptotic behavior of a
difference playing a role in an inequality obtained in this note.
2
General Inequality
We are now in the position to state and prove the following
Theorem. Let A1 , A2 , . . . , An be n fixed points (n ≥ 3) in an arbitrary
unitary space X. Fix the entire number k such that 2 ≤ k ≤ n − 1.
Let furthermore Gi1 ...ik be the ”centroid” of the k points Ai1 , . . . Aik where
1
1 ≤ i1 < . . . < ik ≤ n, that is Gi1 ...ik = (Ai1 + . . . + Aik ).
k
Then for any point P ∈ X the following inequality is valid:
!
µ
¶Ã
n
X
n−k X
n−2
k
P Gi1 ...ik ≤
P Aj + nP G .
k−2
k − 1 j=1
1≤i <...<i ≤n
1
k
(Here G denotes the centroid of all n points A1 , A2 , . . . , An . Furthermore,
AB is the distance between A and B, that is AB = kA − Bk.)
The Hlawka–Djoković Inequality and Points in Unitary Spaces
69
In the spirit of the before-given proof this inequality is an immediate consequence of the Djoković–generalization of the original Hlawka–inequality,
that is
°!
° n
µ
¶Ã
n
°X °
X
n−k X
n−2
°
°
xj °
kxi1 + . . . + xik k ≤
kxj k + °
°
°
k−2
k−1
1≤i1 <...<ik ≤n
j=1
j=1
where x1 , . . . , xn ∈ X. (See [4] and also the referential source [7], p. 522–
523.)
Remark. The special case k = 2 of this inequality was originally proved
by D. Adamović in [1]. It was reconsidered by M. Bencze in [2] where he
also gave several applications of it.
3
Some Applications
(1). For X = C and P = reiϕ , r ≥ 0, and Aj = rj eiϕj , rj ≥ 0, j = 1, . . . , n,
there holds
v
u
n
n
X
X
X
u
2
2
tk r +
rij rim cos(ϕij − ϕim ) − 2kr
rij cos(ϕ − ϕij ) ≤
1≤i1 <...<ik ≤n
≤
µ
j=1
n−2
k−2
¶Ã
n
n − k Xq 2
r + rj2 − 2rrj cos(ϕ − ϕj ) +
k − 1 j=1
v
u
n
X
X
u
+ tn2 r2 + rj2 + 2 + 2
j=1
j=1
rj rm cos(ϕj − ϕm ) − 2nr
i≤j<m≤n
n
X
j=1

rj cos(ϕ − ϕj )
• If all involved points are on the unit–circle this inequality reduces to
v
u
k
X
X
X
u
tk(k + 1) + 2
cos(ϕ − ϕij ) ≤
cos(ϕij − ϕim ) − 2k
1≤i1 <...<ik ≤n
1≤j<m≤n
j=1
70
Walther Janous
≤
µ
µ
¶Ã
¶¯
k ¯
ϕ − ϕj ¯¯
n − k X ¯¯
n−2
sin
2
¯+
k−2
k − 1 j=1 ¯
2
v
u
u
+ tn(n + 1) + 2
X
cos(ϕj − ϕm ) − 2n
n
X
j=1
i≤j<m≤n

cos(ϕ − ϕj ) .
• For P = 0 and all the Aj ′ s on the unit–circle we get the inequality
s
X
X
k+2
cos(ϕij − ϕim ) ≤
1≤i1 <...<ik ≤n
≤
µ
n−2
k−2
¶
i≤j<m≤n

s
X
n − k n + n + 2
cos(ϕj − ϕm )
k−1
1≤j<m≤n

(2). Let I = [a, b] be an interval on the real axis and consider
s the vector
Z b
space X = C[a, b] of all continuous functions on I with kf k =
(f (x))2 dx.
a
For P = f ∈ C[a, b] and Aj = fj ∈ C[a, b], j = 1, . . . , n, there holds the
”master–inequality”
v
!2
uZ b à k
u
X
X
t
f (x) − kf (x) dx ≤
ij
1≤i1 <...<ik ≤n
≤
µ
n−2
k−2
¶
a

j=1
n
X
n
−
k

k − 1 j=1
s
Z
b
(fj (x) − f (x))2 dx +
a
v
!2 
uZ b à n
u
X
fj (x) − nf (x) dx .
+t
a
j=1
This general integral inequality allows specifications in very many directions.
Of specific interest are, of course, orthogonal polynomials.
We investigate now the cases of Legendre, Laguerre and Hermite polynomials.
The Hlawka–Djoković Inequality and Points in Unitary Spaces
71
• I = [−1, 1] and Legendre polynomials Pn .
We let fj = Pj−1 , j = 1, . . . , n. Due to
Z
1
Pr (x)Ps (x)dx =
−1
2
δrs ,
2s + 1
r, s ≥ 0, there holds
v
!
u k
Z b
Z bÃX
k
X
uX 2
t
+ k 2 (f (x))2 dx−2k
Pij (x) f (x)dx ≤
2i
−
1
j
a
a j=1
1≤i <...<i ≤n j=1
1
k

s
µ
¶
Z b
Z b
n
X
2
n − 2 n − k
2
+ (f (x)) dx − 2 Pj (x)f (x)dx +
≤
k−2
k − 1 j=1 2j − 1 a
a
v

à n
!
u n
Z
Z
b
b
X
X
u
2
+t
Pj (x) f (x)dx .
(f (x))2 dx − 2n
+ n2
2j
−
1
a
a
j=1
j=1
If we put in this inequality f = PN , where N ≥ n, then
v
u
k
X
X
u k2
1
t
+
≤
2N + 1 j=1 2ij − 1
1≤i <...<i ≤n
1
k


v
uX
¶
µ
n
n r
2
X
u
1
1
1
n
n − 2 n − k
.
≤
+
+t
+
k−2
k − 1 j=1 2j − 1 2N + 1
2j
−
1
2N
+
1
j=1
Letting N → ∞ yields
X
v
u k
uX
t
1≤i1 <...<ik ≤n
≤
µ
n−2
k−2
¶

n − k
k−1
j=1
n r
X
j=1
1
≤
2ij − 1
1
2j − 1
v
uX
u n
+t
j=1

1 
.
2j − 1
If we set here k = n − 1 then it follows
v
uX
n r
n r
X
X
p
u n
1
1
1
hn −
≤
+t
+ (n − 2) hn
(∗)
2j − 1
2j − 1
2j − 1
j=1
j=1
j=1
72
Walther Janous
where hn =
n
X
k=1
1
.
2j − 1
x
• I = [0, ∞) and Laguerre functions Λn (x) = e 2 Ln (x), where Ln are
the Laguerre polynomials.
∞
X
Λr (x)dx = (s!)2 δrs ,
Letting fj = Λj−1 , j = 1, . . . , n, and noting
0
r, s ≥ 0, we get for f = ΛN , where N ≥ n:
v
u
k
X
X
u
tk 2 (N !)2 +
(ij !)2 ≤
1≤i1 <...<ik ≤n
≤
µ
n−2
k−2
¶

n − k
k−1
n−1
X
j=0
j=1
v

u
n−1
X
p
u
(N !)2 + (j!)2 + tn2 (N !)2 +
(j!)2  .
j=0
This inequality becomes for k = n − 1
s
s
µ ¶2 X
µ ¶2
q
n−1
n−1
X
j!
j!
(N )
(N )
2
(n − 1) + sn −
(1 +
≤
+(n−2) n2 + sn
N!
N!
j=0
j=0
¶2
n−1 µ
X
j!
where
=
.
N!
j=0
Dividing bothµof these
by N ! and letting N −→ ∞ yields
¶ inequalities
µ ¶
n
n
the identities
=
and (n − 1)n = (n − 1)n, respectively.
k
k
This means that these inequalities are sharp (at least ”at infinity”).
)
s(N
n
x2
• I = (−∞, ∞) and Hermite functions ψn (x) = e− 2 Hn (x), where Hn
are the Hermite polynomials.
Z ∞
ψr (x)ψs (x)dx = 2s s!δrs ,
Proceeding as before we get due to
−∞
r, s ≥ 0:
X
1≤i1 <...<ik
v
u
k
X
u
2ij ij !
tk 2 +
≤
2N N !
j=1
≤n
The Hlawka–Djoković Inequality and Points in Unitary Spaces
≤
µ
n−2
k−2
¶

n − k
k−1
n−1
X
j=0
r
73
v

u
n−1
j
i
X
u
j
2
!
2 ij ! t 2
j
 .
+ n +
N
N
2 N!
2
N
!
j=0
From it similar conclusions as before can be drawn.
As a final
Z π example we let I = [0, π] and fj (x) = sin(jx), j = 1, . . . , n.
π
Then
sin(rx) sin(sx)dx = δrs , r, s ≥ 1.
2
0
This, the master–inequality and

Z π

0,
r=s
r+s
sin(rx) cos(sx)dx =
r((−1) − 1)

, r 6= s
0
s2 − r 2
would enable us to deduce various inequalities for Fourier coefficients
∞
a0 X
(am cos(mx) + bm sin(mx))
+
of a functions f (x) having f (x) =
1
m=1
as its Fourier series. We leave this an exercise for the reader.
4
An Observation
In the light of the following recently published result many of the before
proven results have natural extensions for the case k = n − 2.
Indeed, the following inequality is valid ([8], p. 64).
Let x1 , . . . , xn be elements from a Hilbert space (n ≥ 3) and µ1 ≥
1, . . . , µn ≥ 1. Then
°
°
°
à n
!° n
n
n
n
°
°
° X
°X
X
X
X
°
°
°
°
µ j xj ° .
µi kxi k ≥
µi °xi −
µ i xi ° +
µi − 2 °
°
°
°
°
i=1
5
i=1
i=1
i=1
j=1
An Open Problem
At the end of this note we raise the following question in connection with
inequality (*). Unlike the inequalities obtained from the Hermite and Lagrange polynomials is the one stemming from Legendre polynomials, that
is inequality (*), apparently far from being asymptotically sharp.
74
Walther Janous
Therefore it is natural to find out the correct asymptotic behavior of the
following difference we denote in honor of Alexandru Lupaş by AL(n) =
n q
n q
X
X
1
aj (n) where aj (n) =
1 − aj (n) −
, j = 1, . . . , n.
hn (2j − 1)
j=1
j=1
This is, we are left to find f (n), such that AL(n) = n − f (n) + o(f (n)),
as n → ∞.
References
[1] D.D. Adamović, Généralisation d′ une identité de Hlawka et de l′ inégalité
correspondante, Mat. Vestnik 1 (16) (1964), 39–43.
[2] M. Bencze, Despre inegalitatea lui E. Hlawka, Gaz. Matematică, (seria
A) 6 (1985), 48–50.
[3] M. Chiriţă and R. Constantinescu, Asupra unei inegalităţi care caracterizează funcţiilr convexe, Gaz. Matematică (seria B) 89 (1984), 241–242.
[4] D.Ž. Djoković, Generalizations of Hlawka′ s inequality, Glas. Mat.–Fiz.
Astronom., Ser. II, Društvo Mat. Fiz. Hratske 18 (1963), 169–175.
[5] O. Furdui, Problem 3052, Crux math., 31 (2005), 333.
[6] D.S. Mitrinović et al., Recent Advances in Geometric Inequalities,
Kluwer Acd. Publ., Dordrecht, 1989.
[7] D.S. Mitrinović et al., Classical and New Inequalities in Analysis, Kluwer
Acd. Publ., Dordrecht, 1993.
[8] S. Takahasi, Y. Takahashi and S. Wada, An Extension of Hlawka′ s Inequality, Math. Ineq. Appl., 3 (2000), 63–67.
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E-mail address: walter.janous@tirol.com