Multivariate Hardy and Littlewood inequalities on time scales

Nosheen, Ammara; Nawaz, Aneela; Khan, Khuram Ali; Awan, Khalid Mahmood

doi:10.1016/j.ajmsc.2019.12.003

In the paper we extend some Hardy and Littlewood type inequalities on time scales for the function of $n$ variables. Special cases of obtained results include generalized Wirtinger, Hardy and Littlewood type inequalities.

1. Introduction

The discrete Hardy inequality [8] was proved and published by Hardy himself. It states that if $(c_{n})$ is a sequence of non-negative real numbers which are not identically zero, then for every real number $p > 1$ ⁠, one has that

\sum_{k = 1}^{\infty} {(\frac{c_{1} + c_{2} + c_{3} + \dots + c_{k}}{k})}^{p} < {(\frac{p}{p - 1})}^{p} \sum_{k = 1}^{\infty} c_{k}^{p} .

The classical Hardy inequality [9] states that if $f \geq 0$ and integrable over any finite interval $(0, r)$ and $f^{d}$ is integrable and convergent over $(0, \infty)$ then for $d > 1$ ⁠,

{\int_{0}^{\infty} (\frac{1}{r} \int_{0}^{r} f (τ) d τ)}^{d} d r \leq {(\frac{d}{d - 1})}^{d} \int_{0}^{\infty} f^{d} (r) d r,

(1)

equality holds if and only if $f (r) = 0$ almost everywhere. Hardy inequality (1) has been generalized by Hardy himself in [11], where he exposed that, for any integrable function $f (y) > 0$ on $(0, \infty)$ and $d > 1$ ⁠, the following hold

\int_{0}^{\infty} \frac{1}{y^{n}} {(\int_{y}^{\infty} f (h) d h)}^{d} d y \leq {(\frac{d}{1 - n})}^{d} \int_{0}^{\infty} \frac{1}{y^{n - d}} f^{d} (y) d y, n < 1,

(2)

\int_{0}^{\infty} \frac{1}{y^{n}} {(\int_{0}^{y} f (h) d h)}^{d} d y \leq {(\frac{d}{n - 1})}^{d} \int_{0}^{\infty} \frac{1}{y^{n - d}} f^{d} (y) d y, n > 1.

(3)

Hardy and Littlewood [10] demonstrate the discrete versions of (2) and (3). In particular they proved that if $d > 1$ and $(p_{m})$ is a sequence of non-negative terms then

\begin{array}{l} \sum_{m = 1}^{\infty} \frac{1}{m^{j}} {(\sum_{i = m}^{\infty} p_{i})}^{d} \leq N \sum_{m = 1}^{\infty} \frac{1}{m^{j - d}} p_{m}^{d}, j < 1, \\ \sum_{m = 1}^{\infty} \frac{1}{m^{j}} {(\sum_{i = 1}^{m} p_{i})}^{d} \leq N \sum_{m = 1}^{\infty} \frac{1}{m^{j - d}} p_{m}^{d}, j > 1, \end{array}

where $N$ is a non-negative constant. Time scales calculus [12] was introduced in 1988 by the German mathematician Stefan Hilger, which unifies sums and integrals. Some extension of Hardy type inequalities on time scales can be found in [2–4].

S. H. Saker et al. [13] proved some Hardy and Littlewood type inequalities on time scales in the following form:

Theorem 1.1.

Let $T$ be a time scale with $a \in {(0, \infty)}_{T}$ and $p, q > 0$ such that $p / q \geq 2$ and $γ > 1$ . Furthermore assume that $g$ is a nonnegative and the delta integral $\int_{a}^{\infty} t^{\frac{p}{q} - γ} g^{p / q} (t) Δ t$ exists. Let

Λ (t) = \int_{a}^{t} g (s) Δ s, f o r a n y t \in {[a, \infty]}_{T} .

(4)

Then one gets

\begin{array}{l} \int_{a}^{\infty} \frac{1}{t^{γ}} {(Λ^{σ} (t))}^{p / q} Δ t \leq \frac{2^{\frac{p}{q} - 2} p k^{γ}}{q (γ - 1)} {\int_{a}^{\infty} \frac{1}{t^{γ - \frac{p}{q}}} g^{p / q} (t) Δ t}^{\frac{p}{q}} \\ \times {\int_{a}^{\infty} \frac{Λ^{σ} (t) Λ^{p / q}}{t^{γ}} Δ t}^{\frac{p - q}{p}} \\ + \frac{2^{\frac{p}{q} - 2} p k^{γ}}{q (γ - 1)} \int_{a}^{\infty} \frac{μ^{\frac{p}{q}} - 1}{t^{γ - 1}} g^{p / q} (t) Δ t . \end{array}

Theorem 1.2.

Let $T$ be a time scale with $a \in {(0, \infty)}_{T}$ and $p, q > 0$ such that $p / q \geq 2$ and $γ > 1$ . Furthermore assume that $g$ is a nonnegative function and the delta integral $\int_{a}^{\infty} t^{\frac{p}{q} - γ} g^{p / q} (t) Δ t$ exist. Let $Λ (t)$ be as defined in (4). Then

\int_{a}^{\infty} \frac{1}{t^{γ}} {(Λ^{σ} (t))}^{p / q} Δ t \leq {(\frac{2^{\frac{p}{q} - 1} p k^{γ}}{q (γ - 1)})}^{p / q} \int_{a}^{\infty} \frac{1}{t^{γ - \frac{p}{q}}} g^{p / q} (t) Δ t .

Theorem 1.3.

Let $T$ be a time scale with $a \in {(0, \infty)}_{T}$ and $p, q > 0$ such that $p / q > 1$ and $γ > 1$ . Furthermore assume that $g$ is a nonnegative function and the delta integral $\int_{a}^{\infty} t^{\frac{p}{q} - γ} g^{p / q} (t) Δ t$ exists. Let $Λ (t)$ be as defined in (4). Then

\int_{a}^{\infty} \frac{1}{t^{γ}} {(Λ^{σ} (t))}^{p / q} Δ t \leq {(\frac{p k^{γ}}{q (γ - 1)})}^{p / q} \int_{a}^{\infty} \frac{1}{t^{γ - \frac{p}{q}}} g^{p / q} (t) Δ t .

Theorem 1.4.

Let $T$ be a time scale with $a \in {(0, \infty)}_{T}$ and $p, q > 0$ such that $p / q > 1$ and $γ < 1$ . Furthermore assume that $g$ is a nonnegative and delta integral $\int_{a}^{\infty} {(σ (t))}^{\frac{p}{q} - γ} g^{p / q} (t) Δ t$ exists. Let

Ω (t) = \int_{t}^{\infty} g (s) Δ s, f o r a n y t \in {[a, \infty]}_{T} .

Then one gets

\int_{a}^{\infty} \frac{{(Ω (t))}^{p / q}}{σ^{γ} (t)} \leq {(\frac{p}{q (1 - γ)})}^{p / q} \int_{a}^{\infty} \frac{g^{p / q} (t)}{{(σ (t))}^{γ - \frac{p}{q}}} Δ t .

In this paper we extend results of Theorem 1.1 to Theorem 1.4 for the function of $n$ variables.

2. Preliminaries

In this section, we recall the following concepts from theory of time scales [5,7]. A time scale is an arbitrary, non empty closed subset of real numbers. Set of integers and Cantor set are examples of time scales, while rational numbers, complex numbers and open interval between $0$ and $1$ not time scales. Let $T$ be a time scale, for $t \in T$ ⁠, forward and backward jump operators are defined by

σ (t) : = \inf {a \in T; a > t}, ρ (t) : = \sup {a \in T; a < t},

respectively. The conventions for these operators are $i n f ϕ = s u p T$ and $s u p ϕ = i n f T$ ⁠. If $σ (t) > t$ ⁠, then $t$ is right-scattered and if $ρ (t) < t$ ⁠, then $t$ is left-scattered. Points that are right-scattered and left-scattered at the same time are called isolated points.

If $σ (t) = t$ ⁠, then $t$ is right-dense and if $ρ (t) = t$ ⁠, then $t$ is left-dense. Points that are right-dense and left-dense at the same time are called dense points. The functions $μ : T \to ℝ, ν : T \to ℝ$ defined by $μ (t) = σ (t) - t$ and $ν (t) = t - ρ (t)$ are called forward and backward graininess functions, respectively.

A function $g : T \to ℝ$ is said to be right-dense continuous (rd-continuous) provided $g$ is continuous at right-dense points and at left-dense points in $T$ ⁠, left-hand limits exist and are finite. The set of all such rd-continuous functions is denoted by $C_{r d} (T)$ ⁠. For any function $g : T \to ℝ$ ⁠, the notation $g^{σ} (t)$ denotes $g (σ (t))$ ⁠. The delta derivative (also Hilger derivative) $g^{Δ} (t)$ exists if and only if for every $ϵ > 0$ there exists a neighborhood $U$ of $t$ such that

| g (σ (t)) - g (s) - g^{Δ} (t) (σ (t) - s) | \leq | σ (t) - s |, f o r a l l s, t i n U .

Assume that $h : T \to ℝ$ ⁠, if $H^{Δ} (t) = h (t)$ ⁠, then the Cauchy (delta) integral of $h$ ⁠. defined by

\int_{a}^{t} h (s) Δ s : = H (t) - H (a) .

Integration by parts formula [7, Theorem1.77]:

If $a, b \in T$ and $u, v \in C_{r d} (T)$ ⁠, then

\int_{a}^{b} u (t) v^{Δ} (t) Δ t = {[u (t) v (t)]}_{a}^{b} - \int_{a}^{b} u^{Δ} (t) v^{σ} (t) Δ t .

(5)

Chain rule 1 [7, Theorem 1.90]:

Assume that $f : ℝ \to ℝ$ is continuously differentiable and suppose $g : T \to ℝ$ is delta differentiable. Then $f o g : T \to ℝ$ is delta differentiable and

{(f o g)}^{Δ} (t) = {\int_{0}^{1} f^{'} (g (t) + h μ (t) g^{Δ} (t)) d h} g^{Δ} (t)

(6)

holds.

Chain rule 2 [7, Theorem 1.87]:

If $f$ and $g$ satisfy the conditions of Chain rule 1, Then $f o g : T \to ℝ$ is delta differentiable and there exists $c$ in the real interval $[t, σ (t)]$ such that

{(f o g)}^{Δ} (t) = f^{'} (g (c)) g^{Δ} (t) .

(7)

Hölder’s inequality [7, Theorem 6.13]:

For continuous real-valued functions $g : T \to ℝ$ ⁠, $h : T \to ℝ$ ⁠, let $a, b \in T$ ⁠, $p > 1$ and $\frac{1}{p} + \frac{1}{q} = 1$ ⁠, then

\int_{a}^{b} g (t) h (t) d t = {(\int_{a}^{b} g^{p} (t) d t)}^{1 / p} {(\int_{a}^{b} h^{q} (t) d t)}^{1 / q} .

(8)

Fubini’s Theorem on time scales [6]:

Let $(ψ, M, μ_{Δ})$ and $(Γ, N, λ_{Δ})$ be two finite dimensional time scales measure spaces. If $Λ : ψ \times Γ \to ℝ$ is a $μ_{Δ} \times λ_{Δ}$ -integrable function. The function $ς (t_{2}) = \int_{ψ} Λ (t_{1}, t_{2}) Δ t_{1}$ exists for any $t_{1} \in Γ$ and $ξ (t_{1}) = \int_{Γ} Λ (t_{1}, t_{2}) Δ t_{2}$ exists for $t_{2} \in ψ$ ⁠, then

\int_{ψ} Δ t_{1} \int_{Γ} Λ (t_{1}, t_{2}) Δ t_{2} = \int_{Γ} Δ t_{2} \int_{ψ} Λ (t_{1}, t_{2}) Δ t_{1} .

(9)

We assume throughout that all the functions are non-negative and the integrals considered exist.

In this paper, we use the following notations. We assume that there exists constant $k_{i} > 0$ with

\frac{s_{i}}{σ_{i} (s_{i})} \geq \frac{1}{k_{i}} f o r s_{i} \geq a_{i}, i \in {1, \dots, n} .

(10)

\begin{array}{l} Λ_{k}^{σ_{1} \dots σ_{j}} (t_{1}, \dots, t_{n}) ≐ Λ_{k}^{σ_{1} \dots σ_{j}} ≐ Λ_{k} (σ_{1} (t_{1}), \dots, σ_{j} (t_{j}), t_{j + 1}, \dots, t_{n}), k, j \in {1, \dots, n} \\ \int_{a_{1}}^{\infty} \dots \int_{a_{n}}^{\infty} f (t_{1}, \dots, t_{n}) Δ t_{1}, \dots, Δ t_{n} ≐ \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} f (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} Δ t_{i} . \end{array}

3. Hardy and Littlewood-type inequalities for p/q ≥ 2 and γ > 1

The following inequalities are used to prove next results.

a^{λ} + b^{λ} \leq {(a + b)}^{λ} \leq 2^{λ - 1} (a^{λ} + b^{λ}) f o r a, b \geq 0, λ \geq 1.

(11)

2^{λ - 1} (a^{λ} + b^{λ}) \leq {(a + b)}^{λ} \leq a^{λ} + b^{λ} f o r a, b \geq 0,0 \leq λ \leq 1.

(12)

Theorem 3.1.

Assume $i \in {1, \dots, n}$ , $T_{i}$ is a time scale with $a_{i} \in {(0, \infty)}_{T_{i}}$ and $γ_{i} > 1$ , further assume $g : {[a_{1}, \infty)}_{T_{1}} \times \dots \times {[a_{n}, \infty)}_{T_{n}} \to ℝ_{+}$ is such that the delta integrals

$\int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \prod_{i = 1}^{n} {(t_{i})}^{\frac{p}{q} - γ_{i}} g^{p / q} (t_{1}, \dots, t_{n}) Δ t_{i}$ for any $(t_{1}, \dots, t_{n}) \in {[a_{1}, \infty)}_{T_{1}} \times \dots \times {[a_{n}, \infty)}_{T_{n}}$ exist, define

Λ_{k} (t_{1}, \dots, t_{n}) = \int_{\prod_{j = 1}^{k} a_{j}}^{\infty} g (s_{1}, \dots, s_{n}) \prod_{j = 1}^{k} Δ s_{j}, k \in {1, \dots, n},

(13)

then for $p, q > 0$ and $p / q \geq 2$

\begin{array}{l} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{{(Λ_{n}^{σ_{1} \dots σ_{n}})}^{p / q}}{\prod_{i = 1}^{n} t_{i}^{γ_{i}}} \prod_{i = 1}^{n} Δ t_{i} \\ \leq \sum_{r = 1}^{n} \prod_{j = r + 1}^{n} c_{j} {\tilde{c}}_{r} \int_{\prod_{j = r + 1}^{n} a_{j}}^{\infty} \prod_{j = r + 1}^{n} \frac{{(μ_{j} (t_{j}))}^{(p / q - 1)}}{t_{j}^{γ_{j} - 1}} \int_{\prod_{i = 1}^{r - 1} a_{i}}^{\infty} \prod_{i = 1}^{r - 1} \frac{1}{t_{i}^{γ_{i}}} \times \end{array}

\begin{array}{l} {\int_{a_{r}}^{\infty} \frac{{(Λ_{r - 1}^{σ_{1} \dots σ_{r - 1}})}^{p / q}}{t_{r}^{γ_{r} - p / q}} Δ t_{r}}^{q / p} {(Λ_{r}^{σ_{1} \dots σ_{r}}) {(Λ_{r}^{σ_{1} \dots σ_{r - 1}})}^{p / q} Δ t_{r}}^{\frac{p - q}{p}} \prod_{i = 1}^{r - 1} Δ t_{i} \prod_{j = r + 1}^{n} Δ t_{j} \\ + \prod_{i = 1}^{n} {\tilde{c}}_{i} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \prod_{i = 1}^{n} \frac{{(μ_{i} (t_{i}))}^{(p / q - 1)}}{t_{i}^{γ_{i} - 1}} g^{p / q} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} Δ t_{i} \end{array}

(14)

holds, where

{\tilde{c}}_{r} = c_{r} p / q, c_{r} = \frac{2^{p / q - 2} k_{r}^{γ_{r}}}{γ_{r - 1}} .

Proof.

To prove the result, we use the principle of mathematical induction. For $n = 1$ the statement is true by Theorem 1.1. Let the statement be true for $1 \leq n \leq k$ ⁠.

To prove the result for $n = k + 1$ ⁠. The left-hand side of (14) can be written as,

\int_{\prod_{i = 1}^{k + 1} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{k + 1} t_{i}^{γ_{i}}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k + 1}})}^{p / q} \prod_{i = 1}^{k + 1} Δ t_{i} .

(15)

Denote $\int_{a_{k + 1}}^{\infty} \frac{{(Λ_{k + 1}^{σ_{1} \dots σ_{k + 1}})}^{p / q}}{t_{k + 1}^{γ_{k + 1}}} Δ t_{k + 1} = I_{k + 1}$ ⁠. Apply (5) with $\frac{\partial}{Δ t_{k + 1}} u (t_{k + 1}) = \frac{1}{t_{k + 1}^{γ_{k + 1}}}$ and

$v^{σ_{k + 1}} (t_{k + 1}) = {(Λ_{k + 1}^{σ_{1} \dots σ_{k + 1}})}^{p / q}$ by keeping fix $(t_{1}, \dots, t_{k}) \in {[a_{1}, \infty)}_{T_{1}} \times \dots \times {[a_{k}, \infty)}_{T_{k}}$ ⁠.

I_{k + 1} = [u (t_{k + 1}) ({(Λ_{k + 1}^{σ_{1} \dots σ_{k}})}^{p / q})] |_{a_{k + 1}}^{\infty} \int_{a_{k + 1}}^{\infty} - u (t_{k + 1}) \frac{\partial}{Δ t_{k + 1}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k}})}^{p / q} Δ t_{k + 1},

(16)

where,

u (t_{k + 1}) = \int_{t_{k + 1}}^{\infty} - \frac{1}{s_{k + 1}^{γ_{k + 1}}} Δ s_{k + 1} .

(17)

Use chain rule (6) and the fact that $σ_{k + 1} (s_{k + 1}) \geq s_{k + 1}$ to get

\begin{array}{l} \frac{\partial}{Δ s_{k + 1}} (- \frac{1}{s_{k + 1}^{γ_{k + 1} - 1}}) = (γ_{k + 1} - 1) \int_{0}^{1} {[h_{k + 1} σ_{k + 1} (s_{k + 1}) + (1 - h_{k + 1}) s_{k + 1}]}^{- γ_{k + 1}} d h_{k + 1} \\ \geq \frac{(γ_{k + 1} - 1)}{σ_{k + 1}^{γ_{k + 1}} (s_{k + 1})} . \end{array}

(18)

(10) together with (18) gives

\frac{\partial}{Δ s_{k + 1}} (- \frac{1}{s_{k + 1}^{γ_{k + 1} - 1}}) \geq \frac{(γ_{k + 1} - 1)}{k_{k + 1}^{γ_{k + 1}} s_{k + 1}^{γ_{k + 1}}} .

Therefore

\begin{array}{l} \int_{t_{k + 1}}^{\infty} - \frac{1}{s_{k + 1}^{γ_{k + 1}}} Δ s_{k + 1} \\ \geq \int_{t_{k + 1}}^{\infty} - \frac{k_{k + 1}^{γ_{k + 1}}}{γ_{k + 1} - 1} \frac{\partial}{Δ s_{k + 1}} (- \frac{1}{s_{k + 1}^{γ_{k + 1} - 1}}) Δ s_{k + 1} = - \frac{k_{k + 1}^{γ_{k + 1}}}{γ_{k + 1} - 1} (\frac{1}{t_{k + 1}^{γ_{k + 1} - 1}}) . \end{array}

(19)

(17) together with (19) gives

- u (t_{k + 1}) = - \int_{t_{k + 1}}^{\infty} - \frac{1}{s_{k + 1}^{γ_{k + 1}}} Δ s_{k + 1} \leq \frac{k_{k + 1}^{γ_{k + 1}}}{γ_{k + 1} - 1} (\frac{1}{t_{k + 1}^{γ_{k + 1} - 1}}) .

(20)

From (13), (16), (17), (20), we have (note that $u_{k + 1} (\infty) = 0$ and $Λ_{k + 1} (t_{1}, \dots, t_{k}, a_{k + 1}) = 0$ ⁠)

I_{k + 1} = \frac{k_{k + 1}^{γ_{k + 1}}}{γ_{k + 1} - 1} \int_{a_{k + 1}}^{\infty} \frac{1}{t_{k + 1}^{γ_{k + 1} - 1}} \frac{\partial}{Δ t_{k + 1}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k}})}^{p / q} Δ t_{k + 1} .

(21)

Apply chain rule 1 (6) on the right-hand side of (21)

\begin{array}{l} \frac{\partial}{Δ t_{k + 1}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k}})}^{p / q} \\ = \frac{p}{q} \frac{\partial}{Δ t_{k + 1}} Λ_{k + 1}^{σ_{1} \dots σ_{k}} {\int_{0}^{1} [Λ_{k + 1} + h_{k + 1} μ_{k + 1} (t_{k + 1}) \frac{\partial}{Δ t_{k + 1}} Λ_{k + 1}^{σ_{1} \dots σ_{k}}]}^{\frac{p}{q} - 1} d h_{k + 1} . \end{array}

(22)

Use right part of (11) on the right-hand side of (22),

\begin{array}{l} \frac{\partial}{Δ t_{k + 1}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k}})}^{p / q} \\ \leq \frac{p}{q} 2^{p / q - 2} {(Λ_{k + 1}^{σ_{1} \dots σ_{k}})}^{p / q - 1} \frac{\partial}{Δ t_{k + 1}} (Λ_{k + 1}^{σ_{1} \dots σ_{k}}) \\ + \frac{p}{q} 2^{p / q - 2} {(μ_{k + 1} (t_{k + 1}))}^{p / q - 1} {(\frac{\partial}{Δ t_{k + 1}} Λ_{k + 1}^{σ_{1} \dots σ_{k}})}^{p / q} . \end{array}

(23)

Substitute (23) into (21)

\begin{array}{l} I_{k + 1} \leq \frac{p 2^{p / q - 2} k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)} \int_{a_{k + 1}}^{\infty} \frac{1}{t_{k + 1}^{γ_{k + 1} - 1}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k}})}^{p / q - 1} \frac{\partial}{Δ t_{k + 1}} Λ_{k + 1}^{σ_{1} \dots σ_{k}} Δ t_{k + 1} \\ + \frac{p 2^{p / q - 2} k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)} \int_{a_{k + 1}}^{\infty} \frac{1}{t_{k + 1}^{γ_{k + 1} - 1}} {(μ_{k + 1} (t_{k + 1}))}^{p / q - 1} {(\frac{\partial}{Δ t_{k + 1}} Λ_{k + 1}^{σ_{1} \dots σ_{k}})}^{p / q} Δ t_{k + 1} . \end{array}

(24)

Since

\frac{\partial}{Δ t_{k + 1}} Λ_{k + 1}^{σ_{1} \dots σ_{k}} = Λ_{k}^{σ_{1} \dots σ_{k}} \geq 0.

(25)

Use (25) in (24)

\begin{array}{l} I_{k + 1} \leq \frac{p 2^{p / q - 2} k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)} \int_{a_{k + 1}}^{\infty} \frac{1}{t_{k + 1}^{γ_{k + 1} - 1}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k}})}^{p / q - 1} Λ_{k}^{σ_{1} \dots σ_{k}} Δ t_{k + 1} \\ + \frac{p 2^{p / q - 2} k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)} \int_{a_{k + 1}}^{\infty} \frac{1}{t_{k + 1}^{γ_{k + 1} - 1}} {(μ_{k + 1} (t_{k + 1}))}^{p / q - 1} {(Λ_{k}^{σ_{1} \dots σ_{k}})}^{p / q} Δ t_{k + 1} . \end{array}

(26)

Substitute (26) in (15)

\begin{array}{l} \int_{\prod_{i = 1}^{k + 1} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{k + 1} t_{i}^{γ_{i}}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k + 1}})}^{p / q} \prod_{i = 1}^{k + 1} Δ t_{i} \\ \leq \int_{\prod_{i = 1}^{k} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{k} t_{i}^{γ_{i}}} \frac{p 2^{p / q - 2} k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)} \int_{a_{k + 1}}^{\infty} \frac{1}{t_{k + 1}^{γ_{k + 1} - 1}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k}})}^{p / q - 1} Λ_{k}^{σ_{1} \dots σ_{k}} \prod_{i = 1}^{k} Δ t_{k + 1} Δ t_{i} \\ + \int_{\prod_{i = 1}^{k} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{k} t_{i}^{γ_{i}}} \frac{p 2^{p / q - 2} k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)} \int_{a_{k + 1}}^{\infty} \frac{{(μ_{k + 1} (t_{k + 1}))}^{p / q - 1}}{t_{k + 1}^{γ_{k + 1} - 1}} {(Λ_{k}^{σ_{1} \dots σ_{k}})}^{p / q} \prod_{i = 1}^{k} Δ t_{k + 1} Δ t_{i} . \end{array}

(27)

Exchange integrals on right-hand side of (27) $k$ -times by using (9)

\begin{array}{l} = \frac{p 2^{p / q - 2} k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)} \int_{a_{k + 1}}^{\infty} \frac{1}{t_{k + 1}^{γ_{k + 1} - 1}} \int_{\prod_{i = 1}^{k} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{k} t_{i}^{γ_{i}}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k}})}^{p / q - 1} Λ_{k}^{σ_{1} \dots σ_{k}} \prod_{i = 1}^{k} Δ t_{i} Δ t_{k + 1} \\ + \frac{p 2^{p / q - 2} k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)} {(μ_{k + 1} (t_{k + 1}))}^{p / q - 1} \int_{a_{k + 1}}^{\infty} \frac{1}{t_{k + 1}^{γ_{k + 1} - 1}} \\ \times \int_{\prod_{i = 1}^{k} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{k} t_{i}^{γ_{i}}} {(Λ_{k}^{σ_{1} \dots σ_{k}})}^{p / q} \prod_{i = 1}^{k} Δ t_{i} Δ t_{k + 1} . \end{array}

(28)

Use the induction hypothesis with $Λ_{k}^{σ_{1} \dots σ_{k}}$ in (28) for fixed $t_{k + 1} \in T_{k + 1}$ and again apply (9) $k$ -times to get

\begin{array}{l} = \frac{p 2^{p / q - 2} k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)} \int_{a_{k + 1}}^{\infty} \frac{1}{t_{k + 1}^{γ_{{k + 1}^{- 1}}}} \int_{\prod_{i = 1}^{k} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{k} t_{i}^{γ_{i}}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k}})}^{p / q - 1} Λ_{k}^{σ_{1} \dots σ_{k}} \prod_{i = 1}^{k} Δ t_{i} Δ t_{k + 1} \\ + \frac{p 2^{p / q - 2} k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)} {(μ_{k + 1} (t_{k + 1}))}^{p / q - 1} \int_{a_{k + 1}}^{\infty} \frac{1}{t_{k + 1}^{γ_{k + 1} - 1}} \\ \times \sum_{r = 1}^{k} \prod_{j = r + 1}^{k} c_{j} {\tilde{c}}_{r} \int_{\prod_{j = r + 1}^{k} a_{j}}^{\infty} \prod_{j = r + 1}^{k} \frac{{(μ_{j} (t_{j}))}^{(p / q - 1)}}{t_{j}^{γ_{j} - 1}} \times \\ \int_{\prod_{i = 1}^{r - 1} a_{j}}^{\infty} \prod_{i = 1}^{r - 1} \frac{1}{t_{i}^{γ_{i}}} {\int_{a_{r}}^{\infty} \frac{{(Λ_{r - 1}^{σ_{1} \dots σ_{r - 1}})}^{p / q}}{t_{r}^{γ_{r} - p / q}} Δ t_{r}}^{q / p} \\ \times {(Λ_{r}^{σ_{1} \dots σ_{r}}) {(Λ_{r}^{σ_{1} \dots σ_{r - 1}})}^{p / q} Δ t_{r}}^{\frac{p - q}{p}} \prod_{i = 1}^{r - 1} Δ t_{i} \prod_{j = r + 1}^{k} Δ t_{j} \\ + \prod_{i = 1}^{k} {\tilde{c}}_{i} \int_{\prod_{i = 1}^{k} a_{j}}^{\infty} \prod_{i = 1}^{k} \frac{{(μ_{i} (t_{i}))}^{(p / q - 1)}}{t_{i}^{γ_{i} - 1}} g^{p / q} (t_{1}, \dots, t_{k}) \prod_{i = 1}^{k} Δ t_{i} . \end{array}

Hence

\begin{array}{l} \int_{\prod_{i = 1}^{k + 1} a_{i}}^{\infty} \frac{{(Λ_{n}^{σ_{1} \dots σ_{n}})}^{p / q}}{\prod_{i = 1}^{k + 1} t_{i}^{γ_{i}}} \prod_{i = 1}^{k + 1} Δ t_{i} \\ \leq \sum_{r = 1}^{k + 1} \prod_{j = r + 1}^{k + 1} c_{j} {\tilde{c}}_{r} \int_{\prod_{j = r + 1}^{k + 1} a_{j}}^{\infty} \prod_{j = r + 1}^{k + 1} \frac{{(μ_{j} (t_{j}))}^{(p / q - 1)}}{t_{j}^{γ_{j} - 1}} \int_{\prod_{i = 1}^{r - 1} a_{i}}^{\infty} \prod_{i = 1}^{r - 1} \frac{1}{t_{i}^{γ_{i}}} \times \end{array}

\begin{array}{l} {\int_{a_{r}}^{\infty} \frac{{(Λ_{r - 1}^{σ_{1} \dots σ_{r - 1}})}^{p / q}}{t_{r}^{γ_{r} - p / q}} Δ t_{r}}^{q / p} {Λ_{r}^{σ_{1} \dots σ_{r}} {(Λ_{r}^{σ_{1} \dots σ_{r - 1}})}^{p / q} Δ t_{r}}^{\frac{p - q}{p}} \prod_{i = 1}^{r - 1} Δ t_{i} \prod_{j = r + 1}^{k + 1} Δ t_{j} \\ + \prod_{i = 1}^{k + 1} {\tilde{c}}_{i} \int_{\prod_{i = 1}^{k + 1} a_{i}}^{\infty} \prod_{i = 1}^{k + 1} \frac{{(μ_{i} (t_{i}))}^{(p / q - 1)}}{t_{i}^{γ_{i} - 1}} g^{p / q} (t_{1}, \dots, t_{k + 1}) \prod_{i = 1}^{k + 1} Δ t_{i} . \end{array}

Hence by induction principle, the statement is true $\forall n \in ℕ$ ⁠. □

Theorem 3.2.

Assume $i \in {1, \dots, n}$ , $T_{i}$ is a time scale with $a_{i} \in {(0, \infty)}_{T_{i}}$ and $γ_{i} > 1$ , further assume $g : {[a_{1}, \infty)}_{T_{1}} \times \dots \times {[a_{n}, \infty)}_{T_{n}} \to ℝ_{+}$ is such that the delta integrals $\int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \prod_{i = 1}^{n} {(t_{i})}^{\frac{p}{q} - γ_{i}} g^{p / q} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} Δ t_{i}$ exist. Let $Λ_{k} (t_{1}, \dots, t_{n})$ be defined in (13), then for $p, q > 0$ and $p / q \geq 2$

\begin{array}{l} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} t_{i}^{γ_{i}}} {(Λ_{n}^{σ_{1} \dots σ_{n}})}^{p / q} \prod_{i = 1}^{n} Δ t_{i} \\ \leq {(\frac{p}{q})}^{\frac{n p}{q}} \prod_{i = 1}^{n} {(\frac{2^{\frac{p}{q} - 1} k_{i}^{γ_{i}}}{(γ_{i} - 1)})}^{p / q} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \prod_{i = 1}^{n} \frac{1}{t_{i}^{γ_{i} - p / q}} g^{p / q} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} Δ t_{i}, \end{array}

(29)

holds.

Proof.

To prove the result, we use the principle of mathematical induction. For $n = 1$ the statement is true by Theorem 1.2. Let the statement be true for $1 \leq n \leq k$ .

To prove the result for $n = k + 1$ ⁠. Proceed it as in the proof of Theorem 3.1 up to (21). Apply chain rule 1 (6) on the right-hand side of (21) yields

\begin{array}{l} \frac{\partial}{Δ t_{k + 1}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k}})}^{p / q} \\ = (\frac{p}{q}) \frac{\partial}{Δ t_{k + 1}} Λ_{k + 1}^{σ_{1} \dots σ_{k}} {\int_{0}^{1} [h_{k + 1} Λ_{k + 1}^{σ_{1} \dots σ_{k + 1}} + (1 - h_{k + 1}) Λ_{k + 1}^{σ_{1} \dots σ_{k}}]}^{\frac{p}{q} - 1} d h_{k + 1} . \end{array}

(30)

Use (11) on the right-hand side of (30),

\leq (\frac{p}{q}) 2^{\frac{p}{q} - 2} {(Λ_{k + 1}^{σ_{1} \dots σ_{k + 1}})}^{\frac{p}{q} - 1} \frac{\partial}{Δ t_{k + 1}} Λ_{k + 1}^{σ_{1} \dots σ_{k}} + (\frac{p}{q}) 2^{\frac{p}{q} - 2} {(Λ_{k + 1}^{σ_{1} \dots σ_{k}})}^{\frac{p}{q} - 1} \frac{\partial}{Δ t_{k + 1}} Λ_{k + 1}^{σ_{1} \dots σ_{k}},

use the fact $σ_{k + 1} (t_{k + 1}) \geq t_{k + 1}$

\begin{array}{l} = (\frac{p}{q}) 2^{\frac{p}{q} - 2} {(Λ_{k + 1}^{σ_{1} \dots σ_{k + 1}})}^{\frac{p}{q} - 1} \frac{\partial}{Δ t_{k + 1}} Λ_{k + 1}^{σ_{1} \dots σ_{k}} + (\frac{p}{q}) 2^{\frac{p}{q} - 2} {(Λ_{k + 1}^{σ_{1} \dots σ_{k + 1}})}^{\frac{p}{q} - 1} \frac{\partial}{Δ t_{k + 1}} Λ_{k + 1}^{σ_{1} \dots σ_{k}} \\ = (\frac{p}{q}) 2^{\frac{p}{q} - 1} {(Λ_{k + 1}^{σ_{1} \dots σ_{k + 1}})}^{\frac{p}{q} - 1} \frac{\partial}{Δ t_{k + 1}} Λ_{k + 1}^{σ_{1} \dots σ_{k}} . \end{array}

(31)

Since

\frac{\partial}{Δ t_{k + 1}} Λ_{k + 1}^{σ_{1} \dots σ_{k}} = Λ_{k}^{σ_{1} \dots σ_{k}} \geq 0.

(32)

Use (32) in (31) and substitute in (21) to get

I_{k + 1} \leq \frac{p 2^{\frac{p}{q} - 1} k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)} \int_{a_{k + 1}}^{\infty} \frac{1}{t_{k + 1}^{γ_{k + 1} - 1}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k + 1}})}^{\frac{p}{q} - 1} Λ_{k}^{σ_{1} \dots σ_{k}} Δ t_{k + 1} .

(33)

Apply Hölder’s inequality on the right-hand side of (33) with indices $p / q$ and $p / (p - q)$

I_{k + 1} \leq \frac{p 2^{\frac{p}{q} - 1} k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)} {\int_{a_{k + 1}}^{\infty} {\frac{t_{k + 1}^{γ_{k + 1} (\frac{p - q}{q})}}{t_{k + 1}^{γ_{k + 1} - 1}} Λ_{k}^{σ_{1} \dots σ_{k}}}^{p / q} Δ t_{k + 1}}^{q / p} \times {I_{k + 1}}^{\frac{p - q}{p}} .

After simplification, we get

I_{k + 1} \leq {(\frac{p 2^{\frac{p}{q} - 1} k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)})}^{p / q} \int_{a_{k + 1}}^{\infty} \frac{{(Λ_{k}^{σ_{1} \dots σ_{k}})}^{p / q}}{t_{k + 1}^{- \frac{p}{q} + γ_{k + 1}}} Δ t_{k + 1} .

(34)

Substitute (34) into (15)

\begin{array}{l} \int_{\prod_{i = 1}^{k + 1} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{k + 1} t_{i}^{γ_{i}}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k + 1}})}^{p / q} \prod_{i = 1}^{k + 1} Δ t_{i} \\ \leq \int_{\prod_{i = 1}^{k} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{k} t_{i}^{γ_{i}}} {(\frac{p 2^{\frac{p}{q} - 1} k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)})}^{p / q} \int_{a_{k + 1}}^{\infty} \frac{{(Λ_{k}^{σ_{1} \dots σ_{k}})}^{p / q}}{t_{k + 1}^{- \frac{p}{q} + γ_{k + 1}}} Δ t_{k + 1} . \end{array}

(35)

Exchange integrals on right-hand side of (35) $k$ -times by using (9)

{(\frac{p 2^{\frac{p}{q} - 1} k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)})}^{p / q} \int_{a_{k + 1}}^{\infty} \frac{1}{t_{k + 1}^{- \frac{p}{q} + γ_{k + 1}}} {\int_{\prod_{i = 1}^{k} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{k} t_{i}^{γ_{i}}} {(Λ_{k}^{σ_{1} \dots σ_{k}})}^{p / q} \prod_{i = 1}^{k} Δ t_{i}} Δ t_{k + 1} .

(36)

Use the induction hypothesis for $Λ_{k}^{σ_{1} \dots σ_{k}}$ in (36) for fixed $t_{k + 1} \in T_{k + 1}$ and again apply (9) $k$ times to get

\begin{array}{l} \int_{\prod_{i = 1}^{k + 1} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{k + 1} t_{i}^{γ_{i}}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k + 1}})}^{p / q} \prod_{i = 1}^{k + 1} Δ t_{i} \\ \leq {(\frac{p}{q})}^{\frac{(k + 1) p}{q}} \prod_{i = 1}^{k + 1} {(\frac{2^{\frac{p}{q} - 1} k_{i}^{γ_{i}}}{γ_{i} - 1})}^{p / q} \int_{\prod_{i = 1}^{k + 1} a_{i}}^{\infty} \prod_{i = 1}^{k + 1} \frac{1}{t_{i}^{γ_{i} - p / q}} g^{p / q} (t_{1}, \dots, t_{k + 1}) \prod_{i = 1}^{k} Δ t_{i} . \end{array}

Hence by induction principle, the statement is true $\forall n \in ℕ$ . □

Corollary 3.3.

As a special case of Theorem 3.2, when $T_{1} = \dots = T_{n} = ℝ$ , $p / q = λ > 1$ and $γ_{i} < 1$ , (29) becomes the following Wirtinger type inequality

\begin{array}{l} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} t_{i}^{γ_{i}}} {(G (t_{1}, \dots, t_{n}))}^{λ} \prod_{i = 1}^{n} d t_{i} \\ \leq \prod_{i = 1}^{n} {(\frac{λ 2^{λ - 1}}{1 - γ_{i}})}^{λ} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} t_{1}^{γ_{i} - λ}} (\frac{\partial^{n}}{\partial t_{1} \dots \partial t_{n}} G^{λ} (t_{1}, \dots, t_{n})) \prod_{i = 1}^{n} d t_{i}, \end{array}

where $G (t_{1}, \dots, t_{n}) ≐ \int_{\prod_{i = 1}^{n} a_{i}}^{t_{i}} g (s_{1}, \dots, s_{n}) \prod_{i = 1}^{n} d s_{i}$ .

When $γ_{1} = \dots = γ_{n} = λ > 1$ , we have another Hardy type inequality for function of $n$ -variables

\begin{array}{l} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} t_{i}} {(\int_{\prod_{i = 1}^{n} a_{i}}^{t_{i}} g (s_{1}, \dots, s_{n}) \prod_{i = 1}^{n} d s_{i})}^{λ} \prod_{i = 1}^{n} d t_{i} \\ \leq {(\frac{λ 2^{λ - 1}}{λ - 1})}^{λ} g^{λ} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} d t_{i} . \end{array}

Remark 3.4.

Assume that $T_{1} = \dots = T_{n} = ℕ$ in Theorem 3.2, $p / q = λ > 1$ ⁠, $a_{i} > 1$ ⁠, $γ_{i} > 1$ for $i \in {1, \dots, n}$ ⁠, further assume that $\sum_{m_{1} = 1}^{\infty} \dots \sum_{m_{n} = 1}^{\infty} g^{λ} (m_{1}, \dots, m_{n})$ is convergent. (29) becomes the following discrete Hardy and Littlewood inequality

\begin{array}{l} \sum_{m_{1} = 1}^{\infty} \dots \sum_{m_{n} = 1}^{\infty} \frac{1}{m_{1}^{γ_{1}} \dots m_{n}^{γ_{n}}} {(\sum_{k_{1} = 1}^{m_{1}} \dots \sum_{k_{n} = 1}^{m_{n}} g (k_{1}, \dots, k_{n}))}^{λ} \\ \leq \prod_{i = 1}^{n} {(\frac{2^{λ - 1} λ}{γ_{i} - 1})}^{λ} \sum_{m_{1} = 1}^{\infty} \dots \sum_{m_{n} = 1}^{\infty} \frac{1}{\prod_{i = 1}^{n} m_{i}^{γ_{i} - λ}} g^{λ} (m_{1}, \dots, m_{n}) . \end{array}

4. Hardy and Littlewood-type inequalities for p/q ≥ 1 and γ > 1

Theorem 4.1.

Assume $i \in {1, \dots, n}$ , $T_{i}$ is a time scale with $a_{i} \in {(0, \infty)}_{T_{i}}$ and $γ_{i} < 1$ , further assume $g : {[a_{1}, \infty)}_{T_{1}} \times \dots \times {[a_{n}, \infty)}_{T_{n}} \to ℝ_{+}$ is such that the delta integrals $\int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \prod_{i = 1}^{n} {t_{i}}^{\frac{p}{q} - γ_{i}} g^{p / q} (t_{1}, \dots, t_{n}) {\prod_{i = 1}^{n}}^{} Δ t_{i}$ exist. Let $Λ_{k} (t_{1}, \dots, t_{n})$ be defined in (13), then for $p, q > 0$ and $p / q > 1$

\begin{array}{l} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} t_{i}^{} γ_{i}} {(Λ_{n}^{σ_{1} \dots σ_{n}})}^{p / q} \prod_{i = 1}^{n} Δ t_{i} \\ \leq {(\frac{p}{q})}^{\frac{n p}{q}} \prod_{i = 1}^{n} {(\frac{k_{i}^{γ_{i}}}{γ_{i} - 1})}^{p / q} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \prod_{i = 1}^{n} \frac{1}{t_{i}^{γ_{i} - p / q}} g^{p / q} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} Δ t_{i}, \end{array}

(37)

holds, where $n$ is a positive integer.

Proof.

To prove the result, we use the principle of mathematical induction. For $n = 1$ the statement is true by Theorem 1.3. Let the statement be true for $1 \leq n \leq k$ ⁠.

To prove the result for $n = k + 1$ ⁠. Proceed it as in the proof of Theorem 3.1 up to (21). Apply the chain rule 2 (7) to get

\frac{\partial}{Δ t_{k + 1}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k}})}^{p / q} = \frac{p}{q} {(Λ_{k + 1}^{σ_{1} \dots σ_{k}} (t_{1}, \dots, t_{k}, c_{k + 1}))}^{\frac{p}{q} - 1} \frac{\partial}{Δ t_{k + 1}} Λ_{k + 1}^{σ_{1} \dots σ_{k}},

where $c_{k + 1} \in [t_{k + 1}, σ_{k + 1} (t_{k + 1})]$ ⁠. Since

\frac{\partial}{Δ t_{k + 1}} Λ_{k + 1}^{σ_{1} \dots σ_{k}} ≐ Λ_{k}^{σ_{1} \dots σ_{k}} \geq 0,

and $σ_{k + 1} (t_{k + 1}) \geq c_{k + 1}$ ⁠, one has that

\frac{\partial}{Δ t_{k + 1}} Λ_{k + 1}^{σ_{1} \dots σ_{k}} \leq \frac{p}{q} {(Λ_{k + 1}^{σ_{1} \dots σ_{k + 1}})}^{\frac{p}{q} - 1} Λ_{k}^{σ_{1} \dots σ_{k}} .

(38)

Substitute (38) into (21)

I_{k + 1} \leq \frac{p k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)} \int_{a_{k + 1}}^{\infty} \frac{{(Λ_{k + 1}^{σ_{1} \dots σ_{k + 1}})}^{\frac{p}{q} - 1}}{t_{k + 1}^{γ_{k + 1} - 1}} Λ_{k}^{σ_{1} \dots σ_{k}} Δ t_{k + 1} .

(39)

Apply Hölder’s inequality on the right-hand side of (39) with indices $p / q$ and $p / (p - q)$

I_{k + 1} \leq \frac{p k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)} {\int_{a_{k + 1}}^{\infty} {\frac{t_{k + 1}^{γ_{k + 1} (\frac{p - q}{p})}}{t_{k + 1}^{γ_{k + 1} - 1}} Λ_{k}^{σ_{1}, \dots, σ_{k}}}^{p / q} Δ t_{k + 1}}^{q / p} \times {I_{k + 1}}^{\frac{p - q}{p}} .

After simplification, we get

I_{k + 1} \leq {(\frac{p k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)})}^{p / q} \int_{a_{k + 1}}^{\infty} \frac{{(Λ_{k}^{σ_{1}, \dots, σ_{k}})}^{p / q}}{t_{k + 1}^{- \frac{p}{q} + γ_{k + 1}}} Δ t_{k + 1} .

(40)

Substitute (40) into (15)

\begin{array}{l} \int_{\prod_{i = 1}^{k + 1} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{k + 1} t_{i}^{γ_{i}}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k + 1}})}^{p / q} \prod_{i = 1}^{k + 1} Δ t_{i} \\ \leq \int_{\prod_{i = 1}^{k} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{k} t_{i}^{γ_{i}}} {(\frac{p k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)})}^{p / q} \int_{a_{k + 1}}^{\infty} \frac{{(Λ_{k}^{σ_{1}, \dots, σ_{k}})}^{p / q}}{t_{k + 1}^{γ_{k + 1} - \frac{p}{q}}} Δ t_{k + 1} . \end{array}

(41)

Exchange integrals on right-hand side of (41) $k$ -times by using (9)

= {(\frac{p k_{k + 1}^{γ_{k + 1}}}{q (γ_{k + 1} - 1)})}^{p / q} \int_{a_{k + 1}}^{\infty} \frac{1}{t_{k + 1}^{γ_{k + 1} - \frac{p}{q}}} {\int_{\prod_{i = 1}^{k} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{k} t_{i}^{γ_{i}}} {(Λ_{k}^{σ_{1} \dots σ_{k}})}^{p / q} \prod_{i = 1}^{k} Δ t_{i}} Δ t_{k + 1} .

(42)

Use the induction hypothesis with ${(Λ_{k}^{σ_{1} \dots σ_{k}})}^{p / q}$ in (42) for fixed $t_{k + 1} \in T_{k + 1}$ and again apply (9) $k$ -times to get

\begin{array}{l} \int_{\prod_{i = 1}^{k + 1} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{k + 1} t_{i}^{γ_{i}}} {(Λ_{k + 1}^{σ_{1} \dots σ_{k + 1}})}^{p / q} \prod_{i = 1}^{k + 1} Δ t_{i} \\ \leq {(\frac{p}{q})}^{\frac{(k + 1) p}{q}} \prod_{i = 1}^{k + 1} {(\frac{k_{i}^{γ_{i}}}{γ_{i} - 1})}^{p / q} \int_{\prod_{i = 1}^{k + 1} a_{i}}^{\infty} \prod_{i = 1}^{k + 1} \frac{1}{t_{i}^{γ_{i} - p / q}} g^{p / q} (t_{1}, \dots, t_{k + 1}) \prod_{i = 1}^{k + 1} Δ t_{i} . \end{array}

Hence by induction principle, the statement is true $\forall n \in ℕ$ ⁠. □

Corollary 4.2.

As a special case of Theorem 4.1, when $T_{1} = \dots = T_{n} = ℝ$ , $p / q = λ > 1$ and $γ_{1}, \dots, γ_{n} < 1$ , (37) becomes the following Wirtinger type inequality,

\begin{array}{l} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} t_{i}^{γ_{i}}} G^{λ} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} d t_{i} \\ \leq \prod_{i = 1}^{n} {(\frac{λ}{1 - γ_{i}})}^{λ} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} t_{1}^{γ_{i} - λ}} (\frac{\partial^{n}}{\partial t_{1} \dots \partial t_{n}} G^{λ} (t_{1}, \dots, t_{n})) \prod_{i = 1}^{n} d t_{i}, \end{array}

where $G (t_{1}, \dots, t_{n}) ≐ \int_{\prod_{i = 1}^{n} a_{i}}^{t_{i}} g (s_{1}, \dots, s_{n}) \prod_{i = 1}^{n} Δ s_{i}$ .

When $γ_{1} = \dots = γ_{n} = λ > 1$ , we have the classical Hardy type inequality for function of $n$ -variables

\begin{array}{l} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} t_{i}} {(\int_{\prod_{i = 1}^{n} a_{i}}^{t_{i}} g (s_{1}, \dots, s_{n}) \prod_{i = 1}^{n} d s_{i})}^{λ} \prod_{i = 1}^{n} d t_{i} \\ \leq {(\frac{λ}{λ - 1})}^{λ} g^{λ} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} d t_{i} . \end{array}

Corollary 4.3.

Assume that $T_{1} = \dots = T_{n} = ℕ$ in Theorem 4.1, $p / q = λ > 1$ , $a_{i} > 1$ , $γ_{i} > 1$ for $i \in {1, \dots, n}$ , further assume that ${\sum_{m_{1} = 1}^{\infty} \dots \sum_{m_{n} = 1}^{\infty} g^{λ} (m_{1}, \dots, m_{n})}^{}$ is convergent. Note that in this case $\frac{m_{i}}{σ_{i} (m_{i})} = \frac{m_{i}}{m_{i} + 1}$ therefore $\frac{1}{2} \leq \frac{m_{i}}{m_{i} + 1} \leq 1$ , and we get following discrete Hardy and Littlewood inequality

\begin{array}{l} \sum_{m_{1} = 1}^{\infty} \dots \sum_{m_{n} = 1}^{\infty} \frac{1}{m_{1}^{γ_{1}} \dots m_{n}^{γ_{n}}} {(\sum_{k_{1} = 1}^{m_{1}} \dots \sum_{k_{n} = 1}^{m_{n}} g (k_{1}, \dots, k_{n}))}^{λ} \\ \leq \prod_{i = 1}^{n} {(\frac{2^{λ} λ}{γ_{i} - 1})}^{λ} \sum_{m_{1} = 1}^{\infty} \dots \sum_{m_{n} = 1}^{\infty} \frac{1}{\prod_{i = 1}^{n} m_{i}^{γ_{i} - λ}} g^{λ} (m_{1}, \dots, m_{n}) . \end{array}

Remark 4.4.

Assume $i \in {1, \dots, n}$ ⁠, $T_{i}$ is a time scale with $a_{i} \in {(0, \infty)}_{T_{i}}$ and $γ_{i} < 1$ ⁠, further assume $g : {[a_{1}, \infty)}_{T_{1}} \times \dots \times {[a_{n}, \infty)}_{T_{n}} \to ℝ_{+}$ is such that the delta integrals $\int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \prod_{i = 1}^{n} σ_{i} {(t_{i})}^{\frac{p}{q} - γ_{i}} {(\frac{σ_{i} (t_{i})}{t_{i}})}^{\frac{p}{q} (γ_{i} - 1)}$ $g_{n}^{p / q} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} Δ t_{i}$ exist. Let $Λ_{k} (t_{1}, \dots, t_{n})$ be defined in Theorem 3.1, then for $p, q > 0$ and $p / q > 1$

\begin{array}{l} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{{(Λ_{n}^{σ_{1} \dots σ_{n}})}^{p / q}}{\prod_{i = 1}^{n} {(σ_{i} (t_{i}))}^{γ_{i}}} \prod_{i = 1}^{n} Δ t_{i} \\ \leq {(\frac{p}{q})}^{\frac{n p}{q}} \prod_{i = 1}^{n} {(\frac{1}{γ_{i} - 1})}^{p / q} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{g^{p / q} (t_{1}, \dots, t_{n})}{\prod_{i = 1}^{n} σ^{γ_{i} - \frac{p}{q}} (t_{i})} \prod_{i = 1}^{n} {(\frac{σ_{i} (t_{i})}{t_{i}})}^{\frac{p}{q} (γ_{i} - 1)} \prod_{i = 1}^{n} Δ t_{i}, \end{array}

holds.

Proof. Replace left-hand side of (37) in Theorem 4.1 by

\int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{{(Λ_{n}^{σ_{1} \dots σ_{n}})}^{p / q}}{\prod_{i = 1}^{n} {(σ_{i} (t_{i}))}^{γ_{i}}} \prod_{i = 1}^{n} Δ t_{i},

and proceed as in the proof of Theorem 4.1. □

5. Hardy and Littlewood-type inequalities for p/q ≤ 2 and γ > 1

Theorem 5.1.

Assume $i \in {1, \dots, n}$ , $T_{i}$ is a time scale with $a_{i} \in {(0, \infty)}_{T_{i}}$ and $γ_{i} < 1$ , further assume $g : {[a_{1}, \infty)}_{T_{1}} \times \dots \times {[a_{n}, \infty)}_{T_{n}} \to ℝ_{+}$ is such that the delta integrals $\int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \prod_{i = 1}^{n} {(t_{i})}^{\frac{p}{q} - γ_{i}} g^{p / q} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} Δ t_{i}$ exist. Let $Λ_{k} (t_{1}, \dots, t_{n})$ be defined in (13), then for $p, q > 0$ and $p / q \leq 2$

\begin{array}{l} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} t_{i}^{γ_{i}}} {(Λ_{n}^{σ_{1} \dots σ_{n}})}^{p / q} \prod_{i = 1}^{n} Δ t_{i} \\ \leq {(\frac{p}{q})}^{\frac{n p}{q}} \prod_{i = 1}^{n} {(\frac{2 k_{i}^{γ_{i}}}{(γ_{i} - 1)})}^{p / q} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} t_{i}^{γ_{i} - \frac{p}{q}}} g^{p / q} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} Δ t_{i} . \end{array}

(43)

Proof.

Proceed as in the proof of Theorem 3.2 and apply inequality (12) in (21) to get (43). □

Remark 5.2.

As a special case of Theorem 5.1, when $T_{1} = \dots = T_{n} = ℝ$ ⁠, $p / q = λ > 1$ and $γ_{1}, \dots, γ_{n} < 1$ ⁠, we have the following Hardy type inequality

\begin{array}{l} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} t_{i}^{γ_{i}}} {(\int_{\prod_{i = 1}^{n} a_{i}}^{t_{i}} g (s_{1}, \dots, s_{n}) \prod_{i = 1}^{n} d s_{i})}^{λ} \prod_{i = 1}^{n} d t_{i} \\ \leq \prod_{i = 1}^{n} {(\frac{2 λ}{1 - γ_{i}})}^{λ} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} t_{i}^{γ_{i} - λ}} g^{λ} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} d t_{i} . \end{array}

Remark 5.3.

Assume that $T_{1} = \dots = T_{n} = ℕ$ in Theorem 5.1, $p / q = λ > 1$ ⁠, $a_{i} > 1$ ⁠, $γ_{i} > 1$ for $i \in {1, \dots, n}$ ⁠, further assume that $\sum_{m_{1} = 1}^{\infty} \dots \sum_{m_{n} = 1}^{\infty} g^{λ} (m_{1}, \dots, m_{n})$ is convergent. In this case, (43) becomes the following discrete Hardy and Littlewood inequality

\begin{array}{l} \sum_{m_{1} = 1}^{\infty} \dots \sum_{m_{n} = 1}^{\infty} \frac{1}{m_{1}^{γ_{1}} \dots m_{n}^{γ_{n}}} {(\sum_{k_{1} = 1}^{m_{1}} \dots \sum_{k_{n} = 1}^{m_{n}} g (k_{1}, \dots, k_{n}))}^{λ} \\ \leq \prod_{i = 1}^{n} {(\frac{2 λ}{γ_{i} - 1})}^{λ} \sum_{m_{1} = 1}^{\infty} \dots \sum_{m_{n} = 1}^{\infty} \frac{1}{\prod_{i = 1}^{n} m_{i}^{γ_{i} - λ}} g^{λ} (m_{1}, \dots, m_{n}) . \end{array}

6. Hardy and Littlewood-type inequalities for p/q > 1 and γ < 1

Theorem 6.1.

Assume $i \in {1, \dots, n}$ , $T_{i}$ is a time scale with $a_{i} \in {(0, \infty)}_{T_{i}}$ and $γ_{i} < 1$ , further assume $g : {[a_{1}, \infty)}_{T_{1}} \times \dots \times {[a_{n}, \infty)}_{T_{n}} \to ℝ_{+}$ is such that the delta integrals $\int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \prod_{i = 1}^{n} {(σ_{i} (t_{i}))}^{\frac{p}{q} - γ_{i}} g^{p / q} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} Δ t_{i}$ exist, for any $(t_{1}, \dots, t_{n}) \in {[a_{1}, \infty)}_{T_{1}} \times \dots \times {[a_{n}, \infty)}_{T_{n}}$ , define

Ω_{k} (t_{1}, \dots, t_{n}) = \int_{\prod_{j = 1}^{k} a_{j}}^{t_{j}} g (s_{1}, \dots, s_{n}) \prod_{j = 1}^{k} Δ s_{j}, k \in {1, \dots, n}

(44)

then for $p, q > 0$ and $p / q > 1$

\begin{array}{l} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{Ω_{n}^{p / q} (t_{1}, \dots, t_{n})}{\prod_{i = 1}^{n} σ_{i}^{γ_{i}} (t_{i})} \prod_{i = 1}^{n} Δ t_{i} \\ \leq {(\frac{p}{q})}^{\frac{n p}{q}} \prod_{i = 1}^{n} {(\frac{1}{1 - γ_{i}})}^{p / q} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} {(σ_{i} (t_{i}))}^{γ_{i} - p / q}} g^{p / q} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} Δ t_{i} \end{array}

(45)

holds, where $n$ is any positive integer.

Proof.

To prove the result, we use the principle of mathematical induction. For $n = 1$ the statement is true by Theorem 1.4. Let the statement be true for $1 \leq n \leq k$ ⁠.

To prove the result for $n = k + 1$ ⁠. The left-hand side of (45) can be written as

\int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{Ω_{k + 1}^{p / q} (t_{1}, \dots, t_{k + 1})}{\prod_{i = 1}^{k + 1} σ_{i}^{γ_{i}} (t_{i})} \prod_{i = 1}^{k + 1} Δ t_{i}

(46)

Denote $\int_{a_{k + 1}}^{\infty} \frac{Ω_{k + 1}^{p / q} (t_{1}, \dots, t_{k + 1})}{σ_{k + 1}^{γ_{k + 1}} (t_{k + 1})} Δ t_{k + 1} = I_{k + 1}$ ⁠. Apply (5) with $\frac{\partial}{Δ t_{k + 1}} v (t_{k + 1}) = \frac{1}{σ_{k + 1}^{γ_{k + 1}} (t_{k + 1})}$ and $u (t_{k + 1}) = Ω_{k + 1}^{p / q} (t_{1}, \dots, t_{k + 1})$ ⁠. Thus

\begin{array}{l} I_{k + 1} = v (t_{k + 1}) Ω_{k + 1}^{p / q} (t_{1}, \dots, t_{k + 1}) |_{a_{k + 1}}^{\infty} \\ + \int_{a_{k + 1}}^{\infty} v^{σ_{k + 1}} (t_{k + 1}) (- \frac{\partial}{Δ t_{k + 1}} Ω_{k + 1}^{p / q} (t_{1}, \dots, t_{k + 1})) Δ t_{k + 1}, \end{array}

(47)

where $v (t_{k + 1}) = \int_{a_{k + 1}}^{t_{k + 1}} 1 / σ_{k + 1}^{γ_{k + 1}} (s_{k + 1}) Δ s_{k + 1}$ ⁠. Use chain rule (6) and the fact that $σ_{k + 1} (s_{k + 1}) \geq s_{k + 1}$ to get

\begin{array}{l} \frac{\partial}{Δ s_{k + 1}} (s_{k + 1}^{1 - γ_{k + 1}}) = (1 - γ_{k + 1}) \int_{0}^{1} {[h_{k + 1} σ_{k + 1} (s_{k + 1}) + (1 - h_{k + 1}) s_{k + 1}]}^{- γ_{k + 1}} d h_{k + 1} \\ \geq (1 - γ_{k + 1}) \frac{1}{σ_{k + 1}^{γ_{k + 1}} (s_{k + 1})}, \end{array}

which gives

v^{σ_{k + 1}} (t_{k + 1}) = \int_{a_{k + 1}}^{σ_{k + 1} (t_{k + 1})} \frac{1}{σ_{k + 1}^{γ_{k + 1}} (s_{k + 1})} Δ s_{k + 1} \leq \frac{1}{(1 - γ_{k + 1})} {(σ_{k + 1} (t_{k + 1}))}^{1 - γ_{k + 1}} .

(48)

Combine (47), (48) and use the facts $Ω_{k + 1} (t_{1}, \dots, t_{k}, \infty) = 0$ ⁠, $v (a_{k + 1}) = 0$ to get

I_{k + 1} \leq \frac{1}{(1 - γ_{k + 1})} \int_{a_{k + 1}}^{\infty} \frac{- \frac{\partial}{Δ t_{k + 1}} Ω_{k + 1}^{p / q} (t_{1}, \dots, t_{k + 1})}{{(σ_{k + 1} (t_{k + 1}))}^{γ_{k + 1} - 1}} Δ t_{k + 1} .

(49)

Apply chain rule 2 (7) to find

- \frac{\partial}{Δ t_{k + 1}} Ω_{k + 1}^{p / q} (t_{1}, \dots, t_{k + 1}) = - (\frac{p}{q}) Ω_{k + 1}^{\frac{p}{q} - 1} (t_{1}, \dots, t_{k}, c_{k + 1}) \frac{\partial}{Δ t_{k + 1}} Ω_{k + 1} (t_{1}, \dots, t_{k + 1}),

where, $c_{k + 1} \in [t_{k + 1}, σ_{k + 1} (t_{k + 1})]$ ⁠. Since

\begin{array}{l} \frac{\partial}{Δ t_{k + 1}} Ω_{k + 1} (t_{1}, \dots, t_{k + 1}) = - \int_{\prod_{i = 1}^{k} a_{i}}^{\infty} g (s_{1}, \dots, s_{k}, t_{k + 1}) \prod_{i = 1}^{k} Δ s_{i} \\ ≐ Ω_{k} (t_{1}, \dots, t_{k + 1}) \leq 0, \end{array}

and $c_{k + 1} \geq t_{k + 1}$ ⁠, one has that

- \frac{\partial}{Δ t_{k + 1}} Ω_{k + 1}^{p / q} (t_{1}, \dots, t_{k + 1}) \leq \frac{p}{q} Ω_{k + 1}^{\frac{p}{q} - 1} (t_{1}, \dots, t_{k + 1}) Ω_{k} (t_{1}, \dots, t_{k + 1}) .

(50)

Substitute (50) into (49)

I_{k + 1} \leq \frac{p}{q (1 - γ_{k + 1})} \int_{a_{k + 1}}^{\infty} \frac{Ω_{k + 1}^{\frac{p}{q} - 1} (t_{1}, \dots, t_{k + 1})}{{(σ_{k + 1} (t_{k + 1}))}^{γ_{k + 1} - 1}} Ω_{k} (t_{1}, \dots, t_{k + 1}) Δ t_{k + 1} .

(51)

Apply Hölder’s inequality on the right-hand side of (51) with indices $p / q$ and $p / (p - q)$ to obtain

\begin{array}{l} I_{k + 1} \leq \frac{p}{q (1 - γ_{k + 1})} {[{\int_{a_{k + 1}}^{\infty} [\frac{{(σ_{k + 1}^{γ_{k + 1}} (t_{k + 1}))}^{\frac{p - q}{p}}}{{(σ_{k + 1} (t_{k + 1}))}^{γ_{k + 1} - 1}} Ω_{k} (t_{1}, \dots, t_{k + 1})]}^{p / q} Δ t_{k + 1}]}^{q / p} \\ \times {[I_{k + 1}]}^{\frac{p - q}{p}} . \end{array}

After simplification, we get

I_{k + 1} \leq {(\frac{p}{q (1 - γ_{k + 1})})}^{p / q} \int_{a_{k + 1}}^{\infty} \frac{Ω_{k}^{p / q} (t_{1}, \dots, t_{k + 1})}{{(σ_{k + 1} (t_{k + 1}))}^{- \frac{p}{q} + γ_{k + 1}}} Δ t_{k + 1} .

(52)

Substitute (52) into (46)

\begin{array}{l} \int_{\prod_{i = 1}^{k + 1} a_{i}}^{\infty} \frac{Ω_{k + 1}^{p / q} (t_{1}, \dots, t_{k + 1})}{\prod_{i = 1}^{k + 1} σ_{i}^{γ_{i}} (t_{i})} \prod_{i = 1}^{k + 1} Δ t_{i} \\ \leq \int_{\prod_{i = 1}^{k} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{k} σ_{i}^{γ_{i}} (t_{i})} {(\frac{p}{q (1 - γ_{k + 1})})}^{p / q} \int_{a_{k + 1}}^{\infty} \frac{Ω_{k}^{p / q} (t_{1}, \dots, t_{k + 1})}{{(σ_{k + 1} (t_{k + 1}))}^{- \frac{p}{q} + γ_{k + 1}}} \prod_{i = 1}^{k + 1} Δ t_{i} . \end{array}

(53)

Exchange integrals on right-hand side of (53) $k$ -times by using (9)

\begin{array}{l} = {(\frac{p}{q (1 - γ_{k + 1})})}^{p / q} \int_{a_{k + 1}}^{\infty} \frac{1}{{(σ_{k + 1} (t_{k + 1}))}^{- \frac{p}{q} + γ_{k + 1}}} \\ \times {\int_{\prod_{i = 1}^{k} a_{i}}^{\infty} \frac{Ω_{k}^{p / q} (t_{1}, \dots, t_{k + 1})}{\prod_{i = 1}^{k} σ_{i}^{γ_{i}} (t_{i})} \prod_{i = 1}^{k} Δ t_{i}} Δ t_{k + 1} . \end{array}

(54)

Use the induction hypothesis for $Ω_{k} (t_{1}, \dots, t_{k + 1})$ in (54) instead for $Ω_{k} (t_{1}, \dots, t_{k})$ for fixed $t_{k + 1} \in T_{k + 1}$ and again apply (9) $k$ times to get

\begin{array}{l} \int_{\prod_{i = 1}^{k + 1} a_{i}}^{\infty} \frac{Ω_{k + 1}^{p / q} (t_{1}, \dots, t_{k + 1})}{\prod_{i = 1}^{k + 1} σ_{i}^{γ_{i}} (t_{i})} \prod_{i = 1}^{k + 1} Δ t_{i} \\ \leq {(\frac{p}{q})}^{\frac{(k + 1) p}{q}} \prod_{i = 1}^{k + 1} {(\frac{1}{1 - γ_{i}})}^{p / q} \int_{\prod_{i = 1}^{k + 1} a_{i}}^{\infty} \prod_{i = 1}^{k + 1} \frac{1}{{(σ_{i} (t_{i}))}^{γ_{i} - p / q}} g^{p / q} (t_{1}, \dots, t_{k + 1}) \prod_{i = 1}^{k + 1} Δ t_{i} . \end{array}

Hence by induction principle, the statement is true $\forall n \in ℕ$ ⁠. □

Corollary 6.2.

Under the conditions of Theorem 6.1, we get the following inequality

\begin{array}{l} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} σ_{i}^{γ_{i}} (t_{i})} {(Ω_{n} (σ_{1} (t_{1}), \dots, σ_{n} (t_{n})))}^{p / q} \prod_{i = 1}^{n} Δ t_{i} \\ \leq {(\frac{p}{q})}^{\frac{n p}{q}} \prod_{i = 1}^{n} {(\frac{1}{1 - γ_{i}})}^{p / q} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \prod_{i = 1}^{n} \frac{1}{{(σ_{i} (t_{i}))}^{γ_{i} - p / q}} g^{p / q} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} Δ t_{i} . \end{array}

(55)

Proof.

The fact $\frac{\partial^{n} Ω_{n}}{Δ t_{1} \dots Δ t_{n}} \leq 0$ implies

\begin{array}{l} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} σ_{i}^{γ_{i}} (t_{i})} {(Ω_{n} (σ_{1} (t_{1}), \dots, σ_{n} (t_{n})))}^{p / q} \prod_{i = 1}^{n} Δ t_{i} \\ \leq \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} σ_{i}^{γ_{i}} (t_{i})} {(Ω_{n} (t_{1}, \dots, t_{n}))}^{p / q} \prod_{i = 1}^{n} Δ t_{i} . \end{array}

(56)

Now use (45) in (56) to get (55). □

Remark 6.3.

Consider $T_{1} = \dots = T_{n} = ℝ$ ⁠, $p / q = λ > 1$ and $γ_{1}, \dots, γ_{n} < 1$ ⁠, in Theorem 6.1. Denote $G (t_{1}, \dots, t_{n}) = \int_{\prod_{i = 1}^{n} t_{i}}^{\infty} g (s_{1}, \dots, s_{n}) \prod_{i = 1}^{n} d s_{i}$ ⁠. Thus, (45) takes the form

\begin{array}{l} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} t_{i}^{γ_{i}}} (G^{λ} (t_{1}, \dots, t_{n})) \prod_{i = 1}^{n} d t_{i} \\ \leq \prod_{i = 1}^{n} {(\frac{λ}{1 - γ_{i}})}^{λ} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} {(t_{i})}^{γ_{i} - λ}} \frac{\partial^{n}}{\partial t_{1} \dots \partial t_{n}} G^{λ} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} d t_{i}, \end{array}

which can be considered as a generalization of Wirtinger’s inequality [1].

Remark 6.4.

As a special case of Theorem 6.1, assume that $T_{1} = \dots = T_{n} = ℕ$ ⁠, $p / q = λ > 1$ ⁠, $a_{1} = \dots = a_{n} = 1$ and $γ_{1}, \dots, γ_{n} < 1$ ⁠. In this case (55) becomes the following discrete Hardy and Littlewood inequality

\begin{array}{l} \sum_{m_{1} = 1}^{\infty} \dots \sum_{m_{n} = 1}^{\infty} \frac{1}{\prod_{i = 1}^{n} {(m_{i} + 1)}^{γ_{i}}} {(\sum_{k_{1} = m_{1} + 1}^{\infty} \dots \sum_{k_{n} = m_{n} + 1}^{\infty} g (k_{1}, \dots, k_{n}))}^{λ} \\ \leq \prod_{i = 1}^{n} {(\frac{λ}{1 - γ_{i}})}^{λ} \sum_{m_{1} = 1}^{\infty} \dots \sum_{m_{n} = 1}^{\infty} \frac{1}{\prod_{i = 1}^{n} {(m_{i} + 1)}^{γ_{i} - λ}} g^{λ} (m_{1}, \dots, m_{n}) . \end{array}

7. Hardy and Littlewood-type inequalities for p/q ≤ 2 and γ < 1

Theorem 7.1.

Assume $i \in {1, \dots, n}$ , $T_{i}$ is a time scale with $a_{i} \in {(0, \infty)}_{T_{i}}$ and $γ_{i} < 1$ , further assume $g : {[a_{1}, \infty)}_{T_{1}} \times \dots \times {[a_{n}, \infty)}_{T_{n}} \to ℝ_{+}$ is such that the delta integrals $\int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \prod_{i = 1}^{n} {(σ_{i} (t_{i}))}^{\frac{p}{q} - γ_{i}} g^{p / q} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} Δ t_{i}$ exist, then for $p, q > 0$ and $p / q \leq 2$ . Then

\begin{array}{l} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} σ_{i}^{γ_{i}} (t_{i})} {(\int_{\prod_{i = 1}^{n} t_{i}}^{\infty} g (s_{1}, \dots, s_{n}) \prod_{i = 1}^{n} Δ s_{i})}^{p / q} \prod_{i = 1}^{n} Δ t_{i} \\ \leq {(\frac{p}{q})}^{\frac{n p}{q}} \prod_{i = 1}^{n} {(\frac{2}{1 - γ_{i}})}^{p / q} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \prod_{i = 1}^{n} \frac{1}{{(σ_{i} (t_{i}))}^{γ_{i} - p / q}} g^{p / q} (t_{1}, \dots, t_{n}) \prod_{i = 1}^{n} Δ t_{i} . \end{array}

(57)

Proof: Use (12) and proceed as in the proof of Theorem 6.1 to get (57). □

Remark 7.2.

In Theorem 7.1, when $T_{1} = \dots = T_{n} = ℝ$ ⁠, $p / q = λ > 1$ and $γ_{i} < 1$ ⁠, (57) becomes the following Wirtinger type inequality,

\begin{array}{l} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} t_{i}^{γ_{i}}} {(G (t_{1}, \dots, t_{n}))}^{λ} \prod_{i = 1}^{n} d t_{i} \\ \leq \prod_{i = 1}^{n} {(\frac{2 λ}{1 - γ_{i}})}^{λ} \int_{\prod_{i = 1}^{n} a_{i}}^{\infty} \frac{1}{\prod_{i = 1}^{n} {(t_{i})}^{γ_{i} - λ}} {(\frac{\partial^{n}}{\partial t_{1} \dots \partial t_{n}} G (t_{1}, \dots, t_{n}))}^{λ} \prod_{i = 1}^{n} d t_{i}, \end{array}

where $G (t_{1}, \dots, t_{n}) ≐ \int_{\prod_{i = 1}^{n} t_{i}}^{\infty} g (s_{1}, \dots, s_{n}) \prod_{i = 1}^{n} d s_{i}$ ⁠.

Remark 7.3.

In Theorem 7.1, assume that $T_{1} = \dots = T_{n} = ℕ$ ⁠, $p / q = λ > 1$ ⁠, $a_{1} = \dots = a_{n} = 1$ and $γ_{i} < 1$ ⁠. (57) becomes the following discrete Hardy and Littlewood inequality

\begin{array}{l} \sum_{m_{1} = 1}^{\infty} \dots \sum_{m_{n} = 1}^{\infty} \frac{1}{\prod_{i = 1}^{n} {(m_{i} + 1)}^{γ_{i}}} {(\sum_{k_{1} = m_{1} + 1}^{\infty} \dots \sum_{k_{n} = m_{n} + 1}^{\infty} g (k_{1}, \dots, k_{n}))}^{λ} \\ \leq \prod_{i = 1}^{n} {(\frac{2 λ}{1 - γ_{i}})}^{λ} \sum_{m_{1} = 1}^{\infty} \dots \sum_{m_{n} = 1}^{\infty} \frac{1}{\prod_{i = 1}^{n} {(m_{i} + 1)}^{γ_{i} - λ}} g^{λ} (m_{1}, \dots, m_{n}) . \end{array}

Declaration of Competing Interest: The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper. Funding: This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.The publisher wishes to inform readers that the article “Multivariate Hardy and Littlewood inequalities on time scales” was originally published by the previous publisher of the Arab Journal of Mathematical Sciences and the pagination of this article has been subsequently changed. There has been no change to the content of the article. This change was necessary for the journal to transition from the previous publisher to the new one. The publisher sincerely apologises for any inconvenience caused. To access and cite this article, please use Nosheen, A., Nawaz, A., Khan, K. A., Awan, K. M. (2019), “Multivariate Hardy and Littlewood inequalities on time scales”, Arab Journal of Mathematical Sciences, Vol. 26 No. 1/2, pp. 245-263. The original publication date for this paper was 27/12/2019.

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2019

Ammara Nosheen, Aneela Nawaz, Khuram Ali Khan and Khalid Mahmood Awan

Published in the Arab Journal of Mathematical Sciences. Published by Emerald Publishing Limited. This article is published under the Creative Commons Attribution (CC BY 4.0) license. Anyone may reproduce, distribute, translate and create derivative works of this article (for both commercial and non-commercial purposes), subject to full attribution to the original publication and authors. The full terms of this license may be seen at http://creativecommons.org/licences/by/4.0/legalcode

Multivariate Hardy and Littlewood inequalities on time scales

1. Introduction

2. Preliminaries

3. Hardy and Littlewood-type inequalities for p/q ≥ 2 and γ > 1

4. Hardy and Littlewood-type inequalities for p/q ≥ 1 and γ > 1

5. Hardy and Littlewood-type inequalities for p/q ≤ 2 and γ > 1

6. Hardy and Littlewood-type inequalities for p/q > 1 and γ < 1

7. Hardy and Littlewood-type inequalities for p/q ≤ 2 and γ < 1

References

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Multivariate Hardy and Littlewood inequalities on time scales Open Access

1. Introduction

2. Preliminaries

3. Hardy and Littlewood-type inequalities for p/q ≥ 2 and γ > 1

4. Hardy and Littlewood-type inequalities for p/q ≥ 1 and γ > 1

5. Hardy and Littlewood-type inequalities for p/q ≤ 2 and γ > 1

6. Hardy and Littlewood-type inequalities for p/q > 1 and γ < 1

7. Hardy and Littlewood-type inequalities for p/q ≤ 2 and γ < 1

References

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Multivariate Hardy and Littlewood inequalities on time scales