Throughout the paper, we will assume that the functions in the statements of the theorems are nonnegative and rd-continuous functions and (without mentioning) the integrals in the statements of the theorems are assumed to exist. Now we state and prove the basic lemmas that will be used in the proofs of our main results. The first lemma is adapted from [25, Lemma 2.6, p. 593].
Lemma 3.1
Let
\(\mathbb{T}\)
be a time scale with
\(a, b\in \mathbb{T}\)
and
\(f, g\in \mathrm{C}_{\mathrm{rd}}( [ a,b ] _{\mathbb{T}},\mathbb{R})\). If
\(m\geq 1\), then
$$ \biggl( \int _{a}^{b}f(t) \biggl( \int _{a}^{\sigma ( t ) }g(s) \Delta s \biggr) ^{m} \Delta t \biggr) ^{\frac{1}{m}}\leq \int _{a}^{b}g(s) \biggl( \int _{s}^{b}f(t)\Delta t \biggr) ^{\frac{1}{m}} \Delta s. $$
(3.1)
From now on, we will deal with the following half-linear dynamic equation:
$$ \lambda \bigl( v^{\frac{q}{p}}(x) \bigl( y^{\bigtriangleup }(x) \bigr) ^{\frac{q}{p^{\ast }}} \bigr) ^{\bigtriangleup }+w(x)y^{\frac{q}{p ^{\ast }}}\bigl(\sigma ( x ) \bigr)=0, $$
(3.2)
where \(p^{\ast }\) is the conjugate of p, and the weighted dynamic Hardy-type inequality
$$ \biggl( \int _{a}^{b}w(x) \biggl( \int _{a}^{x}u(s)\,ds \biggr) ^{q}\Delta x \biggr) ^{\frac{1}{q}}\leq C_{L} \biggl( \int _{a}^{b}v(x)u^{p}(x) \Delta x \biggr) ^{\frac{1}{p}} $$
(3.3)
for \(1< p\leq q<\infty \).
Actually, the main question that we will give the affirmative answer to states that the solvability of the dynamic equation (3.2) not only is necessary for the validity of the weighted dynamic Hardy-type inequality (3.3) but also is sufficient. The next result will guarantee the first direction, which emphasizes the need to achieve the equation in order to prove the legitimacy of the inequality. In the rest of the paper, we will assume that the function \(v(x)\) satisfies the condition
$$ \int _{a}^{\infty }v^{-\frac{1}{p-1}}(x)\Delta x=\infty . $$
(3.4)
Lemma 3.2
Let
\(\mathbb{T}\)
be a time scale with
\(a,b\in \mathbb{T}\), \(1< p\leq q< \infty \), \(u\in C_{\mathrm{rd}}([a,b]_{\mathbb{T}}, \mathbb{R})\)
be a nonnegative function, w, v
be positive rd-continuous functions on
\((a,b)_{\mathbb{T}}\),
$$ \int _{a}^{x}v^{1-p^{\ast }}(t)\Delta t< \infty \quad \textit{for }x \in [ a,b ] _{\mathbb{T}}, $$
(3.5)
and there exists a number
\(\lambda >0\)
such that the dynamic equation (3.2) has a positive solution
\(y(x)\). Then the following inequality
$$ \biggl( \int _{a}^{b}w(x) \biggl( \int _{a}^{\sigma ( x ) }f(t) \Delta t \biggr) ^{q} \Delta x \biggr) ^{\frac{1}{q}}\leq C \biggl( \int _{a}^{b}v(x)f^{p} ( x ) \Delta x \biggr) ^{\frac{1}{p}} $$
(3.6)
holds for every positive function
\(f(x) \)
on
\([ a,b ] _{\mathbb{T}} \), with the constant
$$ C=\lambda ^{\frac{1}{q}}. $$
(3.7)
Proof
Suppose that \(y(x)\) is a positive solution of (3.2). By utilizing Lemma 1.2.1 in [24, Lemma 1.2.1, p. 17] and condition (3.4), we see that y satisfies
$$ y(x),y^{\bigtriangleup }(x)>0\quad \text{and}\quad y^{\bigtriangleup \bigtriangleup }(x)< 0 \quad \text{for }x\in [ a,b ] _{ \mathbb{T}}. $$
(3.8)
For, x, \(t\in ( a,b ) _{\mathbb{T}}\) denote
$$\begin{aligned}& \varphi ( x ) :=-\lambda \bigl( v^{\frac{q}{p}}(x) \bigl( y^{\bigtriangleup }(x) \bigr) ^{\frac{q}{p^{\ast }}} \bigr) ^{ \bigtriangleup }, \end{aligned}$$
(3.9)
$$\begin{aligned}& \psi ( t ) :=f^{p} ( t ) \bigl( y^{\bigtriangleup }(t) \bigr) ^{\frac{-p}{p^{\ast }}}. \end{aligned}$$
(3.10)
Then (3.2) yields that \(\varphi ( x ) =w(x)y^{\frac{q}{p ^{\ast }}}(\sigma ( x ) )\) and the time scales Hölder’s inequality together with (3.8) imply that
$$\begin{aligned} \biggl( \int _{a}^{\sigma ( x ) }f(t)\Delta t \biggr) ^{q}w(x) =& \biggl( \int _{a}^{\sigma ( x ) }f(t) \bigl( y ^{\bigtriangleup }(t) \bigr) ^{\frac{-1}{p^{\ast }}} \bigl( y^{ \bigtriangleup }(t) \bigr) ^{\frac{1}{p^{\ast }}}\Delta t \biggr) ^{q}w(x) \\ \leq &w(x) \biggl( \int _{a}^{\sigma ( x ) }\psi ( t ) \Delta t \biggr) ^{\frac{q}{p}} \biggl( \int _{a}^{\sigma ( x ) }y^{\bigtriangleup }(t)\Delta t \biggr) ^{\frac{q}{p ^{\ast }}} \\ =&w(x) \biggl( \int _{a}^{\sigma ( x ) }\psi ( t ) \Delta t \biggr) ^{\frac{q}{p}} \bigl( y\bigl(\sigma ( x ) \bigr)-y(a) \bigr) ^{\frac{q}{p^{\ast }}} \\ \leq &w(x)y^{\frac{q}{p^{\ast }}}\bigl(\sigma ( x ) \bigr) \biggl( \int _{a}^{\sigma ( x ) }\psi ( t ) \Delta t \biggr) ^{\frac{q}{p}} \\ =&\varphi ( x ) \biggl( \int _{a}^{\sigma ( x ) }\psi ( t ) \Delta t \biggr) ^{\frac{q}{p}}. \end{aligned}$$
Integrating from a to b with respect to x and denoting that \(r=q/p\), we get that
$$ \biggl( \int _{a}^{b} \biggl( \int _{a}^{\sigma ( x ) }f(t) \Delta t \biggr) ^{q}w(x)\Delta x \biggr) ^{\frac{1}{r}}\leq \biggl( \int _{a}^{b}\varphi ( x ) \biggl( \int _{a}^{\sigma ( x ) }\psi ( t ) \Delta t \biggr) ^{r}\Delta x \biggr) ^{\frac{1}{r}}. $$
Applying the time scales Minkowski’s inequality (3.1), we have that
$$ \biggl( \int _{a}^{b} \biggl( \int _{a}^{\sigma ( x ) }f(t) \Delta t \biggr) ^{q}w(x)\Delta x \biggr) ^{\frac{1}{r}}\leq \int _{a} ^{b}\psi ( t ) \biggl( \int _{t}^{b}\varphi ( x ) \Delta x \biggr) ^{\frac{1}{r}}\Delta t. $$
(3.11)
Using (3.9) to estimate the inner integral on the right-hand side yields that
$$\begin{aligned} \int _{t}^{b}\varphi ( x ) \Delta x =&-\lambda \int _{t} ^{b} \bigl( v^{\frac{q}{p}}(x) \bigl( y^{\bigtriangleup }(x) \bigr) ^{\frac{q}{p^{\ast }}} \bigr) ^{\bigtriangleup }\Delta x \\ =&\lambda v^{\frac{q}{p}}(x) \bigl( y^{\bigtriangleup }(x) \bigr) ^{\frac{q}{p^{\ast }}} \big\vert _{b}^{t} \\ =&\lambda \bigl( v^{\frac{q}{p}}(t) \bigl( y^{\bigtriangleup }(t) \bigr) ^{\frac{q}{p^{\ast }}}-v^{\frac{q}{p}}(b) \bigl( y^{\bigtriangleup }(b) \bigr) ^{\frac{q}{p^{\ast }}} \bigr) \\ \leq &\lambda v^{\frac{q}{p}}(t) \bigl( y^{\bigtriangleup }(t) \bigr) ^{\frac{q}{p^{\ast }}}, \end{aligned}$$
which leads directly to
$$ \biggl( \int _{t}^{b}\varphi ( x ) \Delta x \biggr) ^{ \frac{1}{r}}\leq \lambda ^{\frac{p}{q}}v(t) \bigl( y^{\bigtriangleup }(t) \bigr) ^{\frac{p}{p^{\ast }}}. $$
Substituting this estimate in (3.11) and using (3.10), we have that
$$\begin{aligned} \biggl( \int _{a}^{b} \biggl( \int _{a}^{\sigma ( x ) }f(t) \Delta t \biggr) ^{q}w(x)\Delta x \biggr) ^{\frac{p}{q}} \leq & \int _{a}^{b}\psi ( t ) \lambda ^{\frac{p}{q}}v(t) \bigl( y^{ \bigtriangleup }(t) \bigr) ^{\frac{p}{p^{\ast }}}\Delta t \\ =&\lambda ^{\frac{p}{q}} \int _{a}^{b}v(t)f^{p} ( t ) \bigl( y^{\bigtriangleup }(t) \bigr) ^{\frac{-p}{p^{\ast }}} \bigl( y^{ \bigtriangleup }(t) \bigr) ^{\frac{p}{p^{\ast }}}\Delta t \\ =&\lambda ^{\frac{p}{q}} \int _{a}^{b}v(t)f^{p} ( t ) \Delta t. \end{aligned}$$
Finally, taking \(1/p\) power to both sides, we get the required inequality (3.6) with constant C as in (3.7). The proof is complete. □
Now the remaining part, which ensures that our answer to the main question is fully covered, is to prove the other direction, i.e., the sufficient condition, which is the main job of the next Lemmas 3.4–3.5. To prove these lemmas, we need the following auxiliary result, in which we will make use of Riccati-like inequality to get a useful integral inequality in the sequel.
Lemma 3.3
Suppose that
\(y(x)\)
is a positive solution of (3.2) and set
$$ z ( x ) =\frac{y(x)}{y^{\bigtriangleup }(x)}v^{1-p^{\ast }}(x) \quad \textit{for }x\in [ a,b ] _{\mathbb{T}}. $$
Then
\(z ( x ) >0\)
and satisfies the dynamic inequality
$$ z^{\bigtriangleup } ( x ) >\frac{p^{\ast }}{\lambda q}w(x)z ^{\frac{q}{p^{\ast }}+1} ( x ) +v^{1-p^{\ast }}(x). $$
(3.12)
Proof
For convenience, we sometimes skip the argument x in the computations. By using the quotient rule to differentiate
$$ z ( x ) =\frac{y(x)}{y^{\bigtriangleup }(x)}v^{1-p^{\ast }}(x), $$
we get that
$$\begin{aligned} z^{\bigtriangleup } =&\frac{y^{\bigtriangleup } [ y ( \sigma ( x ) ) ( v^{1-p^{\ast }} ) ^{\Delta }+y^{\bigtriangleup }v^{1-p^{\ast }} ] -y^{\bigtriangleup \bigtriangleup } [ yv^{1-p^{\ast }} ] }{y^{\bigtriangleup }(x)y^{\bigtriangleup }(\sigma ( x ) )} \\ =&\frac{y^{\bigtriangleup } [ y ( \sigma ( x ) ) ( 1-p^{\ast } ) v^{-p^{\ast }}+y^{\bigtriangleup }v^{1-p^{\ast }} ] -y^{\bigtriangleup \bigtriangleup } [ yv ^{1-p^{\ast }} ] }{y^{\bigtriangleup }(x)y^{\bigtriangleup }( \sigma ( x ) )} \\ =&\frac{y ( \sigma ( x ) ) y^{\bigtriangleup } ( 1-p^{\ast } ) v^{-p^{\ast }}+y^{\bigtriangleup }y^{ \bigtriangleup }v^{1-p^{\ast }}-y^{\bigtriangleup \bigtriangleup }yv ^{1-p^{\ast }}}{y^{\bigtriangleup }(x)y^{\bigtriangleup }(\sigma ( x ) )} \\ =&\frac{y^{\bigtriangleup }v^{1-p^{\ast }}}{y^{\bigtriangleup }( \sigma ( x ) )}+ \bigl( 1-p^{\ast } \bigr) \frac{y ( \sigma ( x ) ) v^{-p^{\ast }}}{y^{\bigtriangleup }( \sigma ( x ) )}- \frac{yy^{\bigtriangleup \bigtriangleup }v ^{1-p^{\ast }}}{y^{\bigtriangleup }y^{\bigtriangleup }(\sigma ( x ) )}. \end{aligned}$$
From (3.8) it follows that \(y^{\bigtriangleup }(x)>y^{\bigtriangleup }(\sigma ( x ) )\), and then we get that
$$\begin{aligned} z^{\bigtriangleup } >&\frac{y^{\bigtriangleup }(\sigma ( x ) )v^{1-p^{\ast }}}{y^{\bigtriangleup }(\sigma ( x ) )}+ \bigl( 1-p^{\ast } \bigr) \frac{y ( \sigma ( x ) ) v^{-p^{\ast }}}{y^{\bigtriangleup }(\sigma ( x ) )}-\frac{yy^{\bigtriangleup \bigtriangleup }v^{1-p^{\ast }}}{y^{ \bigtriangleup }y^{\bigtriangleup }} \\ =&v^{1-p^{\ast }}+ \bigl( 1-p^{\ast } \bigr) \frac{y ( \sigma ( x ) ) v^{-p^{\ast }}}{y^{\bigtriangleup }(\sigma ( x ) )}- \frac{yy^{\bigtriangleup \bigtriangleup }v^{1-p ^{\ast }}}{ ( y^{\bigtriangleup } ) ^{2}} \\ >&v^{1-p^{\ast }}+ \bigl( 1-p^{\ast } \bigr) v^{-p^{\ast }} \frac{y ( \sigma ( x ) ) }{y^{\bigtriangleup }( \sigma ( x ) )}-v^{1-p^{\ast }}\frac{y ( \sigma ( x ) ) y^{\bigtriangleup \bigtriangleup }}{ ( y^{\bigtriangleup } ) ^{2}}. \end{aligned}$$
(3.13)
For the last inequality, we have used the fact that \(y(x)< y(\sigma ( x ) )\) since \(y^{\bigtriangleup }(x)>0\). But, since
$$ wy^{\frac{q}{p^{\ast }}}\bigl(\sigma ( x ) \bigr)=-\lambda \bigl( v ^{\frac{q}{p}} \bigl( y^{\bigtriangleup } \bigr) ^{\frac{q}{p^{ \ast }}} \bigr) ^{\bigtriangleup }, $$
it follows by using the chain rule (noting that \(y^{\bigtriangleup \bigtriangleup }(x)<0\)) that
$$\begin{aligned} wy^{\frac{q}{p^{\ast }}}\bigl(\sigma ( x ) \bigr) =&-\lambda \bigl( v^{\frac{q}{p}} \bigl( \bigl( y^{\bigtriangleup } \bigr) ^{\frac{q}{p ^{\ast }}} \bigr) ^{\bigtriangleup }+ \bigl( v^{\frac{q}{p}} \bigr) ^{\bigtriangleup } \bigl( y^{\bigtriangleup }\bigl(\sigma ( x ) \bigr) \bigr) ^{\frac{q}{p^{\ast }}} \bigr) \\ < &-\lambda \biggl( v^{\frac{q}{p}}\frac{q}{p^{\ast }} \bigl( y^{ \bigtriangleup } \bigr) ^{\frac{q}{p^{\ast }}-1}y^{\bigtriangleup \bigtriangleup }(x)+\frac{q}{p}v^{\frac{q}{p}-1} \bigl( y^{\bigtriangleup }\bigl(\sigma ( x ) \bigr) \bigr) ^{\frac{q}{p^{\ast }}} \biggr) \\ =&-\lambda \frac{q}{p^{\ast }} \biggl( v^{\frac{q}{p}} \bigl( y^{ \bigtriangleup } \bigr) ^{\frac{q}{p^{\ast }}-1}y^{\bigtriangleup \bigtriangleup }+\frac{p^{\ast }}{p}v^{\frac{q}{p}-1} \bigl( y^{ \bigtriangleup }\bigl(\sigma ( x ) \bigr) \bigr) ^{\frac{q}{p^{ \ast }}} \biggr), \end{aligned}$$
which leads to
$$\begin{aligned} &w\frac{y^{\frac{q}{p^{\ast }}+1}(\sigma ( x ) )}{ ( y^{\bigtriangleup } ) ^{\frac{q}{p^{\ast }}+1}} \\ &\quad < -\lambda \frac{q}{p^{\ast }} \biggl( v^{\frac{q}{p}} \frac{y(\sigma ( x ) )}{ ( y^{\bigtriangleup } ) ^{2}}y^{ \bigtriangleup \bigtriangleup }+\frac{p^{\ast }}{p}v^{\frac{q}{p}-1}y\bigl( \sigma ( x ) \bigr)\frac{ ( y^{\bigtriangleup }(\sigma ( x ) ) ) ^{\frac{q}{p^{\ast }}}}{ ( y^{\bigtriangleup } ) ^{\frac{q}{p^{\ast }}+1}} \biggr) \\ &\quad < -\lambda \frac{q}{p^{\ast }}v^{\frac{q}{p}+p^{\ast }-1} \biggl( v ^{1-p^{\ast }} \frac{y(\sigma ( x ) )}{ ( y^{\bigtriangleup } ) ^{2}}y^{\bigtriangleup \bigtriangleup }+\frac{p^{\ast }}{p}v ^{-p^{\ast }}y\bigl( \sigma ( x ) \bigr)\frac{ ( y^{\bigtriangleup }(\sigma ( x ) ) ) ^{\frac{q}{p^{\ast }}}}{ ( y ^{\bigtriangleup } ) ^{\frac{q}{p^{\ast }}+1}} \biggr) , \end{aligned}$$
and hence,
$$\begin{aligned} &w \biggl( \frac{y(\sigma ( x ) )v^{1-p^{\ast }}}{y^{ \bigtriangleup }} \biggr) ^{\frac{q}{p^{\ast }}+1} \\ &\quad < \lambda \frac{q}{p^{\ast }} \biggl( -v^{1-p^{\ast }} \frac{y(\sigma ( x ) )}{ ( y^{\bigtriangleup } ) ^{2}}y^{ \bigtriangleup \bigtriangleup }+ \bigl( 1-p^{\ast } \bigr) v^{-p^{ \ast }}y\bigl(\sigma ( x ) \bigr)\frac{ ( y^{\bigtriangleup }( \sigma ( x ) ) ) ^{\frac{q}{p^{\ast }}}}{ ( y ^{\bigtriangleup } ) ^{\frac{q}{p^{\ast }}+1}} \biggr) , \end{aligned}$$
and then we get that
$$\begin{aligned} &\frac{p^{\ast }}{\lambda q}w \biggl( \frac{y(\sigma ( x ) )v^{1-p^{\ast }}}{y^{\bigtriangleup }} \biggr) ^{\frac{q}{p^{\ast }}+1} \\ &\quad < \bigl(1-p^{\ast }\bigr)v^{-p^{\ast }}y\bigl(\sigma ( x ) \bigr) \frac{ ( y^{\bigtriangleup }(\sigma ( x ) ) ) ^{\frac{q}{p ^{\ast }}}}{ ( y^{\bigtriangleup } ) ^{\frac{q}{p^{\ast }}+1}}-v ^{1-p^{\ast }}\frac{y(\sigma ( x ) )}{ ( y^{\bigtriangleup } ) ^{2}}y^{\bigtriangleup \bigtriangleup }. \end{aligned}$$
Since \(y^{\bigtriangleup }(x)>y^{\bigtriangleup }(\sigma ( x ) )\) and \((1-p^{\ast })\) is always negative, we obtain that
$$\begin{aligned} &\frac{p^{\ast }}{\lambda q}w \biggl( \frac{y(\sigma ( x ) )v^{1-p^{\ast }}}{y^{\bigtriangleup }} \biggr) ^{\frac{q}{p^{\ast }}+1} \\ &\quad < \bigl(1-p^{\ast }\bigr)v^{-p^{\ast }}y\bigl(\sigma ( x ) \bigr) \frac{ ( y^{\bigtriangleup }(\sigma ( x ) ) ) ^{\frac{q}{p ^{\ast }}}}{ ( y^{\bigtriangleup }(\sigma ( x ) ) ) ^{\frac{q}{p^{\ast }}+1}}-v^{1-p^{\ast }}\frac{y(\sigma ( x ) )}{ ( y^{\bigtriangleup } ) ^{2}}y^{ \bigtriangleup \bigtriangleup } \\ &\quad =\bigl(1-p^{\ast }\bigr)v^{-p^{\ast }}\frac{y(\sigma ( x ) )}{y ^{\bigtriangleup }(\sigma ( x ) )}-v^{1-p^{\ast }} \frac{y( \sigma ( x ) )}{ ( y^{\bigtriangleup } ) ^{2}}y ^{\bigtriangleup \bigtriangleup }. \end{aligned}$$
(3.14)
Since \(y^{\bigtriangleup }(x)>0\), it follows that \(y(x)< y(\sigma ( x ) )\), and hence
$$\begin{aligned} \frac{p^{\ast }}{\lambda q}w \biggl( \frac{y(\sigma ( x ) )v ^{1-p^{\ast }}}{y^{\bigtriangleup }} \biggr) ^{\frac{q}{p^{\ast }}+1} >& \frac{p^{\ast }}{\lambda q}w \biggl( \frac{y(x)v^{1-p^{\ast }}}{y ^{\bigtriangleup }} \biggr) ^{\frac{q}{p^{\ast }}+1} \\ =&\frac{p^{\ast }}{\lambda q}wz^{\frac{q}{p^{\ast }}+1}. \end{aligned}$$
(3.15)
Finally, assembling (3.13), (3.14), and (3.15), we get that
$$ z^{\bigtriangleup }(x)>\frac{p^{\ast }}{\lambda q}w(x)z^{\frac{q}{p ^{\ast }}+1}(x)+v^{1-p^{\ast }}(x), $$
(3.16)
which is the desired inequality (3.12). The proof is complete. □
Lemma 3.4
Let
\(\mathbb{T}\)
be a time scale with
\(a,b\in \mathbb{T}\), \(1< p\leq q< \infty \), \(u\in C_{\mathrm{rd}}([a,b]_{\mathbb{T}}, \mathbb{R})\)
is a nonnegative function, and let
w, v
be positive rd-continuous functions on
\([ a,b ] _{\mathbb{T}}\). Denote
$$ K=\frac{p^{\ast }}{q}\inf_{f}\sup_{a< x< b} \frac{1}{f ( x ) } \int _{a}^{x}w(t) \biggl( f ( t ) + \int _{a}^{t}v^{1-p^{\ast }}(s)\Delta s \biggr) ^{ \frac{q}{p^{\ast }+1}}\Delta t, $$
(3.17)
where the infimum is taken for every positive function
\(f(t)\)
defined on
\([ a,b ] _{\mathbb{T}}\).
-
(i)
If there exists a positive constant
λ
such that the dynamic equation (3.2) has a positive solution
\(y(x)\), then
$$ K\leq \lambda < \infty . $$
(3.18)
-
(ii)
If
\(K<\infty \), then the dynamic equation (3.2) has a positive solution
\(y(x)\)
satisfying (3.8) for every
\(\lambda >K\).
Proof
(i) Suppose that \(y(x)\) is a positive solution of (3.2) which satisfies (3.8) and set
$$ z ( x ) =\frac{y(x)}{y^{\bigtriangleup }(x)}v^{1-p^{\ast }}(x), $$
which leads directly to that \(z ( x ) \) is a positive solution on \([ a,b ] _{\mathbb{T}}\) for the following dynamic inequality:
$$ z^{\bigtriangleup } ( x ) \geq \frac{p^{\ast }}{\lambda q}w(x)z ^{\frac{q}{p^{\ast }}+1} ( x ) +v^{1-p^{\ast }}(x). $$
(3.19)
Since
$$ z ( x ) \geq \int _{a}^{x}z^{\bigtriangleup } ( x ) \Delta x, $$
then we have that
$$ z ( x ) \geq \frac{p^{\ast }}{\lambda q} \int _{a}^{x}w(t)z ^{\frac{q}{p^{\ast }}+1} ( t ) \Delta t+ \int _{a}^{x}v^{1-p ^{\ast }}(t)\Delta t. $$
Now, assume that
$$ f ( x ) =z ( x ) - \int _{a}^{x}v^{1-p^{\ast }}(t) \Delta t, $$
then we get that \(f ( x ) >0\) for \(x\in [ a,b ] _{\mathbb{T}}\), and
$$ \lambda \geq \frac{p^{\ast }}{q}\frac{1}{f ( x ) } \int _{a} ^{x}w(t) \biggl( f ( t ) + \int _{a}^{t}v^{1-p^{\ast }}(s) \Delta s \biggr) ^{\frac{q}{p^{\ast }}+1}\Delta t, $$
which gives the validity of (3.18) according to the definition of K (3.17).
(ii) Assume that \(\lambda >K\). In view of definition (3.17), there is a positive function \(f ( x ) \) such that
$$ f ( x ) \geq \frac{p^{\ast }}{\lambda q} \int _{a}^{x}w(t) \biggl( f ( t ) + \int _{a}^{t}v^{1-p^{\ast }}(s)\Delta s \biggr) ^{\frac{q}{p^{\ast }}+1} \Delta t. $$
(3.20)
We will formulate a solution for problem (3.2)–(3.8) as follows. First, define for \(n\in \mathbb{N}\) the following sequence \(\{ z_{n}(x) \} \) of functions:
$$\begin{aligned} \begin{aligned} &z_{0}(x) =f ( x ) + \int _{a}^{x}v^{1-p^{\ast }}(t)\Delta t, \\ &z_{n+1}(x) =\frac{p^{\ast }}{\lambda q} \int _{a}^{x}w(t)z_{n}^{\frac{q}{p ^{\ast }}+1}(t) \Delta t+ \int _{a}^{x}v^{1-p^{\ast }}(t)\Delta t. \end{aligned} \end{aligned}$$
(3.21)
It is obvious that \(z_{n}(x)>0\) for \(x\in [ a,b ] _{ \mathbb{T}}\), and using (3.20) we get that
$$ \int _{a}^{x}w(t)z_{0}^{\frac{q}{p^{\ast }}+1}(t) \Delta t< \infty $$
(3.22)
and
$$ z_{0}(x)-z_{1}(x)=f ( x ) -\frac{p^{\ast }}{\lambda q} \int _{a}^{x}w(t)z_{0}^{\frac{q}{p^{\ast }}+1}(t) \Delta t>0, $$
which leads us to
$$ z_{n}(x)-z_{n+1}(x)=\frac{p^{\ast }}{\lambda q} \int _{a}^{x}w(t) \bigl( z_{n-1}^{\frac{q}{p^{\ast }}+1}(t)-z_{n}^{\frac{q}{p^{\ast }}+1}(t) \bigr) \Delta t>0. $$
Thus the sequence \(\{ z_{n}(x) \} \) is decreasing on \([ a,b ] _{\mathbb{T}}\) and asserts with the positivity of \(z_{n}(x)\) to the existence of a nonnegative function \(z ( x ) \) on \([ a,b ] _{\mathbb{T}}\) such that
$$ z ( x ) =\lim_{n\rightarrow \infty }z_{n}(x). $$
Now, we obtain from (3.21) that
$$ z ( x ) =\frac{p^{\ast }}{\lambda q} \int _{a}^{x}w(t)z^{\frac{q}{p ^{\ast }}+1}(t)\Delta t+ \int _{a}^{x}v^{1-p^{\ast }}(t)\Delta t. $$
Actually, this formula asserts that \(z ( x ) >0\) belong to \(C_{\mathrm{rd}}( [ a,b ] _{\mathbb{T}},\mathbb{R})\) and satisfies the dynamic inequality (3.19). The proof is complete. □
Remark 3.1
According to Lemma 3.4, we have shown that the number K from (3.17) is finite if and only if there is \(\lambda \in ( 0,\infty ) \) such that problem (3.2), (3.8) is solvable. Consequently, using in addition Lemma 3.2, Theorem 3.1 will be proved if we show that the validity of Hardy’s inequality implies the finiteness of the number K. This will follow from the next assertion.
Lemma 3.5
Let
\(\mathbb{T}\)
be a time scale with
\(a,b\in \mathbb{T}\), \(1< p\leq q< \infty \), \(u\in C_{\mathrm{rd}}([a,b]_{\mathbb{T}}, \mathbb{R})\)
is a nonnegative function, and let
w, v
be positive on
\([ a,b ] _{\mathbb{T}}\). Suppose that
K
is defined by (3.17)
$$ B_{L}=\sup_{a< x< b} \biggl( \int _{x}^{b}w(t)\Delta t \biggr) ^{ \frac{1}{q}} \biggl( \int _{a}^{\sigma ( x ) }v^{1-p^{ \ast }}(t)\Delta t \biggr) ^{\frac{1}{p^{\ast }}}, $$
and let
\(C_{L}\)
be the best possible constant in (3.3). Then
$$ C_{L}\leq K^{\frac{1}{q}}\leq k(p,q)B_{L}, $$
(3.23)
where
$$ k(p,q)= \biggl( 1+\frac{q}{p^{\ast }} \biggr) ^{\frac{1}{q}} \biggl( 1+ \frac{p ^{\ast }}{q} \biggr) ^{\frac{1}{p^{\ast }}}. $$
(3.24)
Proof
First, we prove the left inequality on (3.23) by contradiction. For this purpose, suppose that \(K^{\frac{1}{q}}< C_{L}\) and assume that there exists a constant \(\lambda _{0}\) such that
$$ K^{\frac{1}{q}}< \lambda _{0}< C_{L}, $$
(3.25)
which gives that \(K<\lambda _{0}^{q}\) and problem (3.2)–(3.8) is solvable for \(\lambda = \lambda _{0}^{q}\) (due to Lemma 3.4). Now, Lemma 3.2 implies that
$$ C_{L}\leq \lambda ^{\frac{1}{q}}=\lambda _{0}, $$
which contradicts (3.25). Next, we prove the right inequality on (3.23). Suppose that \(B_{L}<\infty \), then we have that
$$ 0< \int _{t}^{b}w(y)\Delta y< \infty \quad \text{for }t \in ( a,b ) _{\mathbb{T}}, $$
and the function
$$ f ( t ) =sB_{L}^{p^{\ast }} \biggl( \int _{t}^{b}w(y)\Delta y \biggr) ^{-\frac{p^{\ast }}{q}}- \int _{a}^{t}v^{1-p^{\ast }}(y) \Delta y $$
is finite for every \(s\in ( 1,\infty ) \). Moreover, we get that
$$ sB_{L}^{p^{\ast }}>B_{L}^{p^{\ast }}\geq \biggl( \int _{t}^{b}w(y) \Delta y \biggr) ^{\frac{p^{\ast }}{q}} \biggl( \int _{a}^{t}v^{1-p^{ \ast }}(y)\Delta y \biggr) , $$
which gives directly that \(f ( t ) >0\) for \(t\in ( a,b ) _{\mathbb{T}}\). From (3.17), we can write that
$$\begin{aligned} K \leq &\frac{p^{\ast }}{q}\sup_{a< x< b}\frac{\int _{a}^{\sigma ( x ) }w(t) [ sB_{L}^{p^{\ast }} ( \int _{t}^{b}w(y) \Delta y ) ^{-\frac{p^{\ast }}{q}} ] ^{ \frac{q}{p^{\ast }}+1}\Delta t}{sB_{L}^{p^{\ast }} ( \int _{t} ^{b}w(y)\Delta y ) ^{-\frac{p^{\ast }}{q}}-\int _{a}^{t}v^{1-p ^{\ast }}(y)\Delta y} \\ =&\frac{p^{\ast }}{q} \bigl( sB_{L}^{p^{\ast }} \bigr) ^{\frac{q}{p ^{\ast }}+1}\sup_{a< x< b}\frac{h(x)}{sB_{L}^{p^{\ast }}- ( \int _{x}^{b}w(y)\Delta y ) ^{\frac{p^{\ast }}{q}} ( \int _{a} ^{x}v^{1-p^{\ast }}(y)\Delta y ) } \\ \leq &\frac{p^{\ast }}{q} \bigl( sB_{L}^{p^{\ast }} \bigr) ^{\frac{q}{p ^{\ast }}+1}\sup_{a< x< b}\frac{h(x)}{sB_{L}^{p^{\ast }}- ( \int _{x}^{b}w(y)\Delta y ) ^{\frac{p^{\ast }}{q}} ( \int _{a} ^{\sigma ( x ) }v^{1-p^{\ast }}(y)\Delta y ) } \\ \leq &\frac{p^{\ast }}{q}\frac{ ( sB_{L}^{p^{\ast }} ) ^{\frac{q}{p^{\ast }}+1}}{ ( s-1 ) B_{L}^{p^{\ast }}} \sup_{a< x< b}h(x), \end{aligned}$$
(3.26)
where
$$\begin{aligned} h(x) =& \int _{a}^{\sigma ( x ) }w(t) \biggl( \int _{t}^{b}w(y) \Delta y \biggr) ^{-\frac{p^{\ast }}{q} ( \frac{q}{p^{\ast }}+1 ) } \biggl( \int _{x}^{b}w(y)\Delta y \biggr) ^{\frac{p^{\ast }}{q}} \Delta t \\ \leq &\frac{q}{p^{\ast }} \biggl[ 1- \biggl( \int _{a}^{b}w(y)\Delta y \biggr) ^{-\frac{p^{\ast }}{q}} \biggl( \int _{x}^{b}w(y)\Delta y \biggr) ^{\frac{p^{\ast }}{q}} \biggr] \\ \leq &\frac{q}{p^{\ast }}. \end{aligned}$$
(3.27)
If we set
$$ g(s)= \biggl( \frac{s}{s-1} \biggr) ^{\frac{1}{q}}s^{\frac{1}{p^{ \ast }}}\quad \text{for }s\in ( 1,\infty ) , $$
then, using (3.26) and (3.27), we obtain that
$$ K^{\frac{1}{q}}\leq g(s)B_{L}. $$
But, we know that (see [25, Theorem 3.1, p. 594])
$$ k(p,q)=\inf_{1< s< \infty }g(s), $$
which claims the second inequality in (3.23). This completes the proof. □
By combining the above results together (necessary and sufficient conditions), we are ready to state our main result in this paper.
Theorem 3.1
Let
\(\mathbb{T}\)
be a time scale with
\(a,b\in \mathbb{T}\), \(1< p\leq q<\infty \), \(u\in C_{\mathrm{rd}}([a,b]_{ \mathbb{T}}, \mathbb{R})\)
is a nonnegative function, and let
w, v
be positive rd-continuous functions on
\((a,b)_{\mathbb{T}}\). If
$$ \int _{a}^{x}v^{1-p^{\ast }}(t)\Delta t< \infty \quad \textit{for }x \in [ a,b]_{\mathbb{T}}, $$
(3.28)
then inequality (3.3) holds with a finite constant
\(C_{L}\)
if and only if there is a number
\(\lambda >0\)
such that the half-linear dynamic equation (3.2) has a solution
\(y(x)\)
satisfying (3.8).
Remark 3.2
As a special case of Theorem 3.1 (when \(\mathbb{T}= \mathbb{N}\)), we get the following result which connects the discrete Hardy-type inequality with the half-linear difference equation. It is worth to mention here that the next result coincides with the one obtained by Liao [20, Proposition 2.2, p. 812], while there are some parts of Liao’s proof that were essentially based on the idea of the variational principle presented in [7] and [16, p. 181] which we did not rely on in our proof.
Corollary 3.1
Suppose that
\(1< p\leq q<\infty \), w
and
v
are positive sequences on the discrete interval
\(I=\{1,2,\ldots,N\}\)
with
\(N\leq \infty \). If
$$ \sum_{n=1}^{\infty }v_{n}^{\frac{-1}{p-1}}= \infty \quad \textit{and}\quad \sum_{n=1}^{N}v_{n}^{1-p^{\ast }}< \infty \quad \textit{for }n\in I, $$
then the following discrete weighted Hardy-type inequality
$$ \Biggl( \sum_{n=1}^{N}w_{n} \Biggl( \sum_{n=1}^{k}u_{n} \Biggr) ^{q} \Biggr) ^{\frac{1}{q}}\leq C _{1} \Biggl( \sum _{n=1}^{N}v_{n}u_{n}^{p} \Biggr) ^{\frac{1}{p}} $$
holds for an arbitrary non-negative sequence
u
on
I, with a finite constant
\(C_{1}\), if and only if there is a number
\(\lambda >0\)
such that the difference equation
$$ \lambda \Delta \bigl( v_{n}^{\frac{q}{p}} ( \Delta y_{n} ) ^{\frac{q}{p^{\ast }}} \bigr) +w_{n}y_{n+1}^{\frac{q}{p^{\ast }}}=0 $$
has a positive solution
\(y_{n}\)
for
\(n\in I\).
Remark 3.3
As a special case of Theorem 3.1 (when \(\mathbb{T}=q ^{\mathbb{N}_{0}}\)), we get the following result which connects the discrete Hardy-type inequality with the half-linear difference equation. It is worth to mention here that the next result is entirely new and has not been dealt with before to the knowledge of the authors. Assume that
$$ Hx \bigl( q^{k} \bigr) = \textstyle\begin{cases} \sum_{k=1}^{n}q^{k}x ( q^{k} ) ;&n=1,2,\ldots,N, \\ 0 ;&n=0. \end{cases} $$
Corollary 3.2
Suppose that
\(1< p\leq q<\infty \), w
and
v
are positive sequences defined on
\(\mathbb{T}=q^{\mathbb{N}_{0}}\). If
$$ \sum_{k=1}^{N}q^{k}v^{-1/(p-1)} \bigl( q^{k} \bigr) =\infty \quad \textit{and}\quad \sum _{k=1}^{N}q^{k}v^{1-p^{\ast }} \bigl( q^{k} \bigr) < \infty \quad \textit{for }n\in q^{\mathbb{N}_{0}}, $$
then the following discrete weighted Hardy-type inequality
$$ \Biggl( \sum_{k=1}^{N}q^{k}w \bigl( q^{k} \bigr) \Biggl( \sum_{k=1}^{n}q^{k}u \bigl( q^{k} \bigr) \Biggr) ^{q} \Biggr) ^{\frac{1}{q}} \leq C_{2} \Biggl( \sum_{n=1}^{N}q^{k}v \bigl( q ^{k} \bigr) u^{p} \bigl( q^{k} \bigr) \Biggr) ^{\frac{1}{p}} $$
holds for an arbitrary non-negative sequence
u
on
\(q^{\mathbb{N} _{0}}\), with a finite constant
\(C_{1}\), if and only if there is a number
\(\lambda >0\)
such that the following second-order
q-difference equation
$$ w \bigl( q^{k} \bigr) Hy^{\frac{q}{p^{\ast }}} \bigl( q^{k} \bigr) + \lambda \bigl[ v^{\frac{q}{p}} \bigl( q^{k+1} \bigr) y^{\frac{q}{p ^{\ast }}} \bigl( q^{k+1} \bigr) -v^{\frac{q}{p}} \bigl( q^{k} \bigr) y^{\frac{q}{p^{\ast }}} \bigl( q^{k} \bigr) \bigr] =0 $$
has a positive solution
\(y_{n}\)
for
\(n\in I\).
Now, let us conclude this section with some applications that illustrate and clarify the main ideas of the paper. Specifically, we consider the special case \(\mathbb{T}=\mathbb{R}\).
Example 1
By setting \(a=0\), \(b=\infty \), \(p=q\), \(v(x)=x^{p-k}\), \(w ( x ) = ( \frac{ \vert k-1 \vert }{p} ) ^{p}x^{-k}\) with \(k>1\) and \(F ( 0 ) =0\), then the general weighted inequality (3.6) reduces to the following inequality:
$$ \biggl( \int _{0}^{\infty }x^{-k} \biggl( \int _{0}^{x}f(t)\,dt \biggr) ^{p}\,dx \biggr) < \biggl( \frac{p}{ \vert k-1 \vert } \biggr) ^{p} \int _{0}^{\infty }x^{-k} \bigl( xf ( x ) \bigr) ^{p}\,dx $$
due to Hardy [16, Theorem 330, p. 245]. In this case, for the corresponding differential equation of (3.2), we have \(y(x)=x^{\frac{k-1}{p}}\) which satisfies the corresponding conditions (3.8).
Example 2
By setting \(a=0\), \(b=\infty \), \(p=q\), \(v(x)=1\), \(w ( x ) = ( \frac{p-1}{p} ) ^{p}x^{-p}\) with \(p>1\) and \(F ( 0 ) =0\), then the general weighted inequality (3.6) reduces to the following inequality:
$$ \biggl( \int _{0}^{\infty } \biggl( \frac{F(x)}{x} \biggr) ^{p}\,dx \biggr) < \biggl( \frac{p}{p-1} \biggr) ^{p} \int _{0}^{\infty }f^{p} ( x ) \,dx, $$
where \(\int _{0}^{x}f(t)\,dt\), due to Hardy [16, Theorem 327, p. 240]. In this case, for the corresponding differential equation of (3.2), we have \(y(x)=x^{\frac{p-1}{p}}\) which satisfies the corresponding conditions (3.8).
As another application for our main results, we could get the following inequality (see [16, Theorem 256, p. 182]).
Example 3
If \(p>1\), \(y^{\prime }>0\), and \(y(x)=\int _{0}^{x}y^{\prime }(t)\,dt\), then
$$ \biggl( \int _{0}^{\frac{\pi }{2}}y^{p}\,dx \biggr) \leq \frac{1}{p-1} \biggl( \frac{p}{2}\sin \frac{p}{2} \biggr) ^{p} \int _{0}^{\frac{ \pi }{2}} \bigl( y^{{\prime }} \bigr) ^{p}\,dx, $$
where \(y(x)\) is the solution of the equation
$$ x=\frac{p}{2}\sin \frac{p}{2} \int _{0}^{y} \bigl( 1-t^{p} \bigr) ^{ \frac{-1}{p}}\,dt, \quad 0\leq y\leq 1. $$