In this section, we introduce some new definitions and establish new characterization results of compact sets in functional spaces on time scales which will play an important role in studying abstract discontinuous dynamic equations on time scales.
First, let \(\delta_{L}^{+\infty},\delta_{R}^{+\infty}:[T_{0},+\infty )_{\mathbb{T}}\rightarrow\mathbb{R}^{+}\cup\{0\}\). Similar to [20], we can extend the Δ-gauge for \([a,b]_{\mathbb{T}}\) to \([T_{0},+\infty)_{\mathbb{T}}\).
Definition 3.1
We say \(\delta^{+\infty}=(\delta_{L}^{+\infty},\delta_{R}^{+\infty})\) is a Δ-gauge for \([T_{0},+\infty)_{\mathbb{T}}\) provided \(\delta_{L}^{+\infty}(t)>0\) on \((T_{0},+\infty)_{\mathbb{T}}\) and \(\delta_{L}^{+\infty}(T_{0})\geq0\), \(\delta_{R}^{+\infty}(t)>0\) on \([T_{0},+\infty)_{\mathbb{T}}\) and \(\delta_{R}^{+\infty}(t)\geq\mu(t)\) for all \(t\in[T_{0},+\infty)_{\mathbb{T}}\).
For a Δ-gauge, \(\delta^{+\infty}\), we always assume \(\delta_{L}^{+\infty}(T_{0})\geq0\) (we will sometimes not even point this out).
For \(T_{0}\in\mathbb{T}\) and a Banach space \((X, \Vert \cdot \Vert )\), let
$$\begin{aligned} \mathfrak{G} \bigl([T_{0},+\infty)_{\mathbb{T}}, X \bigr) :=& \Bigl\{ x:[T_{0},+\infty)_{\mathbb{T}}\rightarrow X; \\ &{}\lim_{s\rightarrow t^{+}}x(s)=x\bigl(t^{+}\bigr)\mbox{ and } \lim_{s\rightarrow t^{-}}x(s)=x\bigl(t^{-}\bigr)\mbox{ exist and are finite}, \\ &{}s,t< +\infty\mbox{ and } \sup_{t\in[T_{0},+\infty)_{\mathbb{T}}}\bigl\Vert x(t)\bigr\Vert < +\infty \Bigr\} . \end{aligned}$$
Endow \(\mathfrak{G} ([T_{0},+\infty)_{\mathbb{T}}, X )\) with the norm \(\Vert x\Vert _{\infty}=\sup_{t\in[T_{0},+\infty)_{\mathbb{T}}}\Vert x(t)\Vert \).
Lemma 3.1
\((\mathfrak{G} ([T_{0},+\infty)_{\mathbb{T}}, X ),\Vert \cdot \Vert _{\infty} )\)
is a Banach space.
Proof
Let \(\{x_{n}\}\) be an arbitrary Cauchy sequence in \(\mathfrak{G}\), i.e., for any \(\varepsilon>0\), there exists N such that \(n,m>N\) implies
$$ \bigl\Vert x_{n}(t)-x_{m}(t)\bigr\Vert < \varepsilon \quad\mbox{for all }t\in[T_{0},+\infty)_{\mathbb{T}}. $$
(3.1)
Since X is a Banach space, for each \(t\in[T_{0},+\infty)_{\mathbb{T}}\), \(\{x_{n}(t)\}\subset X\) is a Cauchy sequence so \(x_{n}(t)\rightarrow x(t)\). Hence, let \(m\rightarrow+\infty\) in (3.1), and we have
$$\bigl\Vert x_{n}(t)-x(t)\bigr\Vert < \varepsilon \quad\mbox{for all }t\in[T_{0},+\infty)_{\mathbb{T}}. $$
Furthermore, since \(\{x_{n}\}\subset\mathfrak{G}\), for any \(\varepsilon>0\) and \(t_{0}\in[T_{0},+\infty)_{\mathbb{T}}\), there exists \(\delta>0\), \(t\in (t_{0}-\delta_{L}^{+\infty}(t_{0}),t_{0} )_{\mathbb{T}}\) with
$$\bigl\Vert x_{n}(t)-x_{n}\bigl(t_{0}^{+} \bigr)\bigr\Vert < \varepsilon. $$
Hence, we obtain
$$\begin{aligned} \bigl\Vert x(t)-x\bigl(t_{0}^{+}\bigr) \bigr\Vert \leq\bigl\Vert x(t)-x_{n}(t)\bigr\Vert +\bigl\Vert x_{n}(t)-x_{n}\bigl(t_{0}^{+}\bigr) \bigr\Vert +\bigl\Vert x_{n}\bigl(t_{0}^{+} \bigr)-x\bigl(t_{0}^{+}\bigr)\bigr\Vert \leq3\varepsilon. \end{aligned}$$
(3.2)
Similarly, for \(t\in (t_{0},t_{0}+\delta_{R}^{+\infty}(t_{0}) )_{\mathbb{T}}\), we also obtain
$$\begin{aligned} \bigl\Vert x(t)-x\bigl(t_{0}^{-}\bigr) \bigr\Vert \leq\bigl\Vert x(t_{0})-x_{n}(t)\bigr\Vert +\bigl\Vert x_{n}(t)-x_{n}\bigl(t_{0}^{-} \bigr)\bigr\Vert +\bigl\Vert x_{n}\bigl(t_{0}^{-} \bigr)-x\bigl(t_{0}^{-}\bigr)\bigr\Vert \leq3\varepsilon. \end{aligned}$$
(3.3)
Therefore, from (3.2) and (3.3), we obtain \(x\in\mathfrak{G}\). Hence, \(\mathfrak{G}\) is a Banach space. □
In the following, we will introduce the definition of a partition \(\mathcal{P}\) for \([T_{0},+\infty)_{\mathbb{T}}\).
Definition 3.2
A partition \(\mathcal{P}\) for \([T_{0},+\infty)_{\mathbb{T}}\) is a division of \([T_{0},+\infty)_{\mathbb{T}}\) denoted by
$$\mathcal{P}= \bigl\{ T_{0}=t_{0}^{\mathcal{P}}\leq \eta_{1}\leq t_{1}^{\mathcal{P}}\leq\cdots\leq t_{n-1}^{\mathcal {P}}\leq\eta_{n}\leq t_{n}^{\mathcal{P}} \leq\cdots< \cdots \bigr\} $$
with \(t_{i}^{\mathcal{P}}>t_{i-1}^{\mathcal{P}}\) for \(i=1,2,\ldots\) and \(t_{i},\eta_{i}\in\mathbb{T}\). We call the points \(\eta_{i}\) tag points and the points \(t_{i}\) end points.
Definition 3.3
If \(\delta^{+\infty}\) is a Δ-gauge for \([T_{0},+\infty)_{\mathbb{T}}\), then we say a partition \(\mathcal {P}\) is \(\delta^{+\infty}\)-fine if
$$\eta_{i}-\delta_{L}^{+\infty}(\eta_{i}) \leq t_{i-1}^{\mathcal {P}}< t_{i}^{\mathcal{P}}\leq \eta_{i}+\delta_{R}^{+\infty}(\eta_{i}) $$
for \(i=1,2,\ldots\) .
Remark 3.1
From Definition 3.3, one can observe that if a partition \(\mathcal{P}\) is \(\delta^{+\infty}\)-fine for \([T_{0},+\infty)_{\mathbb{T}}\), then for any closed interval \([a,b]_{\mathbb{T}}\subset[T_{0},+\infty)_{\mathbb{T}}\), there must exist a δ-fine partition \(\mathcal{P}^{*}\) and \(\mathcal {P}^{*}\subset\mathcal{P}\).
Definition 3.4
A set \(\mathfrak{A}\subset\mathfrak{G} ([T_{0},+\infty)_{\mathbb {T}},X )\) is called uniformly equi-regulated, if it has the following property: for every \(\varepsilon>0\) and \(t_{0}\in[T_{0},+\infty)_{\mathbb{T}}\), there is a \(\delta^{+\infty}=(\delta_{L}^{+\infty},\delta_{R}^{+\infty})\) such that
-
(a)
If \(x\in\mathfrak{A}\), \(t^{\prime}\in[T_{0},+\infty )_{\mathbb{T}}\) and \(t_{0}-\delta_{L}^{+\infty}(t_{0})< t^{\prime}< t_{0}\), then \(\Vert x(t_{0}^{-})-x(t^{\prime})\Vert <\varepsilon\).
-
(b)
If \(x\in\mathfrak{A}\), \(t^{\prime\prime}\in[T_{0},+\infty )_{\mathbb{T}}\) and \(t_{0}< t^{\prime\prime}< t_{0}+\delta_{R}^{+\infty}(t_{0})\), then \(\Vert x(t_{0}^{+})-x(t^{\prime\prime})\Vert <\varepsilon\).
From Definition 3.4, we obtain the following theorem.
Theorem 3.1
A set
\(\mathfrak{A}\subset\mathfrak{G} ([T_{0},+\infty)_{\mathbb {T}},X )\)
is uniformly equi-regulated, if and only if, for every
\(\varepsilon>0\), there is a
\(\delta^{+\infty}\)-fine partition
\(\mathcal{P}\):
$$T_{0}=t_{0}^{\mathcal{P}}< t_{1}^{\mathcal{P}}< t_{2}^{\mathcal {P}}< \cdots< t_{n}^{\mathcal{P}}< \cdots< \cdots $$
such that
$$ \bigl\Vert x\bigl(t^{\prime}\bigr)-x(t)\bigr\Vert \leq \varepsilon, $$
(3.4)
for every
\(x\in\mathfrak{A}\)
and
\([t,t^{\prime}]_{\mathbb{T}}\subset(t_{j-1}^{\mathcal {P}},t_{j}^{\mathcal{P}})_{\mathbb{T}}, j=1,2,\ldots\) .
Proof
Let \(\varepsilon>0\) be given and let
$$\mathfrak{D}= \bigl\{ \xi;\xi\in(T_{0},+\infty)_{\mathbb{T}} \bigr\} $$
such that there is a partition \(\mathcal{P}\):
$$T_{0}=t_{0}^{\mathcal{P}}< t_{1}^{\mathcal {P}}< \cdots< t_{k}^{\mathcal{P}}=\xi $$
for which (3.4) holds with \(j=1,2,\ldots,k\).
(i) If \(\mathfrak{A}\subset\mathfrak{G} ([T_{0},+\infty)_{\mathbb {T}},X )\) is uniformly equi-regulated, then there is a \(\delta_{R}^{+\infty}(T_{0})>0\) such that
$$\bigl\Vert x(t)-x\bigl(T_{0}^{+}\bigr)\bigr\Vert < \frac{\varepsilon}{2}, $$
for every \(x\in\mathfrak{A}\) and \(t\in (T_{0},T_{0}+\delta_{R}^{+\infty}(T_{0}) )_{\mathbb{T}}\). Denote \(\xi_{1}=T_{0}+\delta_{R}^{+\infty}(T_{0}), T_{0}=t_{0}^{\mathcal {P}}< t_{1}^{\mathcal{P}}=\xi_{1}\). Thus, for \([t,t^{\prime}]_{\mathbb{T}}\subset(T_{0},\xi_{1})_{\mathbb{T}}\) and \(x\in\mathfrak{A}\), the inequalities
$$\bigl\Vert x(t)-x\bigl(t^{\prime}\bigr)\bigr\Vert \leq\bigl\Vert x(t)-x\bigl(T_{0}^{+}\bigr)\bigr\Vert +\bigl\Vert x \bigl(t^{\prime}\bigr)-x\bigl(T_{0}^{+}\bigr)\bigr\Vert \leq\varepsilon, $$
holds and we have \(\xi_{1}\in\mathfrak{D}\).
Let \(\xi_{2}>\xi_{1}>T_{0}\). Since \(x\in\mathfrak{A}\), then there is a \(\delta_{L}^{+\infty}(\xi_{2})\) such that
$$\bigl\Vert x\bigl(\xi_{2}^{-}\bigr)-x(t)\bigr\Vert < \frac{\varepsilon}{2} $$
for every \(x\in\mathfrak{A}\) and \(t\in (\xi_{2}-\delta_{L}^{+\infty}(\xi_{2}),\xi_{2} )_{\mathbb{T}}\cap[T_{0},+\infty)_{\mathbb{T}}\).
Let \(\tilde{\xi}\in (\xi_{2}-\delta_{L}(\xi_{2}),\xi_{2} )_{\mathbb{T}}\) and a partition \(T_{0}=t_{0}^{\mathcal{P}}< t_{1}^{\mathcal {P}}< t_{2}^{\mathcal{P}}<\cdots<t_{k}^{\mathcal{P}}=\tilde{\xi}\) be such that (3.4) holds with \(j=1,2,\ldots,k\). Denote \(t_{k+1}^{\mathcal{P}}=\xi_{2}\). Then for \([t,t^{\prime}]_{\mathbb{T}}\subset(t_{k}^{\mathcal {P}},t_{k+1}^{\mathcal{P}})_{\mathbb{T}}\) and \(x\in\mathfrak{A}\), we have
$$\bigl\Vert x(t)-x\bigl(t^{\prime}\bigr)\bigr\Vert \leq\bigl\Vert x(t)-x\bigl(\xi_{2}^{-}\bigr)\bigr\Vert +\bigl\Vert x \bigl(t^{\prime}\bigr)-x\bigl(\xi _{2}^{-}\bigr)\bigr\Vert \leq\varepsilon $$
which implies \(\xi_{2}\in\mathfrak{D}\). Thus we use the same argument as before to find that \(\xi_{i}\in(\xi_{i-1},+\infty)_{\mathbb{T}}\) such that \(\xi_{i}\in\mathfrak{D}, i=1,2,\ldots\) . Hence \(\xi_{\infty}=\sup\mathfrak{D}=+\infty\) and we are finished.
(ii) Reciprocally, for any given \(\varepsilon>0\), there is a \(\delta^{+\infty}\)-fine partition \(\mathcal{P}\): \(T_{0}=t_{0}^{\mathcal{P}}< t_{1}^{\mathcal{P}}< t_{2}^{\mathcal {P}}<\cdots<t_{k}^{\mathcal{P}}<\cdots<\cdots\) such that
$$ \bigl\Vert x(t)-x\bigl(t^{\prime}\bigr)\bigr\Vert \leq \varepsilon, $$
(3.5)
for every \(x\in\mathfrak{A}\) and \([t,t^{\prime}]_{\mathbb{T}}\subset(t_{j-1}^{\mathcal {P}},t_{j}^{\mathcal{P}})_{\mathbb{T}}, j=1,2,\ldots\) .
Let \(\eta_{j}\) be a tag of \((t_{j-1},t_{j})_{\mathbb{T}}\). Since this partition is \(\delta^{+\infty}\)-fine, we have \((t_{j-1}^{\mathcal{P}},t_{j}^{\mathcal {P}})_{\mathbb{T}}\subset (\eta_{j}-\delta_{L}^{+\infty}(\eta _{j}),\eta_{j}+\delta_{R}^{+\infty}(\eta_{j}) )_{\mathbb{T}}\). Therefore, the inequality (3.5) holds, for \(t,t^{\prime}\in (\eta_{j}-\delta_{L}^{+\infty}(\eta_{j}),\eta _{j}+\delta_{R}^{+\infty}(\eta_{j}) )_{\mathbb{T}}\). Taking \(t=\eta_{j}^{-}\) and \(t^{\prime}\in (\eta_{j}-\delta_{L}^{+\infty}(\eta_{j}),\eta_{j} ]_{\mathbb{T}}\), then the inequality (3.5) remains true. Also, if \(t=\eta_{j}^{+}\) and \(t^{\prime}\in [\eta_{j},\eta_{j}+\delta_{R}^{+\infty}(\eta_{j}) )_{\mathbb{T}}\), the inequality (3.5) is fulfilled. Then, from Definition 3.4, it follows that \(\mathfrak{A}\) is uniformly equi-regulated. This completes the proof. □
Definition 3.5
Let \(\mathfrak{A}\subset\mathfrak{G} ([T_{0},+\infty)_{\mathbb {T}},X )\). We say \(\mathfrak{A}\) is uniformly Cauchy if for any \(\varepsilon>0\), there exist \(T_{1}\in(T_{0},+\infty)_{\mathbb{T}}\) and a \(\delta^{+\infty}=(\delta_{L}^{+\infty},\delta_{R}^{+\infty})\)-fine partition \(\mathcal{P}\):
$$t_{0}^{\mathcal{P}}=T_{1}< t_{1}^{\mathcal{P}}< t_{2}^{\mathcal {P}}< \cdots< t_{n}^{\mathcal{P}}< \cdots< \cdots $$
such that:
-
(a)
If \(x\in\mathfrak{A}\), \(t^{\prime},t_{0}\in[T_{1},+\infty )_{\mathbb{T}}\) and \(t_{0}-\delta_{L}^{+\infty}(t_{0})< t^{\prime}< t_{0}\), then \(\Vert x(t_{0}^{-})-x(t^{\prime})\Vert <\varepsilon\).
-
(b)
If \(x\in\mathfrak{A}\), \(t^{\prime\prime},t_{0}\in[T_{1},+\infty )_{\mathbb{T}}\) and \(t_{0}< t^{\prime\prime}< t_{0}+\delta_{R}^{+\infty}(t_{0})\), then \(\Vert x(t_{0}^{+})-x(t^{\prime\prime})\Vert <\varepsilon\).
-
(c)
If \(x\in\mathfrak{A}\), \(t^{\prime}\in[a^{\prime},b^{\prime}]_{\mathbb{T}}\subset(t_{j-1}^{\mathcal {P}},t_{j}^{\mathcal {P}})_{\mathbb{T}}, t^{\prime\prime}\in[a^{\prime\prime},b^{\prime\prime}]_{\mathbb{T}}\subset (t_{i-1}^{\mathcal {P}},t_{i}^{\mathcal{P}})_{\mathbb{T}}\), \(i,j=1,2,\ldots\) , then \(\Vert x(t^{\prime})-x(t^{\prime\prime})\Vert <\varepsilon\).
Remark 3.2
From Definition 3.5, one can observe that if \(\mathfrak{A}\subset\mathfrak{G} ([T_{0},+\infty)_{\mathbb {T}},X )\) is uniformly Cauchy, then there exists \(T_{1}\in(T_{0},+\infty)_{\mathbb{T}}\) such that \(\mathfrak{A}\) is uniformly equi-regulated on \([T_{1},+\infty)_{\mathbb{T}}\).
Theorem 3.2
Assume that a set
\(\mathfrak{A}\subset\mathfrak{G} ([T_{0},+\infty)_{\mathbb {T}},X )\)
is uniformly equi-regulated and uniformly Cauchy, and for any
\(t\in[T_{0},+\infty)_{\mathbb{T}}\), there is a number
\(\beta_{t}\)
such that, for
\(x\in\mathfrak{A}\),
$$ \bigl\Vert x(t)-x\bigl(t^{-}\bigr)\bigr\Vert \leq \beta_{t},\qquad \bigl\Vert x\bigl(t^{+}\bigr)-x(t)\bigr\Vert \leq\beta_{t},\quad t\in[T_{0},+\infty)_{\mathbb{T}}. $$
(3.6)
Then there is a constant
\(K>0\)
such that
\(\Vert x(t)-x(T_{0})\Vert \leq K\), for every
\(x\in\mathfrak{A}\)
and
\(t\in[T_{0},+\infty)_{\mathbb{T}}\).
Proof
Since \(\mathfrak{A}\) is uniformly Cauchy, according to Definition 3.5, there exists \(T_{1}\in(T_{0},+\infty)_{\mathbb{T}}\) such that
$$ \bigl\Vert x(t)-x(T_{1})\bigr\Vert < 1,\quad t \in[T_{1},+\infty)_{\mathbb{T}}. $$
(3.7)
Let \(\mathfrak{C}\) be the set of all \(\tau\in(T_{0},T_{1}]_{\mathbb{T}}\) such that there exists \(K_{\tau}>0\) such that
$$\bigl\Vert x(t)-x(T_{0})\bigr\Vert \leq K_{\tau}, $$
for any \(x\in\mathfrak{A}\) and \(t\in[T_{0},\tau]_{\mathbb{T}}\).
Since \(\mathfrak{A}\) is uniformly equi-regulated, there is a \(\delta_{R}^{+\infty}(T_{0})\) such that
$$ \bigl\Vert x(t)-x\bigl(T_{0}^{+}\bigr)\bigr\Vert \leq1, $$
for every \(x\in\mathfrak{A}\) and \(t\in (T_{0},T_{0}+\delta_{R}^{+\infty}(t_{0}) ]_{\mathbb{T}}\). This fact together with the hypothesis implies that
$$\bigl\Vert x(t)-x(T_{0})\bigr\Vert \leq\bigl\Vert x(t)-x \bigl(T_{0}^{+}\bigr)\bigr\Vert +\bigl\Vert x \bigl(T_{0}^{+}\bigr)-x(T_{0})\bigr\Vert \leq1+\beta_{T_{0}}:=K_{T_{0}+\delta^{+\infty}}, $$
for every \(t\in (T_{0},T_{0}+\delta_{R}(T_{0}) ]_{\mathbb{T}}\) and \(x\in\mathfrak{A}\). Hence, \((T_{0},T_{0}+\delta_{R}(T_{0}) ]_{\mathbb{T}}\subset \mathfrak{C}\).
Denote \(\tau_{0}=\sup\mathfrak{C}\). As a consequence of the uniformly equi-regulatedness of \(\mathfrak{A}\), there is a \(\delta_{L}^{+\infty}(\tau_{0})>0\) such that \(\Vert x(t)-x(\tau_{0}^{-})\Vert \leq1\) for \(x\in\mathfrak{A}\) and \(t\in [\tau_{0}-\delta_{L}^{+\infty}(\tau_{0}),\tau_{0} )_{\mathbb{T}}\).
Let \(\tau\in\mathfrak{C}\cap [\tau_{0}-\delta_{L}^{+\infty}(\tau _{0}),\tau_{0} )_{\mathbb{T}}\). Then
$$\bigl\Vert x(t)-x(T_{0})\bigr\Vert \leq\bigl\Vert x(t)-x\bigl( \tau_{0}^{-}\bigr)\bigr\Vert +\bigl\Vert x\bigl(\tau _{0}^{-}\bigr)-x(\tau)\bigr\Vert +\bigl\Vert x( \tau)-x(T_{0})\bigr\Vert \leq1+1+K_{\tau}=2+K_{\tau}, $$
for every \(x\in\mathfrak{A}\) and \(t\in(\tau,\tau_{0})_{\mathbb{T}}\). Also,
$$\bigl\Vert x\bigl(\tau_{0}^{-}\bigr)-x(T_{0}) \bigr\Vert \leq\bigl\Vert x\bigl(\tau_{0}^{-}\bigr)-x( \tau)\bigr\Vert +\bigl\Vert x(\tau)-x(T_{0})\bigr\Vert \leq 1+K_{\tau}. $$
These inequalities and this hypothesis imply that
$$ \bigl\Vert x(\tau_{0})-x(T_{0})\bigr\Vert \leq\bigl\Vert x(\tau_{0})-x\bigl(\tau_{0}^{-} \bigr)\bigr\Vert +\bigl\Vert x\bigl(\tau _{0}^{-} \bigr)-x(T_{0})\bigr\Vert \leq\beta_{\tau_{0}}+1+K_{\tau}. $$
(3.8)
Thus \(\tau_{0}\in\mathfrak{C}\), where \(K_{\tau_{0}}=\beta_{\tau_{0}}+1+K_{\tau}\).
If \(\tau_{0}< T_{1}\), then, since \(\mathfrak{A}\) is uniformly equi-regulated, there is a \(\delta_{R}^{+\infty}(\tau_{0})>0\) such that
$$\bigl\Vert x(t)-x\bigl(\tau_{0}^{+}\bigr)\bigr\Vert \leq1, \quad\mbox{for any }x\in\mathfrak{A} \mbox{ and } t\in \bigl(\tau_{0}, \tau_{0}+\delta_{R}(\tau_{0}) \bigr]_{\mathbb{T}}, $$
which implies
$$\begin{aligned} \bigl\Vert x(t)-x(T_{0})\bigr\Vert \leq&\bigl\Vert x(t)-x \bigl(\tau_{0}^{+}\bigr)\bigr\Vert +\bigl\Vert x\bigl( \tau _{0}^{+}\bigr)-x(\tau_{0})\bigr\Vert + \bigl\Vert x(\tau_{0})-x(T_{0})\bigr\Vert \\ \leq&1+\beta_{\tau_{0}}+K_{\tau_{0}}=K_{\tau_{0}+\delta_{R}^{+\infty }(\tau_{0})}, \end{aligned}$$
for \(t\in (\tau_{0},\tau_{0}+\delta_{R}^{+\infty}(\tau_{0}) ]_{\mathbb{T}}\) and \(x\in\mathfrak{A}\). Thus \(\tau_{0}+\delta_{R}^{+\infty}(\tau_{0})\in\mathfrak{C}\), which contradicts the fact that \(\tau_{0}=\sup\mathfrak{C}\). Therefore, \(\tau_{0}=T_{1}\). Hence, by (3.8), we have
$$ \bigl\Vert x(T_{1})-x(T_{0})\bigr\Vert \leq K_{\tau_{0}}. $$
Combining with (3.7), we have
$$\bigl\Vert x(t)-x(T_{0})\bigr\Vert \leq\bigl\Vert x(t)-x(T_{1})\bigr\Vert +\bigl\Vert x(T_{0})-x(T_{1}) \bigr\Vert \leq1+K_{\tau_{0}}, $$
for \(t\in(T_{1},+\infty)_{\mathbb{T}}\). Then we can get the desired result. □
Now, we give some sufficient conditions to guarantee that \(\mathfrak{A}\) is relatively compact in \(\mathfrak{G} ([T_{0},+\infty)_{\mathbb{T}},X )\).
Theorem 3.3
Let
\(\mathfrak{A}\subset\mathfrak{G} ([T_{0},+\infty)_{\mathbb {T}},X )\)
be uniformly equi-regulated and uniformly Cauchy, for every
\(t\in[T_{0},+\infty)_{\mathbb{T}}\), and let the set
\(\{x(t);x\in\mathfrak{A}\}\)
be relatively compact in
X. Then the set
\(\mathfrak{A}\)
is relatively compact in
\(\mathfrak{G} ([T_{0},+\infty)_{\mathbb{T}},X )\).
Proof
Since \(\mathfrak{A}\) is uniformly Cauchy, \(\forall x\in\mathfrak{A}\), by (a), (b) from Definition 3.5, for any \(\varepsilon>0\), there exists \(T_{1}\in(T_{0},+\infty)_{\mathbb{T}}\), there is a \(\delta^{+\infty}=(\delta_{L}^{+\infty},\delta_{R}^{+\infty})\)-fine partition \(\mathcal{P}_{1}\):
$$t_{0}^{\mathcal{P}_{1}}=T_{1}< t_{1}^{\mathcal{P}_{1}}< t_{2}^{\mathcal {P}_{1}}< \cdots< t_{n}^{\mathcal{P}_{1}}< \cdots< \cdots $$
such that
$$\begin{aligned}& \bigl\Vert x\bigl(t_{j}^{\mathcal{P}_{1}+}\bigr)-x \bigl(t^{\prime}\bigr)\bigr\Vert < \varepsilon \quad\mbox{for } t^{\prime}\in \bigl(t_{j}^{\mathcal{P}_{1}},t_{j}^{\mathcal{P}_{1}}+ \delta _{R}^{+\infty}\bigl(t_{j}^{\mathcal{P}_{1}}\bigr) \bigr]_{\mathbb{T}}, \end{aligned}$$
(3.9)
$$\begin{aligned}& \bigl\Vert x\bigl(t_{j}^{\mathcal{P}_{1}-}\bigr)-x \bigl(t^{\prime\prime}\bigr)\bigr\Vert < \varepsilon \quad\mbox{for } t^{\prime\prime}\in \bigl[t_{j}^{\mathcal{P}_{1}}-\delta_{L}^{+\infty } \bigl(t_{j}^{\mathcal{P}_{1}}\bigr),t_{j}^{\mathcal{P}_{1}} \bigr)_{\mathbb{T}}, \end{aligned}$$
(3.10)
$$\begin{aligned}& \bigl\Vert x\bigl(T_{1}^{+}\bigr)-x\bigl(t^{\prime\prime\prime} \bigr)\bigr\Vert < \varepsilon \quad\mbox{for } t^{\prime\prime\prime}\in \bigl(T_{1},T_{1}+\delta_{R}^{+\infty}(T_{1}) \bigr]_{\mathbb{T}}, \end{aligned}$$
(3.11)
for each \(j=0,1,2,\ldots\) . From (c) in Definition 3.5, we have
$$ \bigl\Vert x \bigl(T_{1}+\delta_{R}^{+\infty}(T_{1}) \bigr)-x(t) \bigr\Vert < \varepsilon \quad\mbox{for } t\in \bigl[t_{j-1}^{\mathcal {P}_{1}}+ \delta_{R}^{+\infty}\bigl(t_{j-1}^{\mathcal {P}_{1}} \bigr),t_{j}^{\mathcal{P}_{1}}-\delta_{L}^{+\infty} \bigl(t_{j}^{\mathcal {P}_{1}}\bigr) \bigr]_{\mathbb{T}} $$
(3.12)
and
$$ \bigl\Vert x\bigl(t^{\prime\prime\prime}\bigr)-x\bigl(t^{\prime} \bigr)\bigr\Vert < \varepsilon. $$
(3.13)
From (3.10) and (3.12), we have
$$ \begin{aligned}[b] \bigl\Vert x\bigl(T_{1}+\delta_{R}^{+\infty}(T_{1}) \bigr)-x\bigl(t_{j}^{\mathcal {P}_{1}-}\bigr) \bigr\Vert \leq{}& \bigl\Vert x \bigl(T_{1}+\delta_{R}^{+\infty }(T_{1}) \bigr)-x \bigl(t_{j}^{\mathcal {P}_{1}}-\delta_{L}^{+\infty} \bigl(t_{j}^{\mathcal {P}_{1}}\bigr) \bigr) \bigr\Vert \\ &{}+ \bigl\Vert x\bigl(t_{j}^{\mathcal{P}_{1}-}\bigr)-x \bigl(t_{j}^{\mathcal {P}_{1}}-\delta_{L}^{+\infty} \bigl(t_{j}^{\mathcal{P}_{1}}\bigr) \bigr) \bigr\Vert < 2\varepsilon, \end{aligned} $$
(3.14)
for each \(j=0,1,2,\ldots\) . Similarly, from (3.9), (3.11), and (3.13), we also have
$$\begin{aligned} \bigl\Vert x\bigl(T_{1}^{+}\bigr)-x \bigl(t_{j}^{\mathcal {P}_{1}+}\bigr)\bigr\Vert \leq&\bigl\Vert x \bigl(T_{1}^{+}\bigr)-x\bigl(t^{\prime\prime\prime}\bigr)\bigr\Vert +\bigl\Vert x\bigl(t_{j}^{\mathcal {P}_{1}+}\bigr)-x \bigl(t^{\prime}\bigr)\bigr\Vert \\ &{}+\bigl\Vert x\bigl(t^{\prime\prime\prime}\bigr)-x\bigl(t^{\prime}\bigr) \bigr\Vert < 3\varepsilon, \end{aligned}$$
(3.15)
for each \(j=0,1,2,\ldots\) . Hence, from (3.12), we obtain
$$ \bigl\Vert x \bigl(T_{1}+\delta_{R}^{+\infty}(T_{1}) \bigr)-x(t) \bigr\Vert < 4\varepsilon,\quad t\in \bigl[t_{j-1}^{\mathcal {P}_{1}}+ \delta_{R}^{+\infty}\bigl(t_{j-1}^{\mathcal {P}_{1}} \bigr),t_{j}^{\mathcal{P}_{1}}-\delta_{L}^{+\infty} \bigl(t_{j}^{\mathcal {P}_{1}}\bigr) \bigr]_{\mathbb{T}}. $$
(3.16)
From (3.10) and (3.14), we get
$$\begin{aligned} \bigl\Vert x \bigl(T_{1}+\delta_{R}^{+\infty}(T_{1}) \bigr)-x(t) \bigr\Vert =&\bigl\Vert x\bigl(t_{j}^{\mathcal {P}_{1}-} \bigr)-x(t)\bigr\Vert + \bigl\Vert x\bigl(T_{1}+ \delta_{R}^{+\infty}(T_{1}) \bigr)-x \bigl(t_{j}^{\mathcal {P}_{1}-}\bigr) \bigr\Vert \\ < &4\varepsilon,\quad t\in \bigl[t_{j}^{\mathcal {P}_{1}}-\delta_{L}^{+\infty} \bigl(t_{j}^{\mathcal{P}_{1}}\bigr), t_{j}^{\mathcal {P}_{1}} \bigr)_{\mathbb{T}}. \end{aligned}$$
(3.17)
Similarly, from (3.11) and (3.15), we also obtain
$$\begin{aligned}[b] \bigl\Vert x\bigl(t_{j}^{\mathcal {P}_{1}+}\bigr)-x(t) \bigr\Vert &= \bigl\Vert x\bigl(T_{1}^{+}\bigr)-x(t)\bigr\Vert +\bigl\Vert x\bigl(T_{1}^{+}\bigr)-x\bigl(t_{j}^{\mathcal {P}_{1}+} \bigr)\bigr\Vert \\ &< 4\varepsilon, \quad t\in \bigl(t_{j}^{\mathcal{P}_{1}},t_{j}^{\mathcal {P}_{1}}+ \delta_{R}^{+\infty}\bigl(t_{j}^{\mathcal {P}_{1}}\bigr) \bigr]_{\mathbb{T}}, \end{aligned} $$
so we obtain
$$ \bigl\Vert x\bigl(T_{1}^{+}\bigr)-x(t) \bigr\Vert < 4\varepsilon, \quad t\in \bigl(t_{j}^{\mathcal {P}_{1}},t_{j}^{\mathcal{P}_{1}}+ \delta_{R}^{+\infty}\bigl(t_{j}^{\mathcal {P}_{1}}\bigr) \bigr]_{\mathbb{T}}. $$
(3.18)
Further, since \(\mathfrak{A}\) is uniformly equi-regulated on \([T_{0},+\infty)_{\mathbb{T}}\), given \(\varepsilon>0\), there is a \(\delta^{+\infty}\)-fine partition \(\mathcal{P}_{2}\):
$$t_{0}^{\mathcal{P}_{2}}=T_{0}< t_{1}^{\mathcal {P}_{2}}< \cdots< t_{K}^{\mathcal{P}_{2}}=T_{1}, $$
such that
$$\bigl\Vert x\bigl(t^{\prime}\bigr)-x(t)\bigr\Vert < \varepsilon, $$
for every \([t,t^{\prime}]_{\mathbb{T}}\subset(t_{j-1}^{\mathcal {P}_{2}},t_{j}^{\mathcal {P}_{2}})_{\mathbb{T}}, j\in\{1,2,\ldots,K\}\). Obviously, \(\mathcal {P}=\mathcal{P}_{1}\cup\mathcal{P}_{2}\) is a \(\delta^{+\infty}\)-fine partition for \([T_{0},+\infty)_{\mathbb{T}}\).
Let \(\{x_{n};n\in\mathbb{N}\}\) be a given sequence. By assumption, the set
$$\bigl\{ x_{n}\bigl(t_{j}^{\mathcal {P}_{2}} \bigr),x_{n} \bigl(T_{1}+\delta_{R}^{+\infty}(T_{1}) \bigr),x_{n}\bigl(T_{1}^{+}\bigr),n\in\mathbb{N} \bigr\} $$
is relatively compact in X for every \(j=0,1,2,\ldots,K\). Therefore, we can find a subsequence of indices \(\{n_{k};k\in\mathbb{N}\}\subset\{n;n\in\mathbb{N}\}\) such that the set
$$\bigl\{ x_{n_{k}}\bigl(t_{j}^{\mathcal {P}_{2}} \bigr),x_{n_{k}} \bigl(T_{1}+\delta_{R}^{+\infty}(T_{1}) \bigr),x_{n_{k}}\bigl(T_{1}^{+}\bigr),k\in \mathbb{N} \bigr\} $$
is also relatively compact in X for every \(j=0,1,\ldots,K\). Using this fact, we can find the elements \(\{y_{j};j=0,1,2,\ldots,K,K+1,K+2\}\subset X\) such that \(y_{j}=\lim_{k\rightarrow\infty}x_{n_{k}}(t_{j}^{\mathcal {P}_{2}})\), \(y_{K+1}=\lim_{k\rightarrow\infty}x_{n_{k}} (T_{1}+\delta _{R}^{+\infty}(T_{1}) )\) and \(y_{K+2}=\lim_{k\rightarrow\infty}x_{n_{k}}(T_{1}^{+})\). Therefore, there exists \(N\in\mathbb{N}\) such that, for every \(k>N\), we have
$$\bigl\Vert x_{n_{k}}\bigl(t_{j}^{\mathcal{P}_{2}} \bigr)-y_{j}\bigr\Vert < \frac{\varepsilon}{4}. $$
Let \(q>k\), and then
$$\bigl\Vert x_{n_{q}}\bigl(t_{j}^{\mathcal{P}_{2}} \bigr)-y_{j}\bigr\Vert < \frac{\varepsilon}{4}, $$
for any \(j=0,1,\ldots,K\). Similarly, we also have
$$\begin{aligned}& \bigl\Vert x_{n_{q}} \bigl(T_{1}+\delta_{R}^{+\infty}(T_{1}) \bigr)-y_{K+1} \bigr\Vert < \frac{\varepsilon}{4}, \\& \bigl\Vert x_{n_{q}}\bigl(T_{1}^{+} \bigr)-y_{K+2} \bigr\Vert < \frac{\varepsilon}{4}. \end{aligned}$$
Now, let \(t\in[T_{0},T_{1}]_{\mathbb{T}}\) and \(q\in\mathbb{N}\) such that \(q>k\). Then \(t=t_{j}^{\mathcal{P}_{2}}\) for some \(j\in\{0,1,\ldots,K\}\) and, in this case, we have
$$\bigl\Vert x_{n_{k}}(t)-x_{n_{q}}(t)\bigr\Vert \leq\bigl\Vert x_{n_{k}}\bigl(t_{j}^{\mathcal {P}_{2}} \bigr)-y_{j}\bigr\Vert +\bigl\Vert x_{n_{q}} \bigl(t_{j}^{\mathcal {P}_{2}}\bigr)-y_{j}\bigr\Vert < \frac{\varepsilon}{2}, $$
or \(t\in(t_{j-1}^{\mathcal{P}_{2}},t_{j}^{\mathcal {P}_{2}})_{\mathbb{T}}\) for some \(j\in\{1,2,\ldots,K\}\) and, in this case, we have
$$\begin{aligned} \bigl\Vert x_{n_{k}}(t)-x_{n_{q}}(t)\bigr\Vert \leq&\bigl\Vert x_{n_{k}}(t)-x_{n_{q}}\bigl(t_{j-1}^{\mathcal {P}_{2}} \bigr)\bigr\Vert +\bigl\Vert x_{n_{q}}(t)-x_{n_{k}} \bigl(t_{j-1}^{\mathcal {P}_{2}}\bigr)\bigr\Vert +\bigl\Vert x_{n_{q}}\bigl(t_{j-1}^{\mathcal {P}_{2}}\bigr)-x_{n_{k}} \bigl(t_{j-1}^{\mathcal{P}_{2}}\bigr)\bigr\Vert \\ < &\frac{\varepsilon}{4}+\frac{\varepsilon}{4}+\frac{\varepsilon }{2}=\varepsilon. \end{aligned}$$
Moreover, from (3.16), let \(t\in[T_{1},+\infty)_{\mathbb{T}}\), we also obtain
$$\begin{aligned} \bigl\Vert x_{n_{k}}(t)-x_{n_{q}}(t)\bigr\Vert \leq& \bigl\Vert x_{n_{k}}(t)-x_{n_{q}} \bigl(T_{1}+ \delta_{R}^{+\infty}(T_{1}) \bigr) \bigr\Vert + \bigl\Vert x_{n_{q}}(t)-x_{n_{k}} \bigl(T_{1}+ \delta_{R}^{+\infty}(T_{1}) \bigr) \bigr\Vert \\ &{}+ \bigl\Vert x_{n_{q}} \bigl(T_{1}+ \delta_{R}^{+\infty}(T_{1}) \bigr)-x_{n_{k}} \bigl(T_{1}+\delta_{R}^{+\infty}(T_{1}) \bigr) \bigr\Vert \\ < &\frac{\varepsilon}{4}+\frac{\varepsilon}{4}+\frac{\varepsilon }{2}< \varepsilon, \end{aligned}$$
for \(t\in [t_{j-1}^{\mathcal {P}_{1}}+\delta_{R}^{+\infty}(t_{j-1}^{\mathcal {P}_{1}}),t_{j}^{\mathcal{P}_{1}}-\delta_{L}^{+\infty}(t_{j}^{\mathcal {P}_{1}}) ]_{\mathbb{T}}, j\in\{1,2,\ldots\}\).
Similarly, from (3.17), we also obtain
$$\begin{aligned} \bigl\Vert x_{n_{k}}(t)-x_{n_{q}}(t)\bigr\Vert \leq& \bigl\Vert x_{n_{k}}(t)-x_{n_{q}} \bigl(T_{1}+ \delta_{R}^{+\infty}(T_{1}) \bigr) \bigr\Vert + \bigl\Vert x_{n_{q}}(t)-x_{n_{k}} \bigl(T_{1}+ \delta_{R}^{+\infty}(T_{1}) \bigr) \bigr\Vert \\ &{}+ \bigl\Vert x_{n_{q}} \bigl(T_{1}+ \delta_{R}^{+\infty}(T_{1}) \bigr)-x_{n_{k}} \bigl(T_{1}+\delta_{R}^{+\infty}(T_{1}) \bigr) \bigr\Vert \\ < &\frac{\varepsilon}{4}+\frac{\varepsilon}{4}+\frac{\varepsilon }{2}< \varepsilon, \end{aligned}$$
for \(t\in [t_{j}^{\mathcal {P}_{1}}-\delta_{L}^{+\infty}(t_{j}^{\mathcal{P}_{1}}),t_{j}^{\mathcal {P}_{1}} )_{\mathbb{T}}, j\in\{0,1,2,\ldots\}\).
From (3.18), we have
$$\begin{aligned} \bigl\Vert x_{n_{k}}(t)-x_{n_{q}}(t)\bigr\Vert \leq& \bigl\Vert x_{n_{k}}(t)-x_{n_{q}}\bigl(T_{1}^{+} \bigr) \bigr\Vert + \bigl\Vert x_{n_{q}}(t)-x_{n_{k}} \bigl(T_{1}^{+}\bigr) \bigr\Vert + \bigl\Vert x_{n_{q}}\bigl(T_{1}^{+} \bigr)-x_{n_{k}}\bigl(T_{1}^{+}\bigr) \bigr\Vert \\ < &\frac{\varepsilon}{4}+\frac{\varepsilon}{4}+\frac{\varepsilon }{2}< \varepsilon, \end{aligned}$$
for \(t\in (t_{j}^{\mathcal{P}_{1}},t_{j}^{\mathcal {P}_{1}}+\delta_{R}^{+\infty}(t_{j}^{\mathcal {P}_{1}}) ]_{\mathbb{T}}, j\in\{0,1,2,\ldots\}\).
Thus, for \(t\in[T_{0},+\infty)_{\mathbb{T}}\), the sequence
$$\bigl\{ x_{n_{k}}(t);k\in\mathbb{N} \bigr\} \subset X $$
is a Cauchy sequence. Since X is complete, \(\lim_{k\rightarrow\infty}x_{n_{k}}(t)\) exists. Hence, any \(\{x_{n}\}\subset\mathfrak{A}\) has a convergent subsequence which means that \(\mathfrak{A}\) is a relatively compact set. The proof is complete. □
In the following, let \(X=\mathbb{R}^{n}\); we will give some sufficient conditions to guarantee that \(\mathfrak{A}\subset\mathfrak{G} ([T_{0},+\infty)_{\mathbb {T}},\mathbb{R}^{n} )\) is relatively compact.
Theorem 3.4
Let a set
\(\mathfrak{A}\subset\mathfrak{G} ([T_{0},+\infty)_{\mathbb {T}},\mathbb{R}^{n} )\). If
\(\mathfrak{A}\)
is relatively compact in the sup-norm topology, then it is uniformly equi-regulated. If
\(\mathfrak{A}\)
is uniformly equi-regulated and uniformly Cauchy, satisfying (3.6), then
\(\mathfrak{A}\)
is relatively compact in
\(\mathfrak{G} ([T_{0},+\infty)_{\mathbb{T}},\mathbb{R}^{n} )\).
Proof
A subset A of a Banach space X is relatively compact if and only if it is totally bounded, i.e., for every \(\varepsilon>0\), there is a finite ε-net F for A, i.e., such a subset \(F=\{x_{1},x_{2},\ldots,x_{k}\}\) of X that, for every \(x\in A\), there is \(x_{n}\in F\) satisfying \(\Vert x-x_{n}\Vert \leq\varepsilon\).
(i) Assume that \(\mathfrak{A}\) is relatively compact. Then it is bounded by a constant C, and evidently (3.6) is satisfied with \(\beta_{t}=2C\) for every \(t\in[T_{0},+\infty)_{\mathbb{T}}\).
Let \(t_{0}\in[T_{0},+\infty)_{\mathbb{T}}\) and ε be given. Let \(\{x_{1},x_{2},\ldots,x_{k}\}\subset\mathfrak{G} ([T_{0},+\infty )_{\mathbb{T}},\mathbb{R}^{n} )\) be a finite \(\varepsilon/3\)-net for \(\mathfrak{A}\). For every \(n=1,2,\ldots,k\), there is a \(\delta_{n}^{+\infty}= (\delta_{L}^{+\infty, n},\delta_{R}^{+\infty, n} )\) such that
$$\bigl\vert x_{n}(t)-x_{n}\bigl(t_{0}^{+} \bigr)\bigr\vert < \frac{\varepsilon}{3}\quad \mbox{for } t\in \bigl(t_{0},t_{0}+ \delta_{R}^{+\infty,n}(t_{0}) \bigr)_{\mathbb {T}} \cap[T_{0},+\infty)_{\mathbb{T}} $$
and
$$\bigl\vert x_{n}\bigl(t_{0}^{-} \bigr)-x_{n}(t)\bigr\vert < \frac{\varepsilon}{3}\quad \mbox{for } t\in \bigl(t_{0}-\delta_{L}^{+\infty,n}(t_{0}),t_{0} \bigr)_{\mathbb {T}}\cap[T_{0},+\infty)_{\mathbb{T}}. $$
Denote \(\delta^{+\infty}= (\min_{1\leq n\leq k} (\delta_{L}^{+\infty,n}(t_{0}) ),\min_{1\leq n\leq k} (\delta_{R}^{+\infty,n}(t_{0}) ) )= (\delta _{L}^{+\infty}(t_{0}),\delta_{R}^{+\infty}(t_{0}) )\).
For arbitrary \(x\in\mathfrak{A}\), we can find \(x_{n}\) such that \(\Vert x-x_{n}\Vert _{\infty}\leq\varepsilon/3\) for every \(t\in (t_{0},t_{0}+\delta_{R}^{+\infty}(t_{0}) )_{\mathbb {T}}\cap[T_{0},+\infty)_{\mathbb{T}}\), and we have the inequality
$$\begin{aligned} \bigl\vert x(t)-x\bigl(t_{0}^{+}\bigr)\bigr\vert \leq &\bigl\vert x(t)-x_{n}(t)\bigr\vert +\bigl\vert x_{n}(t)-x_{n}\bigl(t_{0}^{+}\bigr) \bigr\vert +\bigl\vert x_{n}\bigl(t_{0}^{+} \bigr)-x\bigl(t_{0}^{+}\bigr)\bigr\vert \\ \leq&2\Vert x-x_{n}\Vert _{\infty}+\bigl\vert x_{n}(t)-x_{n}\bigl(t_{0}^{+}\bigr) \bigr\vert < \varepsilon, \end{aligned}$$
and similarly, \(\vert x(t_{0}^{-})-x(t)\vert <\varepsilon\) for \(t\in (t_{0}-\delta_{L}^{+\infty}(t_{0}),t_{0} )_{\mathbb{T}}\).
(ii) Assume that \(\mathfrak{A}\) is uniformly equi-regulated, (3.6) holds, i.e., there exists \(\tilde{\alpha}>0\) such that \(\vert x(T_{0})\vert \leq\tilde{\alpha}\) for every \(x\in\mathfrak{A}\).
From Theorem 3.2, there is K such that \(\vert x(t)-x(T_{0})\vert \leq K\) for any \(x\in\mathfrak{A}\) and \(t\in[T_{0},+\infty)_{\mathbb{T}}\). Hence, \(\vert x(t)\vert \leq \vert x(t)+x(T_{0})\vert +\vert x(T_{0})\vert \leq K+\tilde{\alpha}\). If we denote \(\tilde{\gamma}=K+\tilde{\alpha}\), then \(\Vert x\Vert \leq\tilde{\gamma}\) for \(x\in\mathfrak{A}\). Since \(\mathfrak{A}\) is uniformly bounded in \(\mathbb{R}^{n}\), \(\mathfrak{A}\) is a sequentially compact closed set in \(\mathbb{R}^{n}\), i.e., \(\mathfrak{A}\) is relatively compact in \(\mathbb{R}^{n}\). From Theorem 3.3, one can obtain \(\mathfrak{A}\) is a relatively compact set in \(\mathfrak{G} ([T_{0},+\infty)_{\mathbb{T}},\mathbb{R}^{n} )\). This completes the proof. □
Remark 3.3
If \(\tilde{\mathfrak{A}}\) is uniformly bounded, equi-continuous and uniformly Cauchy, then \(\tilde{\mathfrak{A}}\subset\mathfrak{A}\). Hence, one can observe that Lemma 4 from [21] is just a particular case of Theorem 3.4.
Remark 3.4
For \(a,b\in\mathbb{T}\), let
$$\begin{aligned} \mathfrak{G}_{1} \bigl([a,b]_{\mathbb{T}}, X \bigr) :=& \Bigl\{ x:[a,b]_{\mathbb{T}}\rightarrow X; \\ &{}\lim_{s\rightarrow t^{+}}x(s)=x\bigl(t^{+}\bigr)\mbox{ and } \lim_{s\rightarrow t^{-}}x(s)=x\bigl(t^{-}\bigr)\mbox{ exist and are finite} \Bigr\} , \end{aligned}$$
and according to each \(x\in\mathfrak{G}_{1}\), we can construct the set \(\mathfrak{G}_{2} ([a,+\infty)_{\mathbb{T}}, X )\), satisfying, for each \(\tilde{x}\in\mathfrak{G}_{2}\):
$$ \tilde{x}(t)= \textstyle\begin{cases} x(t), & t\in[a,b]_{\mathbb{T}},\\ x(b^{-}),& t\in(b,+\infty)_{\mathbb{T}}. \end{cases} $$
One can immediately see that \(\mathfrak{G}_{1}\) and \(\mathfrak{G}_{2}\) are topological homeomorphic, i.e., there exists a homeomorphic mapping \(f: \mathfrak{G}_{1}\rightarrow\mathfrak{G}_{2}\) such that \(f(x)=\tilde{x}\) and \(f^{-1}(\tilde{x})=x\). Since for any set \(\mathfrak{A}\subset\mathfrak{G}_{1}\), according to the construction of the set \(\mathfrak{G}_{2}\), one can see that \(f(\mathfrak{A})\) is uniformly Cauchy. If the set \(\{f(x);x\in\mathfrak{A}\}\) is relatively compact in X, then, by Theorem 3.3, the relative compactness of \(f(\mathfrak{A})\) is decided by the uniformly equi-regulatedness of \(f(\mathfrak{A})\). However, for all \(t\in(b,+\infty)_{\mathbb{T}}\), obviously, \(f(\mathfrak{A})\) is uniformly equi-regulated. Thus the uniformly equi-regulatedness of \(f(\mathfrak{A})\) on \([a,+\infty)_{\mathbb{T}}\) is actually decided by the equi-regulatedness of \(\mathfrak{A}\) on \([a,b]_{\mathbb{T}}\). Therefore, if \(\mathfrak{A}\) is equi-regulated on \([a,b]_{\mathbb{T}}\), then \(f(\mathfrak{A})\subset\mathfrak{G}_{2}\) is relatively compact, i.e., \(\mathfrak{A}\) is relatively compact in \(\mathfrak{G}_{1}\).
From Theorem 3.4 and Remark 3.4, we can obtain the following corollaries.
Corollary 3.1
Let
\(X=\mathbb{R}^{n}\). A set
\(\mathfrak{A}\subset\mathfrak{G}_{1}\)
is relatively compact if and only if it is equi-regulated and for every
\(t\in[a,b]_{\mathbb{T}}\), the set
\(\{x(t);x\in\mathfrak{A}\}\)
is bounded in
\(\mathbb{R}^{n}\).
Proof
If \(\mathfrak{A}\subset\mathfrak{G}_{1}\) is relatively compact, then \(\mathfrak{A}\) is totally bounded in \(\mathbb{R}^{n}\), and then for every \(t\in[a,b]_{\mathbb{T}}\), \(\{x(t);x\in\mathfrak{A}\}\) is bounded in \(\mathbb{R}^{n}\). Moreover, by Remark 3.4, there exists a homeomorphic mapping \(f: \mathfrak{G}_{1}\rightarrow\mathfrak{G}_{2}\) such that \(f(\mathfrak{A})\subset\mathfrak{G}_{2}\) is relatively compact, which means that \(f(\mathfrak{A})\subset\mathfrak{G}_{2}\) is equi-regulated on \([a,b]_{\mathbb{T}}\) according to Theorem 3.4. Since f is continuous and a one-to-one mapping, \(\mathfrak{A}\) is equi-regulated on \([a,b]_{\mathbb{T}}\).
If \(\mathfrak{A}\subset\mathfrak{G}_{1}\) is equi-regulated and for every \(t\in[a,b]_{\mathbb{T}}\), the set \(\{x(t);x\in\mathfrak{A}\}\) is bounded in \(\mathbb{R}^{n}\), then \(f(\mathfrak{A})\subset\mathfrak{G}_{2}\) is equi-regulated and for every \(t\in[a,+\infty)_{\mathbb{T}}\), the set \(\{f(x); f(x)\in f(\mathfrak{A})\}\) is bounded in \(\mathbb{R}^{n}\), according to Theorem 3.3, \(f(\mathfrak{A})\) is relatively compact in \(\mathfrak{G}_{2}\), i.e., \(\mathfrak{A}\) is relatively compact in \(\mathfrak{G}_{1}\). This completes the proof. □
Corollary 3.2
Let
\(\mathfrak{A}\subset\mathfrak{G}_{1}\)
be equi-regulated, and for every
\(t\in[a,b]_{\mathbb{T}}\), let the set
\(\{x(t);x\in\mathfrak{A}\}\)
be relatively compact in
X. Then the set
\(\mathfrak{A}\)
is relatively compact in
\(\mathfrak{G}_{1} ([a,b]_{\mathbb{T}},X )\).
Proof
From the assumption of this corollary and Remark 3.4, by Theorem 3.3, we can see that the set \(f(\mathfrak{A})\) is relatively compact in \(\mathfrak{G}_{2} ([a,+\infty)_{\mathbb{T}},X )\), i.e., the set \(\mathfrak{A}\) is relatively compact in \(\mathfrak{G}_{1} ([a,b]_{\mathbb{T}},X )\). This completes the proof. □
Remark 3.5
In fact, if we let \(\mathbb{T}=\mathbb{R}\), Corollaries 3.1 and 3.2 can include Corollary 2.4 from [22] and Theorem 1.13 from [23], respectively.
Similarly, for \(\bar{T}_{0}\in\mathbb{T}\) and a Banach space \((X, \Vert \cdot \Vert )\), let
$$\begin{aligned} \mathfrak{G} \bigl((-\infty,\bar{T}_{0} ]_{\mathbb{T}}, X \bigr) :=& \Bigl\{ x:(-\infty,\bar{T}_{0}]_{\mathbb{T}}\rightarrow X; \\ &{}\lim_{s\rightarrow t^{+}}x(s)=x\bigl(t^{+}\bigr)\mbox{ and } \lim_{s\rightarrow t^{-}}x(s)=x\bigl(t^{-}\bigr)\mbox{ exist and are finite}, \\ &{}s,t< +\infty\mbox{ and } \sup_{t\in(-\infty,\bar{T}_{0}]_{\mathbb{T}}}\bigl\Vert x(t)\bigr\Vert < +\infty \Bigr\} . \end{aligned}$$
From Definition 3.1, we can also introduce a \(\delta^{-\infty}\) and extend the Δ-gauge for \([a,b]_{\mathbb{T}}\) to \((-\infty,\bar{T_{0}}]_{\mathbb{T}}\), and then we can repeat the same above discussion, and the following theorems can also be obtained (we omit the proofs).
Theorem 3.5
Assume that a set
\(\mathfrak{A}\subset\mathfrak{G} ((-\infty,\bar{T}_{0} ]_{\mathbb{T}}, X )\)
is uniformly equi-regulated and uniformly Cauchy, and for any
\(t\in(-\infty,\bar{T}_{0}]_{\mathbb{T}}\), there is a number
\(\beta_{t}\)
such that, for
\(x\in\mathfrak{A}\),
$$ \bigl\Vert x(t)-x\bigl(t^{-}\bigr)\bigr\Vert \leq \beta_{t},\qquad \bigl\Vert x\bigl(t^{+}\bigr)-x(t)\bigr\Vert \leq\beta_{t},\quad t\in(-\infty,\bar{T}_{0}]_{\mathbb{T}}. $$
(3.19)
Then there is a constant
\(K>0\)
such that
\(\Vert x(t)-x(\bar{T}_{0})\Vert \leq K\), for every
\(x\in\mathfrak{A}\)
and
\(t\in(-\infty,\bar{T}_{0}]_{\mathbb{T}}\).
Theorem 3.6
Let
\(\mathfrak{A}\subset\mathfrak{G} ((-\infty,\bar{T}_{0} ]_{\mathbb{T}}, X )\)
be uniformly equi-regulated and uniformly Cauchy, for every
\(t\in(-\infty,\bar{T}_{0}]_{\mathbb{T}}\), let the set
\(\{x(t);x\in\mathfrak{A}\}\)
be relatively compact in
X. Then the set
\(\mathfrak{A}\)
is relatively compact in
\(\mathfrak{G} ((-\infty,\bar{T}_{0} ]_{\mathbb{T}}, X )\).
Theorem 3.7
Let a set
\(\mathfrak{A}\subset\mathfrak{G} ((-\infty,\bar{T}_{0} ]_{\mathbb{T}},\mathbb{R}^{n} )\). If
\(\mathfrak{A}\)
is relatively compact in the sup-norm topology, then it is uniformly equi-regulated. If
\(\mathfrak{A}\)
is uniformly equi-regulated and uniformly Cauchy, satisfying (3.19), then
\(\mathfrak{A}\)
is relatively compact in
\(\mathfrak{G} ((-\infty,\bar{T}_{0} ]_{\mathbb{T}},\mathbb {R}^{n} )\).
For the more general case, for a Banach space \((X, \Vert \cdot \Vert )\), let
$$\begin{aligned} \mathfrak{G} \bigl((-\infty,+\infty)_{\mathbb{T}}, X \bigr) :=& \Bigl\{ x:(- \infty,+\infty)_{\mathbb{T}}\rightarrow X; \\ &{}\lim_{s\rightarrow t^{+}}x(s)=x\bigl(t^{+}\bigr)\mbox{ and } \lim_{s\rightarrow t^{-}}x(s)=x\bigl(t^{-}\bigr)\mbox{ exist and are finite}, \\ &{}-\infty< s,t< +\infty\mbox{ and } \sup_{t\in(-\infty,+\infty)_{\mathbb{T}}}\bigl\Vert x(t)\bigr\Vert < +\infty \Bigr\} . \end{aligned}$$
From Definition 3.1, we can also introduce a \(\delta^{\pm\infty}\) and extend the Δ-gauge for \([a,b]_{\mathbb{T}}\) to \((-\infty,+\infty)_{\mathbb{T}}\), since for any \(T_{0}>\bar{T}_{0}\), \((-\infty,+\infty)_{\mathbb{T}}=(-\infty,\bar{T}_{0}]_{\mathbb {T}}\cup[\bar{T}_{0},T_{0}]_{\mathbb{T}}\cup[T_{0},+\infty )_{\mathbb{T}}\), and then we can repeat the above discussion, and the above similar theorems can also be obtained (we omit these similar statements here).
Let
$$\mathfrak{G}_{0}[T_{0},+\infty)_{\mathbb{T}}:= \Bigl\{ x; x\in PC_{rd} \bigl([T_{0},+\infty)_{\mathbb{T}}, \mathbb{R}^{n} \bigr) \mbox{ and } \sup_{t\in[T_{0},+\infty)_{\mathbb{T}}}\bigl\vert x(t)\bigr\vert < +\infty \Bigr\} , $$
where \(PC_{rd} ([T_{0},+\infty)_{\mathbb{T}},\mathbb{R}^{n} )\) is the set formed by all rd-piecewise continuous functions (one can consult Definition 2.4 from [24]). Endow \(\mathfrak{G}_{0}\) with the norm \(\Vert x\Vert =\sup_{t\in[T_{0},+\infty)_{\mathbb{T}}}\vert x(t)\vert \), and note \((\mathfrak{G}_{0},\Vert \cdot \Vert )\) is a Banach space.
Next, we will establish some theorems to guarantee that \(\mathfrak{A}\subset\mathfrak{G}_{0} ([T_{0},+\infty)_{\mathbb {T}},\mathbb{R}^{n} )\) is uniformly Cauchy and uniformly equi-regulated.
Now, we give a new definition called ‘equi-absolutely continuity’ for the function set \(\mathfrak{A}\subset\mathfrak{G}_{0}\), which will be used in the following lemma’s proof.
Definition 3.6
Let \(\mathfrak{A}\subset\mathfrak{G}_{0}\). We say \(\mathfrak{A}\) is equi-absolutely continuous if for any \(f\in\mathfrak{A}\) and \(\varepsilon>0\), there exists \(\delta>0\), such that for any finite mutually disjoint open interval \((x_{i},y_{i})_{\mathbb{T}}\subset[T_{0},+\infty)_{\mathbb{T}} \) (\(i=1,2,\ldots,n\))
$$\sum_{i=1}^{n}(y_{i}-x_{i})< \delta $$
implies
$$\sum_{i=1}^{n} \bigl\vert f(x_{i})-f(y_{i}) \bigr\vert < \varepsilon. $$
Lemma 3.2
Let
\(\mathfrak{A}\subset\mathfrak{G}_{0} ([T_{0},+\infty)_{\mathbb {T}},\mathbb{R}^{n} )\)
be uniformly bounded and
$$\mathscr{Z}= \bigl\{ t\in[T_{0},+\infty)_{\mathbb{T}}: x\textit{ is not }\Delta\textit{-differentiable at }t \bigr\} $$
and
\(\mu_{\Delta}(\mathscr{Z})=0\)
for all
\(x\in\mathfrak{A}\), i.e., x
is Δ-differentiable at
\([T_{0},+\infty)_{\mathbb{T}}\backslash\mathscr{Z}\)
and there exists
\(M>0\)
such that
\(\vert x^{\Delta}(t)\vert \leq M\)
for all
\(x\in\mathfrak{A}\). Then
\(\mathfrak{A}\)
is uniformly equi-regulated, and for any
\(\varepsilon>0\), there exists
\(T_{1}>0\)
such that, for any
\(x\in\mathfrak{A}\), the following is fulfilled:
$$\biggl\vert \int_{[T_{0},+\infty)_{\mathbb{T}}\backslash\mathscr {Z}}x^{\Delta}(s)\Delta s \biggr\vert \leq \varepsilon. $$
Proof
From the condition of the theorem, since x is Δ-differentiable at \([T_{0},+\infty)_{\mathbb{T}}\backslash\mathscr{Z}\) and there exists \(M>0\) such that \(\vert x^{\Delta}(t)\vert \leq M\), according to Corollary 1.68 from [1], we can obtain for all \(t\in[T_{0},+\infty)_{\mathbb{T}}\backslash\mathscr{Z}\), \(x(t)\) satisfies the Lipschitz condition
$$\bigl\vert x(t_{1})-x(t_{2}) \bigr\vert \leq M\vert t_{1}-t_{2}\vert , \quad \forall t_{1},t_{2} \in[T_{0},+\infty)_{\mathbb{T}}\backslash\mathscr{Z}. $$
So for each \(N_{0}\in\mathbb{Z}^{+}\), we can obtain
$$ \sum_{j=1}^{N_{0}}\bigl\vert t_{j}^{1}-t_{j}^{2}\bigr\vert < \frac{\varepsilon }{M}, \quad\bigl(t_{j}^{1},t_{j}^{2} \bigr)_{\mathbb{T}}\subset[T_{0},+\infty )_{\mathbb{T}}, $$
(3.20)
which implies
$$ \sum_{j=1}^{N_{0}}\bigl\vert x\bigl(t_{j}^{1}\bigr)-x\bigl(t_{j}^{2} \bigr)\bigr\vert < M\sum_{j=1}^{N_{0}}\bigl\vert t_{j}^{1}-t_{j}^{2}\bigr\vert < \varepsilon, $$
(3.21)
i.e., \(\mathfrak{A}\) is equi-absolutely continuous on \([T_{0},+\infty)_{\mathbb{T}}\backslash\mathscr{Z}\). Hence, from (3.20) and (3.21), for all \(t\in[T_{0},+\infty)_{\mathbb{T}}\backslash\mathscr{Z}\), \(\mathfrak{A}\) is uniformly equi-regulated. For all \(t_{0}\in\mathscr{Z}\), since \(\mu_{\Delta}(\mathscr{Z})=0\), from (3.20) and (3.21), we can take \(\delta^{+\infty}= (\delta^{+\infty}_{L},\delta^{+\infty }_{R} )\) and \(\delta_{L},\delta_{R}<\frac{\varepsilon}{M}\), so
$$\bigl\vert x\bigl(t_{0}^{+}\bigr)-x(t)\bigr\vert < \sum_{j=1}^{N_{0}}\bigl\vert x \bigl(t_{j}^{1}\bigr)-x\bigl(t_{j}^{2} \bigr)\bigr\vert < \varepsilon \quad\mbox{if } t\in \bigl(t_{0},t_{0}+ \delta_{R}^{+\infty}(t_{0}) \bigr)_{\mathbb{T}} $$
and
$$\bigl\vert x\bigl(t_{0}^{-}\bigr)-x(t)\bigr\vert < \sum_{j=1}^{N_{0}}\bigl\vert x \bigl(t_{j}^{1}\bigr)-x\bigl(t_{j}^{2} \bigr)\bigr\vert < \varepsilon \quad \mbox{if } t\in \bigl(t_{0}- \delta^{+\infty}_{L}(t_{0}),t_{0} \bigr)_{\mathbb{T}}, $$
and thus \(\mathfrak{A}\) is uniformly equi-regulated on \([T_{0},+\infty)_{\mathbb{T}}\). According to Theorem 3.1, for any closed interval \([a_{\mathcal{P}_{1}},b_{\mathcal {P}_{1}}]_{\mathbb{T}}\), there is a \(\delta_{1}^{+\infty}\)-fine partition \(\mathcal{P}_{1}\):
$$a_{\mathcal{P}_{1}}:=T_{0}=t_{0}^{\mathcal{P}_{1}}< t_{1}^{\mathcal {P}_{1}}< t_{2}^{\mathcal{P}_{1}}< \cdots< t_{N}^{\mathcal {P}_{1}}:=b_{\mathcal{P}_{1}} $$
such that
$$ \bigl\vert x\bigl(t^{\prime}\bigr)-x\bigl(t^{\prime\prime}\bigr)\bigr\vert \leq\frac{1}{2}, $$
for every \(x\in\mathfrak{A}\) and \([t^{\prime},t^{\prime\prime}]_{\mathbb{T}}\subset(t_{j-1}^{\mathcal {P}_{1}},t_{j}^{\mathcal{P}_{1}})_{\mathbb{T}}, j=1,2,\ldots,N\). Hence, we have
$$\begin{aligned} \bigl\vert x\bigl(t_{j}^{{\mathcal{P}_{1}}^{-}}\bigr)-x \bigl(t_{j-1}^{{\mathcal {P}_{1}}^{+}}\bigr) \bigr\vert \leq& \bigl\vert x \bigl(t_{j}^{{\mathcal {P}_{1}}^{-}}\bigr)-x\bigl(t^{\prime}\bigr) \bigr\vert +\bigl\vert x\bigl(t^{\prime}\bigr)-x\bigl(t^{\prime\prime}\bigr) \bigr\vert + \bigl\vert x\bigl(t^{\prime\prime}\bigr)-x\bigl(t_{j-1}^{{\mathcal {P}_{1}}^{+}} \bigr) \bigr\vert \leq\frac{3}{2}, \end{aligned}$$
where \(t^{\prime}\in (t_{j}^{{\mathcal {P}_{1}}}-\delta_{L}^{+\infty}(t_{j}^{{\mathcal {P}_{1}}}),t_{j}^{{\mathcal {P}_{1}}} )_{\mathbb{T}}, t^{\prime\prime}\in (t_{j-1}^{{\mathcal {P}_{1}}},t_{j-1}^{{\mathcal {P}_{1}}}+\delta_{R}^{+\infty}(t_{j-1}^{{\mathcal {P}_{1}}}) )_{\mathbb{T}}\).
Similarly, for any closed interval \([a_{\mathcal{P}_{2}},b_{\mathcal {P}_{2}}]_{\mathbb{T}}\), there is a \(\delta_{2}^{+\infty}\)-fine partition \(\mathcal{P}_{2}\):
$$a_{\mathcal{P}_{2}}:=b_{\mathcal{P}_{1}}=t_{0}^{\mathcal {P}_{2}}< t_{1}^{\mathcal{P}_{2}}< t_{2}^{\mathcal {P}_{2}}< \cdots< t_{N}^{\mathcal{P}_{2}}:=b_{\mathcal{P}_{2}} $$
such that
$$ \bigl\vert x\bigl(t^{\prime}\bigr)-x\bigl(t^{\prime\prime}\bigr)\bigr\vert \leq\frac{1}{2^{2}}, $$
for every \(x\in\mathfrak{A}\) and \([t^{\prime},t^{\prime\prime}]_{\mathbb{T}}\subset(t_{j-1}^{\mathcal {P}_{2}},t_{j}^{\mathcal{P}_{2}})_{\mathbb{T}}, j=1,2,\ldots,N\). Hence, we have
$$\begin{aligned} \bigl\vert x\bigl(t_{j}^{{\mathcal{P}_{2}}^{-}}\bigr)-x \bigl(t_{j-1}^{{\mathcal {P}_{2}}^{+}}\bigr) \bigr\vert \leq& \bigl\vert x \bigl(t_{j}^{{\mathcal {P}_{2}}^{-}}\bigr)-x\bigl(t^{\prime}\bigr) \bigr\vert +\bigl\vert x\bigl(t^{\prime}\bigr)-x\bigl(t^{\prime\prime}\bigr) \bigr\vert + \bigl\vert x\bigl(t^{\prime\prime}\bigr)-x\bigl(t_{j-1}^{{\mathcal {P}_{2}}^{+}} \bigr) \bigr\vert \leq\frac{3}{2^{2}}, \end{aligned}$$
where \(t^{\prime}\in (t_{j}^{{\mathcal {P}_{2}}}-\delta_{L}^{+\infty}(t_{j}^{{\mathcal {P}_{2}}}),t_{j}^{{\mathcal {P}_{2}}} )_{\mathbb{T}}, t^{\prime\prime}\in (t_{j-1}^{{\mathcal {P}_{2}}},t_{j-1}^{{\mathcal {P}_{2}}}+\delta_{R}^{+\infty}(t_{j-1}^{{\mathcal {P}_{2}}}) )_{\mathbb{T}}\).
We can repeat the above process, then for each \(i=2,3,\ldots\) , we can see that, for any closed interval \([a_{\mathcal{P}_{i}},b_{\mathcal {P}_{i}}]_{\mathbb{T}}\), there is a \(\delta_{i}^{+\infty}\)-fine partition \(\mathcal{P}_{i}\):
$$a_{\mathcal{P}_{i}}:=b_{\mathcal{P}_{i-1}}=t_{0}^{\mathcal {P}_{i}}< t_{1}^{\mathcal{P}_{i}}< t_{2}^{\mathcal {P}_{i}}< \cdots< t_{N}^{\mathcal{P}_{i}}:=b_{\mathcal{P}_{i}} $$
such that
$$ \bigl\vert x\bigl(t^{\prime}\bigr)-x\bigl(t^{\prime\prime} \bigr)\bigr\vert \leq\frac{1}{2^{i}}, $$
(3.22)
for every \(x\in\mathfrak{A}\) and \([t^{\prime},t^{\prime\prime}]_{\mathbb{T}}\subset(t_{j-1}^{\mathcal {P}_{i}},t_{j}^{\mathcal{P}_{i}})_{\mathbb{T}}\), \(j=1,2,\ldots,N\). Hence, we have
$$\begin{aligned} \bigl\vert x\bigl(t_{j}^{{\mathcal{P}_{i}}^{-}}\bigr)-x \bigl(t_{j-1}^{{\mathcal {P}_{i}}^{+}}\bigr) \bigr\vert \leq& \bigl\vert x \bigl(t_{j}^{{\mathcal {P}_{i}}^{-}}\bigr)-x\bigl(t^{\prime}\bigr) \bigr\vert +\bigl\vert x\bigl(t^{\prime}\bigr)-x\bigl(t^{\prime\prime}\bigr) \bigr\vert + \bigl\vert x\bigl(t^{\prime\prime}\bigr)-x\bigl(t_{j-1}^{{\mathcal {P}_{i}}^{+}} \bigr) \bigr\vert \leq\frac{3}{2^{i}}, \end{aligned}$$
where \(t^{\prime}\in (t_{j}^{{\mathcal {P}_{i}}}-\delta_{L}^{+\infty}(t_{j}^{{\mathcal {P}_{i}}}),t_{j}^{{\mathcal {P}_{i}}} )_{\mathbb{T}}, t^{\prime\prime}\in (t_{j-1}^{{\mathcal {P}_{i}}},t_{j-1}^{{\mathcal {P}_{i}}}+\delta_{R}^{+\infty}(t_{j-1}^{{\mathcal {P}_{i}}}) )_{\mathbb{T}}\). Thus, from (3.22), we obtain
$$ \sum_{i=1}^{+\infty}\sum _{j=1}^{N} \bigl\vert x\bigl(t_{j}^{{\mathcal {P}_{i}}^{-}} \bigr)-x\bigl(t_{j-1}^{{\mathcal {P}_{i}}^{+}}\bigr) \bigr\vert \leq3N\sum _{i=1}^{+\infty}\frac {1}{2^{i}}< +\infty, $$
which implies that, for any \(\varepsilon>0\), there exists \(i_{0}>0\) such that
$$ \sum_{i=i_{0}}^{+\infty}\sum _{j=1}^{N} \bigl\vert x\bigl(t_{j}^{{\mathcal {P}_{i}}^{-}} \bigr)-x\bigl(t_{j-1}^{{\mathcal{P}_{i}}^{+}}\bigr) \bigr\vert \leq\varepsilon, $$
which implies that, for any \(\varepsilon>0\), there exists \(T_{1}\geq t_{0}^{\mathcal{P}_{i_{0}}}\) and a partition
$$\mathcal{P}:=\bigcup_{i=i_{0}}^{+\infty} \mathcal{P}_{i}, $$
i.e., \(t_{0}^{\mathcal{P}_{i_{0}}}=t_{0}^{\mathcal {P}}< t_{1}^{\mathcal{P}_{i_{0}}}=t_{1}^{\mathcal {P}}< t_{2}^{\mathcal{P}_{i_{0}}}=t_{2}^{\mathcal {P}}<\cdots<t_{N}^{\mathcal{P}_{i_{0}}}=t_{N}^{\mathcal {P}}<t_{0}^{\mathcal{P}_{i_{0}+1}}=t_{N+1}^{\mathcal {P}}<t_{1}^{\mathcal{P}_{i_{0}+1}}=t_{N+2}^{\mathcal {P}}<\cdots<\cdots\) such that
$$\sum_{j=1}^{+\infty}\bigl\vert x \bigl(t_{j}^{\mathcal{P^{-}}}\bigr)-x\bigl(t_{j-1}^{\mathcal {P}^{+}} \bigr)\bigr\vert =\sum_{i=i_{0}}^{+\infty}\sum _{j=1}^{N} \bigl\vert x \bigl(t_{j}^{{\mathcal {P}_{i}}^{-}}\bigr)-x\bigl(t_{j-1}^{{\mathcal{P}_{i}}^{+}} \bigr) \bigr\vert \leq\varepsilon. $$
Moreover, since x is Δ-differentiable almost everywhere on \([T_{0},+\infty)_{\mathbb{T}}\), we obtain
$$\begin{aligned} \biggl\vert \int_{[T_{1},+\infty)_{\mathbb{T}}\backslash\mathscr {Z}}x^{\Delta}(s)\Delta s \biggr\vert =&\sum _{j=1}^{+\infty} \biggl\vert \int_{(t_{j-1}^{\mathcal {P}+},t_{j}^{\mathcal {P}-})\backslash\mathscr{Z}}x^{\Delta}(s)\Delta s \biggr\vert \\ =&\sum_{j=1}^{+\infty}\bigl\vert x \bigl(t_{j}^{\mathcal {P^{-}}}\bigr)-x\bigl(t_{j-1}^{\mathcal{P}^{+}} \bigr)\bigr\vert \leq\varepsilon. \end{aligned}$$
This completes the proof. □
Theorem 3.8
Let
\(\mathfrak{A}\subset\mathfrak{G}_{0} ([T_{0},+\infty)_{\mathbb {T}},\mathbb{R}^{n} )\)
be uniformly bounded and
$$\mathscr{Z}= \bigl\{ t\in[T_{0},+\infty)_{\mathbb{T}}: x\textit{ is not }\Delta\textit{-differentiable at }t \bigr\} $$
and
\(\mu_{\Delta}(\mathscr{Z})=0\)
for all
\(x\in\mathfrak{A}\), i.e., x
is Δ-differentiable at
\([T_{0},+\infty)_{\mathbb{T}}\backslash\mathscr{Z}\)
and there exists
\(M>0\)
such that
\(\vert x^{\Delta}(t)\vert \leq M\)
for all
\(x\in\mathfrak{A}\). Then there exists
\(T_{1}>0\)
such that
\(\mathfrak{A}\)
is uniformly Cauchy on
\([T_{1},+\infty)_{\mathbb{T}}\).
Proof
According to Lemma 3.2, for any \(\varepsilon>0\), there exists \(T_{1}>0\) such that, for any \(x\in\mathfrak{A}\), the following is fulfilled:
$$\biggl\vert \int_{[T_{1},+\infty)_{\mathbb{T}}\backslash\mathscr {Z}}x^{\Delta}(s)\Delta s \biggr\vert \leq \varepsilon. $$
Hence, for any \(t_{1},t_{2}\notin\mathscr{Z}, t_{1},t_{2}>T_{1}\), we obtain
$$\bigl\vert x(t_{1})-x(t_{2})\bigr\vert = \biggl\vert \int_{t_{1}}^{t_{2}}x^{\Delta}(s)\Delta s \biggr\vert < \biggl\vert \int_{[T_{1},+\infty)_{\mathbb{T}}\backslash \mathscr{Z}}x^{\Delta}(s)\Delta s \biggr\vert \leq \varepsilon. $$
This completes the proof. □
In the following, we will give the following useful corollaries.
Corollary 3.3
Let
\(\mathfrak{A}\subset\mathfrak{G}_{0} ([T_{0},+\infty)_{\mathbb {T}},\mathbb{R}^{n} )\)
be uniformly bounded and
$$\mathscr{Z}= \bigl\{ t\in[T_{0},+\infty)_{\mathbb{T}}: x\textit{ is not }\Delta\textit{-differentiable at }t \bigr\} $$
and
\(\mu_{\Delta}(\mathscr{Z})=0\)
for all
\(x\in\mathfrak{A}\), and there exists
\(M>0\)
such that
\(\vert x^{\Delta}(t)\vert \leq M\)
for all
\(x\in\mathfrak{A}\). Then
\(\mathcal{A}\)
is relatively compact in
\(\mathfrak{G}_{0}\).
Proof
According to Theorem 3.8, \(\mathfrak{A}\) is uniformly equi-regulated, uniformly Cauchy. Further, since \(\mathfrak{A}\) is uniformly bounded, so it satisfies (3.6), by Theorem 3.4, we get the desired result immediately. □
Let
$$\mathcal{B}\mathcal{C}[T_{0},+\infty)_{\mathbb{T}}:= \Bigl\{ x\in BC\bigl([T_{0},+\infty)_{\mathbb{T}},\mathbb{R}^{n}\bigr) \mbox{ and } \sup_{t\in[T_{0},+\infty)_{\mathbb{T}}}\bigl\vert x(t)\bigr\vert < \infty \Bigr\} , $$
where \(BC([T_{0},+\infty)_{\mathbb{T}},\mathbb{R}^{n})\) denotes the set of all bounded continuous functions on \([T_{0}, +\infty)_{\mathbb{T}}\). Then we can obtain the following corollary.
Corollary 3.4
Let
\(\mathfrak{A}\subset\mathcal{B}\mathcal {C}[T_{0},+\infty)_{\mathbb{T}}\)
be uniformly bounded and for all
\(x\in\mathfrak{A}\), x
is Δ-differentiable and there exists
\(M>0\)
such that
\(\vert x^{\Delta}(t)\vert < M\). Then
\(\mathfrak{A}\)
is relatively compact in
\(\mathcal{B}\mathcal{C}\).
Proof
Since x is Δ-differentiable on \([T_{0},+\infty)_{\mathbb{T}}\), \(\mathfrak{A}\) is equi-absolutely continuous on \([T_{0},+\infty)_{\mathbb{T}}\), for any \(t_{1},t_{2}\in[T_{0},+\infty)_{\mathbb{T}}\), we can obtain
$$\bigl\vert x(t_{1})-x(t_{2})\bigr\vert = \biggl\vert \int_{t_{1}}^{t_{2}}x^{\Delta}(s)\Delta s \biggr\vert \leq2\sup_{t\in[T_{0},+\infty)_{\mathbb{T}}}\bigl\vert x(t)\bigr\vert < 2M_{0}, $$
which means that
$$\biggl\vert \int_{t_{1}}^{+\infty}x^{\Delta}(s)\Delta s \biggr\vert < 2M_{0}, \quad M_{0}\mbox{ is some constant}. $$
Thus, for any \(\varepsilon>0\), there exists \(T_{1}>0\), and we have \(t_{2}^{\prime}>t_{1}^{\prime}>T_{1}\) implies
$$\bigl\vert x\bigl(t_{1}^{\prime}\bigr)-x\bigl(t_{2}^{\prime} \bigr)\bigr\vert = \biggl\vert \int _{t_{1}^{\prime}}^{t_{2}^{\prime}}x^{\Delta}(s)\Delta s \biggr\vert < \varepsilon, $$
i.e., \(\mathfrak{A}\) is uniformly Cauchy. According to Theorem 3.4, \(\mathfrak{A}\) is relatively dense in \(\mathcal {B}\mathcal{C}\). This completes the proof. □
Remark 3.6
Note that if for all \(x\in\mathfrak{A}\subset\mathcal{B}\mathcal {C}\), x has uniformly bounded Δ-derivatives, then one can see that \(\mathfrak{A}\) is equi-absolutely continuous, which will lead to that \(\mathfrak{A}\) is uniformly Cauchy. Hence, the uniformly boundedness of \(\mathfrak{A}\) and the uniformly boundedness of Δ-derivatives functions of \(\mathfrak{A}\) can guarantee \(\mathfrak{A}\) is relatively compact.