In this section, by using the measure of noncompactness and fixed point theorems, we give the existence results of the nonlocal initial value problem (1.1). Here we list the following hypotheses.

- (1)
The ${C}_{0}$ semigroup $T(t)$ generated by *A* is equicontinuous. We denote $N=sup\{\parallel T(t)\parallel ;t\in [0,1]\}$.

- (2)
$g:C([0,1];X)\to X$ is continuous and compact, there exist positive constants *c* and *d* such that $\parallel g(x)\parallel \le c\parallel x\parallel +d$, $\mathrm{\forall}x\in C([0,1];X)$.

- (3)
The multivalued operator $F:[0,1]\times X\to {P}_{c}(X)$ satisfies the hypotheses: the set ${S}_{F,x}=\{v\in {L}^{1}(I,X):v(t)\in F(t,x(t));\text{for a.e.}t\in [0,1]\}$ is nonempty.

$t\to F(t,x)$ is measurable for every $x\in X$;

$x\to F(t,x)$ is u.s.c. for a.e.

$t\in [0,1]$;

- (4)
There exists

$L\in {L}^{1}(0,1;{\mathrm{\Re}}^{+})$ such that for any bounded

$D\subset X$,

$\alpha (F(t,D))\le L(t)\alpha (D)$

for a.e.

$t\in [0,1]$.

- (5)
There exist a function

$m\in {L}^{1}(0,1;{\mathrm{\Re}}^{+})$ and a nondecreasing continuous function

$\mathrm{\Omega}:{\mathrm{\Re}}^{+}\to {\mathrm{\Re}}^{+}$ such that

$\parallel F(t,x)\parallel \le m(t)\mathrm{\Omega}(\parallel x\parallel )$

for all $x\in X$, and a.e. $t\in [0,1]$.

**Remark 3.1** If $dimX<\mathrm{\infty}$, then ${S}_{F,x}\ne \mathrm{\varnothing}$ for each $x\in C([0,1];X)$ (see Lasota and Opial [31]). If $dimX=\mathrm{\infty}$ and $x\in C([0,1];X)$, then the set ${S}_{F,x}$ is nonempty if and only if the function $Y:[0,1]\to \mathrm{\Re}$ defined by $Y(t)=inf\{\parallel v\parallel :v\in F(t,x(t))\}$ belongs to ${L}^{1}(0,1;{\mathrm{\Re}}^{+})$ (see Hu and Papageorgiou [32]).

The following lemma plays a crucial role in the proof of the main theorem.

**Lemma 3.2** [26]

*Under assumptions* (3)-(5), *if we consider sequences* ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}\subset C([0,1];X)$ *and* ${\{{v}_{n}\}}_{n=1}^{\mathrm{\infty}}\subset {L}^{1}(0,1;X)$, *where* ${v}_{n}\in {S}_{F,{x}_{n}}$, *such that* ${x}_{n}\to x$, ${v}_{n}\rightharpoonup v$, *then* $v\in {S}_{F,x}$.

Now we give the existence results under the above hypotheses.

**Theorem 3.3** *If* (1)-(5)

*are satisfied*,

*then there is at least one mild solution for* (1.1)

*provided that there exists a constant* *R* *with* ${\int}_{0}^{1}m(s)\phantom{\rule{0.2em}{0ex}}ds<{\int}_{N(cR+d)}^{R}\frac{1}{N\mathrm{\Omega}(s)}\phantom{\rule{0.2em}{0ex}}ds.$

(3.1)

*Proof* Define the operator

$\mathrm{\Gamma}:C([0,1];X)\to C([0,1];X)$ by

$(\mathrm{\Gamma}x)(t)=\{y(t)\in C([0,1];X):y(t)=T(t)g(x)+{\int}_{0}^{t}T(t-s)v(s)\phantom{\rule{0.2em}{0ex}}ds;v\in {S}_{F,x}\}.$

We shall show that the multivalued Γ has at least one fixed point. The fixed point is then a mild solution of the problem (1.1).

- (1)
We contract a bounded, convex, closed and compact set $W\subset C([0,1];X)$ such that Γ maps *W* into itself.

In view of (3.1), we know there exists a constant

$\eta >0$ such that

${\int}_{0}^{1}m(s)\phantom{\rule{0.2em}{0ex}}ds<{\int}_{{T}_{0}+\eta}^{R}\frac{1}{N\mathrm{\Omega}(s)}\phantom{\rule{0.2em}{0ex}}ds,$

(3.2)

where ${T}_{0}=N(cR+d)$.

Then there exists a positive integer

*K* such that

${\int}_{{T}_{0}+\eta}^{{T}_{0}+K\eta}\frac{1}{N\mathrm{\Omega}(s)}\phantom{\rule{0.2em}{0ex}}ds<{\int}_{0}^{1}m(s)\phantom{\rule{0.2em}{0ex}}ds\le {\int}_{{T}_{0}+\eta}^{{T}_{0}+(K+1)\eta}\frac{1}{N\mathrm{\Omega}(s)}\phantom{\rule{0.2em}{0ex}}ds.$

(3.3)

Hence, we get

$0={t}_{0}<{t}_{1}<{t}_{2}<\cdots <{t}_{K-1}<{t}_{K}=1$ such that

$\begin{array}{c}{\int}_{0}^{{t}_{1}}m(s)\phantom{\rule{0.2em}{0ex}}ds={\int}_{{T}_{0}+\eta}^{{T}_{0}+2\eta}\frac{1}{N\mathrm{\Omega}(s)}\phantom{\rule{0.2em}{0ex}}ds,\hfill \\ {\int}_{{t}_{1}}^{{t}_{2}}m(s)\phantom{\rule{0.2em}{0ex}}ds={\int}_{{T}_{0}+2\eta}^{{T}_{0}+3\eta}\frac{1}{N\mathrm{\Omega}(s)}\phantom{\rule{0.2em}{0ex}}ds,\hfill \\ \cdots ,\hfill \\ {\int}_{{t}_{K-2}}^{{t}_{K-1}}m(s)\phantom{\rule{0.2em}{0ex}}ds={\int}_{{T}_{0}+(K-1)\eta}^{{T}_{0}+K\eta}\frac{1}{N\mathrm{\Omega}(s)}\phantom{\rule{0.2em}{0ex}}ds,\hfill \\ {\int}_{{t}_{K-1}}^{1}m(s)\phantom{\rule{0.2em}{0ex}}ds\le {\int}_{{T}_{0}+K\eta}^{{T}_{0}+(K+1)\eta}\frac{1}{N\mathrm{\Omega}(s)}\phantom{\rule{0.2em}{0ex}}ds.\hfill \end{array}$

We denote ${W}_{0}=\{x\in C([0,1];X),{\parallel x(t)\parallel}_{i}=sup\{\parallel x(t)\parallel :t\in [{t}_{i-1},{t}_{i}]\}\le {T}_{0}+i\eta ,i=1,2,\dots ,K\}$, then ${W}_{0}\subseteq C([0,1];X)$ is nonempty, bounded, closed and convex.

For any

$x\in {W}_{0}$, we have

$(\mathrm{\Gamma}x)(t)=\{y(t):y(t)=T(t)g(x)+{\int}_{0}^{t}T(t-s)v(s)\phantom{\rule{0.2em}{0ex}}ds;v(t)\in {S}_{F,x}\}.$

Therefore

$\begin{array}{rcl}\parallel y(t)\parallel & \le & \parallel T(t)g(x)\parallel +\parallel {\int}_{0}^{t}T(t-s)v(s)\phantom{\rule{0.2em}{0ex}}ds\parallel \le N(c\parallel x\parallel +d)+N{\int}_{0}^{t}m(s)\mathrm{\Omega}\left(\parallel x(s)\parallel \right)\phantom{\rule{0.2em}{0ex}}ds\\ \le & N(c({T}_{0}+K\eta )+d)+N{\int}_{0}^{t}m(s)\mathrm{\Omega}\left(\parallel x(s)\parallel \right)\phantom{\rule{0.2em}{0ex}}ds\\ \le & N(cR+d)+N{\int}_{0}^{t}m(s)\mathrm{\Omega}\left(\parallel x(s)\parallel \right)\phantom{\rule{0.2em}{0ex}}ds\le {T}_{0}+N{\int}_{0}^{t}m(s)\mathrm{\Omega}\left(\parallel x(s)\parallel \right)\phantom{\rule{0.2em}{0ex}}ds,\end{array}$

and

$\begin{array}{rcl}{\parallel y\parallel}_{i}& =& sup\{\parallel y(t)\parallel :t\in [{t}_{i-1},{t}_{i}]\}\\ \le & sup\{{T}_{0}+N{\int}_{0}^{t}m(s)\mathrm{\Omega}\left(\parallel x(s)\parallel \right)\phantom{\rule{0.2em}{0ex}}ds:t\in [{t}_{i-1},{t}_{i}]\}\\ \le & {T}_{0}+N{\int}_{0}^{{t}_{i}}m(s)\mathrm{\Omega}\left(\parallel x(s)\parallel \right)\phantom{\rule{0.2em}{0ex}}ds\\ \le & {T}_{0}+N[{\int}_{0}^{{t}_{1}}m(s)\mathrm{\Omega}\left(\parallel x(s)\parallel \right)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{{t}_{1}}^{{t}_{2}}m(s)\mathrm{\Omega}\left(\parallel x(s)\parallel \right)\phantom{\rule{0.2em}{0ex}}ds\\ +\cdots +{\int}_{{t}_{i-1}}^{{t}_{i}}m(s)\mathrm{\Omega}\left(\parallel x(s)\parallel \right)\phantom{\rule{0.2em}{0ex}}ds]\\ \le & {T}_{0}+N[{\int}_{0}^{{t}_{1}}m(s)\phantom{\rule{0.2em}{0ex}}ds\mathrm{\Omega}({T}_{0}+\eta )+{\int}_{{t}_{1}}^{{t}_{2}}m(s)\phantom{\rule{0.2em}{0ex}}ds\mathrm{\Omega}({T}_{0}+2\eta )\\ +\cdots +{\int}_{{t}_{i-1}}^{{t}_{i}}m(s)\phantom{\rule{0.2em}{0ex}}ds\mathrm{\Omega}({T}_{0}+i\eta )]\\ \le & {T}_{0}+N[{\int}_{{T}_{0}+\eta}^{{T}_{0}+2\eta}\frac{1}{N\mathrm{\Omega}(s)}\phantom{\rule{0.2em}{0ex}}ds\mathrm{\Omega}({T}_{0}+\eta )+{\int}_{{T}_{0}+2\eta}^{{T}_{0}+3\eta}\frac{1}{N\mathrm{\Omega}(s)}\phantom{\rule{0.2em}{0ex}}ds\mathrm{\Omega}({T}_{0}+2\eta )\\ +\cdots +{\int}_{{T}_{0}+i\eta}^{{T}_{0}+(i+1)\eta}\frac{1}{N\mathrm{\Omega}(s)}\phantom{\rule{0.2em}{0ex}}ds\mathrm{\Omega}({T}_{0}+i\eta )]\\ \le & {T}_{0}+i\eta ,\end{array}$

which implies $\mathrm{\Gamma}:{W}_{0}\to {2}^{{W}_{0}}$ is a bounded operator.

Define

${W}_{1}=\overline{\mathit{conv}}\mathrm{\Gamma}({W}_{0})$, where

$\overline{\mathit{conv}}$ means the closure of the convex hull in

$C([0,1];X)$. Then

${W}_{1}\subset C([0,1];X)$ is nonempty bounded closed convex on

$[0,1]$ with

${W}_{1}\subseteq {W}_{0}$. Let

${W}_{n+1}=\overline{\mathit{conv}}\mathrm{\Gamma}({W}_{n})$ for all

$n\ge 1$. Similarly to the above discussions, we know that

${W}_{n+1}\subseteq {W}_{n}$ for

$n=1,2,\dots $ as

${W}_{1}\subseteq {W}_{0}$ and

${W}_{1},{W}_{2},\dots $ are both nonempty, closed, bounded and convex. Thus

${\{{W}_{n}\}}_{n=1}^{+\mathrm{\infty}}$ is a decreasing sequence consisting of subsets of

$C([0,1];X)$. Moreover, set

$W=\bigcap _{n=1}^{+\mathrm{\infty}}{W}_{n},$

then *W* is a convex, closed and bounded subset of $C([0,1];X)$ and $\mathrm{\Gamma}(W)\subseteq W$.

Now, we claim that

*W* is nonempty and compact in

$C([0,1];X)$. To do so, from Lemma 2.6, we know for arbitrary given

$\u03f5>0$, there exist sequences

${\{{v}_{n}\}}_{n=1}^{+\mathrm{\infty}}\subset {S}_{F,{W}_{n}}$ such that

$\begin{array}{rcl}\alpha ({W}_{n+1}(t))& =& \alpha ((\mathrm{\Gamma}{W}_{n})(t))\\ \le & 2\alpha ({\int}_{0}^{t}T(t-s){v}_{n}{(s)}_{n=1}^{\mathrm{\infty}}\phantom{\rule{0.2em}{0ex}}ds)+\epsilon \\ \le & 4{\int}_{0}^{t}\alpha (T(t-s){v}_{n}{(s)}_{n=1}^{\mathrm{\infty}})\phantom{\rule{0.2em}{0ex}}ds+\epsilon \\ \le & 4N{\int}_{0}^{t}\alpha \left({v}_{n}{(s)}_{n=1}^{\mathrm{\infty}}\right)\phantom{\rule{0.2em}{0ex}}ds+\epsilon \\ \le & 4N{\int}_{0}^{t}\alpha (F(s,{W}_{n}(s))\phantom{\rule{0.2em}{0ex}}ds+\epsilon \\ \le & 4N{\int}_{0}^{t}L(s)\alpha ({W}_{n}(s))\phantom{\rule{0.2em}{0ex}}ds+\epsilon .\end{array}$

Since this is true for arbitrary

$\u03f5>0$, we have

$\alpha ({W}_{n+1}(t))\le 4N{\int}_{0}^{t}L(s)\alpha ({W}_{n}(s))\phantom{\rule{0.2em}{0ex}}ds.$

Because

${W}_{n}$ is decreasing for

*n*, we can define

$\mu (t)=\underset{n\to +\mathrm{\infty}}{lim}\alpha (({W}_{n})(t)).$

Let

$n\to +\mathrm{\infty}$, we have

$\mu (t)\le 4N{\int}_{0}^{t}L(s)\mu (s)\phantom{\rule{0.2em}{0ex}}ds.$

It implies that

$\mu (t)=0$ for all

$t\in [0,1]$. By Lemma 2.2, we know that

${lim}_{n\to +\mathrm{\infty}}\alpha ({W}_{n})=0$. Using Lemma 2.3, we obtain

$W={\bigcap}_{n=1}^{+\mathrm{\infty}}{W}_{n}$ is nonempty and compact in

$C([0,1];X)$.

- (2)
We shall show that Γ is closed on *W* with closed convex values. It is very easy to see that Γ has convex values.

Let us now verity that

$graph(\mathrm{\Gamma})$ is closed. Let

${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}\subset W$ with

${x}_{n}\to x$ in

$C([0,1];X)$, and

${y}_{n}\in \mathrm{\Gamma}{x}_{n}$ with

${y}_{n}\to y$ in

$C([0,1];X)$. Moreover, let

${\{{v}_{n}\}}_{n=1}^{\mathrm{\infty}}\subset {L}^{1}(0,1;X)$ be a sequence such that

${v}_{n}\in {S}_{F,{x}_{n}}$ for any

$n\ge 1$, and

${y}_{n}(t)=T(t)g({x}_{n})+{\int}_{0}^{t}T(t-s){v}_{n}(s)\phantom{\rule{0.2em}{0ex}}ds.$

As ${x}_{n}\to x$ in $C([0,1];X)$, we know that ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is a bounded set of $C([0,1];X)$, we denote ${R}_{x}=sup\{\parallel {x}_{n}\parallel :n=1,2,\dots \}$.

From hypothesis (5), we obtain

$\parallel {v}_{n}(t)\parallel \le \parallel F(t,{x}_{n}(t)\parallel \le m(t)\mathrm{\Omega}(\parallel {x}_{n}\parallel )\le m(t)\mathrm{\Omega}({R}_{x}).$

Then we have the set ${\{{v}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is integrably bounded for a.e. $t\in [0,1]$.

From hypothesis (4), we know

$\alpha \left({\{{v}_{n}(t)\}}_{n=1}^{\mathrm{\infty}}\right)\le \alpha \left(F(t,{\{{x}_{n}(t)\}}_{n=1}^{\mathrm{\infty}})\right)\le L(t)\alpha \left({\{{x}_{n}(t)\}}_{n=1}^{\mathrm{\infty}}\right)=0$

for a.e. $t\in [0,1]$. Then the set ${\{{v}_{n}(t)\}}_{n=1}^{\mathrm{\infty}}$ is relatively compact for a.e. $t\in [0,1]$.

So, the set

${\{{v}_{n}(t)\}}_{n=1}^{\mathrm{\infty}}$ is semicompact. By applying Lemma 2.7, it yields that

${\{{v}_{n}(t)\}}_{n=1}^{\mathrm{\infty}}$ is weakly compact in

${L}^{1}(0,1;X)$. We get that there exists

$v\in {L}^{1}(0,1;X)$ such that

${v}_{n}\rightharpoonup v$. Therefore, we infer that

${\int}_{0}^{t}T(t-s){v}_{n}(s)\phantom{\rule{0.2em}{0ex}}ds\to {\int}_{0}^{t}T(t-s)v(s)\phantom{\rule{0.2em}{0ex}}ds.$

Further, we have

${y}_{n}(t)\to T(t)g(x)+{\int}_{0}^{t}T(t-s)v(s)\phantom{\rule{0.2em}{0ex}}ds,$

and hence

$y(t)=T(t)g(x)+{\int}_{0}^{t}T(t-s)v(s)\phantom{\rule{0.2em}{0ex}}ds.$

By Lemma 3.2, it implies that

$v\in {S}_{F,x}$,

*i.e.*,

$y\in \mathrm{\Gamma}(x)$. Therefore

$graph(\mathrm{\Gamma})$ is closed. And hence Γ has closed values on

*W*.

- (4)

Since $\overline{\mathrm{\Gamma}W}\subseteq W$ is compact, *W* is closed and $graph(\mathrm{\Gamma})$ is closed, we can come to the conclusion that Γ is u.s.c. (see [30]).

Finally, due to fixed point Lemma 2.9, Γ has at least one point $x\in \mathrm{\Gamma}(x)$, and *x* is a mild solution to the semilinear evolution differential inclusions with the nonlocal conditions (1.1). Thus the proof is complete. □

**Remark 3.4** In [8–12] the authors discuss the nonlocal initial value problem (1.1) when $T(t)$ is compact. In [14] the existence of mild solutions of the differential inclusions (1.1) is proved when A generates an equicontinuous semigroup and Banach space *X* is separable and uniformly smooth. In this paper, by using a new method, we prove the operator Γ maps compact set *W* into itself. We do not impose any restriction on the coefficient $L(t)$, and we only require $T(t)$ to be an equicontinuous semigroup. So, Theorem 3.3 generalizes and improves the related results in [8–12, 14].

**Theorem 3.5** [15]

*If* (1)-(5)

*are satisfied*,

*then there is at least one mild solution for* (1.1)

*provided that there exists a constant* $R>0$ *such that* $N(cR+d)+N{\int}_{0}^{1}m(s)\phantom{\rule{0.2em}{0ex}}ds\mathrm{\Omega}(R)\le R.$

(3.4)

*Proof* In view of (3.4), we get

${\int}_{0}^{1}m(s)\phantom{\rule{0.2em}{0ex}}ds\le \frac{R-N(cR+d)}{N\mathrm{\Omega}(R)}<{\int}_{N(cR+d)}^{R}\frac{1}{N\mathrm{\Omega}(s)}\phantom{\rule{0.2em}{0ex}}ds.$

From Theorem 3.3, the nonlocal initial value problem (1.1) has at least one mild solution. □

**Remark 3.6** If

$N=1$,

$c=\frac{1}{3}$,

$d=0$,

$\mathrm{\Omega}(x)=x$ and

${\int}_{0}^{1}m(s)\phantom{\rule{0.2em}{0ex}}ds=1$. We cannot obtain a constant

*R* such that

By using Theorem 3.5, we do not know whether or not equation (

1.1) has a mild solution. But we know there exists a constant

$R=1$ such that

${\int}_{0}^{1}m(s)\phantom{\rule{0.2em}{0ex}}ds=1<ln3={\int}_{N(cR+d)}^{R}\frac{1}{N\mathrm{\Omega}(s)}\phantom{\rule{0.2em}{0ex}}ds.$

So, Theorem 3.3 is better than Theorem 3.5.

**Theorem 3.7** [12]

*If* (1)-(5)

*are satisfied and* $\parallel g(x)\parallel \le d$,

*then there is at least one mild solution for* (1.1)

*provided that* ${\int}_{0}^{1}m(s)\phantom{\rule{0.2em}{0ex}}ds<{\int}_{d}^{+\mathrm{\infty}}\frac{1}{N\mathrm{\Omega}(s)}\phantom{\rule{0.2em}{0ex}}ds.$

(3.5)

*Proof* In view of (3.5), we get there exists a constant

*R* such that

${\int}_{0}^{1}m(s)\phantom{\rule{0.2em}{0ex}}ds<{\int}_{0R+d}^{R}\frac{1}{N\mathrm{\Omega}(s)}\phantom{\rule{0.2em}{0ex}}ds.$

By Theorem 3.3, we complete the proof of this theorem. □

Next, we give the existence result for (1.1) when *g* is Lipschitz continuous.

We suppose that:

- (6)
There exists a constant $c\in {\mathrm{\Re}}^{+}$ such that $\parallel g(u)-g(v)\parallel \le c\parallel u-v\parallel $ for all $u,v\in C([0,1];X)$. Therefore, $\parallel g(x)\parallel \le c\parallel x\parallel +d$, where $d=\parallel g(0)\parallel $.

**Theorem 3.8** *If* (1)

*and* (3)-(6)

*are satisfied and* $Nc+4N{\int}_{0}^{1}L(s)\phantom{\rule{0.2em}{0ex}}ds<1,$

*then there is at least one mild solution for* (1.1)

*provided that there exists a constant* *R* *satisfying* ${\int}_{0}^{1}m(s)\phantom{\rule{0.2em}{0ex}}ds<{\int}_{{T}_{0}}^{+\mathrm{\infty}}\frac{1}{N\mathrm{\Omega}(s)}\phantom{\rule{0.2em}{0ex}}ds.$

(3.6)

*Proof* With the same arguments as given in the first portion of the proof of Theorem 3.3, we know $\mathrm{\Gamma}:{W}_{0}\to {2}^{{W}_{0}}$ is a bounded map with convex values and is closed on ${W}_{0}$.

Now, we prove the values of Γ are compact in $C([0,1];X)$.

Let

$x\in C([0,1];X)$ and

${y}_{n}\in \mathrm{\Gamma}(x)$. To prove that

$\mathrm{\Gamma}(x)$ is compact, we have to show that

${y}_{n}$ has a subsequence converging to a point

$y\in \mathrm{\Gamma}(x)$. We have

${v}_{n}\in {S}_{F,x}$ such that

${y}_{n}(t)=T(t)g(x)+{\int}_{0}^{t}T(t-s){v}_{n}(s)\phantom{\rule{0.2em}{0ex}}ds.$

From hypothesis (5), we obtain

$\parallel {v}_{n}(t)\parallel \le \parallel F(t,x(t))\parallel \le m(t)\mathrm{\Omega}(\parallel x\parallel ).$

Then we have the set ${\{{v}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is integrably bounded for a.e. $t\in [0,1]$.

From hypothesis (4), we know

$\alpha \left({\{{v}_{n}(t)\}}_{n=1}^{\mathrm{\infty}}\right)\le \alpha \left(F(t,x(t))\right)\le L(t)\alpha (x(t))=0$

for a.e. $t\in [0,1]$. Then the set ${\{{v}_{n}(t)\}}_{n=1}^{\mathrm{\infty}}$ is relatively compact for a.e. $t\in [0,1]$.

So, the set

${\{{v}_{n}(t)\}}_{n=1}^{\mathrm{\infty}}$ is semicompact. By applying Lemma 2.7, it yields that

${\{{v}_{n}(t)\}}_{n=1}^{\mathrm{\infty}}$ is weakly compact in

${L}^{1}(0,1;X)$. We get that there exists

$v\in {L}^{1}(0,1;X)$ such that

${v}_{n}\rightharpoonup v$. Therefore, we infer that

${\int}_{0}^{t}T(t-s){v}_{n}(s)\phantom{\rule{0.2em}{0ex}}ds\to {\int}_{0}^{t}T(t-s)v(s)\phantom{\rule{0.2em}{0ex}}ds,$

and

$\underset{n\to +\mathrm{\infty}}{lim}{y}_{n}(t)=T(t)g(x)+{\int}_{0}^{t}T(t-s)v(s)\phantom{\rule{0.2em}{0ex}}ds=y(t).$

By Lemma 3.2, it implies that $v\in {S}_{F,x}$, *i.e.*,$y\in \mathrm{\Gamma}(x)$. Therefore Γ has compact values.

Next, we prove Γ is an

*α* contraction map. For any

$B\subseteq {W}_{0}$, we have

$\begin{array}{rcl}\alpha ((\mathrm{\Gamma}B)(t))& =& \alpha (T(t)g(B)+{\int}_{0}^{t}T(t-s){S}_{F,B}\phantom{\rule{0.2em}{0ex}}ds)\\ \le & Nc\alpha (B)+\alpha ({\int}_{0}^{t}T(t-s){S}_{F,B}\phantom{\rule{0.2em}{0ex}}ds).\end{array}$

From Lemma 2.6, we know for arbitrary given

$\u03f5>0$, there exist sequences

${\{{v}_{n}\}}_{n=1}^{+\mathrm{\infty}}\subset {S}_{F,B}$ such that

$\begin{array}{rcl}\alpha ({\int}_{0}^{t}T(t-s){S}_{F,B}\phantom{\rule{0.2em}{0ex}}ds)& =& 2\alpha ({\int}_{0}^{t}T(t-s){v}_{n}{(s)}_{n=1}^{\mathrm{\infty}}\phantom{\rule{0.2em}{0ex}}ds)+\epsilon \\ \le & 4{\int}_{0}^{t}\alpha (T(t-s){v}_{n}{(s)}_{n=1}^{\mathrm{\infty}})\phantom{\rule{0.2em}{0ex}}ds+\epsilon \\ \le & 4N{\int}_{0}^{t}\alpha \left({v}_{n}{(s)}_{n=1}^{\mathrm{\infty}}\right)\phantom{\rule{0.2em}{0ex}}ds+\epsilon \\ \le & 4N{\int}_{0}^{t}\alpha \left(F(s,B(s))\right)\phantom{\rule{0.2em}{0ex}}ds+\epsilon \\ \le & 4N{\int}_{0}^{t}L(s)\alpha (B(s))\phantom{\rule{0.2em}{0ex}}ds+\epsilon \\ \le & 4N\alpha (B){\int}_{0}^{1}L(s)\phantom{\rule{0.2em}{0ex}}ds+\epsilon .\end{array}$

Since this is true for arbitrary

$\u03f5>0$, we have

$\alpha ({\int}_{0}^{t}T(t-s){S}_{F,B}\phantom{\rule{0.2em}{0ex}}ds)\le 4N\alpha (B){\int}_{0}^{1}L(s)\phantom{\rule{0.2em}{0ex}}ds.$

Therefore, we obtain

$\alpha (\mathrm{\Gamma}B)\le (Nc+4N{\int}_{0}^{1}L(s)\phantom{\rule{0.2em}{0ex}}ds)\alpha (B).$

Noting $Nc+4N{\int}_{0}^{t}L(s)\phantom{\rule{0.2em}{0ex}}ds<1$, therefore Γ is an *α* contraction map.

Finally, due to Lemma 2.10, Γ has at least one fixed point. This completes the proof. □