Georgian Mathematical Journal
Volume 8 (2001), Number 1, 69-86
WEIGHTED EXPONENTIAL INEQUALITIES
H. P. HEINIG, R. KERMAN, AND M. KRBEC
Abstract. Necessary and sufficient conditions on weight pairs are found for
the validity of a class of weighted exponential inequalities involving certain
classical operators. Among the operators considered are the Hardy averaging
operator and its variants in one and two dimensions, as well as the Laplace
transform. Discrete analogues yield characterizations of weighted forms of
Carleman’s inequality.
2000 Mathematics Subject Classification: 26D15, 26D10, 40A05,
42B25.
Key words and phrases: Hardy’s inequality, Carleman’s inequality,
Laplace transform, weight function.
1. Introduction
It is well known ([10], [3]) that for nonnegative locally integrable functions u,
v, and f defined on R+ and 1 < p < ∞, the inequality
Z∞
0
Ã
x
1Z
u(x)
f (t)1/p dt
x
!p
Z∞
dx ≤
Bpp p(p0 )p−1
0
v(x)f (x) dx
(1.1)
0
holds, if and only if
ÃZ∞
Bpp ≡ sup
s>0
!ÃZs
x−p u(x) dx
s
!p−1
1−p0
v(x)
dx
< ∞.
0
Here and throughout, p0 denotes the conjugate index of p defined by p0 = p/(p−
1). Moreover, the inequality (1.1), as well as those in the sequel, are interpreted
in the sense that, if the right side is finite, so is the left side, and the inequality
holds.
Now, if we assume f is positive a.e., then, since
Ã
x
1Z
f (t)1/p dt
lim
p→∞ x
0
!p
Ã
x
1Z
= exp
log f (t) dt
x
0
c Heldermann Verlag www.heldermann.de
ISSN 1072-947X / $8.00 / °
!
70
H. P. HEINIG, R. KERMAN, AND M. KRBEC
(cf. [15], p. 344; 5c) it follows from Fatou’s lemma and (1.1) that
Z∞
0
Ã
x
!
Ã
∞
!p
x
Z
1Z
1Z
u(x) exp
log f (t) dt dx ≤ lim inf u(x)
f (t)1/p dt
p→∞
x
x
0
0
Z∞
≤C
dx
0
(1.2)
v(x)f (x) dx,
0
provided lim Bpp p(p0 )p−1 = C. This limit exists if u(x) = v(x) = xλ , λ ≥ 0, and
p→∞
also if u(x) = v(x) is non-increasing. However, such a result for more general
weights seems not to be known.
In this paper we consider the operator K defined by
Z∞
(Kf )(x) =
k(x, y)f (y) dy,
f > 0,
0
where k(x, y) ≥ 0 satisfies certain mild restrictions. Conditions on positive, locally integrable weights u, v are given for which a weighted exponential inequality of the form (1.4) below holds. Here, locally integrable refers to integrating
on intervals (a, b), 0 < a < b < ∞. For a number of operators, specifically, the
Laplace transform and the general averaging operator
Zx
(Pβ f )(x) = βx
−β
tβ−1 f (t) dt,
β > 0,
(1.3)
0
we characterize the weight functions u and v for which the exponential inequality
ÃZ∞
!1/q
q
u(x)(exp(K log f )(x)) dx
0
ÃZ∞
!1/p
p
≤C
v(x)f (x) dx
(1.4)
0
holds, where 0 < p ≤ q < ∞ or 0 < q < p < ∞, p > 1 (Theorem 2.2). The basic
result of Heinig, Kerman and Krbec was given in [7], where some of the specific
cases were announced and then generalized for the wide range of parameters p
and q as above in [8]. The original result in [6] has in fact triggerred a number
of papers, for example [11], [12]. In particular, in [11], the inequality (1.4) was
obtained for β = 1 and 0 < p ≤ q < ∞. A survey of earlier results can be found
in [7]; we recall here various generalizations of Carleman’s inequality in [9], [6],
[5], [16].
Note that in the sequel all functions are assumed to be measurable, χE denotes
the characteristic function of a set E and constants C may be different at
different occurrences.
WEIGHTED EXPONENTIAL INEQUALITIES
71
2. Exponential Integral Inequalities
Let K be defined by
Z∞
k(x, y)f (y) dy
(Kf )(x) =
0
where f is a positive function and k(x, y) ≥ 0 satisfies
(i) k(λxλy) = λ−1 k(x, y), λ > 0 (k is homogeneous of degree −1);
(ii)
R∞
0
k(1, t) dt = 1;
Ã
(iii) exp −
R∞
!
k(1, t) log t dt = A,
for some constant A.
0
Theorem 2.1. Suppose K is the integral operator defined above with kernel
K satisfying (i), (ii) and (iii). Let u and v be positive weight functions and set
w(x) = u(x) (exp[(K log 1/v)(x)])q/p .
Given exponents p, q > 0, there holds
ÃZ∞
!1/q
q
u(x) (exp[(K log f )(x)]) dx
ÃZ∞
!1/p
p
≤C
.
v(x)f (x) dx
0
(2.1)
0
with C > 0 independent of f > 0 if
(a)
Z∞ "
sup
y>0
0
yk(x, y)
x
#q/p
w(x) dx < ∞
(2.2)
when 0 < p ≤ q; and
(b)
Z∞"Z∞
0
0
yk(x, y)
w(x) dx
x
#r
dy < ∞,
where 1/r = 1 − q/p,
(2.3)
when 0 < q < p.
Proof. Let s > 0 and g = (vf p )1/s . Then (2.1) is equivalent to
ÃZ∞
!p/(qs)
w(x) (exp[(K log g)(x)])
qs/p
dx
ÃZ∞
≤C
0
!1/s
s
g(x) dx
.
(2.4)
0
The change of variable y = xt, the conditions (i)–(iii) and finally Jensen’s
inequality yield for the left side of (2.4),
ÃZ∞
!p/(qs)
qs/p
w(x) (exp [(K log g)(x)])
0
dx
72
H. P. HEINIG, R. KERMAN, AND M. KRBEC
Ã
ÃZ∞
=
"Z∞
w(x) exp
0
=A
0
"Z∞
w(x) exp
0
ÃZ∞
≤A
0
!p/(qs)
dx
#
!qs/p
k(1, t) log tg(xt) dt
0
ÃZ∞
w(x)
!qs/p
k(1, t) log g(xt) dt
Ã
ÃZ∞
#
dx
!qs/p
k(1, t)t log g(xt) dt
!p/(qs)
!p/(qs)
dx
.
0
Setting y = xt in the last inner integral and invoking (2.4) again, we have that
(2.1) is a consequence of
ÃZ∞
!p/(qs)
f
w(x)[(Kg)(x)]
qs/p
dx
!1/s
ÃZ∞
s
≤C
0
g(x) dx
,
(2.5)
0
e
f is the integral operator with the kernel k(x,
where K
y) = yk(x, y)/x. According
to [2], Lemma 2.3 (a), one has (2.5) with s = 1 when 0 < p ≤ q if and only if
Z∞
e
[k(x,
y)]q/p w(x) dx = sup
sup
y>0
y>0
0
Z∞ "
0
yk(x, y)
x
#q/p
w(x) dx < ∞,
and (2.5) with s = p/q when 0 < q < p if and only if
Z∞ÃZ∞
0
!r
e
k(x,
y)w(x) dx
Z∞ÃZ∞
dy =
0
0
0
yk(x, y)
w(x) dx
x
!r
dy < ∞,
where 1/r = 1 − q/p (by the duality counterpart to [2], Lemma 2.3 (b)).
The next result yields a complete characterization of the weights in (2.1) for
the operator Pβ with the kernel kβ (x, y) = βx−β y β−1 χ(0,x) , x, y > 0, defined by
(1.3).
Theorem 2.2. Let β >R 0 and suppose u, v are positive locally integrable
functions on (0, ∞) with 0a tβ−1 v(t)−1 dt < ∞ for all a > 0. Set w(x) =
u(x) (exp[Pβ log(1/v)(x)])q/p . Then, a necessary and sufficient condition that
the inequality
ÃZ∞
!1/q
q
u(x) (exp[(Pβ log f )(x)]) dx
0
ÃZ∞
!1/p
p
≤C
v(x)f (x) dx
(2.6)
0
holds with C > 0 independent of f > 0 is
(a)
Z∞ Ã
sup
y>0
when 0 < p ≤ q < ∞; and
y
yβ
xβ+1
!q/p
w(x) dx < ∞,
(2.7)
WEIGHTED EXPONENTIAL INEQUALITIES
73
(b)
Z∞ÃZ∞ β
y w(x)
xβ+1
y
0
!r
dx
dy < ∞,
1/r = 1 − q/p,
(2.8)
when 0 < q < p < ∞.
Proof. In view of the preceding theorem we need only prove the necessity. Suppose first that (2.5) – which is equivalent to (2.6) – holds with 0 < p ≤ q,
f = P , and s = 1. Given fixed y > 0, define
K
β
h(t) = e1−1/β βtβ−1 y −β χ(0,y) (t) + βt1−β y β χ(y,∞) (t),
t > 0.
Then, for x > y,
Zy
³
´
tβ−1 log e1−1/β βy −β tβ−1 dt
−β
(Pβ log h)(x) = βx
0
Zx
+ βx
−β
³
´
tβ−1 log y β t1−β dt
y
=−
µ ¶
2 y β
β
x
− (1 − β) log x + β log y + 1 +
1
+ log β.
β
Substituting this in (2.6) and restricting the integral on the left side to (y, ∞),
we obtain
C1
!q/p
Z∞ Ã β
y w(x)
y
dx
xβ+1
ÃZ∞
"
q/p
≤ C1
w(x)
y
#
"
2q
q
(yx)β exp − (1 + β) log x
exp −
βp
p
"
#
q
exp β log y dx
p
Z∞
³
´
h(x) dx = C 1 + e1−1/β ,
≤C
0
where C1 = β exp(1 + 1/β) > 0, so that (2.7) holds.
Now, let wn (t) = w(t)χ(1/n,n) (t) and define h by
ÃZ∞
βqr/p
hn (t) = t
t
Then
Zy
wn (s)
ds
sβ+1

ÃZ∞
xβ−1 log xβqr/p
(Pβ log hn )(y) = βy −β
0
x
!qr/p
.
wn (t)
dt
tβ+1
!qr/p 
 dx
#
!p/q
74
H. P. HEINIG, R. KERMAN, AND M. KRBEC
"Z∞
#
y
y
wn (t)
βqr −β Z β−1
β 2 qr −β Z β−1
x
log x dx +
x
log
y
y
dt dx
=
β+1
p
p
t
x
0
β 2 qr −β
≥
y
p
Ã
yβ
yβ
log y − 2
β
β
!
0
"Z∞
#
Zy
βqr
wn (x)
−β
+
log
dx y
xβ−1 dx
β+1
p
x
y
0
Ã ∞
!qr/p 
Z
qr
w
(x)
n
.
≥ log y βqr/p −
+ log 
dx
β+1
p
y
x
Substituting in the equivalent formulation of (2.6) in which f is replaced by
v −1/p h and v by w exp[(q/p)Pβ log v], we get
ÃZ∞
Z∞
−qr/p
e
wn (y)y
βqr/p
y
0
wn (x)
dx
xβ+1
!qr/p
ÃZ∞Ã
≤C
y
Z∞
β
y
0
wn (x)
dx
xβ+1
!r
!q/p
dy
.
Now, qr/p = r − 1 and βqr/p + β = βr, so the left side of (2.5) equals
̰
!r
∞
e−qr/p Z βqr/p+β+1 d Z wn (x)
y
dx dy
−
r
dy y xβ+1
0
e−qr/p
=
r
Ã
! Z∞à Z∞
!r
βqr
w
(x)
n
+β
dx dy
yβ
p
xβ+1
y
0
Hence,
Z∞Ã
Z∞
y
0
β
y
wn (x)
dx
xβ+1
!r
Ã
dy ≤
reqr/p
·C
(βqr/p) + β
!(q/p)−1
,
and, letting n → ∞, (2.8) follows by Fatou’s lemma.
Remark 2.3. a) For q = p = 1, Theorem 2.2 a) was given in [7]. The special
case β = 1 in Theorems 2.2. a) and b) was proved in [11] and [12], respectively.
The proof of the sufficiency of the respective conditions use arguments different
from ours.
b) If (u, v) = (xδ , xγ ), δ, γ ∈ R where q(1 + γ) = p(1 + δ), 0 < p ≤ q < ∞,
then (2.6) holds. This follows at once if one observes that w(x) = eqγ/(βp) xδ+qγ/p
and (see (2.7))
ÃZ∞
sup saβ/p
s>0
!1/q
tδ−qγ/p−q(1+aβ)/p dt
< ∞.
s
But this supremum is finite if q(1 + γ) = p(1 + δ) and p(δ + 1) < q(1 + γ + aβ),
which is clearly the case.
WEIGHTED EXPONENTIAL INEQUALITIES
75
A result similar to that of Theorem 2.2 holds also for the operator Qβ defined
by
Z∞
(Qβ f )(x) = βx
β
y −β−1 f (y) dy,
β > 0.
x
Corollary 2.4. Let u, v and f be as in Theorem 2.2, and
³
h
i´q/p
e
w(x)
= x−2 u(1/x) exp Pβ (t2 /v(1/t))
Then the inequality (2.6) with Pβ replaced by Qβ is satisfied in the range 0 <
e holds.
p ≤ q < ∞, if and only if (2.7), with w replaced by w,
In the case 0 < q < p < ∞, (2.6) with Pβ replaced by Qβ holds, if and only
if, (2.8) with w replaced by we is satisfied.
Proof. Observe that the change of variable y = 1/t in the definition of Qβ shows
that (Qβ f )(x) = (Pβ fe)(1/x), where fe(x) = f (1/x). Hence Theorem 2.2 and
some obvious changes of variables yields
ÃZ∞
!1/q
q
u(x) (exp[Qβ log f (x)]) dx
ÃZ∞
=
0
³
!1/q
´q
u(x) exp[Pβ log fe(1/x)]
dx
0
and
ÃZ∞
³
´q
x u(1/x) exp[Pβ log fe(x)]
−2
!1/q
dx
ÃZ∞
!1/p
x v(1/x)fe(x) dx
−2
≤C
0
p
0
ÃZ∞
!1/p
p
=C
v(x)f (x) dx
,
0
and we are done.
When p = q = 1 and β = 1, the following equivalent statements hold for
P = P1 :
Proposition 2.5. Let u, v and f be positive functions and set
w(x) = u(x) exp(P log(1/v))(x),
x > 0.
Then, the following statements are equivalent:
Z∞
Z∞
v(x)f (x) dx;
u(x) exp(P log f )(x) dx ≤ C1
(2.9)
0
0
−1
(P w)(x) + α (Qα w)(x) ≤ C2 ;
P w ∈ L∞ ;
(2.10)
(2.11)
Z∞
Z∞
w(x)(P f
1/p p
f (x) dx,
) (x) dx ≤ C3
0
where C3 remains bounded as p → ∞.
0
p > 1,
(2.12)
76
H. P. HEINIG, R. KERMAN, AND M. KRBEC
For the proof see [7].
In our next result we consider the Laplace transform L defined by
Z∞
e−xy f (y) dy,
(Lf )(x) =
x > 0.
0
A characterization of the weights for which (2.9) holds is given next. Our result
generalizes Theorem 7 of [6].
Theorem 2.6. Let u, v and f be positive functions. Then
Z∞
0
·
¸
∞
Z
1
u(x) exp (L log f )(1/x) dx ≤ C v(x)f (x) dx
x
(2.13)
0
if and only if
Z∞
e−tx w(1/x) dx < ∞,
sup t
t>0
"
where w(x) = u(x) exp x
−1
(2.14)
0
#
Rx −y/x
e
log(1/v(y)) dy .
0
Proof. Let k(x, y) = x−1 e−y/x , then x−1 (Lf )(x−1 ) = (Kf )(x) satisfies the conditions of Theorem 2.1 (observe that
R∞ −y
e log y dy exists, so (iii) is satisfied)
0
whence the sufficiency part follows.
Conversely, if g = vf , then (2.13) is equivalent to
Z∞
Z∞
w(x) exp(K log g)(x) dx ≤ C
0
g(x) dx
(2.15)
0
where (Kh)(x) = (1/x)(Lh)(1/x). Now, let g(x) = gt (x) = t−1 e−x/t , t > 0
fixed. Then, (K log gt )(x) = −(x/t) − log t and substituting this into (2.15) we
obtain that
Z∞
(Kw)(t) = t
−1
w(x)e−x/t dx ≤ C.
0
Next, define K1 by
Z∞
e−x/y y −2 h(y) dy.
(K1 h)(x) = x
0
Then,
Z∞
e−x/y y −2 dy = C.
(K1 (Kw))(x) ≤ (K1 C)(x) = Cx
0
WEIGHTED EXPONENTIAL INEQUALITIES
77
The proof will be complete if we can show that there exists a constant C1 > 0
such that (K1 w)(x) ≤ C1 (K1 (Kw))(x). But,
Z∞
e−x/y (Kw)(y)
(K1 (Kw))(x) = x
0
ÃZ∞
Z∞
−x/y −3
=x
e
y
0
!
e
y
dy dt
!
−(x+t)s
w(t)
0
0
0
0
Z∞
=x
w(t) dt dy
ÃZ∞
Z∞
=x
−t/y
−(x+t)/y −3
w(t)
0
!
e
ÃZ∞
Z∞
=x
dy
y2
e
s ds dt
0
w(t) dt
+ x/t)2
t2 (1
∞
e Z w(t) −x/t
≥ x
e
dt
4
t2
0
e
= (K1 w)(x).
4
Here the last inequality follows from e−s ≤ 4e−1 /(1 + s)2 , s > 0.
The characterizations of the weights for the averaging operator between
weighted Lp –spaces in higher dimensions are known only in two dimensions
(cf. [13]). By contrast, the corresponding characterization of weights for the
exponential inequality of the averaging operator in higher dimensions carries
over from the one dimensional case in a straightforward manner.
Below we give the two dimensional result only. However the higher dimensional results carries over in the same way.
We shall write x = (x1 , x2 ), R2+ = (0, ∞) × (0, ∞), dx = dx1 dx2 , with the
two dimensional averaging operator being given by
x1x2
1 ZZ
(P f )(x) =
f (y) dy.
x1 x2
2
0 0
Theorem 2.7. Let u, v and f be positive functions on R2+ . Then
Z
Z
u(x) exp(P 2 log f )(x) dx ≤ C
R2+
v(x)f (x) dx
(2.16)
R2+
if and only if for any α1 , α2 > 0,
Z∞Z∞
sup
y1 >0,y2 >0
y1α1 y2α2
−(1+α1 ) −(1+α2 )
x2
w(x) dx
x1
y1 y2
=A<∞
(2.17)
78
H. P. HEINIG, R. KERMAN, AND M. KRBEC
where w(x) = u(x) exp(P 2 log(1/v))(x).
Proof. The argument that (2.17) is sufficient for (2.16) follows well known lines.
Thus, let g = f v. Then (2.16) is equivalent to
Z
Z
2
w(x) exp(P log g)(x) dx ≤ C
R2+
g(x) dx
(2.18)
R2+
and an obvious substitution gives
ÃZ1Z1
Z
w(x) exp
!
g(x) dx.
log g(x1 t1 , x2 t2 ) dt dx ≤ C
0 0
R2+
Z
R2+
But for α1 > 0, α2 > 0,
Z1Z1
e
−(α1 +α2 )
log (tα1 1 tα2 2 ) dt
= exp
0 0
and the left side of (2.18) becomes
ÃZ1Z1
Z
e
α1 +α2
!
log(tα1 1 tα2 2 g(x1 t1 , x2 t2 )) dt
w(x) exp
dx
0 0
R2+
Z1Z1
Z
≤e
α1 +α2
tα1 1 tα2 2 g(x1 t1 , x2 t2 ) dt dx,
w(x)
R2+
0 0
by Jensen’s inequality. Changing variables (twice) and interchanging the order
of integration the latter is equal to
Z
eα1 +α2

Z∞Z∞
g(y) y1α1 y2α2
R2+

−(1+α1 ) −(1+α2 )
x2
w(x)x1
dx dy,
y1y2
so that (2.17) yields (2.18) with C = eα1 +α2 A.
We now show that (2.16) implies (2.17).
For fixed t1 > 0, t2 > 0 substitute
−1
−1
g(x) =t−1
1 t2 χ(0,t1 ) χ(0,t2 ) (x2 ) + t1 χ(0,t1 ) (x1 )
+
+
e−(1+α2 ) tα2 2
χ(t2 ,∞) (x2 )
2
x1+α
2
e−(1+α1 ) tα1 1
−1
t2 χ(0,t2 ) (x2 )
χ(t1 ,∞) (x1 )
1
x1+α
1
e−(2+α1 +α2 ) tα1 1 tα2 2
χ(t1 ,∞) (x1 )χ(t2 ,∞) (x2 )
1 1+α2
x1+α
x2
1
in (2.16). The right side of (2.16) is
h
C 1 + e−1−α2 /α2 + e−1−α1 /α1 + e−2−α1 −α2 /(α1 α2 )
i
WEIGHTED EXPONENTIAL INEQUALITIES
79
Let us write the left side in the form
Zt1Zt2
Z
Zt1Z∞
=
+
Z∞Z∞
+
+
0 t2
0 0
R2+
Z∞Zt2
t1 0
≡ I1 + I2 + I3 + I4 .
t1 t2
Each integral is positive, so, if we show that
Z∞Z∞
I4 =
tα1 1 tα2 2
−(1+α1 ) −(1+α2 )
x2
w(x)x1
dx,
t1 t2
the result will follow. We have
Z∞Z∞
I4 =
t1 t2
Z∞Z∞
=

0 0
·
w(x) exp
t1 t2

ÃZt1Zt2 Zt1Zx2 Zx1Zt2 Zx1Zx2!
1
log g(y) dy  dx
w(x) exp 
+
+
+
x1 x2
t1 0
0 t2
t1 t2
−1
log(t−1
1 t2 )
x1 x2
t1 t2
¸


!
à −1
x2
t1 Z
t1 e−(1+α2 ) tα2 2

exp
dy2 
log
x1 x2
y21+α2
t2
"
à −1
!
#
x1
t2 Z
t2 e−(1+α1 ) tα1 1
dy1
exp
log
x1 x2
y11+α1
t1
"
!
#
Ã
x1x2
1 ZZ
e−(2+α1 +α2 ) tα1 1 tα2 2
dy dx
exp
log
x1 x2
y11+α1 y21+α2
t1 t2
Z∞Z∞
=
w(x)E1 (x)E2 (x)E3 (x)E4 (x) dx.
t1 t2
Clearly,
·
¸
t1 t2
t1 t2
E1 (x) = exp −
log t1 −
log t2 .
x1 x2
x1 x2
An elementary calculation shows that
·
E2 (x) = exp −
t1 t2
α2 t1
t1
log t1 +
log t1 +
log t2
x1
x1 x2
x1
#
t1 t2
(1 + α2 )t1
+
log t2 −
log x2
x1 x2
x1
so that
"
#
t1
α2 t1
(1 + α2 )t1
E1 (x)E2 (x) = exp − log t1 +
log t2 −
log x2 .
x1
x1
x1
80
H. P. HEINIG, R. KERMAN, AND M. KRBEC
Further calculation – elementary again, but rather tedious so we omit details –
yields
·
t2
t1 t2
α1 t2
log t2 +
log t2 +
log t1
x2
x1 x2
x2
#
t1 t2
(1 + α1 )t2
+
log t1 −
log x1 ,
x1 x2
x2
·
t1
t1 t2
t2
tα1 1 tα2 2
α1 t2
log t1 +
log t1 −
log t1 +
log t2
E4 (x) = 1+α1 1+α2 exp −
x2
x1
x1 x2
x2
x1 x2
#
α1 t1
t1 t2
(1 + α1 )t2
(1 + α2 )t1
−
log t2 −
log t2 +
log x1 +
log x2
x1
x1 x2
x2
x1
E3 (x) = exp −
and
"
#
tα1 1 tα2 2
t1
α2 t1
(1 + α2 )t1
E3 (x)E4 (x) = 1+α1 1+α2 exp
log t1 −
log t2 +
log x2 ,
x1
x1
x1
x1 x2
thus
tα1 1 tα2 2
E1 (x)E2 (x)E3 (x)E4 (x) = 1+α1 1+α2 .
x1 x2
This proves the theorem.
3. The Discrete case
It is not unexpected that integral estimates have corresponding discrete analogues. In this section we give discrete versions of Theorem 2.2 (with β = 1) and
Theorem 2.7, which yield two–weight generalizations of a weighted Carleman
inequality.
∞
∞
Theorem 3.1. Let {un }∞
n=1 , {vn }n=1 , {an }n=1 be sequences of positive numbers and set
Ã
"
#!q/p
n
1X
log(1/vk )
wn = un exp
n k=1
,
n = 1, 2, . . . .
(i) If 0 < p ≤ q < ∞, then there is a constant C > 0 such that the inequality
̰
X
!1/q
q/n
≤C
un (a1 a2 . . . an )
n=1
̰
X
!1/p
vn apn
(3.1)
n=1
if and only if for any a > 0
a/p
sup m
m≥1
à ∞
X
n=m
wn
n(1+a)q/p
!1/q
< ∞.
(3.2)
WEIGHTED EXPONENTIAL INEQUALITIES
81
(ii) If 1 < q < p < ∞, then the inequality (3.1) holds, provided the condition
∞
P
Ã
∞
P
m=1 n=m
!r/q
0
n−q(a+1) wn
m(a+1/q )r < ∞ is satisfied, where a > 0 and
1/r = 1/q − 1/p.
Proof. (i) Let bpn = vn apn , n = 1, 2 . . . . Then, (3.1) is equivalent to
̰
X
Ã
"
n
1X
log bk
exp
n k=1
wn
n=1
#!q !1/q
≤C
̰
X
!1/p
bpn
.
(3.3)
n=1
For any p > 0, if we write γ = p/α where α > 1, then, 0 < p = γα ≤ q < ∞,
or 1 < α ≤ q/γ. With ck = bγk the inequality (3.3) is equivalent to
̰
X
Ã
"
n
1X
log ck
exp
n k=1
wn
n=1
#!q/γ !γ/q
≤C
γ
̰
X
!1/α
cαn
.
(3.4)
n=1
To prove (3.4) define f (t) = ck , if k −1 < t ≤ k, k = 1, 2 . . . , and zero otherwise.
Then, since
is
R1
log y dy = −1, Jensen’s inequality shows that the left side of (3.4)
0
̰
X
n=1
#!q/γ !γ/q
n Zk
1X
log f (t) dt
exp
n k=1
Ã
wn
=
"
̰
X
k−1
Ã
"
Ã
"Z1
wn
n=1
=
̰
X
0
wn exp
̰
X
=e
Ã
"Z1
wn exp
=e
log(yf (ny)) dy
0
wn
!q/γ !γ/q
yf (ny) dy
0
wn n
ÃZn
−2q/γ
n=1
à n
̰
X
X wn
=e
n=1
#!q/γ !γ/q
ÃZ1
n=1
̰
X
(t = ny)
0
n=1
̰
X
#!q/γ !γ/q
log f (ny) dy
n=1
≤e
#!q/γ !γ/q
n
1Z
exp
log f (t) dt
n
n2q/γ
!q/γ !γ/q
f (t) dt
0
!q/γ !γ/q
.
kck
k=1
Applying the discrete version of the weighted Hardy inequality (cf. [1], [4]), the
last expression is dominated by
̰
X
n=1
!1/α
k
−α
(k α cαk )
=
̰
X
n=1
!1/α
cαk
82
H. P. HEINIG, R. KERMAN, AND M. KRBEC
provided in case 0 < p ≤ q < ∞
à ∞
X
wn
sup
2q/γ
m≥1 n=m n
!γ/q à ∞
X
!γ/q0
k
−α(1−α0 )
n=m
1+1/α0
= sup m
à ∞
X
m≥1
n=m

!p/(qα)
wn
n2qα/p
= sup m(2α/p)−(1/p)
à ∞
X
wn
!1/q p/α

2qα/p
n=m n
m≥1
< ∞.
But with a = 2α − 1 > 0 this becomes (3.2).
(ii) In case 1 < q < p < ∞ we see that (just as in the previous case) the left
side of (3.3) is not larger than
̰
X
e
wn n
−q(α+1)
à n
X
n=1
!q !1/q
α
k bk
,
α > 0,
k=1
and again by the discrete form of Hardy’s inequality ([15], [16]) this is dominated
by
̰
X
!1/p
k
−αp
p
α
=
(k bk )
!
̰
X p 1/p
bk
k=1
k=1
whenever
∞
X
à ∞
X
m=1 n=m
!r/q à ∞
X
wn
nq(α+1)
But since
!r/q0
n
αp0
0
mαp < ∞.
n=1
Ã
αp0
1
αp0 αp0
r
+
+
−
(αp0 + 1) 0 + αp0 = r
q
q0
q0
q
p
Ã
!
!
Ã
!
1
1
= r αp + 0 − αp0 + α = r α + 0 ,
q
q
0
this is the condition given in (ii) with a = α.
To prove necessity of part (i) let bpn = 1/k if n ≤ k and e−1−α n−α−1 k α if
n > k, α > 0 and k fixed, n = 1, . . . , in (3.3) Clearly the right side of (3.3) is
finite in this case and we can write the left side in the form
à k
X
+
n=1
∞
X
!1/q
= (S1 + S2 )1/q .
n=k+1
Now,
k
X

"
#q
n
k
X
1X
−q/p


S1 =
wn exp
log(k − 1/p)
=k
wn
n j=1
n=1
n=1
WEIGHTED EXPONENTIAL INEQUALITIES
83
and

"
Ã
#q/p 
#!q/p
n
−(1+α) α
X
1
e
k
exp

log
α+1
"
k
1X
S2 =
wn exp
log(1/k) 
n
j=1
n=k+1
∞
X
Ã
∞
X
=
"
−k
exp
log k
n
wn
n=k+1
Ã
#!q/p Ã
n j=k+1
j
#!q/p
"
³
´
n−k−1
exp
log e−(1+α) k α
n
"
n
α+1 X
× exp −
log j
n j=k+1
×
#!q/p
.
But the Mean Value Theorem shows that
n
X
n+1
X
log j =
j=k+1
=
=
j=k+2
n
X
n
X
j log(j − 1) −
j log j
j=k+1
j[log(j − 1) − log j] − (k + 1) log k + (n + 1) log n
j=k+1
n
X
−j
+ (n + 1) log n − (k + 1) log k
0
j=k+1 cj
m where j − 1 ≤ c0j ≤ j. Hence
n
X
n
X
log j ≤
j=k+1
(−1) + (n + 1) log n − (k + 1) log k
j=k+1
so
∞
X
S2 ≥
Ã
wn
n=k+1
Ã
"
k
exp − log k
n
"
#!q/p
³
e−(1+α) k α
#!q/p Ã
k+1
× exp
(1 + α)
n
´q/p
"
α(k + 1)
exp −
log k
n
"
#!q/p
#
(α + 1)
× exp −
(−n + k + 1 + (n + 1) log n − (k + 1) log k)
n
=e
Ã
∞
X
−(1+α)q/p αq/p
k
wn
n=k+1
µ
·
³
´
α+1 q/p
µ
"
k
exp − log k
n
¶
n+1
× exp −(α + 1)
log n
n
× e
Ã
−(1+α)q/p αq/p
≥e
k
"
n=k+1
Ã
wn n
"
α(k + 1)
exp (α + 1)
log k
n
Ã
−(α+1)q/p
"
α(k + 1)
exp −
log k
n
¸¶q/p Ã
k+1
exp −(α + 1)
n
∞
X
#!q/p Ã
!#!q/p
"
Ã
1
k
exp
log
α+1
n
n
!#!q/p
#!q/p
#!q/p
84
since
H. P. HEINIG, R. KERMAN, AND M. KRBEC
k+1
≤ 1. But, since, for k ≥ 1 (n ≥ 2),
n
Ã
1
k
log
α+1
n
n
!
≥−
α+1
α+1
log n ≥ −
log 2,
n
n
it follows that
Ã
"
1
k
exp
log α+1
n
n
#!q/p
"
q
= exp
p
Ã
1
k
log α+1
n
n
!#
≥ 2−q(α+1)/(2p) .
Therefore
S2 ≥ e−(1+α)q/p 2−q(α+1)/(2p) k αq/p
∞
X
wn n−(α+1)q/p ,
n=k+1
and so
Ã
1/q
(S1 + S2 )
−(1+α)/p −(α+1)/(2p)
≥e
k −q/p wk + k αq/p
2
∞
X
wn n−(α+1)q/p
n=k
−k αq/p wk k −(α+1)q/p
Ã
=e
−(1+α)/p −(α+1)/(2p)
2
k
αq/p
´1/q
!1/q
∞
X
wn
n=k
n(α+1)q/p
.
This implies condition (3.2) with a = α and completes the proof of the theorem.
Remark 3.2. i) If un = vn = 1, then wn = 1, n = 1, 2, . . . , and condition (3.2)
is satisfied only if p = q, so in this case (3.1) reduces to Carleman’s inequality
(with C = e).
ii) If p = q = 1 and
sup ma
m≥1
∞
X
n=m
wn
(1+α)
n
= A,
then the constant C in (3.1) satisfies
w−(1+a)/2 A(1 + 3e−(1+a) /(2a)) ≤ C ≤ ea A.
The previous result with p = q = 1 has a higher dimensional analogue. We
state here the two dimensional case and only sketch the proof. Details can be
found in authors’ preprint [8].
Theorem 3.3. For n, m ∈ N let {un,m }, {vn,m }, {an,m } be double sequences
of positive numbers and
!
Ã
wn,m
1 X
= un,m exp
log 1/vk,j .
nm 1≤k≤n
1≤j≤m
WEIGHTED EXPONENTIAL INEQUALITIES
85
Then,
Ã
X
un,m
n,m≥1
!1/(nm)
Y
aj,k
X
≤C
vn,m an,m
(3.5)
n,m≥1
1≤k≤n
1≤j≤m
if and only if for some α1 > 0, α2 > 0,
sup k α1 j α2
k≥1
j≥1
X
n−α1 −1 m−α2 −1 wn,m ≡ A < ∞.
(3.6)
n≥k
m≥j
For the proof observe that the substitution bn,m = an,m vn,m allows us to write
(3.5) in the equivalent form
Ã
X
!
X
1 X
wn,m exp
log bj,k ≤ C
bn,m .
nm 1≤k≤n
n,m≥1
n,m≥1
(3.7)
1≤j≤m
Now, let f (s, t) = bj,k if k − 1 < t ≤ k, j − 1 < s ≤ j, k, j = 1, 2 . . . , and zero
otherwise. Since
Z1Z1
−α1 −α2
e
log (tα1 1 tα2 2 ) dt1 dt2 ,
= exp
0 0
the left side of (3.7) takes the form
!
j
k
1 X Z Z
wn,m exp
log f (s, t) dsdt
nm 1≤k≤n
n,m≥1
X
Ã
1≤j≤m
k−1 j−1
and this is the case dealt with in the previous theorem.
For the converse we substitute into (3.7) bn,m defined by
bn,m =


k −1 j −1




k −1 e−1−α2 j α2 m−α2 −1

j −1 e−1−α1 k α1 n−α1 −1




e−2−α1 −α2 k α1 j α2 n−α1 −1 m−α2 −1
if
if
if
if
n ≤ k,
n ≤ k,
n > k,
n > k,
m ≤ j,
j > m,
j ≤ m,
j ≥ m,
where k, j are fixed. Then, after some calculation one obtains (3.6).
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(Received 9.12.2000)
Authors’ addresses:
H. P. Heinig
Dept. of Mathematics and Statistics
McMaster University
Hamilton, Ontario
Canada L8S 4K1
E-mail: heinig@mcmail.mcmaster.ca
R. Kerman
Dept. of Mathematics, Brock University
St. Catharines, Ontario
Canada L2S 3A1
E-mail: rkerman@spartan.ac.brocku.ca
M. Krbec
Institute of Mathematics
Academy of Sciences of the Czech Republic
Žitná 25, 115 67 Prague 1
Czech Republic
E-mail: krbecm@matsrv.math.cas.cz