**Ahmed A. El-Deeb 1,\*, Samer D. Makharesh <sup>1</sup> and Dumitru Baleanu 2,3,4**


Received: 12 March 2020; Accepted: 26 March 2020; Published: 7 April 2020

**Abstract:** Our work is based on the multiple inequalities illustrated in 2020 by Hamiaz and Abuelela. With the help of a Fenchel-Legendre transform, which is used in various problems involving symmetry, we generalize a number of those inequalities to a general time scale. Besides that, in order to get new results as special cases, we will extend our results to continuous and discrete calculus.

**Keywords:** Hilbert's inequality; Fubini theorem; Fenchel-Legendre transform; time scale

**AMS Subject Classifications:** 26D10; 26D15; 26E70; 34A40

### **1. Introduction**

In 2020, Hamiaz and Abuelela [1] have studied the following discrete inequalities:

**Theorem 1.** *Suppose q*, *p* > 1, *α* > *β* > <sup>1</sup> 2 *and* (*bm*)*<sup>m</sup>* ≥ 0 (*an*)*<sup>n</sup>* ≥ 0 *are sequences of real numbers. Define A<sup>n</sup>* = ∑ *n s*=1 *as* , *B<sup>m</sup>* = ∑ *m t*=1 *bt* . *Then*

$$\begin{aligned} \sum\_{n=1}^{k} \sum\_{m=1}^{r} \frac{A\_n^{2p} B\_m^{2q}}{h(n) + h^\*(m)} &\leqslant \mathcal{C}\_1^\*(p, q) \left( \sum\_{n=1}^{k} (k - n + 1)(a\_n A\_n^{p-1})^2 \right) \\ &\qquad \times \left( \sum\_{m=1}^{r} (r - m + 1)(b\_m B\_m^{q-1})^2 \right) \end{aligned}$$

*and*

$$\begin{split} \sum\_{n=1}^{k} \sum\_{m=1}^{r} \frac{A\_n^p B\_m^q}{\left( |h(n)|^{\frac{1}{2^p}} + |h^\*(m)|^{\frac{1}{2^p}} \right)^n} & \quad \leqslant \sum\_{n=1}^{k} \sum\_{m=1}^{r} \frac{A\_n^p B\_m^q}{\sqrt{h(n) + h^\*(m)}} \\ & \quad \leqslant \mathcal{C}\_2^\*(p, q, k, r) \left( \sum\_{n=1}^{k} (k - n + 1)(a\_n A\_n^{p-1})^2 \right)^{\frac{1}{2}} \\ & \quad \times \left( \sum\_{m=1}^{r} (r - m + 1)(b\_m B\_m^{q-1})^2 \right)^{\frac{1}{2}} \end{split}$$

*unless* (*an*) *or* (*bm*) *is null, where*

$$\mathbf{C}\_1^\*(p,q) = (pq)^2 \text{ and } \mathbf{C}\_2^\*(p,q,r,k) = pq\sqrt{kr}.$$

Hilger [2] suggested time scales theory to unify discrete and continuous analysis. More Hilbert-type inequalities and other types can be seen in [1,3–36], see also [37–53]. For more details on time scales calculus see [54].

We will need the following important relations between calculus on time scales T and either continuous calculus on R or discrete calculus on Z. Note that:

**(***i***)** If T = R, then

$$
\sigma(t) = t, \quad \mu(t) = 0, \quad f^{\Delta}(t) = f'(t), \quad \int\_{a}^{b} f(t) \Delta t = \int\_{a}^{b} f(t) dt. \tag{1}
$$

**(***ii***)** If T = Z, then

$$
\sigma(t) = t + 1, \quad \mu(t) = 1, \quad f^{\Delta}(t) = f(t+1) - f(t), \quad \int\_{a}^{b} f(t) \Delta t = \sum\_{t=a}^{b-1} f(t). \tag{2}
$$

Next is Hölder's and Jensen's inequality:

**Lemma 1** ([19])**.** *Let a*, *<sup>b</sup>* <sup>∈</sup> <sup>T</sup> *and f , g* <sup>∈</sup> *<sup>C</sup>rd*([*a*, *<sup>b</sup>*]T, [0, <sup>∞</sup>))*. If p, q* <sup>&</sup>gt; <sup>1</sup> *with* <sup>1</sup> *<sup>p</sup>* <sup>+</sup> <sup>1</sup> *<sup>q</sup>* = 1*, then*

$$\int\_{a}^{b} f(t)g(t)\Delta t \le \left[\int\_{a}^{b} f^p(t)\Delta t\right]^{\frac{1}{p}} \left[\int\_{a}^{b} g^q(t)\Delta t\right]^{\frac{1}{q}}.$$

**Lemma 2** ( [19])**.** *Let a, b* ∈ T *and c*˘*,* ˘*<sup>d</sup>* <sup>∈</sup> <sup>R</sup>*. Assume that <sup>g</sup>* <sup>∈</sup> *<sup>C</sup>rd* [*a*, *b*]T, [*c*˘, ˘*d*] *and <sup>r</sup>* <sup>∈</sup> *<sup>C</sup>rd* [*a*, *b*]T, R *are nonnegative with* R *<sup>b</sup> a <sup>r</sup>*(*t*)∆*<sup>t</sup>* <sup>&</sup>gt; <sup>0</sup>*. If* <sup>Φ</sup>˘ <sup>∈</sup> *<sup>C</sup>rd* (*c*˘, ˘*d*), R *be a convex function, then*

$$
\Phi\left(\frac{\int\_a^b g(t)r(t)\Delta t}{\int\_a^b r(t)\Delta t}\right) \leqslant \frac{\int\_a^b r(t)\Phi(g(t))\Delta t}{\int\_a^b r(t)\Delta t}.
$$

Now, we present the Fenchel-Legendre transform and refer, for example, to [11–13], for more details.

**Definition 1.** *Assuming <sup>h</sup>* : <sup>R</sup>*<sup>n</sup>* −→ <sup>R</sup> ∪ {+∞} *is a function: <sup>h</sup>* <sup>6</sup>= +<sup>∞</sup> *i.e.,* Dom(*h*) = {*<sup>x</sup>* <sup>∈</sup> <sup>R</sup>*<sup>n</sup>* , |*h*(*x*) < ∞} 6= ∅*. Then the Fenchel-Legendre transform is defined as:*

$$h^\*: \mathbb{R}^n \longrightarrow \mathbb{R} \cup \{ +\infty \}, \ y \longrightarrow h^\*(y) = \sup \{  -h(x), x \in \text{Dom}(h) \}\tag{3}$$

*where* < ., . > *is the scalar product on* R*<sup>n</sup> . The mapping h* −→ *h* ∗ *is often be called the conjugate operation.*

The domain of *h* ∗ is the set of slopes of all the affine functions minorizing the function *h* over R*<sup>n</sup>* . An equivalent formula for (3) is introduced as follows:

**Corollary 1.** *Assuming h* : <sup>R</sup>*<sup>n</sup>* −→ <sup>R</sup> *is differentiable, strictly convex and 1-coercive function. Then*

$$h^\*(y) = (\nabla h)^{-1}(y)> - h((\nabla h)^{-1}(y)),\tag{4}$$

∀ *y* ∈ Dom(*h* ∗ )*, where* < ., . > *denotes the scalar product on* R*<sup>n</sup> .* **Lemma 3** ( [13])**.** *Let h be a function and h*∗ *its Fenchel-Legendre transform. Then*

$$ \lessapprox h(x) + h^\*(y),\tag{5}$$

*for all x* ∈ Dom(*h*)*, and y* ∈ Dom(*h* ∗ ).

In addition, we will use the following definition and lemma as we will see in the proof of our results:

**Definition 2.** *The function* Φ˘ *is said to be a submultiplicative on* [0, ∞) *if*

$$
\Phi(xy) \lessapprox \Phi(x)\Phi(y), \text{ for all } \ge, y \gg 0. \tag{6}
$$

**Lemma 4** ([20])**.** *Assuming* T *is a time scale with x*, *a* ∈ T *such that x* > *a*. *If f* > 0 *and α*˜ > 1, *then*

$$\left(\int\_{a}^{\sigma(\mathbf{x})} f(\check{\mathbf{r}}) \Delta \check{\mathbf{r}}\right)^{\tilde{\mathbf{a}}} \leqslant \tilde{\mathbf{a}} \int\_{a}^{\sigma(\mathbf{x})} f(\eta) \left(\int\_{a}^{\sigma(\eta)} f(\check{\mathbf{r}}) \Delta \check{\mathbf{r}}\right)^{\tilde{\mathbf{a}}-1} \Delta \eta. \tag{7}$$

Next, we write Fubini's theorem on time scales.

**Lemma 5** (Fubini's Theorem, see [55])**.** *Assume that* (*X*, Σ1, *µ*∆) *and* (*Y*, Σ2, *ν*∆) *are two finite-dimensional time scales measure spaces. Moreover, suppose that f* : *X* × *Y* → R *is a delta integrable function and define the functions*

$$\#\_1(y) = \int\_X f(\mathfrak{x}, y) d\mu\_\Delta(\mathfrak{x}), \quad y \in \Upsilon\_\Delta$$

*and*

$$\mathfrak{H}\_2(\mathfrak{x}) = \int\_X f(\mathfrak{x}, \mathfrak{y}) d\nu\_{\Delta}(\mathfrak{y}) \quad \mathfrak{x} \in \mathrm{X}.$$

*Then π*ˆ <sup>1</sup> *is delta integrable on Y and π*ˆ <sup>2</sup> *is delta integrable on X and*

$$\int\_X d\mu\_\Delta(\mathbf{x}) \int\_Y f(\mathbf{x}, y) d\nu\_\Delta(y) = \int\_Y d\nu\_\Delta(y) \int\_X f(\mathbf{x}, y) d\mu\_\Delta(\mathbf{x}) .$$

In this manuscript, by using Fubini's theorem and the Fenchel-Legendre transform, which is used in various problems involving symmetry, we extend the discrete results proved in [1] on time scales. We start from the inequalities treated in the Theorem 1. Our results can be applied to give more general forms of some previously proved inequalities through substituting *h* and *h* ∗ by suitable functions as we will see in the following two sections.

The following section contains our main results.

#### **2. Main Results**

We start by establishing the following useful inequality:

**Lemma 6.** *Assume x and y* <sup>∈</sup> <sup>R</sup> *such that x* <sup>+</sup> *<sup>y</sup>* <sup>&</sup>gt; 1, *then for <sup>γ</sup>* <sup>&</sup>gt; 0, *and <sup>α</sup>* <sup>&</sup>gt; *<sup>β</sup>* <sup>&</sup>gt; <sup>1</sup> 2 *, we get*

$$(\varkappa + y)^{\frac{1}{\gamma}} \leqslant \left( |\varkappa|^{\frac{1}{2\overline{\mu}}} + |y|^{\frac{1}{2\overline{\mu}}} \right)^{\frac{2a}{\gamma}}.\tag{8}$$

**Proof.** For *x* + *y* > 1 and *<sup>α</sup> <sup>β</sup>* > 1, we have

$$(\mathbf{x} + \mathbf{y})^{\frac{1}{2}} \lessapprox \left[ (\mathbf{x} + \mathbf{y})^{\frac{1}{2}} \right]^{\frac{\kappa}{\overline{\rho}}} = \left[ (\mathbf{x} + \mathbf{y})^{\frac{1}{2\overline{\rho}}} \right]^{\kappa} \lessapprox \left[ (|\mathbf{x}| + |\mathbf{y}|)^{\frac{1}{2\overline{\rho}}} \right]^{\kappa}. \tag{9}$$

From |*x*| + |*y*| 1 *n* 6 |*x*| 1 *<sup>n</sup>* + |*y*| 1 *<sup>n</sup>* , for all *n* > 1. Thus, from (9), and since 2*β* > 1, we obtain:

$$(x+y)^{\frac{1}{2}} \lesssim \left[ (|x|+|y|)^{\frac{1}{2\tilde{\rho}}} \right]^{\alpha} \lesssim \left[ |x|^{\frac{1}{2\tilde{\rho}}} + |y|^{\frac{1}{2\tilde{\rho}}} \right]^{\alpha}.\tag{10}$$

Now, since *γ* > 0, by taking the power <sup>1</sup>/*<sup>γ</sup>* for both sides of (10), we get:

$$(x+y)^{\frac{1}{\gamma}} \leqslant \left( |x|^{\frac{1}{2\beta}} + |y|^{\frac{1}{2\beta}} \right)^{\frac{2\alpha}{\gamma}}.$$

This proves our claim.

In the next theorems, we will let *p* > 1, *q* > 1 and <sup>1</sup> *<sup>q</sup>* <sup>+</sup> <sup>1</sup> *<sup>p</sup>* = 1.

**Theorem 2.** *Let* T *be a time scale with L* > 1, *K* > 1 *and s*, *t*, *t*0, *x*, *y* ∈ T. *Assume a*(*τ*ˇ) ≥ 0 *and b*(*τ*ˇ) ≥ 0 *are right-dense continuous functions on the time scales intervals* [*t*0, *x*]<sup>T</sup> *and* [*t*0, *y*]<sup>T</sup> *respectively and define*

$$A(s) := \int\_{t\_0}^{s} a(\mathfrak{k}) \Delta \mathfrak{k}, \text{ and } B(t) := \int\_{t\_0}^{t} b(\mathfrak{k}) \Delta \mathfrak{k}.$$

*then for σ*(*s*) ∈ [*t*0, *x*]<sup>T</sup> *and σ*(*t*) ∈ [*t*0, *y*]T, *we have that*

Z *x t*0 Z *y t*0 *A qK*(*σ*(*s*))*B qL*(*σ*(*t*)) |*h*(*σ*(*s*) − *t*0)| 1 <sup>2</sup>*<sup>β</sup>* + |*h* <sup>∗</sup>(*σ*(*t*) − *t*0)| 1 2*β* <sup>2</sup>*q<sup>α</sup> p* ∆*s*∆*t* 6 *C*1(*L*, *K*, *q*) Z *x t*0 (*σ*(*x*) <sup>−</sup> *<sup>σ</sup>*(*s*)) *a*(*s*)*A K*−1 (*σ*(*s*) *q* ∆*s* × Z *y t*0 (*σ*(*y*) <sup>−</sup> *<sup>σ</sup>*(*t*)) *b*(*t*)*B L*−1 (*σ*(*t*) *q* ∆*t* (11)

*and*

$$\begin{split} &\int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{A^K(\sigma(s))B^L(\sigma(t))}{\left(|h(\sigma(s)-t\_0)|^{\frac{1}{2\tilde{\eta}}} + |h^\*(\sigma(t)-t\_0)|^{\frac{1}{2\tilde{\eta}}}\right)^{\frac{2\tilde{\alpha}}{p}}} \Delta s \Delta t \\ &\leq \mathcal{C}\_2(L, K, p) \left(\int\_{t\_0}^{\chi} (\sigma(x) - \sigma(s)) \left(A^{K-1}(\sigma(s))a(s)\right)^q \Delta s\right)^{\frac{1}{q}} \\ &\times \left(\int\_{t\_0}^{y} (\sigma(y) - \sigma(t)) \left(B^{L-1}(\sigma(t))b(t)\right)^q \Delta t\right)^{\frac{1}{q}} \end{split} \tag{12}$$

*where C*1(*L*, *K*, *q*) = (*KL*) *q and C*2(*L*, *K*, *p*) = *KL*(*x* − *t*0) 1 *<sup>p</sup>* (*y* − *t*0) 1 *p* .

**Proof.** By using the inequality (7), we obtain

$$A^K(\sigma(s)) \lessapprox K \int\_{t\_0}^{\sigma(s)} a(\eta) A^{K-1}(\sigma(\eta)\Delta \eta) \tag{13}$$

$$\mathcal{B}^L(\sigma(t)) \lessapprox L \int\_{t\_0}^{\sigma(t)} b(\eta) \mathcal{B}^{L-1}(\sigma(\eta)) \Delta \eta. \tag{14}$$

We use Lemma 1. Then from (13), we get

$$A^K(\sigma(s)) \leqslant K(\sigma(s) - t\_0)^{\frac{1}{p}} \left( \int\_{t\_0}^{\sigma(s)} \left( a(\eta) A^{K-1}(\sigma(\eta)) \right)^q \Delta \eta \right)^{\frac{1}{q}}.\tag{15}$$

We use Lemma 1. Then from (14), we also have

$$B^L(\sigma(t)) \lessapprox L(\sigma(t) - t\_0)^{\frac{1}{p}} \left( \int\_{t\_0}^{\sigma(t)} \{b(\eta)B^{L-1}(\sigma(\eta))\}^q \Delta \eta \right)^{\frac{1}{q}}.\tag{16}$$

From (15) and (16), we get

$$A^K(\sigma(s))B^L(\sigma(t)) \ll KL(\sigma(s) - t\_0)^{\frac{1}{p}}(\sigma(t) - t\_0)^{\frac{1}{p}}$$

$$\times \left(\int\_{t\_0}^{\sigma(s)} \left(a(\eta)A^{K-1}(\sigma(\eta))\right)^q \Delta \eta \right)^{\frac{1}{q}}$$

$$\times \left(\int\_{t\_0}^{\sigma(t)} \left(b(\eta)B^{L-1}(\sigma(\eta))\right)^q \Delta \eta \right)^{\frac{1}{q}}.\tag{17}$$

From inequality (17), we have

$$\begin{split} A^{qK}(\sigma(\mathbf{s})) \mathcal{B}^{qL}(\sigma(t)) &\leqslant (KL)^{q} (\sigma(\mathbf{s}) - t\_{0})^{\frac{q}{p}} (\sigma(t) - t\_{0})^{\frac{q}{p}} \\ &\qquad \times \left( \int\_{t\_{0}}^{\sigma(\mathbf{s})} \left( a(\eta) A^{K-1} (\sigma(\eta)) \right)^{q} \Delta \eta \right) \\ &\qquad \times \left( \int\_{t\_{0}}^{\sigma(\mathbf{t})} \left( b(\eta) \mathcal{B}^{L-1} (\sigma(\eta)) \right)^{q} \Delta \eta \right). \end{split} \tag{18}$$

Using Lemma 3 in (17) and (18) gives

$$A^K(\sigma(s))B^L(\sigma(t)) \lessapprox KL\left(h(\sigma(s) - t\_0) + h^\*(\sigma(t) - t\_0)\right)^{\frac{1}{q}}$$

$$\times \left(\int\_{t\_0}^{\sigma(s)} \left(a(\eta)A^{K-1}(\sigma(\eta))\right)^q \Delta \eta\right)^{\frac{1}{q}}$$

$$\times \left(\int\_{t\_0}^{\sigma(t)} \left(b(\eta)B^{L-1}(\sigma(\eta))\right)^q \Delta \eta\right)^{\frac{1}{q}}.\tag{19}$$

$$\begin{split} A^{qK}(\sigma(s))\mathcal{B}^{qL}(\sigma(t)) &\leqslant (KL)^{q} \Big( h(\sigma(s)-t\_{0}) + h^{\*}(\sigma(t)-t\_{0}) \Big)^{\frac{q}{p}} \\ &\qquad \times \Big( \int\_{t\_{0}}^{\sigma(s)} \Big( a(\eta)A^{K-1}(\sigma(\eta))^{q}\Delta\eta \Big) \\ &\qquad \times \Big( \int\_{t\_{0}}^{\sigma(t)} \Big( b(\eta)B^{L-1}(\sigma(\eta))^{q}\Delta\eta \Big) . \end{split} \tag{20}$$

Using Lemma 6 in (19) and (20) gives

$$\begin{split} A^K(\sigma(s))B^L(\sigma(t)) \leqslant & KL\left( |h(\sigma(s)-t\_0)|^{\frac{1}{2\bar{\eta}}} + |h^\*(\sigma(t)-t\_0)|^{\frac{1}{2\bar{\eta}}} \right)^{\frac{2\bar{\alpha}}{\bar{\eta}}} \\ & \times \left( \int\_{t\_0}^{\sigma(s)} \left( a(\eta)A^{K-1}(\sigma(\eta)) \right)^q \Delta \eta \right)^{\frac{1}{q}} \\ & \times \left( \int\_{t\_0}^{\sigma(t)} \left( b(\eta)B^{L-1}(\sigma(\eta)) \right)^q \Delta \eta \right)^{\frac{1}{q}} \end{split} \tag{21}$$

$$\begin{split} A^{qK}(\sigma(\mathbf{s})) \mathcal{B}^{qL}(\sigma(t)) &\leqslant (KL)^{q} \Big( |h(\sigma(\mathbf{s}) - t\_{0})|^{\frac{1}{2\theta}} + |h^{\*}(\sigma(t) - t\_{0})|^{\frac{1}{2\theta}} \Big)^{\frac{2\mu}{\theta}} \\ &\qquad \times \Big( \int\_{t\_{0}}^{\sigma(\mathbf{s})} \Big( a(\eta)A^{K-1}(\sigma(\eta))^{q} \Delta \eta \Big) \\ &\qquad \times \Big( \int\_{t\_{0}}^{\sigma(t)} \Big( b(\eta)B^{L-1}(\sigma(\eta))^{q} \Delta \eta \Big). \end{split} \tag{22}$$

Dividing both sides of (21) and (22) by |*h*(*σ*(*s*) − *t*0)| 1 <sup>2</sup>*<sup>β</sup>* + |*h* ∗ (*σ*(*t*) − *t*0)| 1 2*β* <sup>2</sup>*<sup>α</sup> p* and |*h*(*σ*(*s*) − *t*0)| 1 <sup>2</sup>*<sup>β</sup>* + |*h* ∗ (*σ*(*t*) − *t*0)| 1 2*β* <sup>2</sup>*q<sup>α</sup> p* respectively, we get that

$$\frac{A^{K}(\sigma(s))B^{L}(\sigma(t))}{\left(|h(\sigma(s)-t\_{0})|^{\frac{1}{2\overline{\rho}}}+|h^{\*}(\sigma(t)-t\_{0})|^{\frac{1}{2\overline{\rho}}}\right)^{\frac{2\alpha}{\overline{\rho}}}} \leqslant KL\left(\int\_{t\_{0}}^{\sigma(s)} \left(a(\eta)A^{K-1}(\sigma(\eta))^{q}\Delta\eta\right)^{\frac{1}{q}}\right)^{\frac{1}{q}}$$

$$\times \left(\int\_{t\_{0}}^{\sigma(t)} \left(b(\eta)B^{L-1}(\sigma(\eta))\right)^{q}\Delta\eta\right)^{\frac{1}{q}}.\tag{23}$$

$$\frac{A^{qK}(\sigma(s))B^{qL}(\sigma(t))}{\left(|h(\sigma(s)-t\_0)|^{\frac{1}{2^p}} + |h^\*(\sigma(t)-t\_0)|^{\frac{1}{2^p}}\right)^{\frac{2\mu}{p}}} \lesssim (KL)^{\mathfrak{q}} \left(\int\_{t\_0}^{\sigma(s)} \left(a(\eta)A^{K-1}(\sigma(\eta))^{\mathfrak{q}}\Delta\eta\right) \right)$$

$$\times \left(\int\_{t\_0}^{\sigma(t)} \left(b(\eta)B^{L-1}(\sigma(\eta))\right)^{\mathfrak{q}} \Delta\eta\right). \tag{24}$$

From (23) by using Lemma 1 we obtain

$$\begin{split} &\int\_{t\_0}^{\infty} \int\_{t\_0}^{y} \frac{A^K(\sigma(s)) \mathcal{B}^L(\sigma(t))}{\left( |h(\sigma(s) - t\_0)|^{\frac{1}{2\tilde{p}}} + |h^\*(\sigma(t) - t\_0)|^{\frac{2\tilde{p}}{\tilde{p}}} \right)^{\frac{2\tilde{p}}{\tilde{p}}}} \Delta s \Delta t \\ &\leq KL(x - t\_0)^{\frac{1}{\tilde{p}}} (y - t\_0)^{\frac{1}{\tilde{p}}} \int\_{t\_0}^{x} \left( \int\_{t\_0}^{\sigma(s)} \left( a(\eta) A^{K-1} (\sigma(\eta))^q \Delta \eta \right) \Delta s \right)^{\frac{1}{q}} \\ &\times \int\_{t\_0}^{y} \left( \int\_{t\_0}^{\sigma(t)} \left( b(\eta) B^{L-1} (\sigma(\eta))^q \Delta \eta \right) \Delta t \right)^{\frac{1}{q}}. \end{split}$$

*Symmetry* **2020**, *12*, 582

From (24), we get

$$\begin{split} \int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{A^{qK}(\sigma(s))B^{qL}(\sigma(t))}{\left(|h(\sigma(s)-t\_0)|^{\frac{1}{2\beta}} + |h^\*(\sigma(t)-t\_0)|^{\frac{1}{2\beta}}\right)^{\frac{2q\kappa}{p}}} \Delta s \Delta t \\ \lesssim (KL)^q \int\_{t\_0}^{\chi} \left(\int\_{t\_0}^{\sigma(s)} \left(a(\eta)A^{K-1}(\sigma(\eta))\right)^q \Delta \eta\right) \Delta s \Big) \\ \qquad \times \int\_{t\_0}^{y} \left(\int\_{t\_0}^{\sigma(t)} \left(b(\eta)B^{L-1}(\sigma(\eta))\right)^q \Delta \eta\right) \Delta t \Big). \end{split} \tag{26}$$

Applying Fubini's Theorem on (25) and (26) gives

$$\begin{split} &\int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{A^K(\sigma(s))B^L(\sigma(t))}{\left( |h(\sigma(s)-t\_0)|^{\frac{1}{2\beta}} + |h^\*(\sigma(t)-t\_0)|^{\frac{1}{2\beta}} \right)^{\frac{2\alpha}{p}}} \Delta s \Delta t \\ &\leq KL(\mathbf{x}-t\_0)^{\frac{1}{p}}(y-t\_0)^{\frac{1}{p}} \left( \int\_{t\_0}^{\chi} (\mathbf{x}-\sigma(s)) \left( a(s)A^{K-1}(\sigma(s)) \right)^q \Delta s \right)^{\frac{1}{q}} \\ &\times \left( \int\_{t\_0}^{y} (y-\sigma(t)) \left( b(t)B^{L-1}(\sigma(t)) \right)^q \Delta t \right)^{\frac{1}{q}} . \end{split}$$

$$\begin{split} &\int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{A^{q\mathbf{K}}(\sigma(\mathbf{s})) B^{q\mathbf{L}}(\sigma(t))}{\left( |h(\sigma(\mathbf{s}) - t\_0)|^{\frac{1}{2\mathfrak{F}}} + |h^\*(\sigma(t) - t\_0)|^{\frac{1}{2\mathfrak{F}}} \right)^{\frac{2q\mathbf{a}}{p}}} \Delta \mathbf{s} \Delta t \\ &\lesssim (KL)^{q} \left( \int\_{t\_0}^{\chi} (\mathbf{x} - \sigma(\mathbf{s})) \left( a(\mathbf{s}) A^{K-1}(\sigma(\mathbf{s})) \right)^{q} \right) \Delta t \right) \\ &\times \left( \int\_{t\_0}^{\mathcal{Y}} (y - \sigma(t)) \left( b(t) B^{L-1}(\sigma(t)) \right)^{q} \Delta t \right) . \end{split}$$

Using the facts *σ*(*x*) > *x*, *σ*(*y*) > *y* yields

$$\begin{split} &\int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{A^K(\sigma(s)) \mathcal{B}^L(\sigma(t))}{\left( |h(\sigma(s) - t\_0)|^{\frac{1}{2\tilde{\rho}}} + |h^\*(\sigma(t) - t\_0)|^{\frac{1}{2\tilde{\rho}}} \right)^{\frac{2\tilde{\rho}}{\tilde{p}}}} ds dt \\ &\leq C\_2(L, K, p) \left( \int\_{t\_0}^{\chi} (\sigma(\mathbf{x}) - \sigma(\mathbf{s})) \left( a(\mathbf{s}) A^{K-1}(\sigma(\mathbf{s})) \right)^q \Delta \mathbf{s} \right)^{\frac{1}{q}} \\ &\times \left( \int\_{t\_0}^{y} (\sigma(y) - \sigma(t)) \left( b(t) B^{L-1}(\sigma(t)) \right)^q \Delta t \right)^{\frac{1}{q}} . \end{split}$$

$$\begin{split} &\int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{A^{qK}(\sigma(s))B^{qL}(\sigma(t))}{\left(|h(\sigma(s)-t\_0)|^{\frac{1}{2\beta}} + |h^\*(\sigma(t)-t\_0)|^{\frac{1}{2\beta}}\right)^{\frac{2q\alpha}{p}}} \Delta s \Delta t \\ &\leq C\_1(L, K, q) \left(\int\_{t\_0}^{\chi} (\sigma(\chi)-\sigma(s)) \left(a(s)A^{K-1}(\sigma(s))\right)^2 \Delta s\right), \\ &\quad \times \left(\int\_{t\_0}^{y} (\sigma(y)-\sigma(t)) \left(b(t)B^{L-1}(\sigma(t))\right)^q \Delta t\right). \end{split}$$

#### This completes the proof.

**Theorem 3.** *Let a*(*τ*ˇ), *b*(*η*), *A*(*s*) *and B*(*t*) *be defined as in Theorem 2, thus*

$$\begin{aligned} &\int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{A^q(\sigma(s))B^q(\sigma(t))}{\left( |h(\sigma(s) - t\_0)|^{\frac{1}{2\beta}} + |h^\*(\sigma(t) - t\_0)|^{\frac{1}{2\beta}} \right)^{\frac{2q\alpha}{p}}} \Delta s \Delta t \\ &\leq \left( \int\_{t\_0}^{\chi} (\sigma(\chi) - \sigma(s)) a^q(s) \Delta s \right) \left( \int\_{t\_0}^{y} (\sigma(y) - \sigma(t)) b^q(t) \Delta t \right)^{\frac{1}{2\beta}} \end{aligned}$$

*and*

$$\begin{split} &\int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{A(\sigma(s)) \mathcal{B}(\sigma(t))}{\left( |h(\sigma(s) - t\_0)|^{\frac{1}{2\overline{\rho}}} + |h^\*(\sigma(t) - t\_0)|^{\frac{1}{2\overline{\rho}}} \right)^{\frac{2\kappa}{p}}} \Delta s \Delta t \\ &\leq (\chi - t\_0)^{\frac{1}{p}} (y - t\_0)^{\frac{1}{p}} \left( \int\_{t\_0}^{\chi} (\sigma(\mathbf{x}) - \sigma(s)) a^q(s) \Delta s \right)^{\frac{1}{q}} \left( \int\_{t\_0}^{y} (\sigma(y) - \sigma(t)) b^q(t) \Delta t \right)^{\frac{1}{q}}. \end{split}$$

**Proof.** Put *K* = *L* = 1 in (11) and (12). This completes the proof.

In Theorem 2, if we choose T = R, then we have relation (1) and the next results:

**Corollary 2.** *If a*(*s*) ≥ 0*, b*(*t*) ≥ 0*. Define A*(*s*) := R *s* 0 *a*(*η*)*dη and B*(*t*) := R *t* 0 *b*(*η*)*dη, then*

$$\begin{aligned} &\int\_0^\chi \int\_0^y \frac{A^{qK}(s)B^{qL}(t)}{\left(|h(s)|^{\frac{1}{2\tilde{p}}} + |h^\*(t)|^{\frac{1}{2\tilde{p}}}\right)^{\frac{2q\kappa}{p}}} ds dt \\ &\leq \mathcal{C}\_1(L, K, q) \left(\int\_0^\chi (\varkappa - s) \left(a(s)A^{K-1}(s)\right)^q ds\right), \\ &\qquad \times \left(\int\_0^y (y-t) \left(b(t)B^{L-1}(t)\right)^q dt\right). \end{aligned}$$

*and*

$$\begin{aligned} &\int\_0^\chi \int\_0^y \frac{A^K(s)B^L(t)}{\left(|h(s)|^{\frac{1}{2\beta}} + |h^\*(t)|^{\frac{1}{2\beta}}\right)^{\frac{2\alpha}{p}}} ds dt \\ &\leq \mathcal{C}\_3(L, K, p) \left(\int\_0^x (x - s) \left(A^{K-1}(s)a(s)\right)^q ds\right)^{\frac{1}{q}} \\ &\times \left(\int\_0^y (y - t) \left(B^{L-1}(t)b(t)\right)^q dt\right)^{\frac{1}{q}} \end{aligned}$$

*where*

$$\mathcal{C}\_3(L, K, p) = KL(xy)^{\frac{1}{p}}.$$

In Theorem 2, if we chose T = Z, then we get (2), and the next result:

**Corollary 3.** *If a*(*n*) ≥ *and b*(*m*) ≥ 0*. Define*

$$A(n) = \sum\_{s=0}^{n} a(s), \quad B(m) = \sum\_{k=0}^{m} b(k).$$

*Symmetry* **2020**, *12*, 582

*Then*

$$\begin{aligned} \sum\_{n=1}^{N} \sum\_{m=1}^{M} \frac{A^{qL}(n)B^{qK}(m)}{\left(|h(n+1)|^{\frac{1}{2p}} + |h^\*(m+1)|^{\frac{1}{2p}}\right)^{\frac{2m}{p}}} & \leqslant \mathbb{C}\_1(K, L, q) \left(\sum\_{n=1}^{N} (N+1 - (n+1))(a(n)A^{L-1}(n))^q\right)^{\frac{1}{q}}\\ & \qquad \times \left(\sum\_{m=1}^{M} (M+1 - (m+1))(b(m)B^{L-1}(m))^q\right)^{\frac{1}{q}} \end{aligned}$$

*and*

$$\begin{aligned} \sum\_{n=1}^N \sum\_{m=1}^M \frac{A^L(n)B^K(m)}{\left( |h(n+1)|^{\frac{1}{2p}} + |h^\*(m+1)|^{\frac{1}{2p}} \right)^{\frac{2n}{p}}} & \leqslant \mathcal{C}\_4(K, L, p) \left( \sum\_{n=1}^N (N+1 - (n+1))(a(n)A^{L-1}(n))^q \right)^{\frac{1}{q}} \\ & \qquad \times \left( \sum\_{m=1}^M (M+1 - (m+1))(b(m)B^{L-1}(m))^q \right)^{\frac{1}{q}} \end{aligned}$$

*where*

$$\mathbb{C}\_4(K, L, p) = KL(NM)^{\frac{1}{p}}.$$

**Remark 1.** *Taking p* = *q* = 2 *in Corollary 3 gives the result due to Hamiaz and Abuelela ([1], Theorem 3).*

**Corollary 4.** *With the hypotheses of Theorem 2 we have:*

$$\begin{split} &\int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{A^{q\mathbf{K}}(\sigma(s))B^{q\mathbf{L}}(\sigma(t))}{\left(|h(\sigma(s)-t\_0)|^{\frac{1}{2\beta}} + |h^\*(\sigma(t)-t\_0)|^{\frac{1}{2\beta}}\right)^{\frac{2q\alpha}{p}}} \Delta s \Delta t \\ &\lesssim C\_1(L, K, q) \left\{ h \left( \int\_{t\_0}^{\chi} (\sigma(\chi)-\sigma(s)) \left(a(s)A^{K-1}(\sigma(s))\right)^q \Delta s \right) \right\} \\ &\quad + h^\* \left( \int\_{t\_0}^{y} (\sigma(y)-\sigma(t)) \left(b(t)B^{L-1}(\sigma(t))\right)^q \Delta t \right) \right\} \end{split}$$

*and*

$$\begin{split} &\int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{A^K(\sigma(s))B^L(\sigma(t))}{\left(|h(\sigma(s)-t\_0)|^{\frac{1}{2\tilde{p}}} + |h^\*(\sigma(t)-t\_0)|^{\frac{1}{2\tilde{p}}}\right)^{\frac{2\tilde{q}}{p}}} \Delta s \Delta t \\ &\leq C\_2(L\_\*K,p) \left\{ h \left(\int\_{t\_0}^{\chi} (\sigma(\mathbf{x})-\sigma(s)) \left(A^{K-1}(\sigma(s))a(s)\right)^q \Delta s \right) \right\} \\ &\quad + h^\* \left(\int\_{t\_0}^{y} (\sigma(y)-\sigma(t)) \left(B^{L-1}(\sigma(t))b(t)\right)^q \Delta t \right) \right\}^{\frac{1}{q}}. \end{split}$$

**Proof.** Using the Fenchel-Young inequality (5) in (11) and (12). This proves the claim.

**Theorem 4.** *Assuming the time scale* T *with s*, *t*, *t*0, *x*, *y* ∈ T, *A*(*s*) *and B*(*t*) *are defined as in Theorem 2. Suppose f*(*τ*ˇ) ≥ 0 *and g*(*η*) ≥ 0 *are right-dense continuous functions on* [*t*0, *x*]<sup>T</sup> *and* [*t*0, *y*]<sup>T</sup> *respectively. Suppose that* <sup>Φ</sup>˘ <sup>≥</sup> <sup>0</sup> *and* <sup>Ψ</sup>˘ <sup>≥</sup> <sup>0</sup> *are convex, and submultiplicative functions on* [0, <sup>∞</sup>). *Furthermore assume that*

$$F(\mathbf{s}) := \int\_{t\_0}^{\mathbf{s}} f(\mathbf{t}) \Delta \mathbf{t} \,, \; \text{and} \; \; G(t) := \int\_{t\_0}^{t} \mathbf{g}(\eta) \Delta \eta \,. \tag{27}$$

*then for σ*(*s*) ∈ [*t*0, *x*]<sup>T</sup> *and σ*(*t*) ∈ [*t*0, *y*]T, *we have that*

$$\int\_{t\_0}^{x} \int\_{t\_0}^{y} \frac{\Phi(A^{\sigma}(s)) \Psi(B^{\sigma}(t))}{\left( |h(\sigma(s) - t\_0)|^{\frac{1}{2\overline{\rho}}} + |h^\*(\sigma(t) - t\_0)|^{\frac{1}{2\overline{\rho}}} \right)^{\frac{2\sigma}{\overline{p}}}} \Delta s \Delta t$$

$$\leq M\_1(\rho) \left( \int\_{t\_0}^{x} (\sigma(\mathbf{x}) - \sigma(s)) \left( f(s) \Phi\left(\frac{a(s)}{f(s)}\right) \right)^q \Delta s \right)^{\frac{1}{q}}$$

$$\times \left( \int\_{t\_0}^{y} (\sigma(y) - \sigma(t)) \left( g(t) \Psi\left(\frac{b(t)}{g(t)}\right) \right)^q \Delta t \right)^{\frac{1}{q}} \tag{28}$$

.

*where*

$$M\_1(p) = \left\{ \int\_{t\_0}^{\chi} \left( \frac{\varPhi(F^{\sigma}(s))}{F^{\sigma}(s)} \right)^p \Delta s \right\}^{\frac{1}{p}} \left\{ \int\_{t\_0}^{y} \left( \frac{\varPsi(G^{\sigma}(t))}{G^{\sigma}(t)} \right)^p \Delta t \right\}^{\frac{1}{p}}$$

**Proof.** From the properties of Φ˘ and using (2), we obtain

$$\begin{split} \label{E} \Phi(A^{\tau}(s)) &= \quad \Phi\left(\frac{F(\sigma(s)) \int\_{t\_{0}}^{\sigma(s)} f(\check{\mathbf{r}}) \frac{a(\check{\mathbf{r}})}{f(\check{\mathbf{t}})} \Delta \check{\mathbf{r}}}{\int\_{t\_{0}}^{\sigma(s)} f(\check{\mathbf{r}}) \Delta \check{\mathbf{r}}}\right) \\ &\leqslant \quad \Phi(F(\sigma(s)) \Phi\left(\frac{\int\_{t\_{0}}^{\sigma(s)} f(\check{\mathbf{r}}) \frac{a(\check{\mathbf{r}})}{f(\check{\mathbf{r}})} \Delta \check{\mathbf{r}}}{\int\_{t\_{0}}^{\sigma(s)} f(\check{\mathbf{r}}) \Delta \check{\mathbf{r}}}\right) \\ &\leqslant \quad \frac{\Phi(F(\sigma(s))}{F(\sigma(s))} \int\_{t\_{0}}^{\sigma(s)} f(\check{\mathbf{r}}) \Phi\left(\frac{a(\check{\mathbf{r}})}{f(\check{\mathbf{r}})} \right) \Delta \check{\mathbf{r}}. \end{split} \tag{29}$$

Using (1) in (29), we see that

$$\Phi(A^{\sigma}(s)) \ll \frac{\breve{\Phi}(F^{\sigma}(s))}{F^{\sigma}(s)} (\sigma(s) - t\_0)^{\frac{1}{p}} \left( \int\_{t\_0}^{\sigma(s)} \left( f(\dddot{\tau}) \dot{\Phi} \left[ \frac{a(\dddot{\tau})}{f(\dddot{\tau})} \right] \right)^{q} \Delta \dddot{\tau} \right)^{\frac{1}{q}}.\tag{30}$$

In addition, from the convexity and submultiplicative property of Ψ˘ , we get by using (2) and (1):

$$\Psi(\mathcal{B}^{\sigma}(t)) \leqslant \frac{\Psi(G^{\sigma}(t))}{G^{\sigma}(t)} (\sigma(t) - t\_0)^{\frac{1}{p}} \left( \int\_{t\_0}^{\sigma(t)} \left( g(\eta) \Psi \left[ \frac{b(\eta)}{g(\eta)} \right] \right)^q \Delta \eta \right)^{\frac{1}{q}}.\tag{31}$$

From (30) and (31), we have

$$\begin{split} \left| \Phi(A^{\sigma}(s)) \Psi(B^{\sigma}(t)) \right| &\leqslant \left( \sigma(s) - t\_{0} \right)^{\frac{1}{p}} (\sigma(t) - t\_{0})^{\frac{1}{p}} \left( \frac{\left\langle \Phi(F^{\sigma}(s)) \right\rangle}{F^{\sigma}(s)} \left( \int\_{t\_{0}}^{\sigma(s)} \left( f(\mathbf{t}) \Phi \left[ \frac{a(\mathbf{t})}{f(\mathbf{t})} \right] \right)^{q} \Delta \mathbf{t} \right)^{\frac{1}{q}} \\ &\times \left( \frac{\Psi(G^{\sigma}(t))}{G^{\sigma}(t)} \left( \int\_{t\_{0}}^{\sigma(t)} \left( g(\eta) \Psi \left[ \frac{b(\eta)}{g(\eta)} \right] \right)^{q} \Delta \eta \right)^{\frac{1}{q}} \end{split} \tag{32}$$

Using (5) on (*σ*(*s*) − *t*0) 1 *<sup>p</sup>* (*σ*(*t*) − *t*0) 1 *<sup>p</sup>* gives:

$$
\begin{split}
\langle \Phi(\mathcal{A}^{\sigma}(s)) \Psi(\mathcal{B}^{\sigma}(t)) \rangle &\quad \leqslant \quad \left( h(\sigma(s) - t\_{0}) + h^{\*}(\sigma(t) - t\_{0}) \right)^{\frac{1}{p}} \Big{(} \frac{\langle \Phi(\mathcal{F}^{\sigma}(s)) \rangle}{\mathcal{F}^{\sigma}(s)} \left( \int\_{t\_{0}}^{\sigma(s)} \left( f(\tau) \Phi \Big{[} \frac{a(\tau)}{f(\tilde{\tau})} \Big] \right)^{q} \Delta \tilde{\tau} \right)^{\frac{1}{q}} \\ &\quad \times \Big{(} \frac{\langle \Psi(\mathcal{G}^{\sigma}(t)) \rangle}{\mathcal{G}^{\sigma}(t)} \left( \int\_{t\_{0}}^{\sigma(t)} \left( g(\eta) \Psi \Big{[} \frac{b(\eta)}{g(\eta)} \Big] \right)^{q} \Delta \eta \Big{)}^{\frac{1}{q}}.
\end{split}
\tag{33}
$$

Applying Lemma 6 on the right hand side of (33), we see that

$$
\Phi(A^{\sigma}(s))\Psi(B^{\sigma}(t)) \leqslant \left( |h(\sigma(s) - t\_0)|^{\frac{1}{2p}} + |h^\*(\sigma(t) - t\_0)|^{\frac{1}{2p}} \right)^{\frac{2s}{p}}
$$

$$
\times \left( \frac{\Phi(F^{\sigma}(s))}{F^{\sigma}(s)} \left( \int\_{t\_0}^{\sigma(s)} \left( f(\mathbf{t}) \Phi\left[\frac{a(\mathbf{t})}{f(\mathbf{t})}\right] \right)^q \Delta \mathbf{t} \right)^{\frac{1}{q}} \right.
$$

$$
\times \left( \frac{\Psi(G^{\sigma}(t))}{G^{\sigma}(t)} \left( \int\_{t\_0}^{\sigma(t)} \left( g(\eta) \Psi\left[\frac{b(\eta)}{g(\eta)} \right] \right)^q \Delta \eta \right)^{\frac{1}{q}} . \tag{34}
$$

From (34), we have

$$\frac{\Phi(A^{\sigma}(s))\Psi(B^{\sigma}(t))}{\left(|h(\sigma(s)-t\_{0})|^{\frac{1}{2\overline{\rho}}}+|h^{\*}(\sigma(t)-t\_{0})|^{\frac{1}{2\overline{\rho}}}\right)^{\frac{2\mu}{p}}} \leq \left(\frac{\Phi(F^{\sigma}(s))}{F^{\sigma}(s)}\right) \left(\int\_{t\_{0}}^{\sigma(s)} \left(f(\vec{\tau})\Phi\left[\frac{a(\vec{\tau})}{f(\vec{\tau})}\right]\right)^{q} \Delta\vec{\tau}\right)^{\frac{1}{q}}$$

$$\times \left(\frac{\Psi(G^{\sigma}(t))}{G^{\sigma}(t)} \left(\int\_{t\_{0}}^{\sigma(t)} \left(g(\eta)\Psi\left[\frac{b(\eta)}{g(\eta)}\right]\right)^{q} \Delta\eta\right)^{\frac{1}{q}}\right.\tag{35}$$

From (35), we obtain

$$\int\_{t\_0}^{\infty} \int\_{t\_0}^{y} \frac{\Phi(A^{\sigma}(s)) \Psi(\mathcal{B}^{\sigma}(t))}{\left( |h(\sigma(s) - t\_0)|^{\frac{1}{2\tilde{\sigma}}} + |h^\*(\sigma(t) - t\_0)|^{\frac{1}{2\tilde{\sigma}}} \right)^{\frac{2\tilde{\sigma}}{\tilde{p}}}} d\Delta t$$

$$\leq \int\_{t\_0}^{\infty} \left( \frac{\Phi(F^{\sigma}(s))}{F^{\sigma}(s)} \left( \int\_{t\_0}^{\sigma(s)} \left( f(\mathbf{t}) \Phi \left[ \frac{a(\mathbf{t})}{f(\mathbf{t})} \right] \right)^q \Delta \mathbf{t} \right)^{\frac{1}{q}} \Delta \mathbf{s}$$

$$\times \int\_{t\_0}^{y} \left( \frac{\Psi(G^{\sigma}(t))}{G^{\sigma}(t)} \left( \int\_{t\_0}^{\sigma(t)} \left( g(\eta) \Psi \left[ \frac{b(\eta)}{g(\eta)} \right] \right)^q \Delta \eta \right)^{\frac{1}{q}} \Delta t. \tag{36}$$

From (36), by using (1), we have

$$\int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{\Phi(A^{\sigma}(s)) \Psi(B^{\sigma}(t))}{\left( |h(\sigma(s) - t\_0)|^{\frac{1}{2\theta}} + |h^\*(\sigma(t) - t\_0)|^{\frac{1}{2\theta}} \right)^{\frac{2\theta}{\theta}}} \Delta s \Delta t$$

$$\lesssim \left\{ \int\_{t\_0}^{\chi} \left( \frac{\Phi(F^{\sigma}(s))}{F^{\sigma}(s)} \right)^p \Delta s \right\}^{\frac{1}{p}} \left( \int\_{t\_0}^{\chi} \int\_{t\_0}^{\sigma(s)} \left( f(\tau) \Phi\left[\frac{a(\tau)}{f(\tau)}\right] \right)^q \Delta t \Delta s \right)^{\frac{1}{q}}$$

$$\times \left\{ \int\_{t\_0}^{y} \left( \frac{\Psi(G^{\sigma}(t))}{G^{\sigma}(t)} \right)^p \Delta t \right\}^{\frac{1}{p}} \left( \int\_{t\_0}^{y} \int\_{t\_0}^{\sigma(t)} \left( g(\eta) \Psi\left[\frac{b(\eta)}{g(\eta)}\right] \right)^q \Delta \eta \Delta t \right)^{\frac{1}{q}}.\tag{37}$$

From (37), by using (5), we obtain

$$\begin{split} &\int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{\dot{\Phi}(A^{\sigma}(s)) \dot{\Psi}(B^{\sigma}(t))}{\left( |h(\sigma(s) - t\_0|^{\frac{1}{2\beta}} + |h^\*(\sigma(t) - t\_0)|^{\frac{1}{2\beta}} \right)^{\frac{2\alpha}{p}}} \Delta s \Delta t \\ &\leq M\_1(p) \left( \int\_{t\_0}^{\chi} (x - \sigma(s)) \left( f(s) \Phi\left[\frac{a(s)}{f(s)}\right] \right)^q \Delta s \right)^{\frac{1}{q}} \\ &\qquad \times \left( \int\_{t\_0}^{y} (y - \sigma(t)) \left( g(t) \Psi\left[\frac{b(t)}{g(t)}\right] \right)^q \Delta t \right)^{\frac{1}{q}}. \end{split}$$

By using the facts *σ*(*x*) > *x* and *σ*(*y*) > *y*, we obtain

$$\begin{split} &\int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{\Phi(A^{\sigma}(s)) \Psi(B^{\sigma}(t))}{\left( |h(\sigma(s) - t\_0)|^{\frac{1}{2\tilde{\rho}}} + |h^\*(\sigma(t) - t\_0)|^{\frac{1}{2\tilde{\rho}}} \right)^{\frac{2\tilde{\rho}}{\tilde{p}}}} \Delta s \Delta t \\ &\leq M\_1(p) \left( \int\_{t\_0}^{\chi} (\sigma(\mathbf{x}) - \sigma(\mathbf{s})) \left( f(\mathbf{s}) \Phi \left[ \frac{a(\mathbf{s})}{f(\mathbf{s})} \right] \right)^q \Delta s \right)^{\frac{1}{q}} \\ &\times \left( \int\_{t\_0}^{y} (\sigma(y) - \sigma(t)) \left( g(t) \Psi \left[ \frac{b(t)}{g(t)} \right] \right)^q \Delta t \right)^{\frac{1}{q}} \end{split}$$

where

$$M\_1(p) = \left\{ \int\_{t\_0}^{\chi} \left( \frac{\not\Phi(F^{\sigma}(s))}{F^{\sigma}(s)} \right)^p \Delta s \right\}^{\frac{1}{p}} \left\{ \int\_{t\_0}^{y} \left( \frac{\varPsi(G^{\sigma}(t))}{G^{\sigma}(t)} \right)^p \Delta t \right\}^{\frac{1}{p}}.$$

This completes the proof.

In Theorem 4, taking T = R, we have (1) and the result:

**Corollary 5.** *Assume that a*(*s*) ≥ 0, *b*(*t*) ≥ 0, *f*(*τ*ˇ) ≥ 0 *and g*(*η*) ≥ 0*, we define*

$$A(\mathbf{s}) := \int\_0^\mathbf{s} a(\eta) d\eta,\quad \mathbf{B}(t) := \int\_0^t b(\eta) d\eta,\text{ } \mathbf{F}(\mathbf{s}) := \int\_0^\mathbf{s} f(\check{\mathbf{r}}) d\check{\mathbf{r}},\text{ and }\mathbf{G}(t) := \int\_0^t \mathbf{g}(\eta) d\eta.$$

*Then*

$$\begin{split} \int\_{0}^{\mathbf{x}} \int\_{0}^{\mathbf{y}} \frac{\mathsf{\dot{\Phi}}(A(s)\varPsi(B(t)) )}{\left( |h(s)|^{\frac{1}{2\overline{\rho}}} + |h^\*(t)|^{\frac{1}{2\overline{\rho}}} \right)^{\frac{2\overline{\mathbf{\dot{s}}}}{\overline{p}}}} ds dt & \quad \leqslant M\_2(p) \left( \int\_{0}^{\mathbf{x}} (\mathbf{x} - s) \left( f(s) \varPhi \left( \frac{a(s)}{f(s)} \right) \right)^q ds \right)^{\frac{1}{q}} \\ & \qquad \qquad \qquad \times \left( \int\_{0}^{\mathbf{y}} (y - t) \left( g(t) \varPsi \left( \frac{b(t)}{g(t)} \right) \right)^q dt \right)^{\frac{1}{q}} \end{split}$$

*where*

$$M\_2(p) = \left\{ \int\_0^\chi \left( \frac{\Phi(F(s))}{F(s)} \right)^p ds \right\}^{\frac{1}{p}} \left\{ \int\_0^y \left( \frac{\Psi(G(t))}{G(t)} \right)^p dt \right\}^{\frac{1}{p}}.$$

In Theorem 4, taking T = Z, gives (2) and the result:

**Corollary 6.** *Assume that a*(*n*) ≥ 0*, b*(*m*) ≥ 0*, f*(*n*) ≥ 0*, g*(*m*) ≥ 0 *are sequences of real numbers. Define*

$$A(n) = \sum\_{s=0}^{n} a(s), \text{ } B(m) = \sum\_{k=0}^{m} b(k), \text{ } F(n) = \sum\_{s=0}^{n} f(s) \text{ } and \text{ } G(m) = \sum\_{k=0}^{m} g(k).$$

*Then*

$$\begin{aligned} \sum\_{n=1}^N \sum\_{m=1}^M \frac{\Phi(A(n)) \Psi(B(m))}{\left( |h(n+1)|^{\frac{1}{2\beta}} + |h^\*(m+1)|^{\frac{1}{2\beta}} \right)^{\frac{2\alpha}{p}}} & \leqslant M\_3(p) \left\{ \sum\_{n=1}^N (N+1 - (n+1)) \left( f(n) \Phi\left[ \frac{a(n)}{f(n)} \right] \right)^q \right\}^{\frac{1}{q}} \\ & \qquad \times \left\{ \sum\_{m=1}^M (M+1 - (m+1)) \left( g(m) \Psi\left[ \frac{b(m)}{g(m)} \right] \right)^q \right\}^{\frac{1}{q}} \end{aligned}$$

*Symmetry* **2020**, *12*, 582

*where*

$$M\_3(p) = \left\{ \sum\_{n=1}^N \left( \frac{\Phi(F(n))}{F(n)} \right)^p \right\}^{\frac{1}{p}} \left\{ \sum\_{m=1}^M \left( \frac{\Psi(G(m))}{G(m)} \right)^p \right\}^{\frac{1}{p}}$$

**Remark 2.** *In Corollary 6, if p* = *q* = 2 *we get the result due to Hamiaz and Abuelela ([1], Theorem 5).*

**Corollary 7.** *Under the hypotheses of Theorem 4 the following inequality hold:*

$$\begin{split} &\int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{\mathsf{\dot{\Phi}}(A^{\sigma}(s)) \dot{\Psi}(B^{\sigma}(t))}{\left( |h(\sigma(s) - t\_0)|^{\frac{1}{2\mathfrak{\dot{\sigma}}}} + |h^\*(\sigma(t) - t\_0)|^{\frac{1}{2\mathfrak{\dot{\sigma}}}} \right)^{\frac{2\mathfrak{\dot{\sigma}}}{p}}} \Delta s \Delta t \\ &\leq M\_1(p) \Big[ h \Big( \int\_{t\_0}^{\chi} (\sigma(x) - \sigma(s)) \Big( f(s) \Phi \Big( \frac{a(s)}{f(s)} \Big) \Big)^{q} \Delta s \Big) \\ &\quad + h^\* \Big( \int\_{t\_0}^{y} (\sigma(y) - \sigma(t)) \Big( g(t) \Psi \Big( \frac{b(t)}{g(t)} \Big) \Big)^{q} \Delta t \Big) \Big]^{\frac{1}{q}}. \end{split}$$

**Proof.** Using (5) in (28). This proves our claim.

**Lemma 7.** *With hypotheses of Theorem 4, we get:*

$$\begin{split} & \int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{\Phi(A^{\sigma}(s))^2 \Psi(B^{\sigma}(t))^2}{\left(h(\sigma(s) - t\_0) + h^\*(\sigma(t) - t\_0)\right)^4} \Delta s \Delta t \\ & \quad \lesssim M\_4 \left\{ \int\_{t\_0}^{\chi} (\sigma(\mathbf{x}) - \sigma(s)) \left(f(s) \Phi\left[\frac{a(s)}{f(s)}\right]\right)^4 \Delta s \right\}^{\frac{1}{2}} \left\{ \int\_{t\_0}^{y} (\sigma(t) - \sigma(t)) \left(g(t) \Psi\left[\frac{b(t)}{g(t)}\right]\right)^4 \Delta t \right\}^{\frac{1}{2}} (38) \end{split}$$

*where*

$$M\_{4} = \left\{ \int\_{t\_{0}}^{\chi} \left( \frac{\not{\Phi}(F^{\sigma}(s))^{4}}{(F^{\sigma}(s))^{4}} \right) (\sigma(s) - t\_{0}) \Delta s \right\}^{\frac{1}{2}} \left\{ \int\_{t\_{0}}^{\mathcal{Y}} \left( \frac{\not{\Psi}(G^{\sigma}(t))^{4}}{(G^{\sigma}(t))^{4}} \right) (\sigma(t) - t\_{0}) \Delta t \right\}^{\frac{1}{2}}.\tag{39}$$

**Proof.** From (30) and (31) and by using Fenchel-Young inequality with *p* = *q* = 2 we have

$$\begin{split} &\quad \dot{\Phi}(A^{\sigma}(s))^{2}\Psi(\mathcal{B}^{\sigma}(t))^{2} \\ &\leqslant \left(h(\sigma(s)-t\_{0})+h^{\*}(\sigma(t)-t\_{0})\right)\left(\frac{\dot{\Phi}(F^{\sigma}(s))^{2}}{(F^{\sigma}(s))^{2}}\left(\int\_{t\_{0}}^{\sigma(s)}\left(f(\tau)\Phi\left[\frac{a(\tau)}{f(\tau)}\right]\right)^{2}\Delta\tau\right) \\ &\qquad \times \left(\frac{\Psi(G^{\sigma}(t))^{2}}{(G^{\sigma}(t))^{2}}\left(\int\_{t\_{0}}^{\sigma(t)}\left(g(\eta)\Psi\left[\frac{b(\eta)}{g(\eta)}\right]\right)^{2}\Delta\eta\right). \end{split} \tag{40}$$

From (40), by using (1) with *p* = *q* = 2, we obtain

Z *x t*0 Z *y t*0 Φ˘ (*A σ* (*s*))2Ψ˘ (*B σ* (*t*))<sup>2</sup> *h*(*σ*(*s*) − *t*0) + *h* <sup>∗</sup>(*σ*(*t*) − *t*0) ∆*s*∆*<sup>t</sup>* 6 Z *x t*0 Φ˘ (*F σ* (*s*))<sup>2</sup> (*F <sup>σ</sup>*(*s*))<sup>2</sup> Z *σ*(*s*) *t*0 *f*(*τ*ˇ)Φ˘ *a*(*τ*ˇ) *f*(*τ*ˇ) <sup>2</sup> ∆*τ*ˇ ∆*s* × Z *y t*0 Ψ˘ (*G σ* (*t*))<sup>2</sup> (*Gσ*(*t*))<sup>2</sup> Z *σ*(*t*) *t*0 *g*(*η*)Ψ˘ *b*(*η*) *g*(*η*) <sup>2</sup> ∆*η* ∆*t* 6 Z *x t*0 Φ˘ (*F σ* (*s*))<sup>2</sup> (*F <sup>σ</sup>*(*s*))<sup>2</sup> (*σ*(*s*) − *t*0) 1 2 Z *σ*(*s*) *t*0 *f*(*τ*ˇ)Φ˘ *a*(*τ*ˇ) *f*(*τ*ˇ) <sup>4</sup> ∆*τ*ˇ 1 2 ∆*s* × Z *y t*0 Ψ˘ (*G σ* (*t*))<sup>2</sup> (*Gσ*(*t*))<sup>2</sup> (*σ*(*t*) − *t*0) 1 2 Z *σ*(*t*) *t*0 *g*(*η*)Ψ˘ *b*(*η*) *g*(*η*) <sup>4</sup> ∆*η* 1 2 ∆*t* 6 Z *x t*0 Φ˘ (*F σ* (*s*))<sup>4</sup> (*F <sup>σ</sup>*(*s*))<sup>4</sup> (*σ*(*s*) − *t*0)∆*s* 1 2 Z *x t*0 Z *σ*(*s*) *t*0 *f*(*τ*ˇ)Φ˘ *a*(*τ*ˇ) *f*(*τ*ˇ) <sup>4</sup> ∆*τ*ˇ ∆*s* 1 2 × Z *y t*0 Ψ˘ (*G σ* (*t*))<sup>4</sup> (*Gσ*(*t*))<sup>4</sup> (*σ*(*t*) − *t*0)∆*t* 1 2 Z *y t*0 Z *σ*(*t*) *t*0 *g*(*η*)Ψ˘ *b*(*η*) *g*(*η*) <sup>4</sup> ∆*η* ∆*t* 1 2 . (41)

Applying (5) on (41), we obtain

$$\begin{split} &\int\_{t\_0}^{\mathcal{X}} \int\_{t\_0}^{y} \frac{\tilde{\Phi}(A^{\sigma}(s))^2 \tilde{\Psi}(B^{\sigma}(t))^2}{\left(h(\sigma(s)-t\_0) + h^\*(\sigma(t)-t\_0)\right)} \Delta s \Delta t \\ &\leq \left\{\int\_{t\_0}^{\mathcal{X}} \left(\frac{\tilde{\Phi}(F^{\sigma}(s))^4}{(F^{\sigma}(s))^4}\right) (\sigma(s)-t\_0) \Delta s\right\}^{\frac{1}{2}} \left\{\int\_{t\_0}^{\mathcal{X}} (x-\sigma(s)) \left(f(s)\Phi\left[\frac{a(s)}{f(s)}\right]\right)^4 \Delta s\right\}^{\frac{1}{2}} \\ &\times \left\{\int\_{t\_0}^{\mathcal{Y}} \left(\frac{\tilde{\Psi}(G^{\sigma}(t))^4}{(G^{\sigma}(t))^4}\right) (\sigma(t)-t\_0) \Delta t\right\}^{\frac{1}{2}} \left\{\int\_{t\_0}^{\mathcal{Y}} (t-\sigma(t)) \left(g(t)\Psi\left[\frac{b(t)}{g(t)}\right]\right)^4 \Delta t\right\}^{\frac{1}{2}} \\ &= M\_4 \left\{\int\_{t\_0}^{\mathcal{X}} (x-\sigma(s)) \left(f(s)\Phi\left[\frac{a(s)}{f(s)}\right]\right)^4 \Delta s\right\}^{\frac{1}{2}} \left\{\int\_{t\_0}^{\mathcal{Y}} (t-\sigma(t)) \left(g(t)\Psi\left[\frac{b(t)}{g(t)}\right]\right)^4 \Delta t\right\}^{\frac{1}{2}}. \end{split}$$

Since *σ*(*x*) > *x* and *σ*(*y*) > *y*, from the last inequality above, we have

$$\begin{split} &\int\_{t\_0}^{x} \int\_{t\_0}^{y} \frac{\Phi(A^{\sigma}(s))^2 \Psi(B^{\sigma}(t))^2}{\left(h(\sigma(s) - t\_0) + h^\*(\sigma(t) - t\_0)\right)} ds dt \\ &\leqslant M\_4 \left\{ \int\_{t\_0}^{x} (\sigma(x) - \sigma(s)) \left(f(s) \Phi\left[\frac{a(s)}{f(s)}\right]\right)^4 ds\right\}^{\frac{1}{2}} \left\{ \int\_{t\_0}^{y} (\sigma(t) - \sigma(t)) \left(g(t) \Psi\left[\frac{b(t)}{g(t)}\right]\right)^4 \Delta t \right\}^{\frac{1}{2}} \end{split}$$

where *M*<sup>4</sup> defined as in (39). This proves our claim.

**Theorem 5.** *Assume the time scale* T *with t*, *s*, *x*0, *t*0, *y* ∈ T. *Suppose that b*(*τ*ˇ) ≥ 0 *and a*(*τ*ˇ) ≥ 0 *are right-dense continuous functions on* [*t*0, *<sup>y</sup>*]<sup>T</sup> *and* [*t*0, *<sup>x</sup>*]T*. Let <sup>G</sup>*, *<sup>F</sup>*, *<sup>g</sup>*, *<sup>f</sup>* , <sup>Ψ</sup>˘ *and* <sup>Φ</sup>˘ *be as assumed in Theorem 4. Furthermore assume that*

$$A(s) := \frac{1}{F(s)} \int\_{t\_0}^{s} a(\mathfrak{k}) f(\mathfrak{k}) \Delta \mathfrak{k}, \text{ and } \mathcal{B}(t) := \frac{1}{G(t)} \int\_{t\_0}^{t} b(\eta) g(\eta) \Delta \eta. \tag{42}$$

*then for σ*(*s*) ∈ [*t*0, *x*]<sup>T</sup> *and σ*(*t*) ∈ [*t*0, *y*]T, *we have that*

$$\int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{\Phi(A^{\sigma}(s)) \Psi(\mathcal{B}^{\sigma}(t)) F^{\sigma}(s) \mathcal{G}^{\sigma}(t)}{\left( |h(\sigma(s) - t\_0)|^{\frac{1}{2\tilde{\rho}}} + |h^\*(\sigma(t) - t\_0)|^{\frac{1}{2\tilde{\rho}}} \right)^{\frac{2\tilde{\rho}}{\tilde{p}}}} \Delta s \Delta t$$

$$\leq M\_5(p) \left( \int\_{t\_0}^{\chi} (\sigma(\mathbf{x}) - \sigma(s)) \left( f(s) \Phi \left( a(s) \right) \right)^q \Delta s \right)^{\frac{1}{q}}$$

$$\times \left( \int\_{t\_0}^{y} (\sigma(y) - \sigma(t)) \left( \mathcal{G}(t) \Psi \left( b(t) \right) \right)^q \Delta t \right)^{\frac{1}{q}} \tag{43}$$

*where*

$$M\_5(p) = (x - t\_0)^{\frac{1}{p}} (y - t\_0)^{\frac{1}{p}}.\tag{44}$$

**Proof.** From (42), we see that

$$\Phi(A^{\sigma}(s)) = \Phi\left(\frac{1}{F^{\sigma}(s)} \int\_{t\_0}^{\sigma(s)} f(\check{\tau}) a(\check{\tau}) \Delta \check{\tau}\right). \tag{45}$$

Applying (1) on (45), we obtain

$$
\Phi(A^{\sigma}(s)) \lesssim \frac{(\sigma(s) - t\_0)^{\frac{1}{p}}}{F^{\sigma}(s)} \left( \int\_{t\_0}^{\sigma(s)} \left( f(\dddot{\tau}) \Phi[a(\dddot{\tau})] \right)^q \Delta \dddot{\tau} \right)^{\frac{1}{q}}.\tag{46}
$$

From (46), we get

$$\Phi(A^{\sigma}(s))F^{\sigma}(s) \leqslant (\sigma(s) - t\_0)^{\frac{1}{p}} \left( \int\_{t\_0}^{\sigma(s)} \left( f(\check{\mathbf{r}}) \Phi[a(\check{\mathbf{r}})] \right)^q \Delta \check{\mathbf{r}} \right)^{\frac{1}{q}}.\tag{47}$$

Similarly, we obtain

$$\Psi(\mathcal{B}^{\sigma}(t))\mathcal{G}^{\sigma}(t) \leqslant (\sigma(t) - t\_0)^{\frac{1}{p}} \left( \int\_{t\_0}^{\sigma(t)} \left( g(\eta)\Psi[b(\eta)] \right)^q \Delta \eta \right)^{\frac{1}{q}}.\tag{48}$$

From (47) and (48), we observe that

$$\begin{split} \label{E} \Phi(A^{\sigma}(\mathsf{s})) \dot{\Psi}(\mathcal{B}^{\sigma}(t)) G^{\sigma}(t) F^{\sigma}(\mathsf{s}) &\leqslant \left(\sigma(\mathsf{s}) - t\_{0}\right)^{\frac{1}{p}} (\sigma(t) - t\_{0})^{\frac{1}{p}} \\ &\times \left(\int\_{t\_{0}}^{\sigma(\mathsf{s})} \left(f(\mathsf{t}) \mathsf{\dot{\Phi}}[a(\mathsf{t})]\right)^{q} \Delta \mathsf{\dot{\pi}}\right)^{\frac{1}{q}} \left(\int\_{t\_{0}}^{\sigma(\mathsf{t})} \left(g(\mathsf{\eta}) \mathsf{\dot{\Psi}}[b(\mathsf{\eta})]\right)^{q} \Delta \mathsf{\eta}\right)^{\frac{1}{q}}. \end{split} \tag{49}$$

Applying the Lemma 3 on the term (*σ*(*s*) − *t*0) 1 *<sup>p</sup>* (*σ*(*t*) − *t*0) 1 *p* , gives:

$$\begin{split} \left\| \Phi(A^{\sigma}(s)) \Psi(B^{\sigma}(t)) G^{\sigma}(t) F^{\sigma}(s) \right\| &\quad \leqslant \left( h(\sigma(s) - t\_{0}) + h^{\*}(\sigma(t) - t\_{0}) \right)^{\frac{1}{p}} \left( \int\_{t\_{0}}^{\sigma(s)} \left( f(\mathbf{t}) \Phi[a(\mathbf{t})] \right)^{q} \Delta \mathbf{t} \right)^{\frac{1}{q}} \\ &\quad \times \left( \int\_{t\_{0}}^{\sigma(t)} \left( g(\eta) \Psi[b(\eta)] \right)^{q} \Delta \eta \right)^{\frac{1}{q}}. \end{split} \tag{50}$$

From 6 and (50), we obtain

$$
\Phi(A^{\sigma}(s))\Psi(B^{\sigma}(t))G^{\sigma}(t)F^{\sigma}(s) \leqslant \left(|h(\sigma(s)-t\_{0})|^{\frac{1}{2\overline{\sigma}}} + |h^{\*}(\sigma(t)-t\_{0})|^{\frac{1}{2\overline{\sigma}}}\right)^{\frac{2\alpha}{p}}
$$

$$
\times \left(\int\_{t\_{0}}^{\sigma(s)} \left(f(\mathbf{t})\Phi[a(\mathbf{t})]\right)^{q}\Delta\mathbf{t}\right)^{\frac{1}{q}} \left(\int\_{t\_{0}}^{\sigma(t)} \left(g(\eta)\Psi[b(\eta)]\right)^{q}\Delta\eta\right)^{\frac{1}{q}}.\tag{51}
$$

Dividing both sides of (51) by |*h*(*σ*(*s*) − *t*0)| 1 <sup>2</sup>*<sup>β</sup>* + |*h* ∗ (*σ*(*t*) − *t*0)| 1 2*β* <sup>2</sup>*<sup>α</sup> p* , we get

Φ˘ (*A σ* (*s*))Ψ˘ (*B σ* (*t*))*G σ* (*t*)*F σ* (*s*) |*h*(*σ*(*s*) − *t*0)| 1 <sup>2</sup>*<sup>β</sup>* + |*h* <sup>∗</sup>(*σ*(*t*) − *t*0)| 1 2*β* <sup>2</sup>*<sup>α</sup> p* 6 Z *σ*(*s*) *t*0 *<sup>f</sup>*(*τ*ˇ)Φ˘ [*a*(*τ*ˇ)]*<sup>q</sup>* ∆*τ*ˇ 1 *q* × Z *σ*(*t*) *t*0 *<sup>g</sup>*(*η*)Ψ˘ [*b*(*η*)]*<sup>q</sup>* ∆*η* 1 *q* . (52)

Taking delta-integral for (52), yields:

$$\begin{split} &\int\_{t\_0}^{\infty} \int\_{t\_0}^{y} \frac{\Phi(A^{\sigma}(s))\Psi(B^{\sigma}(t))G^{\sigma}(t)F^{\sigma}(s)}{\left(|h(\sigma(s)-t\_0)|^{\frac{1}{2\beta}} + |h^\*(\sigma(t)-t\_0)|^{\frac{1}{2\beta}}\right)^{\frac{2\kappa}{p}}} \Delta s \Delta t \\ &\lesssim \left(\int\_{t\_0}^{\infty} \left(\int\_{t\_0}^{\sigma(s)} \left(f(\mathbf{\bar{\tau}})\Phi[a(\mathbf{\bar{\tau}})]\right)^q \Delta \mathbf{\bar{\tau}}\right)^{\frac{1}{q}} \Delta s\right) \left(\int\_{t\_0}^{y} \left(\int\_{t\_0}^{\sigma(t)} \left(g(\eta)\Psi[b(\eta)]\right)^q \Delta \eta\right)^{\frac{1}{q}} \Delta t\right). \end{split} \tag{53}$$

Using (1) in (53), yield:

$$\begin{split} &\int\_{t\_{0}}^{\infty} \int\_{t\_{0}}^{y} \frac{\dot{\Phi}(A^{\sigma}(s))\Psi(B^{\sigma}(t))\mathbf{G}^{\sigma}(t)F^{\sigma}(s)}{\left(|h(\sigma(s)-t\_{0})|^{\frac{1}{2^{\beta}}} + |h^{\star}(\sigma(t)-t\_{0})|^{\frac{1}{2^{\beta}}}\right)^{\frac{2\alpha}{\beta}}} \Delta s \Delta t \\ &\leq (x-t\_{0})^{\frac{1}{p}}(y-t\_{0})^{\frac{1}{p}} \left(\int\_{t\_{0}}^{\infty} \left(\int\_{t\_{0}}^{\sigma(s)} \left(f(t)\Phi[a(\tau)]\right)^{q} \Delta \tau\right) \Delta s\right)^{\frac{1}{q}} \\ &\quad \times \left(\int\_{t\_{0}}^{y} \left(\int\_{t\_{0}}^{\sigma(t)} \left(g(\eta)\Psi[b(\eta)]\right)\right)^{q} \Delta \eta\right) \Delta t\right)^{\frac{1}{q}} \\ &= \mathsf{M}\_{5}(p) \left(\int\_{t\_{0}}^{\infty} \left(\int\_{t\_{0}}^{\sigma(s)} \left(f(\tau)\Phi[a(\tau)]\right)^{q} \Delta \tau\right) \Delta s\right)^{\frac{1}{q}} \\ &\quad \times \left(\int\_{t\_{0}}^{y} \left(\int\_{t\_{0}}^{\sigma(t)} \left(g(\eta)\Psi[b(\eta)]\right)^{q} \Delta \eta\right) \Delta t\right)^{\frac{1}{q}}. \end{split} \tag{54}$$

where *M*<sup>5</sup> defined as in (44). From (5) and (54), we get:

Z *x t*0 Z *y t*0 Φ˘ (*A σ* (*s*))Ψ˘ (*B σ* (*t*))*G σ* (*t*)*F σ* (*s*) |*h*(*σ*(*s*) − *t*0)| 1 <sup>2</sup>*<sup>β</sup>* + |*h* <sup>∗</sup>(*σ*(*t*) − *t*0)| 1 2*β* <sup>2</sup>*<sup>α</sup> p* ∆*s*∆*t* = *M*5(*p*) Z *x t*0 (*<sup>x</sup>* <sup>−</sup> *<sup>σ</sup>*(*s*)) *<sup>f</sup>*(*s*)Φ˘ [*a*(*s*)]*<sup>q</sup>* ∆*s* 1 *q* × Z *y t*0 (*<sup>y</sup>* <sup>−</sup> *<sup>σ</sup>*(*t*)) *<sup>g</sup>*(*t*)Ψ˘ [*b*(*t*)]*<sup>q</sup>* ∆*t* 1 *q* .

By using the fact *σ*(*x*) > *x* and *σ*(*y*) > *y*, we obtain

Z *x t*0 Z *y t*0 Φ˘ (*A σ* (*s*))Ψ˘ (*B σ* (*t*))*G σ* (*t*)*F σ* (*s*) |*h*(*σ*(*s*) − *t*0)| 1 <sup>2</sup>*<sup>β</sup>* + |*h* <sup>∗</sup>(*σ*(*t*) − *t*0)| 1 2*β* <sup>2</sup>*<sup>α</sup> p* ∆*s*∆*t* = *M*5(*p*) Z *x t*0 (*σ*(*x*) <sup>−</sup> *<sup>σ</sup>*(*s*)) *<sup>f</sup>*(*s*)Φ˘ [*a*(*s*)]*<sup>q</sup>* ∆*s* 1 *q* × Z *y t*0 (*σ*(*y*) <sup>−</sup> *<sup>σ</sup>*(*t*)) *<sup>g</sup>*(*t*)Ψ˘ [*b*(*t*)]*<sup>q</sup>* ∆*t* 1 *q* .

This completes the proof.

Taking T = R in Theorem 5 with relation (1), we have:

**Corollary 8.** *Assume g*(*t*) ≥ 0*, b*(*t*) ≥ 0, *f*(*s*) ≥ 0*, a*(*s*) ≥ 0*. Define*

$$A(s) := \frac{1}{F(s)} \int\_0^s f(\check{\tau}) a(\check{\tau}) d\check{\tau} \text{ and } \; B(t) := \frac{1}{G(t)} \int\_0^t g(\check{\tau}) b(\check{\tau}) d\check{\tau},$$

$$F(s) := \int\_0^s f(\check{\tau}) d\check{\tau} \text{ and } \; G(t) := \int\_0^t g(\check{\tau}) d\check{\tau}.$$

*Then*

$$\begin{split} \int\_{0}^{\chi} \int\_{0}^{\mathbf{y}} \frac{\tilde{\Phi}(A(s)) \tilde{\Psi}(\mathcal{B}(t)) F(s) G(t)}{\left( |h(s)|^{\frac{1}{2^{\theta}}} + |h^\*(t)|^{\frac{1}{2^{\theta}}} \right)^{\frac{2\theta}{p}}} ds dt & \quad \leqslant M\_{6}(p) \Big( \int\_{0}^{\chi} (\mathbf{x} - \mathbf{s}) \Big( f(s) \tilde{\Phi} \Big( a(s) \Big) \Big)^{q} ds \Big)^{\frac{1}{q}}. \\\\ & \qquad \times \Big( \int\_{0}^{y} (y - t) \Big( g(t) \Psi \Big( b(t) \Big) \Big)^{q} dt \Big)^{\frac{1}{q}} \end{split}$$

*where*

$$M\_6(p) = (\mathfrak{x})^{\frac{1}{p}} (y)^{\frac{1}{p}}$$

Taking T = Z in Theorem 5 with relation (2), gives:

**Corollary 9.** *Assume g*(*n*) ≥ 0*, b*(*n*) ≥ 0, *f*(*n*) ≥ 0*, a*(*n*) ≥ 0*. Define*

$$A(n) := \frac{1}{F(n)} \sum\_{s=0}^{n} f(s)a(s) \text{ and } \; B(m) := \frac{1}{G(m)} \sum\_{k=0}^{m} g(k)b(k).$$

$$F(n) := \sum\_{s=0}^{n} f(s) \text{ and } \mathcal{G}(m) := \sum\_{k=0}^{m} \mathcal{g}(k).$$

*Then*

$$\begin{split} \sum\_{n=1}^{N} \sum\_{m=1}^{M} \frac{\Phi(A(n)) \Psi(B(m)) F(n) G(m)}{\left( |h(n+1)|^{\frac{1}{2^p}} + |h^\*(m+1)|^{\frac{1}{2^p}} \right)^{\frac{2p}{p}}} & \quad \leqslant M\_7(p) \left( \sum\_{n=1}^{N} (N+1-(n+1)) \left( f(n) \Phi\left(a(n) \right) \right) \right)^{q} \cdot \frac{1}{q} \\ & \qquad \qquad \times \left( \sum\_{m=1}^{M} (M+1-(m+1)) \left( g(m) \Psi\left(b(m) \right) \right) \right)^{\frac{1}{q}} \\ \dots \end{split}$$

*where*

$$M\_7(p) = (NM)^{\frac{1}{p}}.$$

**Remark 3.** *In Corollary 9, if p* = *q* = 2 *we get the result due to Hamiaz and Abuelela ([1], Theorem 7).* **Corollary 10.** *With the hypotheses of Theorem 5, we get:*

$$\begin{split} &\int\_{t\_0}^{\chi} \int\_{t\_0}^{y} \frac{\mathsf{\dot{\Phi}}(A^{\sigma}(s)) \overleftarrow{\Psi}(B^{\sigma}(t)) F^{\sigma}(s) G^{\sigma}(t)}{\left( |h(\sigma(s) - t\_0)|^{\frac{1}{2\tilde{p}}} + |h^\*(\sigma(t) - t\_0)|^{\frac{1}{2\tilde{p}}} \right)^{\frac{2\tilde{a}}{p}}} \Delta s \Delta t \\ &\leq M\_5(p) \left\{ h \left( \int\_{t\_0}^{\chi} (\sigma(\mathbf{x}) - \sigma(s)) \left( f(s) \Phi \left( a(s) \right) \right)^q \Delta s \right) \right\} \\ &\quad + h^\* \left( \int\_{t\_0}^{y} (\sigma(y) - \sigma(t)) \left( g(t) \Psi \left( b(t) \right) \right)^q \Delta t \right) \right\}^{\frac{1}{q}}. \end{split}$$

*where M*<sup>5</sup> *defined as in* (44)*.*

**Proof.** We apply the Fenchel-Young inequality (5) in (43). This completes the proof.

#### **3. Some Applications**

We can apply our inequalities to obtain different formulas of Hilbert-type inequalities by suggesting *h* ∗ (*y*) and *h*(*x*) by some functions:

$$\text{In (12), as a special case, if we take } h(\mathbf{x}) = \frac{\mathbf{x}^2}{2} \text{, we have } h^\*(\mathbf{x}) = \frac{\mathbf{x}^2}{2} \text{ see [12], we get}$$

$$\begin{split} &\int\_{t\_{0}}^{\mathbf{x}} \int\_{t\_{0}}^{\mathbf{y}} \frac{A^{K}(\sigma(\mathbf{s})) \mathcal{B}^{L}(\sigma(t))}{\left( |h(\sigma(\mathbf{s}) - t\_{0})|^{\frac{1}{\mathfrak{N}}} + |h^{\*}(\sigma(t) - t\_{0})|^{\frac{1}{\mathfrak{N}}} \right)^{\frac{2\mathfrak{s}}{\mathfrak{p}}}} \frac{\Delta\mathbf{s}\Delta t}{\mathfrak{p}} \\ &= \int\_{t\_{0}}^{\mathbf{x}} \int\_{t\_{0}}^{\mathbf{y}} \frac{A^{K}(\sigma(\mathbf{s})) \mathcal{B}^{L}(\sigma(t))}{\left( (\sigma(\mathbf{s}) - t\_{0})\big)^{\frac{1}{\mathfrak{N}}} + (\sigma(t) - t\_{0})^{\frac{1}{\mathfrak{N}}} \right)^{\frac{2\mathfrak{s}}{\mathfrak{p}}}} \Delta\mathbf{s}\Delta t \\ &\leq \left( \frac{1}{2} \right)^{\frac{\mathfrak{N}}{\mathfrak{p}}} \mathsf{C}\_{2}(L, K, p) \left( \int\_{t\_{0}}^{\mathbf{x}} (\sigma(\mathbf{x}) - \sigma(\mathbf{s})) \Big( A^{K-1}(\sigma(\mathbf{s})) a(\mathbf{s}) \Big)^{q} \Delta\mathbf{s} \right)^{\frac{1}{q}} \\ &\qquad \times \left( \int\_{t\_{0}}^{\mathbf{y}} (\sigma(\mathbf{y}) - \sigma(\mathbf{t})) \Big( B^{L-1}(\sigma(\mathbf{t})) b(\mathbf{t}) \Big)^{q} \Delta\mathbf{t} \right)^{\frac{1}{q}}. \end{split}$$

where *C*2(*L*, *K*, *p*) defined as in Theorem 2. Consequently, for *α* = *β* = 1, inequality (55) produces

$$\begin{split} &\int\_{t\_0}^{\mathcal{X}} \int\_{t\_0}^{\mathcal{Y}} \frac{A^K(\sigma(s))B^L(\sigma(t))}{\left( (\sigma(s) - t\_0) \right) + (\sigma(t) - t\_0)} \Big\prime^{\frac{2}{\tilde{p}}} \\ &\leqslant \left( \frac{1}{2} \right)^{\frac{1}{\tilde{p}}} \mathcal{C}\_2(L, K, p) \left( \int\_{t\_0}^{\mathcal{X}} (\sigma(\mathbf{x}) - \sigma(s)) \left( A^{K-1}(\sigma(s))a(s) \right)^q \Delta s \right)^{\frac{1}{q}} \\ &\qquad \times \left( \int\_{t\_0}^{\mathcal{Y}} (\sigma(y) - \sigma(t)) \left( B^{L-1}(\sigma(t))b(t) \right)^q \Delta t \right)^{\frac{1}{q}}. \end{split} \tag{56}$$

On the other hand if we take *h*(*n*) = *<sup>n</sup> r r* ,*r* > 1, then *h* ∗ (*m*) = *<sup>m</sup><sup>k</sup> <sup>k</sup>* where <sup>1</sup> *<sup>r</sup>* <sup>+</sup> <sup>1</sup> *<sup>k</sup>* = 1 and *n*, *m* R+, then (12) gives

$$\begin{split} &\int\_{t\_{0}}^{\infty} \int\_{t\_{0}}^{y} \frac{A^{K}(\sigma(s))\mathcal{B}^{L}(\sigma(t))}{\left(|h(\sigma(s)-t\_{0})|^{\frac{1}{2\overline{p}}} + |h^{\*}(\sigma(t)-t\_{0})|^{\frac{1}{2\overline{p}}}\right)^{\frac{2\alpha}{p}}} d\Delta t \\ &= \int\_{t\_{0}}^{\infty} \int\_{t\_{0}}^{y} \frac{A^{K}(\sigma(s))\mathcal{B}^{L}(\sigma(t))}{\left((k(\sigma(s)-t\_{0})^{r})^{\frac{1}{2\overline{p}}} + (r(\sigma(t)-t\_{0})^{k})^{\frac{1}{2\overline{p}}}\right)^{\frac{2\alpha}{p}}} d\Delta t \\ &\leq \left(\frac{1}{r\overline{k}}\right)^{\frac{4}{p\overline{p}}} \mathsf{C}\_{2}(L,\mathsf{K},p) \left(\int\_{t\_{0}}^{\infty} (\sigma(\mathbf{x})-\sigma(\mathbf{s})) \left(A^{K-1}(\sigma(\mathbf{s}))a(\mathbf{s})\right)^{q} d\mathbf{s}\right)^{\frac{1}{q}} \\ &\qquad \times \left(\int\_{t\_{0}}^{y} (\sigma(y)-\sigma(t)) \left(\mathcal{B}^{L-1}(\sigma(t))b(t)\right)^{q} d\mathbf{t}\right)^{\frac{1}{q}}. \end{split}$$

Clearly, when *β* = <sup>1</sup> 2*α* , the inequality (57) becomes

$$\begin{split} &\int\_{t\_0}^{\infty} \int\_{t\_0}^{y} \frac{A^K(\sigma(s))B^L(\sigma(t))}{\left((k(\sigma(s)-t\_0)^r)^a + (r(\sigma(t)-t\_0)^k)^a\right)^{\frac{2p}{p}}} \Delta s \Delta t \\ &\leqslant \left(\frac{1}{rk}\right)^{\frac{2q^2}{p}} \mathbb{C}\_2(L, K, p) \left(\int\_{t\_0}^{\chi} (\sigma(x)-\sigma(s)) \left(A^{K-1}(\sigma(s))a(s)\right)^q \Delta s\right)^{\frac{1}{q}} \\ &\qquad \times \left(\int\_{t\_0}^{y} (\sigma(y)-\sigma(t)) \left(B^{L-1}(\sigma(t))b(t)\right)^q \Delta t\right)^{\frac{1}{q}}. \end{split}$$

If *β* = *α* = 1. From (57), we get

$$\begin{split} &\int\_{t\_{0}}^{\chi} \int\_{t\_{0}}^{y} \frac{A^{K}(\sigma(s)) \mathbf{B}^{L}(\sigma(t))}{\left( (k(\sigma(s) - t\_{0})^{r})^{\frac{1}{2}} + (r(\sigma(t) - t\_{0})^{k})^{\frac{1}{2}} \right)^{\frac{2}{p}}} \Delta s \Delta t \\ &\leq \left( \frac{1}{r\lambda} \right)^{\frac{1}{p}} \mathbf{C}\_{2}(L, \mathbf{K}, p) \left( \int\_{t\_{0}}^{\chi} (\sigma(\mathbf{x}) - \sigma(\mathbf{s})) \left( A^{K-1}(\sigma(\mathbf{s})) a(\mathbf{s}) \right)^{q} \Delta \mathbf{s} \right)^{\frac{1}{q}} \\ &\times \left( \int\_{t\_{0}}^{y} (\sigma(y) - \sigma(t)) \left( B^{L-1}(\sigma(t)) b(t) \right)^{q} \Delta t \right)^{\frac{1}{q}}. \end{split}$$

### **4. Conclusions**

In this paper, with the help of a Fenchel-Legendre transform, which is used in various problems involving symmetry, we generalized a number of Hilbert-type inequalities to a general time scale. Besides that, in order to obtain some new inequalities as special cases, we also extended our inequalities to discrete and continuous calculus. In the future, we can generalize these inequalities in a different way by using other mathematical tools.

**Author Contributions:** All authors have contributed equally. All authors have read and agreed to the published version of the manuscript.

**Funding:** The authors declare that they have received no funding from any funding body.

**Conflicts of Interest:** The authors declare that they have no competing interests.

### **References**


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