This section is devoted to state and prove our main results.
Theorem 3.1
Assume \(1\leq k<\infty \). Furthermore, assume that ħ is nonincreasing and \(\int _{\varsigma _{0}}^{\infty }\varphi (\varsigma )\hbar ^{k}( \varsigma )\Delta \varsigma <\infty \). Suppose that there is a constant \(D>0\) with
$$ \int _{\varsigma }^{\infty } \frac{\varphi (s)}{(\sigma (s)-\varsigma _{0})^{k}}\Delta s\leq \frac{D}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \int _{\varsigma _{0}}^{ \sigma ( \varsigma ) }\varphi (s)\Delta s \quad \textit{for all } \varsigma \in (\varsigma _{0},\infty )_{\mathbb{T}}. $$
(18)
Then
$$ \int _{\varsigma _{0}}^{\infty }\varphi (\varsigma ) \biggl( \frac{1}{\sigma (\varsigma )-\varsigma _{0}}\int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x)\Delta x \biggr) ^{k} \Delta \varsigma \leq k^{k} (D+1)^{k} \int _{\varsigma _{0}}^{ \infty }\varphi (\varsigma )\hbar ^{k}(\varsigma )\Delta \varsigma . $$
(19)
Proof
Suppose that (18) holds. Apply Lemma 2.1 with \(z=\hbar \), and we have that
$$ \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar ( x ) \Delta x \biggr) ^{k}\leq k \int _{\varsigma _{0}}^{\sigma ( \varsigma )}\hbar (x) \biggl( \int _{\varsigma _{0}}^{\sigma (x)} \hbar (\tau )\Delta \tau \biggr) ^{k-1}\Delta x. $$
(20)
Substituting (20) into the left-hand side of (19), we get that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \\& \quad \leq k \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \biggl[ \int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x) \biggl( \int _{\varsigma _{0}}^{\sigma (x)}\hbar (\tau )\Delta \tau \biggr) ^{k-1} \Delta x \biggr] \Delta \varsigma . \end{aligned}$$
(21)
Applying Lemma 2.2, with
$$ \varphi (x)=\hbar (x) \biggl( \int _{\varsigma _{0}}^{\sigma (x)} \hbar (\tau )\Delta \tau \biggr) ^{k-1}\quad \text{and}\quad \psi (\varsigma )= \varphi (\varsigma )/ \bigl( \sigma (\varsigma )-\varsigma _{0} \bigr)^{k}, $$
on the right hand side of (21), we see that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \\& \quad \leq k \int _{\varsigma _{0}}^{\infty }\hbar (\varsigma ) \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x)\Delta x \biggr) ^{k-1} \biggl( \int _{\varsigma }^{\infty } \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) \Delta \varsigma \\& \quad = k \int _{\varsigma _{0}}^{\infty } \bigl(\sigma (\varsigma )- \varsigma _{0} \bigr)^{k-1} \biggl( \int _{\varsigma }^{\infty }\frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}} \Delta x \biggr) \hbar (\varsigma ) \biggl( \frac{\int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x)\Delta x}{\sigma (\varsigma )-\varsigma _{0}} \biggr) ^{k-1}\Delta \varsigma . \end{aligned}$$
(22)
Using the additive property of integrals [5, Theorem 1.77(iv)] on time scales, we have for \(\varsigma \in (\varsigma _{0},\infty )_{\mathbb{T}}\) that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \bigl(\sigma (s)- \varsigma _{0} \bigr)^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) { \Delta s} \\& \quad = \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \bigl( \sigma (s)-\varsigma _{0} \bigr)^{k-1} \biggl( \int _{s}^{\sigma (\varsigma )} \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) { \Delta s} \\& \qquad {} + \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \bigl( \sigma (s)-\varsigma _{0} \bigr)^{k-1} \biggl( \int _{\sigma (\varsigma )}^{ \infty }\frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) { \Delta s} \\& \quad \leq \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \bigl(\sigma (s)-\varsigma _{0} \bigr)^{k-1} \biggl( \int _{s}^{\sigma ( \varsigma )}\frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}} \Delta x \biggr) { \Delta s} \\& \qquad {} + \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \bigl( \sigma (s)-\varsigma _{0} \bigr)^{k-1} \biggl( \int _{\varsigma }^{\infty } \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) { \Delta s} \\& \quad = \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \bigl( \sigma (s)-\varsigma _{0} \bigr)^{k-1} \biggl( \int _{s}^{\sigma (\varsigma )} \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) { \Delta s} \\& \qquad {} + \biggl( \int _{\varsigma }^{\infty } \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) \int _{\varsigma _{0}}^{\sigma ( \varsigma ) }{ \bigl( \sigma (s)-\varsigma _{0} \bigr)}^{k-1}{ \Delta s}. \end{aligned}$$
(23)
Substituting (18) into the right-hand side of (23), we see that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \bigl(\sigma (s)- \varsigma _{0} \bigr)^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) { \Delta s} \\& \quad \leq \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \bigl( \sigma (s)-\varsigma _{0} \bigr)^{k-1} \biggl( \int _{s}^{\sigma (\varsigma )} \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) { \Delta s} \\& \qquad {} +\frac{D}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \biggl( \int _{ \varsigma _{0}}^{\sigma ( \varsigma ) }\varphi (x) \Delta x \biggr) \int _{\varsigma _{0}}^{\sigma ( \varsigma ) }{ \bigl(\sigma (s)- \varsigma _{0} \bigr)}^{k-1}\Delta s. \end{aligned}$$
(24)
Applying integration by parts formula (8) on the term
$$ \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \bigl(\sigma (s)- \varsigma _{0} \bigr)^{k-1} \biggl( \int _{s}^{\sigma (\varsigma )} \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) { \Delta s}, $$
with
$$ \Re (s)=\Upsilon (s)= \int _{s}^{\sigma (\varsigma )} \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x, \quad \text{and}\quad \upsilon ^{\Delta }(s)={ \bigl(\sigma (s)-\varsigma _{0} \bigr)}^{k-1}, $$
we get that
$$ \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \bigl(\sigma (s)- \varsigma _{0} \bigr)^{k-1} \biggl( \int _{s}^{\sigma (\varsigma )} \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) { \Delta s=} \Upsilon (s)\upsilon (s)\vert _{ \varsigma _{0}}^{\sigma ( \varsigma ) }- \int _{ \varsigma _{0}}^{\sigma ( \varsigma ) }\Upsilon ^{ \Delta }(s) \upsilon ^{\sigma }(s)\Delta s, $$
where \(\upsilon (s)=\int _{\varsigma _{0}}^{s}{ (}\sigma (x)- \varsigma _{0})^{k-1}\Delta x\). Using \(\Upsilon ^{\sigma }(\varsigma )=0\) and \(\upsilon (\varsigma _{0})=0\), we have that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \bigl(\sigma (s)- \varsigma _{0} \bigr)^{k-1} \biggl( \int _{s}^{\sigma (\varsigma )} \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) \Delta s \\& \quad = \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \bigl[ -\Upsilon ^{\Delta }(s) \bigr] \upsilon ^{\sigma }(s) \Delta s \\& \quad = \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \frac{\varphi (s)}{(\sigma (s)-\varsigma _{0})^{k}} \biggl( \int _{ \varsigma _{0}}^{\sigma ( s ) } \bigl(\sigma (x)-\varsigma _{0} \bigr)^{k-1} \Delta x \biggr) \Delta s. \end{aligned}$$
(25)
Substituting (25) into the right-hand side of (24), we get that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \bigl(\sigma (s)- \varsigma _{0} \bigr)^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) { \Delta s} \\& \quad \leq \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \frac{\varphi (s)}{(\sigma (s)-\varsigma _{0})^{k}} \biggl( \int _{ \varsigma _{0}}^{\sigma (s)}{ \bigl(\sigma (x)-\varsigma _{0} \bigr)}^{k-1}\Delta x \biggr) { \Delta s} \\& \qquad {} + \frac{D}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \biggl( \int _{ \varsigma _{0}}^{\sigma ( \varsigma ) }\varphi (x) \Delta x \biggr) \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \bigl(\sigma (s)-\varsigma _{0} \bigr)^{k-1}\Delta s. \end{aligned}$$
(26)
Since
$$ \int _{\varsigma _{0}}^{\sigma (\varsigma )} \bigl(\sigma (x)-\varsigma _{0} \bigr)^{k-1} \Delta x\leq \bigl(\sigma (\varsigma )- \varsigma _{0} \bigr)^{k}, $$
(27)
inequality (26) becomes
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma ( \varsigma ) } \bigl(\sigma (s)- \varsigma _{0} \bigr)^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) { \Delta s}\leq \int _{\varsigma _{0}}^{\sigma ( \varsigma ) }\varphi (s)\Delta s \\& \qquad {} +D \int _{\varsigma _{0}}^{\sigma ( \varsigma )}\varphi (x)\Delta x \\& \quad =(D+1) \int _{\varsigma _{0}}^{\sigma ( \varsigma ) }\varphi (s)\Delta s. \end{aligned}$$
(28)
Applying Lemma 2.2 on the term
$$ \int _{\varsigma _{0}}^{\infty } \bigl(\sigma (s)-\varsigma _{0} \bigr)^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) { \Delta s}, $$
we see that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \bigl(\sigma (s)-\varsigma _{0} \bigr)^{k-1} \biggl( \int _{s}^{\infty }\frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}} \Delta x \biggr) { \Delta s} \\& \quad = \int _{\varsigma _{0}}^{\infty } \frac{\varphi (s)}{(\sigma (s)-\varsigma _{0})^{k}} \biggl( \int _{\varsigma _{0}}^{\sigma ( s ) } \bigl(\sigma (x)- \varsigma _{0} \bigr)^{k-1}\Delta x \biggr) \Delta s. \end{aligned}$$
(29)
From (27) and (29), we have
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \bigl(\sigma (s)-\varsigma _{0} \bigr)^{k-1} \biggl( \int _{s}^{\infty }\frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}} \Delta x \biggr) { \Delta s} \\& \quad \leq \int _{\varsigma _{0}}^{\infty }\varphi (x)\Delta x \leq (D+1) \int _{\varsigma _{0}}^{\infty }\varphi (x)\Delta x. \end{aligned}$$
(30)
Putting \(\psi (x)=1\) in Lemma 2.4, we see that the function
$$ \hbar (\varsigma ) \biggl( \frac{1}{\sigma (\varsigma )-\varsigma _{0}} \int _{\varsigma _{0}}^{ \sigma (\varsigma )}\hbar (x)\Delta x \biggr) ^{k-1} $$
is nonincreasing. Now, by applying Lemma 2.3, with
$$ \upsilon (\varsigma )= \bigl(\sigma (\varsigma )-\varsigma _{0} \bigr)^{k-1} \biggl( \int _{\varsigma }^{\infty } \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) \quad \text{and}\quad \phi (\varsigma )= ( D+1 ) \varphi (\varsigma ), $$
and
$$ z(\varsigma )=\hbar (\varsigma ) \biggl( \frac{1}{\sigma (\varsigma )-\varsigma _{0}} \int _{\varsigma _{0}}^{ \sigma (\varsigma )}\hbar (x)\Delta x \biggr) ^{k-1}, $$
we have from (28) and (30) that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \bigl(\sigma (\varsigma )- \varsigma _{0} \bigr)^{k-1} \biggl( \int _{\varsigma }^{\infty } \frac{\varphi (x)}{(\sigma (x)-\varsigma _{0})^{k}}\Delta x \biggr) \hbar (\varsigma ) \biggl( \frac{1}{\sigma (\varsigma )-\varsigma _{0}} \int _{\varsigma _{0}}^{ \sigma (\varsigma )}\hbar (x)\Delta x \biggr) ^{k-1}\Delta \varsigma \\& \quad \leq ( D+1 ) \int _{\varsigma _{0}}^{\infty } \varphi (\varsigma )\hbar ( \varsigma ) \biggl[ \frac{\int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x)\Delta x}{\sigma (\varsigma )-\varsigma _{0}} \biggr] ^{k-1}{ \Delta \varsigma }. \end{aligned}$$
(31)
Substituting (31) into the right-hand side of (22), we see that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \\& \quad \leq k ( D+1 ) \int _{ \varsigma _{0}}^{\infty }\varphi (\varsigma )\hbar ( \varsigma ) \biggl[ \frac{1}{ ( \sigma (\varsigma )-\varsigma _{0} ) } \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x)\Delta x \biggr) \biggr] ^{k-1}{ \Delta \varsigma } \\& \quad =k ( D+1 ) \int _{\varsigma _{0}}^{\infty } \varphi ^{\frac{1}{k}}( \varsigma )\hbar (\varsigma ) \biggl[ \frac{\varphi ^{\frac{1}{k}}(\varsigma )}{ ( \sigma (\varsigma )-\varsigma _{0} ) } \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x)\Delta x \biggr) \biggr] ^{k-1}{ \Delta \varsigma }. \end{aligned}$$
(32)
Applying Hölder’s inequality (11) on the term
$$ \int _{\varsigma _{0}}^{\infty } \bigl[ \varphi ^{\frac{1}{k}}( \varsigma )\hbar (\varsigma ) \bigr] \biggl[ \frac{\varphi ^{\frac{1}{k}}(\varsigma )}{\sigma (\varsigma )-\varsigma _{0}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x)\Delta x \biggr) \biggr] ^{k-1}\Delta \varsigma , $$
with indices k and \(k/(k-1)\), we get that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \bigl[ \varphi ^{\frac{1}{k}}( \varsigma )\hbar (\varsigma ) \bigr] \biggl[ \frac{\varphi ^{\frac{1}{k}}(\varsigma )}{\sigma (\varsigma )-\varsigma _{0}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x)\Delta x \biggr) \biggr] ^{k-1}{ \Delta \varsigma } \\& \quad \leq \biggl( \int _{\varsigma _{0}}^{\infty }\varphi ( \varsigma )\hbar ^{k}(\varsigma )\Delta \varsigma \biggr) ^{ \frac{1}{k}} \biggl[ \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \biggl( \int _{\varsigma _{0}}^{ \sigma (\varsigma )}\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \biggr] ^{\frac{k-1}{k}}. \end{aligned}$$
(33)
Finally substituting (33) into (32), we have that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \\& \quad \leq k ( D+1 ) \biggl( \int _{\varsigma _{0}}^{\infty } \varphi (\varsigma )\hbar ^{k}(\varsigma )\Delta \varsigma \biggr) ^{ \frac{1}{k}} \biggl[ \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x)\Delta x \biggr) ^{k} \Delta \varsigma \biggr] ^{\frac{k-1}{k}}. \end{aligned}$$
This implies that
$$ \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \leq k^{k} ( D+1 ) ^{k} \int _{ \varsigma _{0}}^{\infty }\varphi (\varsigma )\hbar ^{k}(\varsigma ) \Delta \varsigma , $$
which is (19). The proof is complete. □
Remark 3.1
In Theorem 3.1 we could replace ħ is rd-continuous with ħ is integrable.
Remark 3.2
Suppose that ħ is integrable and also assume that
$$ \int _{\varsigma _{0}}^{\infty }\varphi (\varsigma ) \biggl( \frac{1}{\sigma (\varsigma )-\varsigma _{0}}\int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x)\Delta x \biggr) ^{k} \Delta \varsigma \leq C \int _{\varsigma _{0}}^{\infty }\varphi ( \varsigma )\hbar ^{k}(\varsigma )\Delta \varsigma $$
(34)
holds for some constant \(C>0\). Then (34) holds when
$$ \hbar (x)= \textstyle\begin{cases} 1 ,& \text{if }x\in {}[ {\varsigma _{0}} ,\sigma (s) ]_{\mathbb{T}}, \\ 0,&\text{if }x\notin {}[ \varsigma _{0} , \sigma (s) ]_{\mathbb{T}}, \end{cases}$$
for any fixed \(s\in (\varsigma _{0},\infty )_{\mathbb{T}}\). For this ħ, the left-hand side of (34) becomes
$$\begin{aligned} \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma =& \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \biggl( \int _{\varsigma _{0}}^{\sigma (s)}\Delta x \biggr) ^{k} \Delta \varsigma \\ =& \bigl(\sigma (s)-\varsigma _{0} \bigr)^{k} \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \Delta \varsigma \\ \geq & \bigl(\sigma (s)-\varsigma _{0} \bigr)^{k} \int _{s}^{\infty } \frac{\varphi (\varsigma )}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \Delta \varsigma , \end{aligned}$$
(35)
and the right-hand side of (34) now becomes
$$ C \int _{\varsigma _{0}}^{\infty }\varphi (\varsigma )\hbar ^{k}( \varsigma )\Delta \varsigma =C \int _{\varsigma _{0}}^{\sigma (s)} \varphi (\varsigma )\Delta \varsigma . $$
(36)
From (35) and (36), we obtain
$$ \bigl(\sigma (s)-\varsigma _{0} \bigr)^{k} \int _{s}^{\infty } \frac{\varphi (\varsigma )}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \Delta \varsigma \leq C \int _{\varsigma _{0}}^{\sigma (s)}\varphi ( \varsigma )\Delta \varsigma , $$
and then
$$ \int _{s}^{\infty } \frac{\varphi (\varsigma )}{(\sigma (\varsigma )-\varsigma _{0})^{k}} \Delta \varsigma \leq \frac{C}{(\sigma (s)-\varsigma _{0})^{k}} \int _{ \varsigma _{0}}^{\sigma (s)}\varphi (s)\Delta s. $$
For particular cases of Theorem 3.1, we have the following results.
Remark 3.3
In the case when \(\mathbb{T}=\mathbb{R}\), inequality (19) in Theorem 3.1 reduces to the continuous inequality (1) of Ariňo and Muckenhoupt [3].
Remark 3.4
In the case when \(\mathbb{T}=\mathbb{N}\), inequality (19) in Theorem 3.1 reduces to inequality (2) of Bennett and Gross–Erdmann [4].
Remark 3.5
If \(\varphi (\varsigma )=1\), \(k>1\), and \(\frac{\sigma (\varsigma )-\varsigma _{0}}{\varsigma -\varsigma _{0}}\leq K\) for \(\varsigma \in (\varsigma _{0},\infty )_{\mathbb{T}}\), we see that inequality (18) holds with
$$ D=\frac{K^{k-1}}{k-1}. $$
(37)
Applying the chain rule formula (10), we see that
$$ - \bigl( ( s-\varsigma _{0} ) ^{-k+1} \bigr) ^{\Delta } \geq ( k-1 ) \bigl( \sigma (s)-\varsigma _{0} \bigr) ^{-k} , $$
and therefore
$$\begin{aligned} \int _{\varsigma }^{\infty } \frac{1}{ ( \sigma (s)-\varsigma _{0} ) ^{k}}\Delta s \leq &\frac{1}{k-1} \int _{\varsigma }^{\infty } \biggl( \frac{-1}{ ( s-\varsigma _{0} ) ^{k-1}} \biggr) ^{\Delta }\Delta s \\ \leq &\frac{K^{k-1}}{k-1} \frac{1}{ ( \sigma (\varsigma )-\varsigma _{0} ) ^{k-1}}. \end{aligned}$$
Thus we get the inequality
$$ \int _{\varsigma _{0}}^{\infty } \biggl( \frac{1}{\sigma (\varsigma )-\varsigma _{0}} \int _{\varsigma _{0}}^{ \sigma (\varsigma )}\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \leq k^{k} \biggl( \frac{K^{k-1}}{k-1}+1 \biggr) ^{k} \int _{\varsigma _{0}}^{\infty }\hbar ^{k}(\varsigma ) \Delta \varsigma . $$
(38)
Remark 3.6
In the case when \(\mathbb{T}=\mathbb{R}\) and \(\varsigma _{0}=0\), then \(K=1\) in the previous remark and from (38) we have
$$ \int _{0}^{\infty } \biggl( \frac{1}{\varsigma } \int _{0}^{\varsigma } \hbar (x)\,dx \biggr) ^{k}\,d\varsigma \leq \biggl( \frac{k^{2}}{k-1} \biggr) ^{k} \int _{0}^{\infty }\hbar ^{k}(\varsigma )\,d\varsigma , $$
which is a Hardy-type inequality with constant \(( k^{2}/(k-1) ) ^{k}\) (see [9]).
Theorem 3.2
Assume that \(k\geq 1\) and \(c>0\). Furthermore, assume that ħ is nonincreasing and
$$ \int _{\varsigma _{0}}^{\infty }\varphi (\varsigma )\hbar ^{k}( \varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{k-c}\Delta \varsigma < \infty . $$
Suppose that there is a constant \(D>0\) with
$$ \int _{\varsigma }^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}}\Delta x\leq \frac{D}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{k}}\int _{\varsigma _{0}}^{\sigma ( \varsigma ) }\varphi (x) \bigl[ \Upsilon ^{\sigma }(x) \bigr] ^{k-c}\Delta x \quad \textit{for all } \varsigma \in (\varsigma _{0},\infty )_{\mathbb{T}}; $$
(39)
here \(\Upsilon (\varsigma )=\int _{\varsigma _{0}}^{\varsigma }\psi (x) \Delta x\). Then
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (x) \hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \\& \quad \leq k^{k} (D+1)^{k} \int _{\varsigma _{0}}^{\infty }\varphi (\varsigma )\hbar ^{k}( \varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{k-c} \Delta \varsigma . \end{aligned}$$
(40)
Proof
Suppose that (39) holds. Apply Lemma 2.1 with \(z=\psi \hbar \), and we get that
$$ \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (x)\hbar (x) \Delta x \biggr) ^{k} \leq k \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (x)\hbar (x) \biggl[ \int _{\varsigma _{0}}^{\sigma (x)}\psi (\tau )\hbar (\tau ) \Delta \tau \biggr] ^{k-1}\Delta x. $$
(41)
Substituting (41) into the left-hand side of (40), we get that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )} \psi (x)\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \\& \quad \leq k \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (x)\hbar (x) \biggl( \int _{\varsigma _{0}}^{\sigma (x)}\psi (\tau )\hbar (\tau ) \Delta \tau \biggr) ^{k-1}\Delta x \biggr) \Delta \varsigma . \end{aligned}$$
(42)
Applying Lemma 2.2 on the term
$$ \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (x) \hbar (x) \biggl( \int _{\varsigma _{0}}^{\sigma (x)}\psi (\tau ) \hbar (\tau )\Delta \tau \biggr) ^{k-1}\Delta x \biggr) \Delta \varsigma , $$
we have that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )} \psi (x)\hbar (x) \biggl( \int _{\varsigma _{0}}^{\sigma (x)}\psi ( \tau )\hbar (\tau )\Delta \tau \biggr) ^{k-1}\Delta x \biggr) \Delta \varsigma \\& \quad = \int _{\varsigma _{0}}^{\infty }\psi (\varsigma )\hbar (\varsigma ) \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau ) \hbar (\tau )\Delta \tau \biggr) ^{k-1} \biggl( \int _{\varsigma }^{ \infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}}\Delta x \biggr) \Delta \varsigma . \end{aligned}$$
(43)
Substituting (43) into (42), we obtain
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (x)\hbar (x) \Delta x \biggr) ^{k}\Delta \varsigma \\& \quad \leq k \int _{\varsigma _{0}}^{\infty }\psi (\varsigma ) \hbar (\varsigma ) \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )} \psi (\tau )\hbar (\tau )\Delta \tau \biggr) ^{k-1} \biggl( \int _{ \varsigma }^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}} \Delta x \biggr) \Delta \varsigma \\& \quad =k \int _{\varsigma _{0}}^{\infty }\psi (\varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{k-1} \biggl( \int _{ \varsigma }^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}} \Delta x \biggr) \hbar (\varsigma ) \biggl( \frac{\int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau }{\Upsilon ^{\sigma }(\varsigma )} \biggr) ^{k-1} \Delta \varsigma . \end{aligned}$$
(44)
Using the additive property of integrals [5, Theorem 1.77(iv)] on time scales, we see for any \(\varsigma \in (\varsigma _{0},\infty )_{\mathbb{T}}\) that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}} \Delta x \biggr) \Delta s \\& \quad = \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-1} \biggl( \int _{s}^{\sigma ( \varsigma )} \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}} \Delta x+ \int _{\sigma (\varsigma )}^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}} \Delta x \biggr) \Delta s \\& \quad \leq \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-1} \biggl( \int _{s}^{\sigma ( \varsigma )} \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}} \Delta x \biggr) \Delta s \\& \qquad {} + \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-1} \biggl( \int _{\varsigma }^{ \infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}} \Delta x \biggr) \Delta s \\& \quad = \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-1} \biggl( \int _{s}^{\sigma ( \varsigma )} \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}} \Delta x \biggr) \Delta s \\& \qquad {} + \biggl( \int _{\varsigma }^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}}\Delta x \biggr) \int _{\varsigma _{0}}^{\sigma ( \varsigma )}\psi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-1} \Delta s. \end{aligned}$$
(45)
Integrating the term
$$ \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl( \Upsilon ^{ \sigma }(s) \bigr) ^{k-1} \biggl[ \int _{s}^{\sigma (\varsigma )} \frac{\varphi (x)}{ ( \Upsilon ^{\sigma }(x) ) ^{c}} \Delta x \biggr] \Delta s, $$
using the parts formula (8) with
$$ \Re (s)=\Phi (s)= \int _{s}^{\sigma (\varsigma )} \frac{\varphi (x)}{ ( \Upsilon ^{\sigma }(x) ) ^{c}} \Delta x\quad \text{and}\quad \omega ^{\Delta }(s)=\psi (s) \bigl( \Upsilon ^{ \sigma }(s) \bigr) ^{k-1}, $$
we have that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k-1} \biggl[ \int _{s}^{\sigma ( \varsigma )} \frac{\varphi (x)}{ ( \Upsilon ^{\sigma }(x) ) ^{c}} \Delta x \biggr] \Delta s \\& \quad = \Phi (s)\omega (s)\vert _{\varsigma _{0}}^{\sigma ( \varsigma )}- \int _{\varsigma _{0}}^{\sigma (\varsigma )}\Phi ^{ \Delta }(s)\omega ^{\sigma }(s)\Delta s, \end{aligned}$$
where \(\omega (s)=\int _{\varsigma _{0}}^{s}\psi (x) ( \Upsilon ^{ \sigma }(x) ) ^{k-1}\Delta x\). Using \(\Phi ^{\sigma }(\varsigma )=\omega (\varsigma _{0})=0\), we get that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k-1} \biggl[ \int _{s}^{\sigma ( \varsigma )} \frac{\varphi (x)}{ ( \Upsilon ^{\sigma }(x) ) ^{c}} \Delta x \biggr] \Delta s \\& \quad = \int _{\varsigma _{0}}^{\sigma (\varsigma )} \bigl[ -\Phi ^{ \Delta }(s) \bigr] \omega ^{\sigma }(s)\Delta s \\& \quad = \int _{\varsigma _{0}}^{\sigma (\varsigma )} \frac{\varphi (s)}{ ( \Upsilon ^{\sigma }(s) ) ^{c}} \biggl( \int _{\varsigma _{0}}^{\sigma (s)}\psi (x) \bigl( \Upsilon ^{\sigma }(x) \bigr) ^{k-1}\Delta x \biggr) \Delta s. \end{aligned}$$
(46)
Substituting (46) into (45), we have that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}} \Delta x \biggr) \Delta s \\& \quad \leq \int _{\varsigma _{0}}^{\sigma (\varsigma )} \frac{\varphi (s)}{ ( \Upsilon ^{\sigma }(s) ) ^{c}} \biggl( \int _{\varsigma _{0}}^{\sigma (s)}\psi (x) \bigl( \Upsilon ^{\sigma }(x) \bigr) ^{k-1}\Delta x \biggr) \Delta s \\& \qquad {} + \biggl( \int _{\varsigma }^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}}\Delta x \biggr) \int _{\varsigma _{0}}^{\sigma ( \varsigma )}\psi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-1} \Delta s. \end{aligned}$$
(47)
Note
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (x) \bigl( \Upsilon ^{\sigma }(x) \bigr) ^{k-1}\Delta x \\& \quad = \int _{\varsigma _{0}}^{\varsigma }\psi (x) \bigl( \Upsilon ^{ \sigma }(x) \bigr) ^{k-1}\Delta x+ \int _{\varsigma }^{\sigma ( \varsigma )}\psi (x) \bigl( \Upsilon ^{\sigma }(x) \bigr) ^{k-1} \Delta x \\& \quad = \int _{\varsigma _{0}}^{\varsigma }\psi (x) \bigl( \Upsilon ^{ \sigma }(x) \bigr) ^{k-1}\Delta x+\mu (\varsigma )\psi ( \varsigma ) \bigl( \Upsilon ^{\sigma }(\varsigma ) \bigr) ^{k-1} \\& \quad = \int _{\varsigma _{0}}^{\varsigma }\psi (x) \bigl( \Upsilon ^{ \sigma }(x) \bigr) ^{k-1}\Delta x+ \bigl( \Upsilon ^{\sigma }( \varsigma ) \bigr) ^{k-1} \int _{\varsigma }^{\sigma (\varsigma )} \psi (x)\Delta x. \end{aligned}$$
(48)
Since \(\Upsilon ^{\Delta }(\varsigma )=\psi (\varsigma )\geq 0\) and σ is an increasing function, we have (for \(x\leq \varsigma \)) that \(\sigma (x)\leq \sigma (\varsigma )\), and then \(\Upsilon ^{\sigma }(x)\leq \Upsilon ^{\sigma }(\varsigma )\), and then \(( \Upsilon ^{\sigma }(x) ) ^{k-1}\leq ( \Upsilon ^{ \sigma }(\varsigma ) ) ^{k-1}\). Therefore
$$ \int _{\varsigma _{0}}^{\varsigma }\psi (x) \bigl( \Upsilon ^{\sigma }(x) \bigr) ^{k-1}\Delta x\leq \bigl( \Upsilon ^{\sigma }(\varsigma ) \bigr) ^{k-1} \int _{\varsigma _{0}}^{\varsigma }\psi (x)\Delta x. $$
(49)
Substituting (49) into the right-hand side of (48), we have that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (x) \bigl( \Upsilon ^{\sigma }(x) \bigr) ^{k-1}\Delta x \\& \quad \leq \bigl( \Upsilon ^{\sigma }(\varsigma ) \bigr) ^{k-1} \int _{ \varsigma _{0}}^{\varsigma }\psi (x)\Delta x+ \bigl( \Upsilon ^{ \sigma }(\varsigma ) \bigr) ^{k-1} \int _{\varsigma }^{\sigma ( \varsigma )}\psi (x)\Delta x \\& \quad = \bigl( \Upsilon ^{\sigma }(\varsigma ) \bigr) ^{k-1} \int _{ \varsigma _{0}}^{\sigma (\varsigma )}\psi (x)\Delta x= \bigl( \Upsilon ^{\sigma }(\varsigma ) \bigr) ^{k}. \end{aligned}$$
(50)
Substituting (50) into (47), we obtain
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}} \Delta x \biggr) \Delta s \\& \quad \leq \int _{\varsigma _{0}}^{\sigma (\varsigma )}\varphi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-c}\Delta s+ \biggl( \int _{\varsigma }^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}}\Delta x \biggr) \bigl( \Upsilon ^{\sigma }(\varsigma ) \bigr) ^{k}. \end{aligned}$$
(51)
Substituting (39) into (51), we get that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}} \Delta x \biggr) \Delta s \\& \quad \leq \int _{\varsigma _{0}}^{\sigma (\varsigma )}\varphi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-c}\Delta s+D \int _{\varsigma _{0}}^{\sigma ( \varsigma ) }\varphi (x) \bigl[ \Upsilon ^{\sigma }(x) \bigr] ^{k-c}\Delta x \\& \quad = ( D+1 ) \int _{\varsigma _{0}}^{\sigma (\varsigma )} \varphi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-c}\Delta s. \end{aligned}$$
(52)
Applying Lemma 2.2 on the term
$$ \int _{\varsigma _{0}}^{\infty }\psi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}} \Delta x \biggr) \Delta s, $$
we have
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty }\psi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}} \Delta x \biggr) \Delta s \\& \quad = \int _{\varsigma _{0}}^{\infty } \frac{\varphi (s)}{ [ \Upsilon ^{\sigma }(s) ] ^{c}} \biggl( \int _{\varsigma _{0}}^{\sigma (s)}\psi (x) \bigl[ \Upsilon ^{\sigma }(x) \bigr] ^{k-1}\Delta x \biggr) \Delta s. \end{aligned}$$
(53)
From (50) and (53), we see that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty }\psi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}} \Delta x \biggr) \Delta s \\& \quad \leq \int _{\varsigma _{0}}^{\infty }\varphi (s) \bigl[ \Upsilon ^{ \sigma }(s) \bigr] ^{k-c}\Delta s \\& \quad \leq ( D+1 ) \int _{\varsigma _{0}}^{\infty }\varphi (s) \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k-c}\Delta s. \end{aligned}$$
(54)
Since from Lemma 2.4, the function
$$ \hbar (\varsigma ) \biggl( \frac{1}{\Upsilon ^{\sigma }(\varsigma )} \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau ) \Delta \tau \biggr) ^{k-1} $$
is nonincreasing, then by applying Lemma 2.3 on the term
$$ \int _{\varsigma _{0}}^{\infty } \biggl[ \psi (\varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{k-1} \biggl( \int _{ \varsigma }^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}} \Delta x \biggr) \biggr] \hbar (\varsigma ) \biggl( \frac{\int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau }{\Upsilon ^{\sigma }(\varsigma )} \biggr) ^{k-1}\Delta \varsigma , $$
with
$$ \upsilon (\varsigma )=\psi (\varsigma ) \bigl[ \Upsilon ^{\sigma }( \varsigma ) \bigr] ^{k-1} \biggl( \int _{\varsigma }^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}}\Delta x \biggr) ,\qquad \phi (\varsigma )= ( D+1 ) \varphi (\varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{k-c}, $$
and
$$ z(\varsigma )=\hbar (\varsigma ) \biggl( \frac{\int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau }{\Upsilon ^{\sigma }(\varsigma )} \biggr) ^{k-1}, $$
we get from (52) and (54) that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \biggl[ \psi (\varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{k-1} \biggl( \int _{\varsigma }^{\infty } \frac{\varphi (x)}{ [ \Upsilon ^{\sigma }(x) ] ^{c}} \Delta x \biggr) \biggr] \hbar (\varsigma ) \biggl( \frac{\int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau }{\Upsilon ^{\sigma }(\varsigma )} \biggr) ^{k-1}\Delta \varsigma \\& \quad \leq ( D+1 ) \int _{\varsigma _{0}}^{\infty } \varphi (\varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{k-c} \hbar (\varsigma ) \biggl( \frac{\int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau }{\Upsilon ^{\sigma }(\varsigma )} \biggr) ^{k-1}\Delta \varsigma \\& \quad = ( D+1 ) \int _{\varsigma _{0}}^{\infty } \varphi (\varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{1-c} \hbar (\varsigma ) \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )} \psi (\tau )\hbar (\tau )\Delta \tau \biggr) ^{k-1}\Delta \varsigma . \end{aligned}$$
(55)
Substituting (55) into the right-hand side of (44), we have that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )} \psi (x)\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \\& \quad \leq k ( D+1 ) \int _{\varsigma _{0}}^{\infty }\varphi ( \varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{1-c} \hbar (\varsigma ) \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )} \psi (\tau )\hbar (\tau )\Delta \tau \biggr) ^{k-1}\Delta \varsigma \\& \quad = k ( D+1 ) \int _{\varsigma _{0}}^{\infty }\varphi ^{ \frac{k-1}{k}}(\varsigma ) \frac{ ( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau ) ^{k-1}}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c ( \frac{k-1}{k} ) }}\varphi ^{\frac{1}{k}}(\varsigma )\hbar (\varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{\frac{k-c}{k}}\Delta \varsigma . \end{aligned}$$
(56)
Applying Hölder’s inequality (11) on the term
$$ \int _{\varsigma _{0}}^{\infty } \biggl[ \varphi ^{\frac{k-1}{k}}( \varsigma ) \frac{ ( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau ) ^{k-1}}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c ( \frac{k-1}{k} ) }} \biggr] \bigl[ \varphi ^{\frac{1}{k}}(\varsigma )\hbar (\varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{\frac{k-c}{k}} \bigr] \Delta \varsigma , $$
with indices k and \(k/(k-1)\), we see that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \biggl[ \varphi ^{\frac{k-1}{k}}( \varsigma ) \frac{ ( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau ) ^{k-1}}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c ( \frac{k-1}{k} ) }} \biggr] \bigl[ \varphi ^{\frac{1}{k}}(\varsigma )\hbar (\varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{\frac{k-c}{k}} \bigr] \Delta \varsigma \\& \quad \leq \biggl( \int _{\varsigma _{0}}^{\infty } \biggl[ \varphi ^{ \frac{k-1}{k}}( \varsigma ) \frac{ ( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau ) ^{k-1}}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c ( \frac{k-1}{k} ) }} \biggr] ^{\frac{k}{k-1}}\Delta \varsigma \biggr) ^{\frac{k-1}{k}} \\& \qquad {} \times \biggl( \int _{\varsigma _{0}}^{\infty } \bigl[ \varphi ^{ \frac{1}{k}}( \varsigma )\hbar (\varsigma ) \bigl[ \Upsilon ^{\sigma }( \varsigma ) \bigr] ^{\frac{k-c}{k}} \bigr] ^{k}\Delta \varsigma \biggr) ^{\frac{1}{k}} \\& \quad = \biggl( \int _{\varsigma _{0}}^{\infty }\varphi (\varsigma ) \frac{ ( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau ) ^{k}}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}} \Delta \varsigma \biggr) ^{\frac{k-1}{k}} \biggl( \int _{\varsigma _{0}}^{ \infty }\varphi (\varsigma )\hbar ^{k}(\varsigma ) \bigl[ \Upsilon ^{ \sigma }(\varsigma ) \bigr] ^{k-c}\Delta \varsigma \biggr) ^{\frac{1}{k}}. \end{aligned}$$
(57)
Finally, substituting (57) into the right-hand side of (56), we obtain
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )} \psi (x)\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \\& \quad \leq k ( D+1 ) \biggl( \int _{\varsigma _{0}}^{\infty } \varphi (\varsigma ) \frac{ ( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau ) ^{k}}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}}\Delta \varsigma \biggr) ^{\frac{k-1}{k}} \\& \qquad {}\times \biggl( \int _{\varsigma _{0}}^{\infty }\varphi (\varsigma )\hbar ^{k}( \varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{k-c} \Delta \varsigma \biggr) ^{\frac{1}{k}}, \end{aligned}$$
and then
$$ \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (x) \hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \leq k^{k} ( D+1 ) ^{k} \int _{\varsigma _{0}}^{\infty }\varphi (\varsigma ) \hbar ^{k}(\varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{k-c} \Delta \varsigma , $$
which is (40). The proof is complete. □
Remark 3.7
In Theorem 3.2 we could replace ħ is rd-continuous with ħ is integrable.
Remark 3.8
Suppose that ħ is integrable, and also assume that
$$ \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (x) \hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \leq D \int _{ \varsigma _{0}}^{\infty }\varphi (\varsigma )\hbar ^{k}(\varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{k-c}\Delta \varsigma $$
(58)
holds for some constant \(D>0\). Then (58) holds when
$$ \hbar ( x ) = \textstyle\begin{cases} 1,& x\in [ \varsigma _{0},\sigma ( s ) ]_{\mathbb{T}}, \\ 0,&x\notin [\varsigma _{0},\sigma ( s )]_{\mathbb{T}},\end{cases} $$
for any fixed \(s\in (\varsigma _{0},\infty )_{\mathbb{T}}\). For this ħ in (58), we obtain
$$ \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}} \biggl( \int _{\varsigma _{0}}^{\sigma (s)}\psi (x)\Delta x \biggr) ^{k}\Delta \varsigma \leq D \int _{\varsigma _{0}}^{\sigma (s)} \varphi (\varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{k-c}\Delta \varsigma . $$
(59)
Note
$$\begin{aligned} \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}} \biggl( \int _{\varsigma _{0}}^{\sigma (s)}\psi (x)\Delta x \biggr) ^{k}\Delta \varsigma \geq & \int _{s}^{\infty } \frac{\varphi (\varsigma )}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}} \biggl( \int _{\varsigma _{0}}^{\sigma (s)}\psi (x) \Delta x \biggr) ^{k}\Delta \varsigma \\ =& \biggl( \int _{\varsigma _{0}}^{\sigma (s)}\psi (x)\Delta x \biggr) ^{k} \int _{s}^{\infty } \frac{\varphi (\varsigma )}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}}\Delta \varsigma \\ =& \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k} \int _{s}^{\infty } \frac{\varphi (\varsigma )}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}} \Delta \varsigma . \end{aligned}$$
(60)
From (59) and (60), we have that
$$ \int _{s}^{\infty } \frac{\varphi (\varsigma )}{ [ \Upsilon ^{\sigma }(\varsigma ) ] ^{c}}\Delta \varsigma \leq \frac{D}{ [ \Upsilon ^{\sigma }(s) ] ^{k}}\int _{\varsigma _{0}}^{\sigma (s)}\varphi (\varsigma ) \bigl[ \Upsilon ^{\sigma }(\varsigma ) \bigr] ^{k-c}\Delta \varsigma . $$
As a particular case of Theorem 3.2 when \(k=c\), we have the following result.
Theorem 3.3
Assume that \(1\leq k<\infty \). Furthermore, assume that ħ is nonincreasing and \(\int _{\varsigma _{0}}^{\infty }\varphi (\varsigma )\hbar ^{k}( \varsigma )\Delta \varsigma <\infty \). Suppose that there is a constant \(D>0\) with
$$ \int _{\varsigma }^{\infty } \frac{\varphi (x)}{ ( \Upsilon ^{\sigma }(x) ) ^{k}}\Delta x\leq \frac{D}{ ( \Upsilon ^{\sigma }(\varsigma ) ) ^{k}}\int _{\varsigma _{0}}^{\sigma ( \varsigma ) }\varphi (x) \Delta x \quad \textit{for all } \varsigma \in (\varsigma _{0},\infty )_{\mathbb{T}}; $$
(61)
here \(\Upsilon (\varsigma )=\int _{\varsigma _{0}}^{\varsigma }\psi (x) \Delta x\). Then
$$ \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{ ( \Upsilon ^{\sigma }(\varsigma ) ) ^{k}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (x)\hbar (x) \Delta x \biggr) ^{k}\Delta \varsigma \leq k^{k} (D+1)^{k} \int _{ \varsigma _{0}}^{\infty }\varphi (\varsigma )\hbar ^{k}(\varsigma ) \Delta \varsigma . $$
(62)
Theorem 3.4
Assume that \(1\leq k<\infty \) and \(c>0\). Furthermore, assume that ħ is nonincreasing and
$$ \int _{\varsigma _{0}}^{\infty }\varphi (\varsigma ) \frac{ ( \Upsilon ^{\sigma }(\varsigma ) ) ^{k}}{\Theta ^{c}(\varsigma )}\hbar ^{k}(\varsigma )\Delta \varsigma < \infty . $$
Suppose that there is a constant \(D>0\) with
$$ \int _{\varsigma }^{\infty }\frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \leq \frac{D}{ ( \Upsilon ^{\sigma }(\varsigma ) ) ^{k}} \int _{ \varsigma _{0}}^{\sigma (\varsigma )}\varphi (x) \frac{ ( \Upsilon ^{\sigma }(x) ) ^{k}}{\Theta ^{c}(x)} \Delta x \quad \textit{for all } \varsigma \in (\varsigma _{0},\infty )_{ \mathbb{T}}; $$
(63)
here \(\Theta (\varsigma )=\int _{\varsigma }^{\infty }\psi (x)\Delta x\) and \(\Upsilon (\varsigma )=\int _{\varsigma _{0}}^{\varsigma }\psi (x) \Delta x\). Then
$$ \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \biggl( \int _{ \varsigma _{0}}^{\sigma (\varsigma )}\psi (x)\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \leq k^{k} (D+1)^{k} \int _{\varsigma _{0}}^{ \infty }\varphi (\varsigma ) \frac{ ( \Upsilon ^{\sigma }(\varsigma ) ) ^{k}}{\Theta ^{c}(\varsigma )} \hbar ^{k}(\varsigma )\Delta \varsigma . $$
(64)
Proof
Suppose that (63) holds. From (41), the left-hand side of (64) becomes
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \biggl( \int _{ \varsigma _{0}}^{\sigma (\varsigma )}\psi (x)\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \\& \quad \leq k \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \biggl( \int _{ \varsigma _{0}}^{\sigma (\varsigma )}\psi (x)\hbar (x) \biggl( \int _{ \varsigma _{0}}^{\sigma (x)}\psi (\tau )\hbar (\tau )\Delta \tau \biggr) ^{k-1}\Delta x \biggr) \Delta \varsigma . \end{aligned}$$
(65)
Applying Lemma 2.2 on the term
$$ \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \biggl( \int _{ \varsigma _{0}}^{\sigma (\varsigma )}\psi (x)\hbar (x) \biggl( \int _{ \varsigma _{0}}^{\sigma (x)}\psi (\tau )\hbar (\tau )\Delta \tau \biggr) ^{k-1}\Delta x \biggr) \Delta \varsigma , $$
we have that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \biggl[ \int _{ \varsigma _{0}}^{\sigma (\varsigma )}\psi (x)\hbar (x) \biggl( \int _{ \varsigma _{0}}^{\sigma (x)}\psi (\tau )\hbar (\tau )\Delta \tau \biggr) ^{k-1}\Delta x \biggr] \Delta \varsigma \\& \quad = \int _{\varsigma _{0}}^{\infty }\psi (\varsigma )\hbar (\varsigma ) \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau ) \hbar (\tau )\Delta \tau \biggr) ^{k-1} \biggl( \int _{\varsigma }^{ \infty }\frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \Delta \varsigma \\& \quad = \int _{\varsigma _{0}}^{\infty }\psi (\varsigma ) \bigl( \Upsilon ^{ \sigma }(\varsigma ) \bigr) ^{k-1} \biggl( \int _{\varsigma }^{\infty } \frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \hbar (\varsigma ) \biggl( \frac{\int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau }{\Upsilon ^{\sigma }(\varsigma )} \biggr) ^{k-1} \Delta \varsigma . \end{aligned}$$
(66)
Substituting (66) into (65), we see that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \biggl( \int _{ \varsigma _{0}}^{\sigma (\varsigma )}\psi (x)\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \\& \quad \leq k \int _{\varsigma _{0}}^{\infty }\psi (\varsigma ) \bigl( \Upsilon ^{\sigma }(\varsigma ) \bigr) ^{k-1} \biggl( \int _{ \varsigma }^{\infty }\frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \hbar (\varsigma ) \biggl( \frac{\int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau }{\Upsilon ^{\sigma }(\varsigma )} \biggr) ^{k-1} \Delta \varsigma . \end{aligned}$$
(67)
Using the additive property of integrals [5, Theorem 1.77(iv)] on time scales, we obtain for any \(\varsigma \in (\varsigma _{0},\infty )_{\mathbb{T}}\) that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl( \Upsilon ^{ \sigma }(s) \bigr) ^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \Delta s \\& \quad = \int _{ \varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl( \Upsilon ^{ \sigma }(s) \bigr) ^{k-1} \biggl( \int _{s}^{\sigma (\varsigma )} \frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \Delta s \\& \qquad {} + \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k-1} \biggl( \int _{\sigma ( \varsigma )}^{\infty }\frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \Delta s \\& \quad \leq \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k-1} \biggl( \int _{s}^{\sigma ( \varsigma )}\frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \Delta s \\& \qquad {} + \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k-1} \biggl( \int _{\varsigma }^{ \infty }\frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \Delta s. \end{aligned}$$
(68)
Integrating the term
$$ \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k-1} \biggl( \int _{s}^{\sigma (\varsigma )} \frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \Delta s, $$
using the parts formula (8) with
$$ \Re (s)=\Phi (s)= \int _{s}^{\sigma (\varsigma )} \frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x,\quad \text{and}\quad \omega ^{\Delta }(s)=\psi (s) \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k-1}, $$
we have that
$$ \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl( \Upsilon ^{ \sigma }(s) \bigr) ^{k-1} \biggl( \int _{s}^{\sigma (\varsigma )} \frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \Delta s= \Phi (s)\omega (s)\vert _{\varsigma _{0}}^{\sigma (\varsigma )}- \int _{\varsigma _{0}}^{\sigma (\varsigma )}\Phi ^{\Delta }(s)\omega ^{ \sigma }(s)\Delta s, $$
where \(\omega (s)=\int _{\varsigma _{0}}^{s}\psi (x) ( \Upsilon ^{ \sigma }(x) ) ^{k-1}\Delta x\). Using \(\Phi ^{\sigma }(\varsigma )=\omega (\varsigma _{0})=0\), we get that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k-1} \biggl( \int _{s}^{\sigma ( \varsigma )}\frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \Delta s \\& \quad = \int _{\varsigma _{0}}^{\sigma (\varsigma )} \bigl[ -\Phi ^{ \Delta }(s) \bigr] \omega ^{\sigma }(s)\Delta s \\& \quad = \int _{\varsigma _{0}}^{\sigma (\varsigma )} \frac{\varphi (s)}{\Theta ^{c}(s)} \biggl( \int _{\varsigma _{0}}^{ \sigma (s)}\psi (x) \bigl( \Upsilon ^{\sigma }(x) \bigr) ^{k-1} \Delta x \biggr) \Delta s. \end{aligned}$$
(69)
Substituting (69) into (68), we obtain
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \Delta s \\& \quad \leq \int _{\varsigma _{0}}^{\sigma (\varsigma )} \frac{\varphi (s)}{\Theta ^{c}(s)} \biggl( \int _{\varsigma _{0}}^{ \sigma (s)}\psi (x) \bigl( \Upsilon ^{\sigma }(x) \bigr) ^{k-1} \Delta x \biggr) \Delta s \\& \qquad {} + \biggl( \int _{\varsigma }^{\infty } \frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \int _{\varsigma _{0}}^{ \sigma (\varsigma )}\psi (s) \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k-1} \Delta s. \end{aligned}$$
(70)
Substituting (63) into (70), we get that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \Delta s \\& \quad \leq \int _{\varsigma _{0}}^{\sigma (\varsigma )} \frac{\varphi (s)}{\Theta ^{c}(s)} \biggl( \int _{\varsigma _{0}}^{ \sigma (s)}\psi (x) \bigl( \Upsilon ^{\sigma }(x) \bigr) ^{k-1} \Delta x \biggr) \Delta s \\& \qquad {} +\frac{D}{ ( \Upsilon ^{\sigma }(\varsigma ) ) ^{k}} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\varphi (x) \frac{ ( \Upsilon ^{\sigma }(x) ) ^{k}}{\Theta ^{c}(x)} \Delta x \biggr) \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k-1}\Delta s. \end{aligned}$$
(71)
Substituting (50) into (71), we have that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (s) \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \Delta s \\& \quad \leq \int _{\varsigma _{0}}^{\sigma (\varsigma )} \frac{\varphi (s)}{\Theta ^{c}(s)} \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k}\Delta s+D \int _{\varsigma _{0}}^{\sigma (\varsigma )} \varphi (x) \frac{ ( \Upsilon ^{\sigma }(x) ) ^{k}}{\Theta ^{c}(x)} \Delta x \\& \quad = ( D+1 ) \int _{\varsigma _{0}}^{\sigma (\varsigma )} \frac{\varphi (s)}{\Theta ^{c}(s)} \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k}\Delta s. \end{aligned}$$
(72)
Applying Lemma 2.2 on the term
$$ \int _{\varsigma _{0}}^{\infty }\psi (s) \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \Delta s, $$
we observe that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty }\psi (s) \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \Delta s \\& \quad = \int _{ \varsigma _{0}}^{\infty }\frac{\varphi (s)}{\Theta ^{c}(s)} \biggl( \int _{\varsigma _{0}}^{\sigma (s)}\psi (x) \bigl[ \Upsilon ^{ \sigma }(x) \bigr] ^{k-1}\Delta x \biggr) \Delta s. \end{aligned}$$
(73)
Substituting (50) into (73), we see that
$$\begin{aligned} \int _{\varsigma _{0}}^{\infty }\psi (s) \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k-1} \biggl( \int _{s}^{\infty } \frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \Delta s \leq & \int _{ \varsigma _{0}}^{\infty }\frac{\varphi (s)}{\Theta ^{c}(s)} \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k}\Delta s \\ \leq &(D+1) \int _{\varsigma _{0}}^{\infty } \frac{\varphi (s)}{\Theta ^{c}(s)} \bigl[ \Upsilon ^{\sigma }(s) \bigr] ^{k}\Delta s. \end{aligned}$$
(74)
Since from Lemma 2.4 the function
$$ \hbar (\varsigma ) \biggl( \frac{1}{\Upsilon ^{\sigma }(\varsigma )} \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau ) \Delta \tau \biggr) ^{k-1} $$
is nonincreasing, then by applying Lemma 2.3 with
$$ \upsilon (\varsigma )=\psi (\varsigma ) \bigl( \Upsilon ^{\sigma }( \varsigma ) \bigr) ^{k-1} \biggl[ \int _{\varsigma }^{\infty } \frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr] ,\qquad \phi (\varsigma )= ( D+1 ) \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \bigl( \Upsilon ^{ \sigma }(\varsigma ) \bigr) ^{k}, $$
and
$$ z(\varsigma )=\hbar (\varsigma ) \biggl( \frac{\int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau }{\Upsilon ^{\sigma }(\varsigma )} \biggr) ^{k-1}, $$
we obtain from (72) and (74) that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty }\psi (\varsigma ) \bigl( \Upsilon ^{ \sigma }(\varsigma ) \bigr) ^{k-1} \biggl( \int _{\varsigma }^{\infty } \frac{\varphi (x)}{\Theta ^{c}(x)}\Delta x \biggr) \hbar (\varsigma ) \biggl( \frac{\int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau }{\Upsilon ^{\sigma }(\varsigma )} \biggr) ^{k-1} \Delta \varsigma \\& \quad \leq ( D+1 ) \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \bigl( \Upsilon ^{ \sigma }(\varsigma ) \bigr) ^{k}\hbar (\varsigma ) \biggl( \frac{\int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau }{\Upsilon ^{\sigma }(\varsigma )} \biggr) ^{k-1}\Delta \varsigma . \end{aligned}$$
(75)
Substituting (75) into (67), we have that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \biggl( \int _{ \varsigma _{0}}^{\sigma (\varsigma )}\psi (x)\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \\& \quad \leq k ( D+1 ) \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \bigl( \Upsilon ^{ \sigma }(\varsigma ) \bigr) ^{k}\hbar (\varsigma ) \biggl( \frac{\int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau }{\Upsilon ^{\sigma }(\varsigma )} \biggr) ^{k-1}\Delta \varsigma \\& \quad = k ( D+1 ) \int _{\varsigma _{0}}^{\infty } \biggl[ \frac{\varphi ^{\frac{1}{k}}(\varsigma )}{\Theta ^{\frac{c}{k}}(\varsigma )} \Upsilon ^{\sigma }( \varsigma )\hbar (\varsigma ) \biggr] \biggl[ \frac{\varphi ^{\frac{k-1}{k}}(\varsigma )}{\Theta ^{\frac{c ( k-1 ) }{k}}(\varsigma )} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau ) \hbar (\tau )\Delta \tau \biggr) ^{k-1} \biggr] \Delta \varsigma . \end{aligned}$$
(76)
Applying Hölder’s inequality (11) on the term
$$ \int _{\varsigma _{0}}^{\infty } \biggl[ \frac{\varphi ^{\frac{1}{k}}(\varsigma )}{\Theta ^{\frac{c}{k}}(\varsigma )} \Upsilon ^{\sigma }(\varsigma )\hbar (\varsigma ) \biggr] \biggl[ \frac{\varphi ^{\frac{k-1}{k}}(\varsigma )}{\Theta ^{\frac{c ( k-1 ) }{k}}(\varsigma )} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau ) \hbar (\tau )\Delta \tau \biggr) ^{k-1} \biggr] \Delta \varsigma , $$
with indices k and \(k/ ( k-1 )\), we see that
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \biggl[ \frac{\varphi ^{\frac{1}{k}}(\varsigma )}{\Theta ^{\frac{c}{k}}(\varsigma )} \Upsilon ^{\sigma }(\varsigma )\hbar ( \varsigma ) \biggr] \biggl[ \frac{\varphi ^{\frac{k-1}{k}}(\varsigma )}{\Theta ^{\frac{c ( k-1 ) }{k}}(\varsigma )} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau ) \hbar (\tau )\Delta \tau \biggr) ^{k-1} \biggr] \Delta \varsigma \\& \quad \leq \biggl( \int _{\varsigma _{0}}^{\infty } \biggl[ \frac{\varphi ^{\frac{1}{k}}(\varsigma )}{\Theta ^{\frac{c}{k}}(\varsigma )} \Upsilon ^{\sigma }(\varsigma ) \hbar (\varsigma ) \biggr] ^{k} \Delta \varsigma \biggr) ^{ \frac{1}{k}} \\& \qquad {} \times \biggl( \int _{\varsigma _{0}}^{\infty } \biggl[ \frac{\varphi ^{\frac{k-1}{k}}(\varsigma )}{\Theta ^{\frac{c ( k-1 ) }{k}}(\varsigma )} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau ) \hbar (\tau )\Delta \tau \biggr) ^{k-1} \biggr] ^{\frac{k}{k-1}}\Delta \varsigma \biggr) ^{\frac{k-1}{k}} \\& \quad = \biggl( \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \bigl( \Upsilon ^{ \sigma }(\varsigma ) \bigr) ^{k}\hbar ^{k}(\varsigma )\Delta \varsigma \biggr) ^{\frac{1}{k}} \\& \qquad {} \times \biggl( \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \biggl( \int _{\varsigma _{0}}^{\sigma (\varsigma )}\psi (\tau ) \hbar (\tau )\Delta \tau \biggr) ^{k}\Delta \varsigma \biggr) ^{ \frac{k-1}{k}}. \end{aligned}$$
(77)
Finally, substituting (77) into (76), we obtain
$$\begin{aligned}& \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \biggl( \int _{ \varsigma _{0}}^{\sigma (\varsigma )}\psi (x)\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \\& \quad \leq k ( D+1 ) \biggl( \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \bigl( \Upsilon ^{\sigma }(\varsigma ) \bigr) ^{k}\hbar ^{k}(\varsigma )\Delta \varsigma \biggr) ^{ \frac{1}{k}} \\& \qquad {}\times \biggl( \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \biggl( \int _{\varsigma _{0}}^{\sigma ( \varsigma )}\psi (\tau )\hbar (\tau )\Delta \tau \biggr) ^{k}\Delta \varsigma \biggr) ^{\frac{k-1}{k}}. \end{aligned}$$
This implies that
$$ \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \biggl( \int _{ \varsigma _{0}}^{\sigma (\varsigma )}\psi (x)\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \leq k^{k} ( D+1 ) ^{k} \int _{ \varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \bigl( \Upsilon ^{ \sigma }(\varsigma ) \bigr) ^{k}\hbar ^{k}(\varsigma )\Delta \varsigma , $$
which is (64). The proof is complete. □
Remark 3.9
In Theorem 3.4 we could replace ħ is rd-continuous with ħ is integrable.
Remark 3.10
Suppose that ħ is integrable, and also assume that
$$ \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \biggl( \int _{ \varsigma _{0}}^{\sigma (\varsigma )}\psi (x)\hbar (x)\Delta x \biggr) ^{k}\Delta \varsigma \leq D \int _{\varsigma _{0}}^{\infty }\varphi ( \varsigma ) \frac{ ( \Upsilon ^{\sigma }(\varsigma ) ) ^{k}}{\Theta ^{c}(\varsigma )} \hbar ^{k}(\varsigma )\Delta \varsigma $$
(78)
holds for some constant \(D>0\). Then (78) holds when
$$ \hbar ( x ) = \textstyle\begin{cases} 1,&x\in {}[ \varsigma _{0},\sigma ( s ) ]_{ \mathbb{T}}, \\ 0,&x\notin {}[ \varsigma _{0},\sigma ( s ) ]_{\mathbb{T}},\end{cases}$$
for any fixed \(s\in (\varsigma _{0},\infty )_{\mathbb{T}}\). For this ħ, note
$$\begin{aligned} \int _{\varsigma _{0}}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )} \biggl( \int _{ \varsigma _{0}}^{\sigma (s)}\psi (x)\Delta x \biggr) ^{k}\Delta \varsigma \geq & \biggl( \int _{\varsigma _{0}}^{\sigma (s)}\psi (x) \Delta x \biggr) ^{k} \int _{s}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )}\Delta \varsigma \\ =& \bigl( \Upsilon ^{\sigma }(s) \bigr) ^{k} \int _{s}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )}\Delta \varsigma , \end{aligned}$$
so
$$ \int _{s}^{\infty } \frac{\varphi (\varsigma )}{\Theta ^{c}(\varsigma )}\Delta \varsigma \leq \frac{D}{ ( \Upsilon ^{\sigma }(s) ) ^{k}} \int _{\varsigma _{0}}^{ \sigma (s)}\varphi (\varsigma ) \frac{ ( \Upsilon ^{\sigma }(\varsigma ) ) ^{k}}{\Theta ^{c}(\varsigma )} \Delta \varsigma . $$