Exploring Generalized Hardy-Type Inequalities via Nabla Calculus on Time Scales

([32]). Γ is left-dense continuous ($ld—$continuous), if it exhibits continuity at each left-dense point in $\mathbb{T}$ and possesses well-defined right-sided limits at right-dense points within $\mathbb{T}$. The set encompassing all such ld-continuous functions is denoted as ${\mathrm{C}}_{ld}(\mathbb{T},\mathbb{R})$.

([32]). Γ is $\nabla —$differentiable at ς, if it is exists within a neighborhood V of ς and $\exists {\Gamma }^{\nabla }\left(\varsigma \right)\in \mathbb{R}$ is unique (which is referred to as the nabla derivative of Γ at ς), s.t. $\forall ϵ>0,$ $\exists N$ is a neighborhood of ς s.t. $N\subseteq V$ and

([32]). Assuming κ and χ are $\nabla —$differentiable at $s\in \mathbb{T}$, then

([33]). Assuming $\kappa :\mathbb{R}\to \mathbb{R}$ is continuously differentiable, $\chi :\mathbb{T}\to \mathbb{R}$ is continuous and $\nabla —$differentiable, then

([32]). Λ is considered a $\nabla —$antiderivative of κ if ${\Lambda }^{\nabla }\left(s\right)=\kappa \left(s\right)$ true for $s\in \mathbb{T}$. The nabla integral of κ is defined as

([32]). Let ${\delta }_1,{\delta }_2,{\delta }_3\in \mathbb{T},$ $ϵ,\epsilon \in \mathbb{R}$ and $\kappa ,\chi \in {\mathrm{C}}_{ld}(\mathbb{T},\mathbb{R})$. Then

(The integration by parts [32]). If ${\delta }_0,\delta \in \mathbb{T}$ and $\kappa ,\chi \in {\mathrm{C}}_{ld}(\mathbb{T},\mathbb{R})$, then

(Hölder’s inequality [34]). Let ${\delta }_0,d\in \mathbb{T},$ $\kappa ,\xi \in {\mathrm{C}}_{ld}(\mathbb{T},\mathbb{R})$ and $ϵ,\epsilon >1$ such that $1/ϵ+1/\epsilon =1.$ Then,

([35]). $\psi :J_{\mathbb{T}}\to \mathbb{R}$ is termed convex if $\forall \varsigma ,s\in J_{\mathbb{T}}$ and $\alpha \in \left[0,1\right]$ s.t. $\alpha \varsigma +\left(1-\alpha \right)s\in J_{\mathbb{T}}.$

([36]). $\psi :J_{\mathbb{T}}\to {\mathbb{R}}^{+}$ is submultiplicative if

(Jensen’s inequality [34]). Assume ${\varsigma }_0,\varsigma \in \mathbb{T}$ and $\delta ,r\in \mathbb{R}$. If $\Psi :({\delta }_0,\delta )\to \mathbb{R}$ is continuous and convex, $\kappa \in {\mathrm{C}}_{ld}({[{\varsigma }_0,\varsigma ]}_{\mathbb{T}},\mathbb{R})$ and $\phi \in {\mathrm{C}}_{ld}({[{\varsigma }_0,\varsigma ]}_{\mathbb{T}},({\delta }_0,\delta )),$ then

([35]). Suppose $\psi :J_{\mathbb{T}}\to \mathbb{R}$ is a continuous function. If ${\psi }^{\nabla \nabla }$ exists on $J_{\mathbb{T}}$ and ${\psi }^{\nabla \nabla }\ge 0,$ then ψ is convex.

From (23), we find that Let $\varsigma \le s.$ Then, $f\left(\varsigma \right)\le f\left(s\right)$ (as f is a nonnegative and nondecreasing), thus As $\psi \ge 0$ is a nondecreasing, it follows that Using (26) into (24), we get This result corresponds to (22). □

From (23), we see that Applying (20) on ${\int }_{\beta }^s\left[{\left(\varsigma -\beta \right)}^{\lambda \left(h-1\right)/h}f\left(\varsigma \right)\right]{\left(\varsigma -\beta \right)}^{-\lambda \left(h-1\right)/h}\nabla \varsigma ,$ with $h>1,$ $h/\left(h-1\right),$ we obtain Substituting (30) into (29), we obtain Applying (18) on ${\left(\varsigma -\beta \right)}^{1-\lambda }$ for $0<\lambda <1,$ we see that Since $d\le \varsigma$, we see that and then Using(32) in (31), we get Applying (19) on ${\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)}{\xi }^{-h}\left(s\right)\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\lambda \left(h-1\right)}{f}^h\left(\varsigma \right)\nabla \varsigma \right)\nabla s,$ with ${\chi }^{\nabla }\left(s\right)={\left(s-\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)}{\xi }^{-h}\left(s\right)$ and $\kappa \left(s\right)={\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\lambda \left(h-1\right)}{f}^h\left(\varsigma \right)\nabla \varsigma ,$ we observe that where Since ${lim}_{s\to \infty }\chi \left(s\right)=0,$ we have Since the function $\left(\varsigma -\beta \right)/\xi \left(\varsigma \right)$ is non-increasing, we get Using (36) in (35), we find that Applying (18) on ${\left(\varsigma -\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)-h+1},$ we see that where $d\in [\rho \left(\varsigma \right),\varsigma ].$ Since $d\le \varsigma ,$ $h>1$ and $-\left[\lambda \left(h-1\right)+1\right]<0,$ it is evident that and then we get By integrating the previous inequality over $\varsigma$ from $\rho \left(s\right)$ to $\infty ,$ (taking into account that $h>1,$ $0<\lambda <1$ and $\left(1-\lambda \right)\left(h-1\right)-h+1=-\lambda \left(h-1\right)<0$), we derive Using (39) in (37), we obtain Using (40) in (33), we observe that Using (27) in (41), we see that which is (28). □

Note that Applying (21), we find that Substituting (45) into (44), we have that Applying (30) on ${\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{-1}{h}}{\left(\varsigma -\beta \right)}^{\frac{1}{h}}\psi \left(f\left(\varsigma \right)\right)\nabla \varsigma ,$ with $h>1,$ $h/\left(h-1\right),$ we find that Substituting (47) into (46), we find that Applying (18) on ${\left(\varsigma -\beta \right)}^{\frac{-1}{h}+1},$ we find that where $d\in \left[\rho \left(\varsigma \right),\varsigma \right].$ Since $d\le \varsigma$ and $h>1,$ we observe that and then we have from (49) that Integrating (50) over $\varsigma$ from $\beta$ to s, we see that Substituting (51) into (48), we observe that Applying (19) on ${\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{\frac{1}{h}-2}\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{h-1}{h}}{\psi }^h\left(f\left(\varsigma \right)\right)\nabla \varsigma \right)\nabla s,$ with ${\chi }^{\nabla }\left(s\right)={\left(s-\beta \right)}^{\frac{1}{h}-2},$ and $\kappa \left(s\right)={\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{h-1}{h}}{\psi }^h\left(f\left(\varsigma \right)\right)\nabla \varsigma ,$ we see that where Since ${lim}_{s\to \infty }v\left(s\right)=0,$ we have Substituting (54) into (52), we obtain Applying (18) on ${\left(\varsigma -\beta \right)}^{\frac{1}{h}-1},$ we get where $d\in \left[\rho \left(\varsigma \right),\varsigma \right].$ Since $d\le \varsigma$ and $(1/h)-2<0,$ we see that and then we have from (56) that Integrating (57) over $\varsigma$ from $\rho \left(s\right)$ to $\infty ,$ considering that $(1/h)-1<0,$ yields Using (58) in (55), we have Using (42), the last inequality becomes which is (43). □