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

∫_{0}^{∞}u(x)(\frac{1}{x} ∫_{0}^{x} f(t)^{1/p} dt)^{p} dx ≤ B_{p}^{p}p(p′)^{p−1} ∫_{0}^{∞}v(x)f(x) dx  (1.1)

holds, if and only if

B_{p}^{p} ≡ sup_{s>0}(∫_{s}^{∞}x^{−p}u(x) dx)(∫_{0}^{s} v(x)^{1−p′} dx)^{p−1} < ∞.

Here and throughout, p′ denotes the conjugate index of p defined by p′ = 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

lim_{p→∞}(\frac{1}{x} ∫_{0}^{x} f(t)^{1/p} dt)^{p} = exp(\frac{1}{x} ∫_{0}^{x} log f(t) dt)

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(cf. [15], p. 344; 5c) it follows from Fatou’s lemma and (1.1) that

∫_{0}^{∞}u(x) exp(\frac{1}{x} ∫_{0}^{x} log f(t) dt) dx ≤ lim inf_{p→∞} ∫_{0}^{∞}u(x)(\frac{1}{x} ∫_{0}^{x} f(t)^{1/p} dt)^{p} dx

≤ C ∫_{0}^{∞}v(x)f(x) dx, (1.2)

provided lim_{p→∞} B_{p}^{p}p(p′)^{p−1} = C. This limit exists if u(x) = v(x) = x^{λ}, λ ≥ 0, and 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

(Kf)(x) = ∫_{0}^{∞}k(x, y)f(y) dy, f > 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

(P_{β}f)(x) = βx^{−β} ∫_{0}^{x} t^{β−1}f(t) dt,  β > 0,  (1.3)

we characterize the weight functions u and v for which the exponential inequality

(∫_{0}^{∞}u(x)(exp(K log f)(x))^{q} dx)^{1/q} ≤ C(∫_{0}^{∞}v(x)f(x)^{p} dx)^{1/p}  (1.4)

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, χ_{E} denotes the characteristic function of a set E and constants C may be different at different occurrences.

2. Exponential Integral Inequalities

Let K be defined by

(Kf)(x) = ∫_{0}^{∞}k(x, y)f(y) dy

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) ∫_{0}^{∞}k(1, t) dt = 1;

(iii) exp(− ∫_{0}^{∞}k(1, t) log t dt) = A, for some constant A.

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

(∫_{0}^{∞}u(x) (exp[(K log f)(x)])^{q} dx)^{1/q} ≤ C(∫_{0}^{∞}v(x)f(x)^{p} dx)^{1/p}. (2.1)

with C > 0 independent of f > 0 if (a)

sup_{y>0} ∫_{0}^{∞}[\frac{yk(x, y)}{x}]^{q/p} w(x) dx < ∞  (2.2)

when 0 < p ≤ q; and (b)

∫_{0}^{∞}[∫_{0}^{∞} \frac{yk(x, y)}{x}w(x) dx]^{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

(∫_{0}^{∞}w(x) (exp[(K log g)(x)])^{qs/p} dx)^{p/(qs)} ≤ C(∫_{0}^{∞}g(x)^{s} dx)^{1/s}. (2.4)

The change of variable y = xt, the conditions (i)–(iii) and finally Jensen’s inequality yield for the left side of (2.4),

(∫_{0}^{∞}w(x) (exp [(K log g)(x)])^{qs/p} dx)^{p/(qs)}

= (∫_{0}^{∞}w(x)(exp[∫_{0}^{∞}k(1, t) log g(xt)] dt)^{qs/p} dx)^{p/(qs)}

= A(∫_{0}^{∞}w(x)(exp[∫_{0}^{∞}k(1, t) log tg(xt)] dt)^{qs/p} dx)^{p/(qs)}

≤ A(∫_{0}^{∞}w(x)(∫_{0}^{∞}k(1, t)t log g(xt) dt)^{qs/p} dx)^{p/(qs)}.

Setting y = xt in the last inner integral and invoking (2.4) again, we have that (2.1) is a consequence of

(∫_{0}^{∞}w(x)[(K̃g)(x)]^{qs/p} dx)^{p/(qs)} ≤ C(∫_{0}^{∞}g(x)^{s} dx)^{1/s},  (2.5)

where K̃ is the integral operator with the kernel k̃(x, 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

sup_{y>0} ∫_{0}^{∞}[k̃(x, y)]^{q/p}w(x) dx = sup_{y>0} ∫_{0}^{∞}[\frac{yk(x, y)}{x}]^{q/p} w(x) dx < ∞,

and (2.5) with s = p/q when 0 < q < p if and only if

∫_{0}^{∞}(∫_{0}^{∞}k̃(x, y)w(x) dx)^{r} dy = ∫_{0}^{∞}(∫_{0}^{∞} \frac{yk(x, y)}{x}w(x) dx)^{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 P_{β} with the kernel k_{β}(x, y) = βx^{−β}y^{β−1}χ_{(0,x)}, x, y > 0, defined by (1.3).

Theorem 2.2. Let β > 0 and suppose u, v are positive locally integrable functions on (0, ∞) with ∫_{0}^{a} t^{β−1}v(t)^{−1} dt < ∞ for all a > 0. Set w(x) = u(x) (exp[P_{β} log(1/v)(x)])^{q/p}. Then, a necessary and sufficient condition that the inequality

(∫_{0}^{∞}u(x) (exp[(P_{β} log f)(x)])^{q} dx)^{1/q} ≤ C(∫_{0}^{∞}v(x)f(x)^{p} dx)^{1/p} (2.6)

holds with C > 0 independent of f > 0 is (a)

sup_{y>0} ∫_{y}^{∞}(\frac{y^{β}}{x^{β+1}} )^{q/p} w(x) dx < ∞,  (2.7)

when 0 < p ≤ q < ∞; and (b)

∫_{0}^{∞}(∫_{y}^{∞} \frac{y^{β}w(x)}{x^{β+1}} dx)^{r} 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, K̃ = P_{β}, and s = 1. Given fixed y > 0, define

h(t) = e^{1−1/β}βt^{β−1}y^{−β}χ_{(0,y)}(t) + βt^{1−β}y^{β}χ_{(y,∞)}(t), t > 0.

Then, for x > y,

(P_{β} log h)(x) = βx^{−β} ∫_{0}^{y} t^{β−1} log (e^{1−1/β}βy^{−β}t^{β−1}) dt

+ βx^{−β} ∫_{y}^{x} t^{β−1} log (y^{β}t^{1−β}) dt

= −\frac{2}{β} (\frac{y}{x})^{β} − (1 − β) log x + β log y + 1 + \frac{1}{β} + log β.

Substituting this in (2.6) and restricting the integral on the left side to (y, ∞), we obtain

C_{1} ∫_{y}^{∞}(\frac{y^{β}w(x)}{x^{β+1}})^{q/p} dx

≤ C_{1}(∫_{y}^{∞}w(x)^{q/p} exp [−\frac{2}{β} \frac{q}{p} (yx)^{β}] exp [−\frac{q}{p}(1 + β) log x]

exp [\frac{q}{p}β log y] dx)^{p/q}

≤ C ∫_{0}^{∞}h(x) dx = C (1 + e^{1−1/β}) ,

where C_{1} = β exp(1 + 1/β) > 0, so that (2.7) holds.

Now, let w_{n}(t) = w(t)χ_{(1/n,n)}(t) and define h by

h_{n}(t) = t^{βqr/p}(∫_{t}^{∞} \frac{w_{n}(s)}{s^{β+1}} ds)^{qr/p}.

Then

(P_{β} log h_{n})(y) = βy^{−β} ∫_{0}^{y} x^{β−1} log [x^{βqr/p}(∫_{x}^{∞} \frac{w_{n}(t)}{t^{β+1}} dt)^{qr/p}] dx

= \frac{β^{2}qr}{p}y^{−β} ∫_{0}^{y} x^{β−1} log x dx + \frac{βqr}{p}y^{−β} ∫_{0}^{y} x^{β−1} log[∫_{x}^{∞} \frac{w_{n}(t)}{t^{β+1}} dt] dx

≥ \frac{β^{2}qr}{p}y^{−β} (\frac{y^{β}}{β} log y − \frac{y^{β}}{β^{2}} ) + \frac{βqr}{p} log[∫_{y}^{∞} \frac{w_{n}(x)}{x^{β+1}} dx]y^{−β} ∫_{0}^{y} x^{β−1} dx

≥ log y^{βqr/p} − \frac{qr}{p} + log [(∫_{y}^{∞} \frac{w_{n}(x)}{x^{β+1}} dx)^{qr/p}] .

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)P_{β} log v], we get

e^{−qr/p} ∫_{0}^{∞}w_{n}(y)y^{βqr/p}(∫_{y}^{∞} \frac{w_{n}(x)}{x^{β+1}} dx)^{qr/p} ≤ C(∫_{0}^{∞}(y^{β} ∫_{y}^{∞} \frac{w_{n}(x)}{x^{β+1}} dx)^{r} dy)^{q/p}.

Now, qr/p = r − 1 and βqr/p + β = βr, so the left side of (2.5) equals

−\frac{e^{−qr/p}}{r} ∫_{0}^{∞}y^{βqr/p+β+1} \frac{d}{dy} (∫_{y}^{∞} \frac{w_{n}(x)}{x^{β+1}} dx)^{r} dy

= \frac{e^{−qr/p}}{r} (\frac{βqr}{p} + β) ∫_{0}^{∞}(y^{β} ∫_{y}^{∞} \frac{w_{n}(x)}{x^{β+1}} dx)^{r} dy

Hence,

∫_{0}^{∞}(y^{β} ∫_{y}^{∞} \frac{w_{n}(x)}{x^{β+1}} dx)^{r} dy ≤ (\frac{re^{qr/p}}{(βqr/p) + β} · C)^{(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) = (x^{δ}, x^{γ}), 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) = e^{qγ/(βp)}x^{δ+qγ/p} and (see (2.7))

sup_{s>0} s^{aβ/p}(∫_{s}^{∞}t^{δ−qγ/p−q(1+aβ)/p} dt)^{1/q} < ∞.

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.

A result similar to that of Theorem 2.2 holds also for the operator Q_{β} defined by

(Q_{β}f)(x) = βx^{β} ∫_{x}^{∞}y^{−β−1}f(y) dy, β > 0.

Corollary 2.4. Let u, v and f be as in Theorem 2.2, and

w̃(x) = x^{−2}u(1/x) (exp [P_{β}(t^{2}/v(1/t))])^{q/p}

Then the inequality (2.6) with P_{β} replaced by Q_{β} is satisfied in the range 0 < p ≤ q < ∞, if and only if (2.7), with w replaced by w̃, holds.

In the case 0 < q < p < ∞, (2.6) with P_{β} replaced by Q_{β} holds, if and only if, (2.8) with w replaced by w̃ is satisfied.

Proof. Observe that the change of variable y = 1/t in the definition of Q_{β} shows that (Q_{β}f)(x) = (P_{β}f̃)(1/x), where f̃(x) = f(1/x). Hence Theorem 2.2 and some obvious changes of variables yields

(∫_{0}^{∞}u(x) (exp[Q_{β} log f(x)])^{q} dx)^{1/q} = (∫_{0}^{∞}u(x) (exp[P_{β} log f̃(1/x)])^{q} dx)^{1/q}

and

(∫_{0}^{∞}x^{−2}u(1/x) (exp[P_{β} log f̃(x)])^{q} dx)^{1/q} ≤ C(∫_{0}^{∞}x^{−2}v(1/x)f̃(x)^{p} dx)^{1/p}

= C(∫_{0}^{∞}v(x)f(x)^{p} dx)^{1/p},

and we are done.

When p = q = 1 and β = 1, the following equivalent statements hold for P = P_{1}:

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:

∫_{0}^{∞}u(x) exp(P log f)(x) dx ≤ C_{1} ∫_{0}^{∞}v(x)f(x) dx;  (2.9)
(P w)(x) + α^{−1}(Q_{α}w)(x) ≤ C_{2};  (2.10)
P w ∈ L_{∞};  (2.11)
∫_{0}^{∞}w(x)(P f^{1/p})^{p}(x) dx ≤ C_{3} ∫_{0}^{∞}f(x) dx,  p > 1,  (2.12)

where C_{3} remains bounded as p → ∞.

For the proof see [7].

In our next result we consider the Laplace transform L defined by

(Lf)(x) = ∫_{0}^{∞}e^{−xy}f(y) dy, x > 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

∫_{0}^{∞}u(x) exp [\frac{1}{x}(L log f)(1/x)] dx ≤ C ∫_{0}^{∞}v(x)f(x) dx  (2.13)

if and only if

sup_{t>0} t ∫_{0}^{∞}e^{−tx}w(1/x) dx < ∞,  (2.14)

where w(x) = u(x) exp[x^{−1} ∫_{0}^{x} e^{−y/x} log(1/v(y)) dy].

Proof. Let k(x, y) = x^{−1}e^{−y/x}, then x^{−1}(Lf)(x^{−1}) = (Kf)(x) satisfies the con-

ditions of Theorem 2.1 (observe that ∫_{0}^{∞}e^{−y} log y dy exists, so (iii) is satisfied)

whence the sufficiency part follows.

Conversely, if g = vf, then (2.13) is equivalent to

∫_{0}^{∞}w(x) exp(K log g)(x) dx ≤ C ∫_{0}^{∞}g(x) dx  (2.15)

where (Kh)(x) = (1/x)(Lh)(1/x). Now, let g(x) = g_{t}(x) = t^{−1}e^{−x/t}, t > 0 fixed. Then, (K log g_{t})(x) = −(x/t) − log t and substituting this into (2.15) we obtain that

(Kw)(t) = t^{−1} ∫_{0}^{∞}w(x)e^{−x/t} dx ≤ C.

Next, define K_{1} by

(K_{1}h)(x) = x ∫_{0}^{∞}e^{−x/y}y^{−2}h(y) dy.

Then,

(K_{1}(Kw))(x) ≤ (K_{1}C)(x) = Cx ∫_{0}^{∞}e^{−x/y}y^{−2} dy = C.

The proof will be complete if we can show that there exists a constant C_{1} > 0 such that (K_{1}w)(x) ≤ C_{1}(K_{1}(Kw))(x). But,

(K_{1}(Kw))(x) = x ∫_{0}^{∞}e^{−x/y}(Kw)(y) \frac{dy}{y^{2}}

= x ∫_{0}^{∞}e^{−x/y}y^{−3}(∫_{0}^{∞}e^{−t/y}w(t) dt) dy

= x ∫_{0}^{∞}w(t)(∫_{0}^{∞}e^{−(x+t)/y}y^{−3} dy) dt

= x ∫_{0}^{∞}w(t)(∫_{0}^{∞}e^{−(x+t)s}s ds) dt

= x ∫_{0}^{∞} \frac{w(t) dt}{t^{2}(1 + x/t)^{2}}

≥ \frac{e}{4} x ∫_{0}^{∞} \frac{w(t)}{t^{2}} e^{−x/t} dt

= \frac{e}{4}(K_{1}w)(x).

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 L^{p}–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 = (x_{1}, x_{2}), ℝ_{+}^{2} = (0, ∞) × (0, ∞), dx = dx_{1}dx_{2}, with the two dimensional averaging operator being given by

(P^{2}f)(x) = \frac{1}{x_{1}x_{2}} ∫_{0}^{x_{1}} ∫_{0}^{x_{2}}f(y) dy.

Theorem 2.7. Let u, v and f be positive functions on ℝ_{+}^{2}. Then

∫_{ℝ_{+}^{2}} u(x) exp(P^{2} log f)(x) dx ≤ C ∫_{ℝ_{+}^{2}} v(x)f(x) dx  (2.16)

if and only if for any α_{1}, α_{2} > 0,

sup_{y_{1}>0,y_{2}>0} y_{1}^{α_{1}}y_{2}^{α_{2}} ∫_{y_{1}}^{∞} ∫_{y_{2}}^{∞}x_{1}^{−(1+α_{1})}x_{2}^{−(1+α_{2})}w(x) dx = A < ∞  (2.17)

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 = fv. Then (2.16) is equivalent to

∫_{ℝ_{+}^{2}} w(x) exp(P^{2} log g)(x) dx ≤ C ∫_{ℝ_{+}^{2}} g(x) dx  (2.18)

and an obvious substitution gives

∫_{ℝ_{+}^{2}} w(x) exp(∫_{0}^{1}∫_{0}^{1} log g(x_{1}t_{1}, x_{2}t_{2}) dt) dx ≤ C ∫_{ℝ_{+}^{2}} g(x) dx.

But for α_{1} > 0, α_{2} > 0,

e^{−(α_{1}+α_{2})} = exp ∫_{0}^{1}∫_{0}^{1} log (t_{1}^{α_{1}}t_{2}^{α_{2}}) dt

and the left side of (2.18) becomes

e^{α_{1}+α_{2}} ∫_{ℝ_{+}^{2}} w(x) exp(∫_{0}^{1}∫_{0}^{1} log(t_{1}^{α_{1}}t_{2}^{α_{2}}g(x_{1}t_{1}, x_{2}t_{2})) dt) dx

≤ e^{α_{1}+α_{2}} ∫_{ℝ_{+}^{2}} w(x) ∫_{0}^{1}∫_{0}^{1} t_{1}^{α_{1}}t_{2}^{α_{2}}g(x_{1}t_{1}, x_{2}t_{2}) dt dx,

by Jensen’s inequality. Changing variables (twice) and interchanging the order of integration the latter is equal to

e^{α_{1}+α_{2}} ∫_{ℝ_{+}^{2}} g(y) (y_{1}^{α_{1}}y_{2}^{α_{2}} ∫_{y_{1}}^{∞} ∫_{y_{2}}^{∞}w(x)x_{1}^{−(1+α_{1})}x_{2}^{−(1+α_{2})} dx) dy,

so that (2.17) yields (2.18) with C = e^{α_{1}+α_{2}}A.

We now show that (2.16) implies (2.17).

For fixed t_{1} > 0, t_{2} > 0 substitute

g(x) =t_{1}^{−1}t_{2}^{−1}χ_{(0,t_{1})}χ_{(0,t_{2})}(x_{2}) + t_{1}^{−1}χ_{(0,t_{1})}(x_{1})\frac{e^{−(1+α_{2})}t_{2}^{α_{2}}}{x_{2}^{1+α_{2}}} χ_{(t_{2},∞)}(x_{2})

+ t_{2}^{−1}χ_{(0,t_{2})}(x_{2})\frac{e^{−(1+α_{1})}t_{1}^{α_{1}}}{x_{1}^{1+α_{1}}} χ_{(t_{1},∞)}(x_{1})

+ \frac{e^{−(2+α_{1}+α_{2})}t_{1}^{α_{1}}t_{2}^{α_{2}}}{x_{1}^{1+α_{1}}x_{2}^{1+α_{2}}} χ_{(t_{1},∞)}(x_{1})χ_{(t_{2},∞)}(x_{2})

in (2.16). The right side of (2.16) is

C [1 + e^{−1−α_{2}}/α_{2} + e^{−1−α_{1}}/α_{1} + e^{−2−α_{1}−α_{2}}/(α_{1}α_{2})]

WEIGHTED EXPONENTIAL INEQUALITIES 79

Let us write the left side in the form

∫_{ℝ_{+}^{2}} = ∫_{0}^{t_{1}}∫_{0}^{t_{2}} + ∫_{0}^{t_{1}} ∫_{t_{2}}^{∞}+ ∫_{t_{1}}^{∞}∫_{0}^{t_{2}} + ∫_{t_{1}}^{∞} ∫_{t_{2}}^{∞} ≡ I_{1} + I_{2} + I_{3} + I_{4}.

Each integral is positive, so, if we show that

I_{4} = t_{1}^{α_{1}}t_{2}^{α_{2}} ∫_{t_{1}}^{∞} ∫_{t_{2}}^{∞}w(x)x_{1}^{−(1+α_{1})}x_{2}^{−(1+α_{2})} dx,

the result will follow. We have

I_{4} = ∫_{t_{1}}^{∞} ∫_{t_{2}}^{∞}w(x) exp [\frac{1}{x_{1}x_{2}} (∫_{0}^{t_{1}}∫_{0}^{t_{2}} + ∫_{0}^{t_{1}} ∫_{t_{2}}^{x_{2}}+ ∫_{t_{1}}^{x_{1}} ∫_{0}^{t_{2}} + ∫_{t_{1}}^{x_{1}} ∫_{t_{2}}^{x_{2}}) log g(y) dy] dx

= ∫_{t_{1}}^{∞} ∫_{t_{2}}^{∞}w(x) exp [\frac{t_{1}t_{2}}{x_{1}x_{2}} log(t_{1}^{−1}t_{2}^{−1})]

exp [\frac{t_{1}}{x_{1}x_{2}} ∫_{t_{2}}^{x_{2}}log (\frac{t_{1}^{−1}e^{−(1+α_{2})}t_{2}^{α_{2}}}{y_{2}^{1+α_{2}}} ) dy_{2}]

exp[\frac{t_{2}}{x_{1}x_{2}} ∫_{t_{1}}^{x_{1}}log (\frac{t_{2}^{−1}e^{−(1+α_{1})}t_{1}^{α_{1}}}{y_{1}^{1+α_{1}}} ) dy_{1}]

exp[\frac{1}{x_{1}x_{2}} ∫_{t_{1}}^{x_{1}} ∫_{t_{2}}^{x_{2}}log (\frac{e^{−(2+α_{1}+α_{2})}t_{1}^{α_{1}}t_{2}^{α_{2}}}{y_{1}^{1+α_{1}}y_{2}^{1+α_{2}}} ) dy] dx

= ∫_{t_{1}}^{∞} ∫_{t_{2}}^{∞}w(x)E_{1}(x)E_{2}(x)E_{3}(x)E_{4}(x) dx.

Clearly,

E_{1}(x) = exp [−\frac{t_{1}t_{2}}{x_{1}x_{2}} log t_{1} − \frac{t_{1}t_{2}}{x_{1}x_{2}} log t_{2}] .

An elementary calculation shows that

E_{2}(x) = exp [−\frac{t_{1}}{x_{1}} log t_{1} + \frac{t_{1}t_{2}}{x_{1}x_{2}} log t_{1} + \frac{α_{2}t_{1}}{x_{1}} log t_{2}

+\frac{t_{1}t_{2}}{x_{1}x_{2}} log t_{2} − \frac{(1 + α_{2})t_{1}}{x_{1}} log x_{2}]

so that

E_{1}(x)E_{2}(x) = exp [−\frac{t_{1}}{x_{1}} log t_{1} + \frac{α_{2}t_{1}}{x_{1}} log t_{2} − \frac{(1 + α_{2})t_{1}}{x_{1}} log x_{2}] .

Further calculation – elementary again, but rather tedious so we omit details – yields

E_{3}(x) = exp [−\frac{t_{2}}{x_{2}} log t_{2} + \frac{t_{1}t_{2}}{x_{1}x_{2}} log t_{2} + \frac{α_{1}t_{2}}{x_{2}} log t_{1}

+\frac{t_{1}t_{2}}{x_{1}x_{2}} log t_{1} − \frac{(1 + α_{1})t_{2}}{x_{2}} log x_{1}] ,

E_{4}(x) = \frac{t_{1}^{α_{1}}t_{2}^{α_{2}}}{x_{1}^{1+α_{1}}x_{2}^{1+α_{2}}} exp [−\frac{α_{1}t_{2}}{x_{2}} log t_{1} + \frac{t_{1}}{x_{1}} log t_{1} − \frac{t_{1}t_{2}}{x_{1}x_{2}} log t_{1} + \frac{t_{2}}{x_{2}} log t_{2}

−\frac{α_{1}t_{1}}{x_{1}} log t_{2} − \frac{t_{1}t_{2}}{x_{1}x_{2}} log t_{2} + \frac{(1 + α_{1})t_{2}}{x_{2}} log x_{1} + \frac{(1 + α_{2})t_{1}}{x_{1}} log x_{2}]

and

E_{3}(x)E_{4}(x) = \frac{t_{1}^{α_{1}}t_{2}^{α_{2}}}{x_{1}^{1+α_{1}}x_{2}^{1+α_{2}}} exp [\frac{t_{1}}{x_{1}} log t_{1} − \frac{α_{2}t_{1}}{x_{1}} log t_{2} + \frac{(1 + α_{2})t_{1}}{x_{1}} log x_{2}] ,

thus

E_{1}(x)E_{2}(x)E_{3}(x)E_{4}(x) = \frac{t_{1}^{α_{1}}t_{2}^{α_{2}}}{x_{1}^{1+α_{1}}x_{2}^{1+α_{2}}} .

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 {u_{n}}_{n=1}^{∞}, {v_{n}}_{n=1}^{∞}, {a_{n}}_{n=1}^{∞} be sequences of positive numbers and set

w_{n} = (u_{n} exp[\frac{1}{n} ∑_{k=1}^{n} log(1/v_{k})])^{q/p} , n = 1, 2, ….

(i) If 0 < p ≤ q < ∞, then there is a constant C > 0 such that the inequality

(∑_{n=1}^{∞} u_{n}(a_{1}a_{2} … a_{n})^{q/n})^{1/q} ≤ C(∑_{n=1}^{∞} v_{n}a_{n}^{p})^{1/p}  (3.1)

if and only if for any a > 0

sup_{m≥1} m^{a/p}( ∑_{n=m}^{∞} \frac{w_{n}}{n^{(1+a)q/p}} )^{1/q} < ∞.  (3.2)

(ii) If 1 < q < p < ∞, then the inequality (3.1) holds, provided the condition

∑_{m=1}^{∞}( ∑_{n=m}^{∞} n^{−q(a+1)}w_{n})^{r/q}m^{(a+1/q′)r} < ∞ is satisfied, where a > 0 and

1/r = 1/q − 1/p.

Proof. (i) Let b_{n}^{p} = v_{n}a_{n}^{p}, n = 1, 2 . . .. Then, (3.1) is equivalent to

(∑_{n=1}^{∞} w_{n} (exp[\frac{1}{n} ∑_{k=1}^{n} log b_{k}])^{q})^{1/q} ≤ C(∑_{n=1}^{∞} b_{n}^{p})^{1/p}.  (3.3)

For any p > 0, if we write γ = p/α where α > 1, then, 0 < p = γα ≤ q < ∞, or 1 < α ≤ q/γ. With c_{k} = b_{k}^{γ} the inequality (3.3) is equivalent to

(∑_{n=1}^{∞} w_{n} (exp[\frac{1}{n} ∑_{k=1}^{n} log c_{k}])^{q/γ})^{γ/q} ≤ C^{γ}(∑_{n=1}^{∞} c_{n}^{α})^{1/α}.  (3.4)

To prove (3.4) define f(t) = c_{k}, if k−1 < t ≤ k, k = 1, 2 …, and zero otherwise.

Then, since ∫_{0}^{1} log y dy = −1, Jensen’s inequality shows that the left side of (3.4)

is

(∑_{n=1}^{∞} w_{n}(exp[\frac{1}{n} ∑_{k=1}^{n} ∫_{k−1}^{k} log f(t) dt])^{q/γ})^{γ/q}

= (∑_{n=1}^{∞} w_{n}(exp[\frac{1}{n} ∫_{0}^{n} log f(t) dt])^{q/γ})^{γ/q}

= (∑_{n=1}^{∞} w_{n}(exp[∫_{0}^{1} log f(ny) dy])^{q/γ})^{γ/q} (t = ny)

= e(∑_{n=1}^{∞} w_{n}(exp[∫_{0}^{1} log(yf(ny)) dy])^{q/γ})^{γ/q}

≤ e(∑_{n=1}^{∞} w_{n}(∫_{0}^{1} yf(ny) dy)^{q/γ})^{γ/q}

= e(∑_{n=1}^{∞} w_{n}n^{−2q/γ}(∫_{0}^{n} f(t) dt)^{q/γ})^{γ/q}

= e(∑_{n=1}^{∞} \frac{w_{n}}{n^{2q/γ}} (∑_{k=1}^{n} kc_{k})^{q/γ})^{γ/q}.

Applying the discrete version of the weighted Hardy inequality (cf. [1], [4]), the last expression is dominated by

(∑_{n=1}^{∞} k^{−α} (k^{α}c_{k}^{α}))^{1/α} = (∑_{n=1}^{∞} c_{k}^{α})^{1/α}

provided in case 0 < p ≤ q < ∞

sup_{m≥1}( ∑_{n=m}^{∞} \frac{w_{n}}{n^{2q/γ}} )^{γ/q}( ∑_{n=m}^{∞} k^{−α(1−α′)})^{γ/q′}

= sup_{m≥1} m^{1+1/α′}( ∑_{n=m}^{∞} \frac{w_{n}}{n^{2qα/p}} )^{p/(qα)}

= (sup_{m≥1} m^{(2α/p)−(1/p)}( ∑_{n=m}^{∞} \frac{w_{n}}{n^{2qα/p}} )^{1/q})^{p/α}

< ∞.

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

e(∑_{n=1}^{∞} w_{n}n^{−q(α+1)}(∑_{k=1}^{n} k^{α}b_{k})^{q})^{1/q}, α > 0,

and again by the discrete form of Hardy’s inequality ([15], [16]) this is dominated by

(∑_{k=1}^{∞} k^{−αp} (k^{α}b_{k})^{p})^{1/p} = (∑_{k=1}^{∞} b_{k}^{p})^{1/p}

whenever

∑_{m=1}^{∞}( ∑_{n=m}^{∞} \frac{w_{n}}{n^{q(α+1)}} )^{r/q}(∑_{n=1}^{∞} n^{αp′})^{r/q′}m^{αp′} < ∞.

But since

(αp′ + 1)\frac{r}{q′} + αp′ = r (\frac{αp′}{q′} + \frac{1}{q′} + \frac{αp′}{q} − \frac{αp′}{p} )

= r (αp′ + \frac{1}{q′} − αp′ + α) = r (α + \frac{1}{q′} ) ,

this is the condition given in (ii) with a = α.

To prove necessity of part (i) let b_{n}^{p} = 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

(∑_{n=1}^{k} + ∑_{n=k+1}^{∞})^{1/q} = (S_{1} + S_{2})^{1/q}.

Now,

S_{1} = ∑_{n=1}^{k} w_{n} (exp[\frac{1}{n} ∑_{j=1}^{n} log(k − 1/p)])^{q} = k^{−q/p} ∑_{n=1}^{k} w_{n}

WEIGHTED EXPONENTIAL INEQUALITIES 83

and

S_{2} = ∑_{n=k+1}^{∞} w_{n} (exp[\frac{1}{n} ∑_{j=1}^{k} log(1/k)])^{q/p} (exp[\frac{1}{n} ∑_{j=k+1}^{n} log (\frac{e^{−(1+α)}k^{α}}{j^{α+1}} ]))^{q/p}

= ∑_{n=k+1}^{∞} w_{n} (exp [\frac{−k}{n} log k])^{q/p} (exp [\frac{n − k − 1}{n} log (e^{−(1+α)}k^{α})])^{q/p} ×

× (exp[−\frac{α + 1}{n} ∑_{j=k+1}^{n} log j])^{q/p}.

But the Mean Value Theorem shows that

∑_{j=k+1}^{n} log j = ∑_{j=k+2}^{n+1} j log(j − 1) − ∑_{j=k+1}^{n} j log j

= ∑_{j=k+1}^{n} j[log(j − 1) − log j] − (k + 1) log k + (n + 1) log n

= ∑_{j=k+1}^{n} \frac{−j}{c′_{j}} + (n + 1) log n − (k + 1) log k

m where j − 1 ≤ c′_{j} ≤ j. Hence

∑_{j=k+1}^{n} log j ≤ ∑_{j=k+1}^{n}(−1) + (n + 1) log n − (k + 1) log k

so

S_{2} ≥ ∑_{n=k+1}^{∞} w_{n} (exp [−\frac{k}{n} log k])^{q/p} (e^{−(1+α)}k^{α})^{q/p}

× (exp [\frac{k + 1}{n}(1 + α)])^{q/p} (exp [−\frac{α(k + 1)}{n} log k])^{q/p}

× exp [−\frac{(α + 1)}{n}(−n + k + 1 + (n + 1) log n − (k + 1) log k)]

= e^{−(1+α)q/p}k^{αq/p} ∑_{n=k+1}^{∞} w_{n} (exp [−\frac{k}{n} log k])^{q/p} (exp [−\frac{α(k + 1)}{n} log k])^{q/p}

× (exp [−(α + 1) (\frac{n + 1}{n}) log n])^{q/p} (exp [(α + 1)\frac{α(k + 1)}{n} log k])^{q/p}

× (e^{α+1})^{q/p} (exp [−(α + 1) (\frac{k + 1}{n})])^{q/p}

≥ e^{−(1+α)q/p}k^{αq/p} ∑_{n=k+1}^{∞} w_{n}n^{−(α+1)q/p} (exp [\frac{1}{n} log (\frac{k}{n^{α+1}} )])^{q/p}

since \frac{k + 1}{n} ≤ 1. But, since, for k ≥ 1 (n ≥ 2),

\frac{1}{n} log (\frac{k}{n^{α+1}} ) ≥ −\frac{α + 1}{n} log n ≥ −\frac{α + 1}{n} log 2,

it follows that

(exp [\frac{1}{n} log \frac{k}{n^{α+1}} ])^{q/p} = exp [\frac{q}{p} (\frac{1}{n} log \frac{k}{n^{α+1}} )]
≥ 2^{−q(α+1)/(2p)}.

Therefore

S_{2} ≥ e^{−(1+α)q/p}2^{−q(α+1)/(2p)}k^{αq/p} ∑_{n=k+1}^{∞} w_{n}n^{−(α+1)q/p},

and so

(S_{1} + S_{2})^{1/q} ≥ e^{−(1+α)/p}2^{−(α+1)/(2p)} (k^{−q/p}w_{k} + k^{αq/p} ∑_{n=k}^{∞} w_{n}n^{−(α+1)q/p}
−k^{αq/p}w_{k}k^{−(α+1)q/p})^{1/q}

= e^{−(1+α)/p}2^{−(α+1)/(2p)}(k^{αq/p} ∑_{n=k}^{∞} \frac{w_{n}}{n^{(α+1)q/p}} )^{1/q}.

This implies condition (3.2) with a = α and completes the proof of the theorem.

Remark 3.2. i) If u_{n} = v_{n} = 1, then w_{n} = 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_{m≥1} m^{a} ∑_{n=m}^{∞} \frac{w_{n}}{n^{(1+α)}} = A,

then the constant C in (3.1) satisfies

w^{−(1+a)/2}A(1 + 3e^{−(1+a)}/(2a)) ≤ C ≤ e^{a}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 {u_{n,m}}, {v_{n,m}}, {a_{n,m}} be double sequences of positive numbers and

w_{n,m} = u_{n,m} exp(\frac{1}{nm} ∑_{\begin{matrix}1≤k≤n \\ 1≤j≤m \end{matrix}} log 1/v_{k,j}).

Then,

∑_{n,m≥1} u_{n,m}( ∏_{\begin{matrix}1≤k≤n \\ 1≤j≤m \end{matrix}} a_{j,k})^{1/(nm)} ≤ C ∑_{n,m≥1} v_{n,m}a_{n,m}  (3.5)

if and only if for some α_{1} > 0, α_{2} > 0,

sup_{\begin{matrix}k≥1 \\ j≥1 \end{matrix}} k^{α_{1}}j^{α_{2}} ∑_{\begin{matrix}n≥k \\ m≥j \end{matrix}} n^{−α_{1}−1}m^{−α_{2}−1}w_{n,m} ≡ A < ∞.  (3.6)

For the proof observe that the substitution b_{n,m} = a_{n,m}v_{n,m} allows us to write (3.5) in the equivalent form

∑_{n,m≥1} w_{n,m} exp(\frac{1}{nm} ∑_{\begin{matrix}1≤k≤n \\ 1≤j≤m \end{matrix}} log b_{j,k}) ≤ C ∑_{n,m≥1} b_{n,m}.  (3.7)

Now, let f(s, t) = b_{j,k} if k − 1 < t ≤ k, j − 1 < s ≤ j, k, j = 1, 2 …, and zero otherwise. Since

e^{−α_{1}−α_{2}} = exp ∫_{0}^{1}∫_{0}^{1} log (t_{1}^{α_{1}}t_{2}^{α_{2}}) dt_{1}dt_{2},

the left side of (3.7) takes the form

∑_{n,m≥1} w_{n,m} exp(\frac{1}{nm} ∑_{\begin{matrix}1≤k≤n \\ 1≤j≤m \end{matrix}} ∫_{k−1}^{k} ∫_{j−1}^{j} log f(s, t) dsdt)

and this is the case dealt with in the previous theorem.

For the converse we substitute into (3.7) b_{n,m} defined by

b_{n,m} = \begin{cases}k^{−1}j^{−1} & if n ≤ k, m ≤ j, \\ k^{−1}e^{−1−α_{2}}j^{α_{2}}m^{−α_{2}−1} & if n ≤ k, j > m, \\ j^{−1}e^{−1−α_{1}}k^{α_{1}}n^{−α_{1}−1} & if n > k, j ≤ m, \\ e^{−2−α_{1}−α_{2}}k^{α_{1}}j^{α_{2}}n^{−α_{1}−1}m^{−α_{2}−1} & if n > k, j ≥ m, \end{cases}

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