On H\"{o}lder and Minkowski Type Inequalities

We obtain inequalities of H\"{o}lder and Minkowski type with weights generalizing both the case of weights with alternating signs and the classical case of non-negative weights.

In this paper, we intend to give inequalities of Hölder and Minkowski type with more general weights, including both the case of weights with alternating signs and the classical case of nonnegative weights (see, e.g., [12,Section 4.2] and [1,13]). Namely, weights , = 1, . . . , , satisfying the property are considered. We follow proofs in [1] with several changes in order to obtain our results.

Hölder Type Inequalities
In this section, we show that there is not a direct analog of Hölder's inequality in the case of our weights, but one of reverse Hölder's inequalities exists. Note that reverse Hölder's inequalities for nonnegative weights are well studied (see [13]). Theorem 1. Let a and b be nonincreasing such that where = 1, . . . , . If, moreover, ⩾ 0, and , > 1, 1/ + 1/ = 1, then The left hand side of (6) should be read as there exists no positive constant, depending on , , , , , or , which bounds the fraction in (6) from below.
Proof. Applying the Abel transformation, we have where the latter expression is nonnegative since the sequences a and { − } are nonincreasing, and ⩾ 0. The equality holds, for example, if b ≡ .
Proof. We denote the fraction in (6) by . Applying the Abel transformation to the numerator and the denominator of easily yields ⩾ 0. But we prove even more, namely, that there exist no positive constants bounding from below. Following [1], let = (−1) +1 , = 1, . . . , , where is even, and let a = { 1 , 1 , 3 , 3 , . . . , , , . . .} be positive and nondecreasing. The sequence b is arbitrary except such that 2 −1 − 2 = 0 for all = 1, . . . , /2. It follows that Thus cannot be bounded from below by a positive absolute constant or a constant depending on , , maximum or minimum elements of a and b. Now we prove the right hand side of (6). Here denotes the numerator of . First we apply the Abel transformation: By the right hand side of (4) and the Abel transformation where and are arbitrary positive constants. Therefore, (3) after several simplifications gives ) . (12) In the latter expression, { /( )+ /( )} is nondecreasing and { } is nonincreasing, because a and b are nonincreasing. Hence by Lemma 2 It is easily seen that, in order to get the smallest constant in the latter inequality, we must choose / = / . It gives the right hand side of (6). Note that the constant there belongs to (1; ∞).

Remark 3.
From Theorem 1, it is seen that the constant in the right hand side of (6) tends to infinity as → 0 or → 0 (note that this constant is better than in [1]). Now we give an example of sequences confirming this [1]. In Theorem 1 we suppose that From the left hand side of (4) we deduce where 1 − 1/ > 0. Therefore, for a fixed positive 2 the sum in the denominator can be made sufficiently small by an appropriate choice of b. Consequently, can be arbitrarily large. The same is for = 2 +1 → 0.
It is clear that if = = 2 then the constant in the right hand side of (6) equals 2 √ ( ) −1 ⩾ 2. Now we give a more precise constant belonging to [1; ∞) for the case when a and b satisfy several additional conditions. Proposition 4. Let a and b be nonincreasing such that the sequence { / } is monotone and 0 < ⩽ / ⩽ < ∞. If, moreover, ⩾ 0 for = 1, . . . , , then Abstract and Applied Analysis 3 The left hand side of (16) should be read as there exists no positive constant, depending on and , which bounds the fraction in (16) from below.
It is easily seen that equality in (16) holds, for example, if a ≡ b. The fact that the constant in the right hand side of (16) belongs to [1; ∞) is obvious.
From the well-known weighted inequality of arithmetic and geometric means (see, e.g., [14,Chapter 2]) supposing ⩾ 0 and V > 0, we have This is a multivariable version of Young's inequality (3). From this we obtain a multivariable version of Theorem 1 (but with less precise constant).
The left hand side of (20) should be read as there exists no positive constant, depending on , , and , which bounds the fraction in (20) from below.
Proof. Set is the fraction in (20). Nonexistence of a positive constant bounding from below follows from Theorem 1. To prove the right hand side we denote the numerator of by . By the Abel transformation The right hand side of (4) and the Abel transformation yields where it is obvious that Several simplifications give the right hand side of (20).

Minkowski Type Inequalities
In this section we prove precise Minkowski type inequalities with our weights. As we have already mentioned, these generalize both the case of weight with alternating signs and the case of nonnegative weights (see [1]).
Let a be a nonincreasing positive sequence and a function convex on [ ; 1 ] such that (0) = 0. Then the necessary and sufficient condition on weights in order that is 0 ⩽ ⩽ 1, = 1, . . . , .
From this point of view, the sufficient condition ⩾ 0, = 1, . . . , , in Theorems 1 and 6 seems to be quite close to the necessary one.

Further Generalizations
Now we give integral versions of Lemma 2 and Theorems 1 and 6. In what follows, we use the notation and suppose that all functions of are integrable and differentiable on [ ; ].
Using Lemma 8 and integration by parts instead of the Abel transformation, we obtain the following results by essential repeating proofs of Theorems 1 and 6. We emphasize that ( ) may be negative here in contrast to the classical case.
The left hand side of (37) should be read as there exists no positive constant, depending on , , , , , and , which bounds the fraction in (37) from below.
If ( ) is nondecreasing for ∈ [ ; ] (i.e., ( ) is nonnegative), Theorems 9 and 10 give the classical case of nonnegative weights, for which we can put 1 instead of 0 in the left hand sides of (37) and (38) due to Hölder's and Minkowski's inequalities.