Proof. Assume that the assertion is not true, then there exists $t_{0} \in \mathbb {R},\,\epsilon >0$ such that $g(t_{0})\notin \overline {\{f(t):t \in \mathbb {R}\}}$. Without loss of generality, let $t_{0}=0$, then there exists $\epsilon >0$, such that

\[ \|g(0)-f(t)\|\geq 2\epsilon,\text{ for all $t \in \mathbb{R}$}. \]

It follows that

\[ \|h(t)\|=\|f(t)-g(t)\|\geq\|f(t)-g(0)\|-\|g(0)-g(t)\|\geq\epsilon, \]

for all $t \in C_{\epsilon }$, where $C_{\epsilon }$ is given in Lemma 2.3. Hence,

\[ \|h(t-s_{i})\|\geq\epsilon, \]

for each $i \in {1,\,\ldots,\,m}$ and $t \in s_{i}+C_{\epsilon }.$ Now, we define the function $H(t)$ by

\[ H(t)=\sum^{m}_{i=1}\|h(t-s_{i})\|. \]

From the above inequality, we can see that

\[ H(t)\geq\epsilon, \]

for all $t \in \mathbb {R}$. On the other hand, by the translation invariance of $\mathscr {E}(\mathbb {R},\,X)$, we conclude that $h(t-s_{i}) \in \mathscr {E}(\mathbb {R},\,X)$ for all $i \in {1,\,...,\,m}$, and thus $H \in \mathscr {E}(\mathbb {R},\,X)$, which is a contradiction to $H(t)\geq \epsilon$. So, the conclusion is true.

Proof. Assume that $f=g_{1}+h_{1}=g_{2}+h_{2}$ with $g_{i} \in BP_{p,k}(\mathbb {R},\,X)$ and $h_{i} \in \mathscr {E}(\mathbb {R},\,X)$ for $i=1,\,2$. Then we have $0=(g_{1}-g_{2})+(h_{1}-h_{2}) \in PBP_{p,k}(\mathbb {R},\,X)$ with $g_{1}-g_{2} \in BP_{p,k}(\mathbb {R},\,X)$ and $h_{1}-h_{2} \in \mathscr {E}(\mathbb {R},\,X)$. From Lemma 3.1, we obtain $(g_{1}-g_{2})(\mathbb {R})\subseteq {0}$. Hence, we have $g_{1}=g_{2}$ and $h_{1}=h_{2}$.

Proof. Note that $BP_{p,k}(\mathbb {R},\,X)$ and $\mathscr {E}(\mathbb {R},\,X)$ are Banach space equipped with the sup-norm, they are closed to addition and multiplication. Hence, it is obvious that $f_{1}+f_{2}$ and $cf$, for any $c \in \mathbb {R}$ are also in $PBP_{p,k}(\mathbb {R},\,X)$. In view of Lemma 2.1(2) and Lemma 2.5(ii), we can easily deduce that $f_{a}(t):=f(t+a)$ also belongs to $PBP_{p,k}(\mathbb {R},\,X)$ for any $a \in \mathbb {R}$.

Proof. Let $\{f_{n}\}$ be a Cauchy sequence in $PBP_{p,k}(\mathbb {R},\,X)$. We can write $f_{n}=g_{n}+h_{n}$ with $g_{n} \in BP_{p,k}(\mathbb {R},\,X),\, h_{n} \in \mathscr {E}(\mathbb {R},\,X)$. From Lemma 3.1, $\|g_{n}-g_{m}\|_{\infty }\leq \|f_{n}-f_{m}\|_{\infty }$, therefore $\{g_{n}\}$ is a Cauchy sequence in the Banach space $(BP_{p,k}(\mathbb {R},\,X),\,\|\cdot \|_{\infty })$. So, $\{h_{n}\}=\{f_{n}-g_{n}\}$ is also a Cauchy sequence in the Banach space $(\mathscr {E}(\mathbb {R},\,X),\,\|\cdot \|_{\infty })$. Then we can conclude that $\displaystyle \lim _{n\rightarrow \infty }g_{n}=g \in BP_{p,k}(\mathbb {R},\,X)$ and $\displaystyle \lim _{n\rightarrow \infty }h_{n}=h \in \mathscr {E}(R,\,X)$, and finally, we have $\displaystyle \lim _{n\rightarrow \infty }f_{n}=g+h \in PBP_{p,k}(\mathbb {R},\,X)$. The above arguments imply that the space $PBP_{p,k}(\mathbb {R},\,X)$ is a closed subspace of $BC(\mathbb {R},\,X)$, and thus $PBP_{p,k}(\mathbb {R},\,X)$ is also a Banach space equipped with the sup-norm.

Proof. The proof is similar to that of Lemma 3.1, and by the translation invariance of $\mathscr {E}(\mathbb {R},\,X,\,\rho )$ owing to the hypothesis H$_{\rho }$, we can conclude that the assertion is true.

Proof. Taking into account Lemma 3.4, the conclusion can be similarly conducted as that of Proposition 3.1.

Proof. Considering $BP_{p,k}(\mathbb {R},\,X)$ and $\mathscr {E}(\mathbb {R},\,X,\,\rho )$ are Banach spaces equipped with the sup-norm, and closed to addition and multiplication, thus $f_{1}+f_{2}$ and $cf$ are also in $WPBP_{p,k}(\mathbb {R},\,X,\,\rho )$ for any $c \in \mathbb {R}$. For $\rho \in \mathcal {U}_{inv}$, the space $\mathscr {E}(\mathbb {R},\,X,\,\rho )$ is translation invariant by Lemma 2.4(b), and thus $f_{a}(t):=f(t+a)\in WPBP_{p,k}(\mathbb {R},\,X)$ by Lemma 2.1(2).

Proof. The proof can be conducted similarly to that of Lemma 3.3 by taking into account Lemma 3.4 and Proposition 3.2.

Proof. Let $x(t):=\alpha (t)+\beta (t)\in WPBP_{p,k}(\mathbb {R},\,X)$ with $\alpha (t) \in BP_{p,k}(\mathbb {R},\,X)$ and $\beta (t) \in \mathscr {E}(\mathbb {R},\,X,\,\rho )$. The function $f$ can be re-written as

\begin{align*} f(t,x(t))& =g(t,\alpha(t))+f(t,x(t))-g(t,\alpha(t))\\ & =g(t,\alpha(t))+f(t,x(t))-f(t,\alpha(t))+h(t,\alpha(t)). \end{align*}

Define

\[ G(t):=f(t,x(t))-f(t,\alpha(t)),~H(t):=h(t,\alpha(t)). \]

According to (A1) and Lemma 2.2, we have $g(t,\,\alpha (t)) \in BP_{p,k}(\mathbb {R},\,X)$. Now to complete the proof, it is enough to prove that $G(t) \in \mathscr {E}(\mathbb {R},\,X,\,\rho )$ and $H(t) \in \mathscr {E}(\mathbb {R},\,X,\,\rho )$.

We firstly show that $G\in \mathscr {E}(\mathbb {R},\,X,\,\rho )$. Clearly, $f(t,\,x(t))-f(t,\,\alpha (t))\in BC(\mathbb {R},\,X)$, without loss of generality, we assume that $\|f(t,\,x(t))-f(t,\,\alpha (t))\|\leq \mathcal {C}$ for some positive constant $\mathcal {C}$. By the fact that $\beta \in \mathscr {E}(\mathbb {R},\,X,\,\rho )$ and Lemma 2.4, for any sufficiently small $\epsilon >0$, we have

\[ \lim_{r\rightarrow\infty}\frac{1}{\mu(r,\rho)}\int_{M_{r,\epsilon}(\beta)}\rho(t){\rm d}t=0, \]

where $M_{r,\epsilon }(\beta )=\{t \in [-r,\,r]: \|\beta (t)\|\geq \epsilon \}$. Therefore,

\begin{align*} & \displaystyle\frac{1}{\mu(r,\rho)}\displaystyle\int_{[{-}r,r]}\|G(t)\|\rho(t){\rm d}t\\ & \quad=\displaystyle\frac{1}{\mu(r,\rho)}\displaystyle\int_{[{-}r,r]}\|f(t,x(t))-f(t,\alpha(t))\|\rho(t){\rm d}t\\ & \quad=\frac{1}{\mu(r,\rho)}\int_{M_{r,\epsilon}(\beta)} \|f(t,x(t))-f(t,\alpha(t))\|\rho(t){\rm d}t\\ & \qquad +\frac{1}{\mu(r,\rho)}\int_{[{-}r,r] \backslash {M_{r,\epsilon}(\beta)}}\|f(t,x(t))-f(t,\alpha(t))\|\rho(t){\rm d}t\\ & \quad \leq \mathcal{C}\frac{1}{\mu(r,\rho)}\int_{M_{r,\epsilon}(\beta)}\rho(t){\rm d}t +\frac{1}{\mu(r,\rho)}\int_{[{-}r,r]\backslash {M_{r,\epsilon}(\beta)}}\mathcal{L}(t)\|\beta(t)\|\rho(t){\rm d}t\\ & \quad\leq \mathcal{C}\frac{1}{\mu(r,\rho)}\int_{M_{r,\epsilon}(\beta)}\rho(t){\rm d}t +\epsilon\frac{1}{\mu(r,\rho)}\int_{[{-}r,r]}\mathcal{L}(t)\rho(t){\rm d}t. \end{align*}

Taking into account that $\mathcal {L}(\cdot )\in \mathscr {S}(r,\,\rho )$ and $\epsilon$ sufficiently small, we obtain

\[ \lim_{r\to\infty}\frac{1}{\mu(r,\rho)}\int_{[{-}r,r]}\|G(t)\|\rho(t){\rm d}t=0, \]

which shows that $G(\cdot )\in \mathscr {E}(\mathbb {R},\,X,\,\rho )$.

In the following, we show that $H(t) \in \mathscr {E}(\mathbb {R},\,X,\,\rho )$. By the continuity of $\alpha$, we deduce that $\mathfrak {I}:=\alpha ([-r,\,r])$ is a compact set. Note that $H(t):=h(t,\,\alpha (t))$ is continuous on a compact interval $[-r,\,r]$, so it is uniformly continuous on $[-r,\,r]$. Thus, there exist finite open balls $\mathfrak {B}_k$ ($k=1,\,2,\,\cdots,\,m$) with centre $x_{k} \in \mathfrak {I}$ and radius $\delta >0$ small enough such that

\[ \mathfrak{I}\subseteq\bigcup^{m}_{k=1}\mathfrak{B}_{k} \]

and

\[ \|h(t,\alpha(t))-h(t,x_{k})\|<\frac{\epsilon}{2}, \]

for any sufficiently small $\epsilon >0$ with $\alpha (t) \in \mathfrak {B}_{k},\,t \in [-r,\,r]$. Define the open set $\mathcal {O}_{k}=\{t \in [-r,\,r]:\alpha (t) \in \mathfrak {B}_{k}\}$ such that $[-r,\,r]=\bigcup ^{m}_{k=1}\mathcal {O}_{k}$. Let

\[ \mathfrak{E}_{1}=\mathcal {O}_{1},\mathfrak{E}_{k}=\mathcal {O}_{k}\backslash\bigcup^{k-1}_{j=1}\mathcal {O}_{j},~2 \leq k \leq m, \]

then $\mathfrak {E}_{i} \cap \mathfrak {E}_{j}=\emptyset$ if $i \neq j$, $1 \leq i,\,j \leq m$. Thus,

\begin{align*} & \left\{t \in [{-}r,r]:\|h(t,\alpha(t))\|\geq\epsilon\right\}\subseteq\bigcup^{m}_{k=1}\left\{t \in \mathfrak{E}_{k}:\|h(t,\alpha(t))-h(t,x_{k})\|+\|h(t,x_{k})\|\geq\epsilon\right\}\\ & \quad\subseteq\bigcup^{m}_{k=1}(\left\{t \in \mathfrak{E}_{k}:\|h(t,\alpha(t))-h(t,x_{k})\|\geq\dfrac{\epsilon}{2}\right\}\bigcup\left\{t \in \mathfrak{E}_{k}:\|h(t,x_{k})\|\geq\dfrac{\epsilon}{2}\right\}). \end{align*}

Noticing that $\alpha (t) \in \mathfrak {B}_{k}$, so $\left \{t \in \mathfrak {E}_{k}:\|h(t,\,\alpha (t))-h(t,\,x_{k})\|\geq \dfrac {\epsilon }{2}\right \}=\emptyset$ for $1\leq k\leq m$. Hence,

\[ \frac{1}{\mu(r,\rho)}\int_{M_{r,\epsilon}(h(t,\alpha(t))}\rho(t){\rm d}t\leq \sum^{m}_{k=1}\frac{1}{\mu(r,\rho)}\int_{M_{r,\frac{\epsilon}{2}}(h(t,x_{k}))}\rho(t){\rm d}t. \]

Taking into account that $h(t,\,x) \in \mathscr {E}(\mathbb {R}\times X,\,X,\,\rho )$, we obtain

\[ \frac{1}{\mu(r,\rho)}\int_{M_{r,\frac{\epsilon}{2}}(h(t,x_{k}))}\rho(t){\rm d}t\rightarrow 0,\text{ as $r\rightarrow\infty$}, \]

which implies that

\[ \displaystyle\lim_{r\rightarrow\infty}\frac{1}{\mu(r,\rho)}\int_{M_{r,\epsilon}(h(t,\alpha(t)))}\rho(t){\rm d}t=0. \]

Therefore $H(t):=h(t,\,\alpha (t)) \in \mathscr {E}(\mathbb {R},\,X,\,\rho )$. This completes the proof.

Proof. Note that any positive constant $l$ must belong to the set $\mathscr {S}(r)$ or $\mathscr {S}(r,\,\rho )$, above conclusions can be deduced directly via Corollary 3.1 and Theorem 3.1.

Proof. It is only to show that the function $L(t)\in L^{p}(\mathbb {R},\,\mathbb {R}_+)(1\leq p<+\infty )$ also belongs to the set $\mathscr {S}(r)$. In fact, if $L(t)\in L^{1}(\mathbb {R},\,\mathbb {R}_+)$, it is obvious that $L(t)\in \mathscr {S}(r)$. Let $L(t)\in L^{p}(\mathbb {R},\,\mathbb {R}_+)$ with $1< p<\infty$, then we have

\[ \displaystyle\frac{1}{2r}\displaystyle\int_{{-}r}^{r}L(t){\rm d}t \leq\frac{\left(\displaystyle\int_{{-}r}^{r}L^{p}(t){\rm d}t\right)^{\frac{1}{p}}}{(2r)^{\frac{1}{p}}} \leq\frac{\left(\displaystyle\int_{-\infty}^{\infty}L^{p}(t){\rm d}t\right)^{\frac{1}{p}}}{(2r)^{\frac{1}{p}}}\rightarrow 0, ~r\to 0, \]

which also implies that $L(t)\in \mathscr {S}(r)$.

Proof. It is easy to see that if $L(t)\rho (t)\in L^{1}(\mathbb {R},\,\mathbb {R}_+)$, then $L(t)\in \mathscr {S}(r,\,\rho )$. Thus, the conclusion is a direct consequence of Theorem 3.1.

Proof. For each $x(t)=\alpha (t)+\beta (t) \in WPBP_{p,k}(\mathbb {R},\,X,\,\rho )$ with $\alpha (t) \in BP_{p,k}(\mathbb {R},\,X)$ and $\beta (t) \in \mathscr {E}(\mathbb {R},\,X,\,\rho )$, we can re-write the function $f$ in the form:

\begin{align*} f(t,x(t))& =g(t,\alpha(t))+f(t,x(t))-g(t,\alpha(t))\\ & =g(t,\alpha(t))+f(t,x(t))-f(t,\alpha(t))+h(t,\alpha(t)). \end{align*}

Define

\[ \varPhi(t):=f(t,x(t))-f(t,\alpha(t)),\quad H(t):=h(t,\alpha(t)). \]

By (A1) and Lemma 2.2, we conclude that $g(t,\,\alpha (t)) \in BP_{p,k}(\mathbb {R},\,X)$. From the proof of Theorem 3.1, we can also have $H(t) \in \mathscr {E}(\mathbb {R},\,X,\,\rho )$. The remainder is only to show that $\varPhi (t) \in \mathscr {E}(\mathbb {R},\,X,\,\rho )$. In fact, $f(t,\,x(t))-f(t,\,\alpha (t))\in BC(\mathbb {R},\,X)$, and we can assume that $\|\varPhi (t)\|\leq \mathcal {C}$. For $\varPhi \in \mathscr {E}(\mathbb {R},\,X,\,\rho )$, it is enough to show that

\[ \lim_{r\to\infty}\frac{1}{\mu(r,\rho)}\int_{[{-}r,r]}\|\varPhi(t)\|\rho(t){\rm d}t=0. \]

By Lemma 3.4, $\alpha (\mathbb {R})\subset \overline {x(\mathbb {R})}$ is a bounded set. By the condition (A4) with $\mathcal {Q}=\overline {x(\mathbb {R})}$, we see that for each $\epsilon >0$, there exists a constant $\delta >0$ such that for all $t\in \mathbb {R}$,

\[ \|x-\alpha\|\leq \delta \Rightarrow \|f(t,x(t))-f(t,\alpha(t))\|\leq\mathbb{L}(t)\epsilon. \]

Considering $M_{r,\epsilon }(\beta )=\{t\in [-r,\,r]:\|\beta (t)\|>\epsilon \}$, we have

\begin{align*} M_{r,\mathbb{L}(t)\epsilon}(\varPhi) & = M_{r,\mathbb{L}(t)\epsilon}(f(t,x(t))-f(t,\alpha(t)))\\ & \subseteq M_{r,\delta}(x(t)-\alpha(t)) = M_{r,\delta}(\beta). \end{align*}

Hence,

\[ \dfrac{\displaystyle\int_{M_{r,\mathbb{L}(t)\epsilon}(\varPhi)}\rho(t){\rm d}t}{\mu(r,\rho)}\leq \dfrac{\displaystyle\int_{M_{r,\delta}(\beta)}\rho(t){\rm d}t}{\mu(r,\rho)}. \]

In view of $\beta \in \mathscr {E}(\mathbb {R},\,X,\,\rho )$ and Lemma 2.4, for the above-mentioned $\delta$, we have

\[ \lim_{r\to\infty}\dfrac{\displaystyle\displaystyle\int_{M_{r,\delta}(\beta)}\rho(t){\rm d}t}{\mu(r,\rho)}=0, \]

and

(3.1)\begin{equation} \lim_{r\to\infty}\dfrac{\displaystyle\displaystyle\int_{M_{r,\mathbb{L}(t)\epsilon}(\varPhi)}\rho(t){\rm d}t}{\mu(r,\rho)}=0. \end{equation}

Therefore,

\begin{align*} & \displaystyle\frac{1}{\mu(r,\rho)}\displaystyle\displaystyle\int_{[{-}r,r]}\|\varPhi(t)\|\rho(t){\rm d}t\\ & \quad= \displaystyle\frac{1}{\mu(r,\rho)}\displaystyle\int_{M_{r,\mathbb{L}(t)\epsilon}(\varPhi)}\|\varPhi(t)\|\rho(t){\rm d}t +\frac{1}{\mu(r,\rho)}\displaystyle\int_{[{-}r,r]\setminus M_{r,\mathbb{L}(t)\epsilon}(\varPhi)} \|\varPhi(t)\|\rho(t){\rm d}t\\ & \quad\leq \mathcal{C}\frac{\displaystyle\int_{M_{r,\mathbb{L}(t)\epsilon}(\varPhi)}\rho(t){\rm d}t}{\mu(r,\rho)} +\epsilon\frac{1}{\mu(r,\rho)}\displaystyle\int_{[{-}r,r]}\mathbb{L}(t)\rho(t){\rm d}t. \end{align*}

From (3.1) and the fact $\mathbb {L}(\cdot )\in \mathscr {S}(r,\,\rho )$, we can see that for any sufficiently small $\epsilon >0$,

\[ \lim_{r\to\infty}\frac{1}{\mu(r,\rho)}\int_{[{-}r,r]}\|\varPhi(t)\|\rho(t){\rm d}t=0, \]

which implies $\varPhi (t)\in \mathscr {E}(\mathbb {R},\,X,\,\rho )$.

Proof. Note that if the condition (C5) holds, then for any bounded subset $\mathcal {Q}\subseteq \mathbb {X}$ and for each $\epsilon >0$, there exists a constant $\delta = (\epsilon )^{\frac {1}{\vartheta }}$ such that for all $x$, $y\in \mathcal {Q}$ with $\|x-y\|\leq \delta =(\epsilon )^{\frac {1}{\vartheta }}$ and $t\in \mathbb {R}$,

\[ \|f(t,x)-f(t,y)\|\leq \mathbb{L}(t)\|x-y\|^{\vartheta}\leq \mathbb {L}(t)[(\epsilon)^{\frac{1}{\vartheta}}]^{\vartheta}< \mathbb {L}(t)\epsilon. \]

Thus, the conclusions hold true by Theorem 3.2 (respectively Corollary 3.5).

Proof. By the Fubini theorem, we have

\begin{align*} \displaystyle\frac{1}{2r}\displaystyle\int^{r}_{{-}r}\|H_{1}(t)\|{\rm d}t & \leq\frac{1}{2r}\int^{r}_{{-}r}[\int^{t}_{-\infty}\|\mathcal {S}(t-s)\|\|h(s)\|ds]{\rm d}t\\ & \leq\displaystyle\frac{1}{2r}\displaystyle\int^{r}_{{-}r}[\int^{\infty}_{0}\|\mathcal {S}(s)\|\|h(t-s)\|ds]{\rm d}t\\ & =\int^{\infty}_{0}\|\mathcal {S}(s)\|[\frac{1}{2r}\int^{r}_{{-}r}\|h(t-s)\|{\rm d}t]ds. \end{align*}

Owing to $h \in \mathscr {E}(\mathbb {R},\,X)$, it is deduced that

\[ \frac{1}{2r}\int^{r}_{{-}r}\|H_{1}(t)\|{\rm d}t \rightarrow 0, \text{ as }r\rightarrow\infty \]

by Lemma 2.5(ii) and Lebesgue dominated convergence theorem.

Proof. Let $f \in PBP_{p,k}(\mathbb {R},\,X)$ be such that $f(t)=g(t)+h(t)$ with $g \in BP_{p,k}(\mathbb {R},\,X)$ and $h \in \mathscr {E}(\mathbb {R},\,X)$, then

\begin{align*} \displaystyle\int^{t}_{-\infty}\mathcal {S}(t-s)f(s)ds& =\int^{t}_{-\infty}\mathcal {S}(t-s)g(s)ds+\int^{t}_{-\infty}\mathcal {S}(t-s)h(s)ds\\ & :=u_1(t)+ H_1(t). \end{align*}

In view of Lemma 2.1(5) and Lemma 3.7, we have $u_1(t)\in BP_{p,k}(\mathbb {R},\,X)$ and $H_1(t)\in \mathscr {E}(\mathbb {R},\,X)$ respectively. Thus, we deduce that $\displaystyle \int ^{t}_{-\infty }\mathcal {S}(t-s)f(s)ds \in PBP_{p,k}(\mathbb {R},\,X)$.

Proof. The proof can be conducted similarly as Theorem 3.3 in view of Lemma 2.1(5) and Lemma 3.8.

Proof. The assertion (1) can be obtained by [Reference Chang and Ponce7, Theorem 3.2]. The family $\{S(t)\}_{t\geq 0}$ is uniformly exponentially stable via (H1)- (H2) and Lemma 4.1. Thus, the assertion (2) is a consequence of Theorem 3.3, and the assertion (3) is a consequence of Theorem 3.4.

Proof. Define the operator $\varUpsilon :PBP_{p,k}(\mathbb {R},\,X)\to PBP_{p,k}(\mathbb {R},\,X)$ by

(4.3)\begin{equation} (\varUpsilon u)(t):=\displaystyle\int_{-\infty} ^{t} S(t-s) f(s,u(s))\,ds, \quad t \in \mathbb{R} \end{equation}

where $\{S(t)\}_{t\geq 0}$ is given in Lemma 4.1. For each $u\in PBP_{p,k}(\mathbb {R},\,X)$, by (1) in Corollary 3.2, the function $s\mapsto f(s,\,u(s))$ belongs to $PBP_{p,k}(\mathbb {R},\,X)$. It follows from Lemma 4.2(2) that $\varUpsilon u\in PBP_{p,k}(\mathbb {R},\,X)$, which implies that $\varUpsilon$ is well defined. By Lemma 4.1, there exist $\omega >0,\, C>0$ such that $||S(t)||\leq Ce^{-\omega t}$ for all $t\geq 0.$ For $u_1,\,u_2 \in PBP_{p,k}(\mathbb {R},\,X)$ and $t\in \mathbb {R}$, we have

\begin{align*} ||(\varUpsilon u_1)(t) -(\varUpsilon u_2)(t) ||& \leq \displaystyle\int_{-\infty}^{t} ||S(t-s)[f(s,u_1(s))-f(s,u_2(s))]||ds\\ & \leq \displaystyle\int_{-\infty }^{t}l \|S(t-s)\|\cdot\|u_1(s)-u_2(s)\|ds\\ & \leq l\|u_1-u_2\|_{\infty}\int_0^{\infty }\|S{(r)}\|dr\\ & \leq \displaystyle\frac{lC}{\omega}\|u_1-u_2\|_{\infty}. \end{align*}

Thus

\[ ||\varUpsilon u_1 -\varUpsilon u_2 ||_{\infty}\leq\frac{lC}{\omega}\|u_1-u_2\|_{\infty}, \]

which shows that $\varUpsilon$ is a contraction owing to the assumption $0< l<\dfrac {\omega }{C}$. By the Banach fixed point theorem there exists a unique $u\in PBP_{p,k}(\mathbb {R},\,X)$ such that $\varUpsilon u=u$. The proof is ended.

Proof. The proof is analogous to Theorem 4.1 by Corollary 3.2(2) and Lemma 4.2(3).

Proof. Let the operator $\varUpsilon$ be defined as (4.3). For each $u\in PBP_{p,k}(\mathbb {R},\,X)$, according to Corollary 3.3, Lemma 4.1 and Lemma 4.2(2) we have $\varUpsilon \in PBP_{p,k}(\mathbb {R},\,X)$. Thus, $\varUpsilon$ is well defined.

For $L(\cdot )\in L^{p}(\mathbb {R},\,\mathbb {R}_+)$ with $1< p<\infty$, let $\tau (t)=\int _{-\infty }^{t} L^{p}(s)ds$. Define an equivalent norm over $PBP_{p,k}(\mathbb {R},\,X)$ by

\[ \|u\|_{\tau}=\sup_{t\in\mathbb{R}}\left\{e^{-\theta\tau(t)}\|u\|_{\infty}\right\},\quad u\in PBP_{p,k}(\mathbb{R},X), \]

where $\theta >0$, is a constant sufficiently large. Now for each $u,\,v\in PBP_{p,k}(\mathbb {R},\,X)$, we have

\begin{align*} \|(\varUpsilon u)(t)-(\varUpsilon v)(t)\| & \leq \displaystyle\int_{-\infty}^{t}\|S(t-s)[f(s,u(s))-f(s,v(s))]\|ds\\ & \leq C\displaystyle\int_{-\infty}^{t}e^{-\omega(t-s)}L(s)\|u(s)-v(s)\|ds\\ & \leq C\int_{-\infty}^{t}e^{-\omega(t-s)}L(s)e^{\theta\tau(s)}\|u-v\|_{\tau}ds\\ & \leq C\left[\int_{-\infty}^{t} e^{\theta p\tau(s)}L^{p}(s)ds\right]^{\frac{1}{p}} \left[\int_{-\infty}^{t}e^{-\omega \frac{p(t-s)}{p-1}}ds\right]^{\frac{p-1}{p}}\|u-v\|_{\tau}\\ & \leq C(\omega \frac{p}{p-1})^{\frac{1-p}{p}}\left[\int_{-\infty}^{t} e^{\theta p\tau(s)}d\tau(s)\right]^{\frac{1}{p}}\|u-v\|_{\tau}\\ & \leq C(\omega \frac{p}{p-1})^{\frac{1-p}{p}}(p\theta)^{-\frac{1}{p}}e^{\theta\tau(t)}\|u-v\|_{\tau}. \end{align*}

Consequently,

\[ \|\varUpsilon u-\varUpsilon v\|_{\tau}\leq C\left(\omega \frac{p}{p-1}\right)^{\frac{1-p}{p}}(p\theta)^{-\frac{1}{p}}\|u-v\|_{\tau}. \]

We can see that the operator $\varUpsilon$ is a contraction for a sufficiently large $\theta$.

For $L(\cdot )\in L^{1}(\mathbb {R},\,\mathbb {R}_+)$, we have

\begin{align*} \|(\varUpsilon u)(t)-(\varUpsilon v)(t)\| & \leq\displaystyle\int_{-\infty}^{t}\|S(t-s)[f(s,u(s))-f(s,v(s))]\|ds\\ & \leq C\displaystyle\int_{-\infty}^{t}L(s)\|u(s)-v(s)\|ds\\ & \leq C\|u-v\|_{\infty}\int_{-\infty}^{t}L(s)ds. \end{align*}

In the general, we have

\begin{align*} \|(\varUpsilon^{n} u)(t)-(\varUpsilon^{n} v)(t)\| & \leq\displaystyle\frac{C^{n}}{(n-1)!}\|u-v\|_{\infty}\left[\displaystyle\int_{-\infty}^{t}L(s)\left(\int_{-\infty}^{s}L(\sigma)d\sigma\right)^{n-1} ds\right]\\ & \leq \displaystyle\frac{C^{n}}{n!}\|u-v\|_{\infty}\left(\displaystyle\int_{-\infty}^{t}L(s)ds\right)^{n}. \end{align*}

Thus,

\[ \|\varUpsilon^{n}u-\varUpsilon^{n}v\|_{\infty}\leq\displaystyle\frac{(C\|L\|_{L^{1}(\mathbb{R},\mathbb{R}_+)})^{n}}{n!}\|u-v\|_{\infty}. \]

Since $\frac {(C\|L\|_{L^{1}(\mathbb {R},\,\mathbb {R}_+)})^{n}}{n!}<1$ for $n$ sufficiently large, $\varUpsilon$ is also a contraction.

From above arguments and the Banach fixed point theorem, there exists a unique $u\in PBP_{p,k}(\mathbb {R},\,X)$ such that $\varTheta u=u$. The proof is finished.

Proof. For each $u\in WPBP_{p,k}(\mathbb {R},\,X,\,\rho )$, define the operator $\varUpsilon$ as (4.3). We can show that the operator $\varUpsilon$ is well defined via Corollary 3.4, Lemma 4.2(3). Taking into account $L(\cdot )\in L^{1}(\mathbb {R},\,\mathbb {R}_+)$, we can show $\varUpsilon$ is a contraction analogously to the proof of Theorem 4.3 in the case $p=1$. Thus, there exists a unique $u\in WPBP_{p,k}(\mathbb {R},\,X,\,\rho )$ such that $\varUpsilon u=u$ by the Banach fixed point theorem. This ends the proof.

Proof. From arguments in the above, we know that the operator $A$ can generate a strongly continuous resolvent family $\{R_\alpha (t)\}_{t\geq 0}$ satisfying (4.7). Note that $\frac {1}{1+|\omega |t^{\alpha }}\in L^{1}(\mathbb {R}_{+},\,\mathbb {R}_{+})$ for $1<\alpha <2$, thus the resolvent family $\{R_\alpha (t)\}_{t\geq 0}$ is uniformly integrable, and assertions (a), (b) are direct consequences of Theorem 3.3, Theorem 3.4 respectively.

Proof. Define the operator $\varTheta : PBP_{p,k}(\mathbb {R},\,X)\to PBP_{p,k}(\mathbb {R},\,X)$ by

(4.9)\begin{equation} (\varTheta u(t)):=\displaystyle\int_{-\infty}^{t} R_{\alpha}(t-s)f(s,u(s))ds,\quad t\in\mathbb{R}.\end{equation}

For each $u\in PBP_{p,k}(\mathbb {R},\,X)$, by (1) in Corollary 3.2, the function $s\mapsto f(s,\,u(s))$ belongs to $PBP_{p,k}(\mathbb {R},\,X)$. It follows from (a) of Lemma 4.3 that $\varTheta u\in PBP_{p,k}(\mathbb {R},\,X)$, which implies that $\varTheta$ is well defined. Now for $u_1,\,u_2\in PBP_{p,k}(\mathbb {R},\,X)$ and $t\in \mathbb {R}$, we have

\begin{align*} ||(\varTheta u_1)(t) -(\varTheta u_2)(t) ||& \leq \displaystyle\int_{-\infty}^{t} ||R_{\alpha}(t-s)[f(s,u_1(s))-f(s,u_2(s))]||ds\\ & \leq l \displaystyle\int_{-\infty }^{t}\ \|R_{\alpha}(t-s)\|\cdot\|u_1(s)-u_2(s)\|ds\\ & \leq l\|u_1-u_2\|_{\infty}\int_0^{\infty }\|R_{\alpha}(r)\|dr\\ & \leq \dfrac{Kl\pi}{\alpha\sin{(\dfrac{\pi}{\alpha})}}|\omega|^{{-}1/\alpha}\|u_1-u_2\|_{\infty}. \end{align*}

Thus,

\[ ||\varTheta u_1-\varTheta u_2 ||_{\infty}\leq\dfrac{Kl\pi}{\alpha\sin{\left(\dfrac{\pi}{\alpha}\right)}}|\omega|^{{-}1/\alpha}\|u_1-u_2\|_{\infty}, \]

which means that $\varTheta$ is a contraction on $PBP_{p,k}(\mathbb {R},\,X)$ in view of (4.8). So, by the Banach fixed point theorem there exists a unique $u\in PBP_{p,k}(\mathbb {R},\,X)$ such that $\varTheta u=u$. The proof is end.

Proof. Taking into account Corollary 3.2(2) and Lemma 4.3(b), the proof can be conducted similarly as Theorem 4.5.

Proof. Taking into account that $\alpha \in (1,\,2)$ and if $t\geq 1$, then $\|R_{\alpha }(t)\|\leq \frac {K}{|\omega |}$, and if $0\leq t \leq 1,$ then $\|R_{\alpha }(t)\|\leq K$. Thus, $\|R_{\alpha }(t)\|\leq \widetilde {K}:=\max \{K,\, \frac {K}{|\omega |}\}$.

Define the operator $\varTheta$ by (4.9). It is followed by Corollary 3.3 and Lemma 4.3(a) that $\varTheta$ is well defined. For $u,\, v\in PBP_{p,k}(\mathbb {R},\,X)$ and $t\in \mathbb {R}$, we have

\begin{align*} ||(\varTheta u)(t) -(\varTheta v)(t) ||& \leq \displaystyle\int_{-\infty}^{t} ||R_{\alpha}(t-s)[f(s,u(s))-f(s,v(s))]||ds\\ & \leq \widetilde{K}\|u-v\|_{\infty}\displaystyle\int_0^{\infty }\mathfrak{L}(t-\xi)d\xi\\ & =\widetilde{K}\|u-v\|_{\infty}\int_{-\infty}^{t }\mathfrak{L}(s)ds. \end{align*}

Generally, we have

\begin{align*} \|(\varTheta^{n} u)(t) -(\varTheta^{n} v)(t) \|& \leq \| u-v\|_{\infty}\displaystyle\frac{(\widetilde{K})^{n}}{(n-1)!} \left(\displaystyle\int_{-\infty}^{t} \mathfrak{L}(s) \left(\int_{-\infty}^{s} L(\xi)d\xi\right)^{n-1} ds\right) \\ & \leq \|u -v\|_{\infty} \displaystyle\frac{(\widetilde{K})^{n}}{n!} ( \displaystyle\int_{-\infty}^{t} \mathfrak{L}(s)ds )^{n} \\ & \leq \| u-v\|_{\infty}\dfrac{(\|\mathfrak{L}\|_{L^{1}}\widetilde{K})^{n} }{n!} . \end{align*}

Since $\frac {(\|\mathfrak {L}\|_{1}\widetilde {K})^{n} }{n!}<1$ for sufficiently large $n$, by the contraction principle $\varTheta$ admits a unique fixed point $u\in PBP_{p,k}(\mathbb {R},\,X)$. The proof is finished.

Proof. The proof can be conducted similarly as Theorem 4.7 in view of Corollary 3.4 and Lemma 4.3(b).

Proof. We define the linear operator $\varLambda :C_{\unicode{x1D559}}(X)\to C_{\unicode{x1D559}}(X)$ by

\[ (\varLambda u)(t):=\displaystyle\int_{-\infty}^{t}R_{\alpha}(t-s)f(s,u(s))ds,\ t\in\mathbb{R}. \]

We will show that $\varLambda$ has a fixed point in $PBP_{p,k}(\mathbb {R},\,X)$. For the sake of convenience, we divide the proof into several steps.

1. (I) For each $u\in C_{\unicode{x1D559}}(X)$, it follows by conditions (INT) and (F2) that

   \begin{align*} \|(\varLambda u)(t)\|& \leq \displaystyle\int_{-\infty}^{t}\varPhi(t-s)W_{f}(\|u(s)\|)ds\\ & \leq \displaystyle\int_{-\infty}^{t}\varPhi(t-s)W_{f}(\|u\|_{h}h(s))ds. \end{align*}
   Thus the condition (F3) implies that $\varLambda :C_{\unicode{x1D559}}(X)\to C_{\unicode{x1D559}}(X)$.

2. (II) The operator $\varLambda$ is continuous. In fact, for any $\varepsilon >0$, we take $\delta >0$ involved in the condition (F4). If $u,\,v\in C_{\unicode{x1D559}}(X)$ and $\|u-v\|_{\unicode{x1D559}}\leq \delta$, then

   \[ \|(\varLambda u)(t)-(\varLambda v)(t)\|\leq\displaystyle\int_{-\infty}^{t}\varPhi(t-s)\|f(s,u(s))-f(s,v(s))\|ds\leq\varepsilon, \]
   which shows the assertion.

3. (III) We will show that $\varLambda$ is completely continuous. We set $B_{r}(Z)$ for the closed ball with centre at $0$ and radius $r$ in the space $Z$. Let $V=\varLambda (B_{r}(C_{\unicode{x1D559}}(X)))$ and $v=\varLambda (u)$ for $u\in B_{r}(C_{\unicode{x1D559}}(X))$. Firstly, we will prove that $V(t)$ is a relatively compact subset of $X$ for each $t\in \mathbb {R}$. It follows from the condition (F3) that the function $s\rightarrow \varPhi (s)W_{f}(r{\unicode{x1D559}}(t-s))$ is integrable on $\mathbb {R}_+$. Hence, for $\varepsilon >0$, we can choose $a\geq 0$ such that $\int _{a}^{\infty }\varPhi (s)W_{f}(r{\unicode{x1D559}}(t-s))ds\leq \varepsilon$. Since

   \[ v(t)=\displaystyle\int_{0}^{a}R_{\alpha}(s)f(t-s,u(t-s))ds+\int_{a}^{\infty}R_{\alpha}(s)f(t-s,u(t-s))ds \]
   and
   \[ \left\|\displaystyle\int_{a}^{\infty}R_{\alpha}(s)f(t-s,u(t-s))ds\right\|\leq\int_{a}^{\infty}\varPhi(s)W_{f}(r{\unicode{x1D559}}(t-s))ds\leq\varepsilon, \]
   we get $v(t)\in a\overline {co(K)}+B_{\varepsilon }(X)$, where $co(K)$ denotes the convex hull of $K$ and $K=\{R_{\alpha }(s)f(\xi,\,u):0\leq s\leq a,\,\ t-a\leq \xi \leq t,\,\ \|u\|_{\unicode{x1D559}}\leq r\}$. Using the strong continuity of $R_{\alpha }(\cdot )$ and the condition (F6), we infer that $K$ is a relatively compact set, and $V(t)\subseteq a\overline {co(K)}+B_{\varepsilon }(X)$, which establishes our assertion.

   Secondly, we show that the set $V$ is equicontinuous. In fact, we can decompose

   \begin{align*} v(t+s)-v(t)& =\displaystyle\int_{0}^{s}R_{\alpha}(\sigma)f(t+s-\sigma,u(t+s-\sigma))d\sigma\\ & \quad+\displaystyle\int_{0}^{a}[R_{\alpha}(\sigma+s)-R_{\alpha}(\sigma)]f(t-\sigma,u(t-\sigma))d\sigma\\ & \quad+\int_{a}^{\infty}[R_{\alpha}(\sigma+s)-R_{\alpha}(\sigma)]f(t-\sigma,u(t-\sigma))d\sigma. \end{align*}
   For each $\varepsilon >0$, we can choose $a>0$ and $\delta _{1}>0$ such that
   \begin{align*} & \left\|\displaystyle\int_{0}^{s}R_{\alpha}(\sigma)f(t+s-\sigma,u(t+s-\sigma))d\sigma\right.\\ & \qquad\left.+\int_{a}^{\infty}[R_{\alpha}(\sigma+s)-R_{\alpha}(\sigma)]f(t-\sigma,u(t-\sigma))d\sigma\right\|\\ & \quad\leq\int_{0}^{s}\varPhi(\sigma)W_{f}(r{\unicode{x1D559}}(t+s-\sigma))d\sigma\\ & \qquad+\int_{a}^{\infty}[\varPhi(\sigma+s)+\varPhi(\sigma)]W_{f}(r{\unicode{x1D559}}(t-\sigma))d\sigma\\ & \quad\leq \displaystyle\frac{\varepsilon}{2} \end{align*}
   for $s\leq \delta _{1}$. Moreover, since $\{f(t-\sigma,\,u(t-\sigma )):0\leq \sigma \leq a,\,\ u\in B_{r}(C_{\unicode{x1D559}}(X)\}$ is a relatively compact set and $R_{\alpha }(\cdot )$ is strongly continuous, we can choose $\delta _{2}>0$ such that for $s\leq \delta _{2}$
   \[ \|[R_{\alpha}(\sigma+s)-R_{\alpha}(\sigma)]f\left(t-\sigma,u(t-\sigma)\right)\|\leq\frac{\varepsilon}{2a}. \]
   Combining these estimates, we get $\|v(t+s)-v(t)\|\leq \varepsilon$ for $s$ sufficiently small and independent of $u\in B_{r}(C_{\unicode{x1D559}}(X))$.

   Finally, applying the condition (F3), it is easy to see that

   \[ \displaystyle\frac{\|v(t)\|}{{\unicode{x1D559}}(t)}\leq\frac{1}{{\unicode{x1D559}}(t)}\displaystyle\int_{-\infty}^{t}\varPhi(t-s)W_{f}(r{\unicode{x1D559}}(s))ds\to 0,\quad |t|\rightarrow\infty, \]
   and this convergence is independent of $u\in B_{r}(C_{\unicode{x1D559}}(X))$. Hence, by Lemma 4.4, $V$ is a relatively compact set in $C_{\unicode{x1D559}}(X)$.

4. (IV) Let $u^{\lambda }(\cdot )$ be a solution to $u^{\lambda }=\lambda \varLambda (u^{\lambda })$ for some $0<\lambda <1$. We can estimate

   \begin{align*} \|u^{\lambda}(t)\|& =\lambda\left\|\displaystyle\int_{-\infty}^{t}R_{\alpha}(t-s)f(s,u^{\lambda}(s))ds\right\|\\ & \leq\displaystyle\int_{-\infty}^{t}\varPhi(t-s)W_{f}(\|u^{\lambda}\|_{\unicode{x1D559}}{\unicode{x1D559}}(s))ds\\ & \leq\beta(\|u^{\lambda}\|_{\unicode{x1D559}}){\unicode{x1D559}}(t). \end{align*}
   Hence, we have
   \[ \displaystyle\frac{\|u^{\lambda}\|_{\unicode{x1D559}}}{\beta(\|u^{\lambda}\|_{\unicode{x1D559}})}\leq 1 \]
   and combining with the condition (F5), we conclude that the set $\{u^{\lambda }:u^{\lambda }=\lambda \varLambda (u^{\lambda }),\,\ \lambda \in (0,\,1)\}$ is bounded.

5. (V) By Corollary 3.8(1) and Lemma 4.3(a), we have $\varLambda (PBP_{p,k}(\mathbb {R},\,X))\subseteq PBP_{p,k}(\mathbb {R},\,X)$. Since $PBP_{p,k}(\mathbb {R},\,X)$ is a closed subspace of $C_{\unicode{x1D559}}(X)$, we can consider $\varLambda :\overline {PBP_{p,k}(\mathbb {R},\,X)}\to \overline {PBP_{p,k}(\mathbb {R},\,X)}$. By steps (I)–(III), we deduce that this map is completely continuous. Applying Lemma 4.5, we infer that $\varLambda$ has a fixed point $u\in \overline {PBP_{p,k}(\mathbb {R},\,X)}$. Let $\{u_n\}$ be a sequence in $PBP_{p,k}(\mathbb {R},\,X)$ which converges to $u$ by the norm of $C_{\unicode{x1D559}}(X)$, i.e. $\|u_n-u\|_{\unicode{x1D559}}\to 0$ as $n\to \infty$. Then

\begin{align*} & \|\varLambda u^{n}-u\|_\infty=\|\varLambda u^{n}-\varLambda u\|_{\infty}\\ & \quad \leq \sup_{t\in\mathbb{R}}\displaystyle\int_0^{t} \varPhi(t-s)\|f(s,u^{n}(s))-f(s,u(s))\|ds. \end{align*}

It follows from the condition (F4) that $\varLambda u^{n}\to u$, as $n\to \infty$ uniformly in $\mathbb {R}$. Since $\varLambda u^{n}\in PBP_{p,k}(\mathbb {R},\,X)$, we have $u\in PBP_{p,k}(\mathbb {R},\,X)$, i.e. Equation (4.5) admits one mild solution $u\in PBP_{p,k}(\mathbb {R},\,X)$.