0$ such that \begin{equation}\label{3.11} \frac{b}{R}\leq (1-\lambda_0 \|T\|-a). \end{equation} It follows from \eqref{3.11} and the hypothesis that the conditions of Theorem \ref{thm3.4} is satisfied. Next, for the case that $p\in (0, 1)$, it suffices to choose $\lambda_0\geq 0$ with $\lambda_0\|T\|<1$ and $R>0$ such that $aR^p+b\leq (1-\lambda_0\|T\|)R$. This is obvious. \end{proof} \begin{remark} \label{rmk3.3}\rm The fixed point results of section 2 can be applied to study the eigenvalue problems of Krasnosel'skii-type in the critical case, that is, the map $T:M\subset E\to E$ is non-expansive. Their arguments are fully analogous to the discission presented in this section. Hence we omit it. \end{remark} \section{Application to integral equation } In this section, our aim is to present some existence results for the nonlinear integral equation \begin{equation}\label{4.1} u(t)=f(u)+\int_0^T g(s, u(s))ds, \quad u\in C(J, E), \end{equation} where $E$ is a reflexive Banach space and $J=[0, T]$. The integral in \eqref{4.1} is understood to be the Pettis integral. To study \eqref{4.1}, we assume for the remained of this section the following hypotheses are satisfied: \begin{itemize} \item [(H1)] $f:E\to E$ is sequentially weakly continuous and onto; \item [(H2)] $ \|f(x)-f(y)\|\geq h\|x-y\|$, ($h\geq 2$) for all $x, y\in E$; and $f$ maps relatively weakly compact sets into bounded sets and is uniformly continuous on weakly compact sets; \item[(H3)] for any $t\in J$, the map $g_t=g(t, \cdot): E\to E$ is sequentially weakly continuous; \item[(H4)] for each $x\in C(J, E)$, $g(\cdot, x(\cdot))$ is Pettis integrable on $[0, T]$; \item[(H5)] there exist $\alpha\in L^1[0, T]$ and a nondecreasing continuous function $\phi$ from $[0, \infty)$ to $(0, \infty)$ such that $\|g(t, x)\|\leq \alpha(t)\phi(\|x\|)$ for a.e. $t\in [0, T]$ and all $x\in E$. Further, assume that $\int_0^T\alpha(s)ds<\int_{\|f(\theta)\|}^\infty\frac{dr}{\phi(r)}$. \end{itemize} We now state and prove an existence principle for \eqref{4.1}. \begin{theorem} \label{thm4.1} Suppose that the conditions (H1)-(H5) are fulfilled. Then \eqref{4.1} has at least one solution $u\in C(J, E)$. \end{theorem} \begin{proof} Put $$ \beta(t)=\int_{\|f(\theta)\|}^t\frac{dr}{\phi(r)} \quad \text{and} \quad b(t)=(h-1)^{-1}\beta^{-1}\big(\int_0^t\alpha(s)ds\big). $$ Then \begin{equation}\label{4.00} \int_{\|f(\theta)\|}^{(h-1)b(t)}\frac{dr}{\phi(r)}=\int_0^t\alpha(s)ds. \end{equation} It follows from \eqref{4.00} and the final part of (H5) that $b(T)<\infty$. We define the set $$ K=\big\{x\in C(J, E): \|x(t)\|\leq (h-1)b(t) \text{ for all } t\in J \big\}. $$ Then $K$ is a closed, convex and bounded subset of $C(J, E)$. Let us now introduce the nonlinear operators $T$ and $S$ as follows: \begin{gather*} (Tx)(t)=f(x(t))-f(\theta), \\ (Sy)(t)=f(\theta)+\int_0^tg(s, y(s))ds. \end{gather*} The conditions (H1) and (H4) imply that $T$ and $S$ are well defined on $C(J,E)$, respectively. Our idea is to use Theorem \ref{thm2.2} to find the fixed point for the sum $T+S$ in $K$. The proof will be shown in several steps. \noindent\textbf{Step 1:} Prove that $S$ maps $K$ into $K$, $S(K)$ is equicontinuous and relatively weakly compact. For any $y\in K$, we shall show that $Sy\in K$. Let $t\in J$ be fixed. Without loss of generality, we may assume that $(Sy)(t)\neq 0$. In view of the Hahn-Banach theorem there exists $y_t^*\in E^*$ with $\|y_t^*\|=1$ such that $\langle y_t^*, (Sy)(t)\rangle=\|(Sy)(t)\|$. Thus, one can deduce from (H5) and \eqref{4.00} that \begin{equation}\label{4.2} \begin{aligned} \|(Sy)(t)\|&=\langle y_t^*, f(\theta)\rangle+\int_0^t\langle y_t^*,g(s, y(s))\rangle ds\\ &\leq \|f(\theta)\| +\int_0^t\alpha(s)\phi(\|y(s)\|)ds\\ &\leq \|f(\theta)\|+\int_0^t\alpha(s)\phi((h-1)b(s))ds\\ &= \|f(\theta)\|+(h-1)\int_0^tb'(s)ds=(h-1)b(t). \end{aligned} \end{equation} It shows from \eqref{4.2} that $S(K)\subset K$ and hence is bounded. This proves the first claim of Step 1. Next, let $t, s\in J$ with $s\neq t$. We may assume that $(Sy)(t)-(Sy)(s)\neq0$. Then there exists $x_t^*\in E^*$ with $\|x_t^*\|=1$ and $\langle x_t^*, (Sy)(t)-(Sy)(s)\rangle=\|(Sy)(t)-(Sy)(s)\|$. Consequently, \begin{equation}\label{4.3} \begin{aligned} \|(Sy)(t)-(Sy)(s)\| &\leq \int_s^t\alpha(\tau)\phi(\|y(\tau)\|)d\tau\\ &\leq \int_s^t\alpha(\tau)\phi((h-1)b(\tau))d\tau \\ &\leq (h-1)\big|\int_s^tb'(\tau)d\tau\big|=(h-1)|b(t)-b(s)|. \end{aligned} \end{equation} It follows from \eqref{4.3} that $S(K)$ is equicontinuous. The reflexiveness of $E$ implies that $S(K)(t)$ is relatively weakly compact for each $t\in J$, where $S(K)(t)=\{z(t): z\in S(K)\}$. By a known result (see \cite{Ku,Mit}), one can easily get that $S(K)$ is relatively weakly compact in $C(J, E)$. This completes Step 1. \noindent\textbf{Step 2:} Prove that $S:K\to K$ is sequentially weakly continuous. Let $\{x_n\}$ be a sequence in $K$ with $x_n\rightharpoonup x$ in $C(J, E)$, for some $x\in K$. Then $x_n(t)\rightharpoonup x(t)$ in $E$ for all $t\in J$. Fix $t\in (0, T]$. From the item (H3) one sees that $g(t, x_n(t))\rightharpoonup g(t, x(t))$ in $E$. Together with (H5) and the Lebesgue dominated convergence theorem for the Pettis integral yield for each $\varphi \in E^*$ that $$ \langle \varphi, (Sx_n)(t)\rangle \to \langle \varphi, (Sx)(t)\rangle; $$ i.e., $(Sx_n)(t)\rightharpoonup(Sx)(t)$ in $E$. We can do this for each $t\in J$ and notice that $S(K)$ is equicontinuous, and accordingly $Sx_n\rightharpoonup Sx$ by \cite{Mit}. The Step 2 is proved. \noindent\textbf{Step 3:} Prove that the conditions (ii) and (iii) of Theorem \ref{thm2.2} hold. Since $E$ is reflexive and $f$ is continuous on weakly compact sets, it shows that $T$ transforms $C(J,E)$ into itself. This, in conjunction with the first part of (H2), one easily gets that $T: C(J, E)\to C(J, E)$ is expansive with constant $h\geq 2$. For all $x, y\in C(J, E)$, one can see from the first part of (H2) that $$ \|(I-T)x(t)-(I-T)y(t)\|\geq (h-1)\|x(t)-y(t)\|\geq \|x(t)-y(t)\|, $$ where $I$ is identity map. Thus, one has \begin{equation}\label{4.4} \|(I-T)x(t)\|\geq (h-1)\|x(t)\|\geq\|x(t)\|, \quad \forall x\in C(J,E). \end{equation} Assume now that $x=Tx+Sy $ for some $y\in K$. We conclude from \eqref{4.2} and \eqref{4.4} that $$ \|x(t)\|\leq\|(I-T)x(t)\|=\|(Sy)(t)\|\leq (h-1)b(t), $$ which shows that $x\in K$. Therefore, the second part of (iii) in Theorem \ref{thm2.2} is fulfilled. Next, for each $y\in C(J, E)$, we define $T_y:C(J, E)\to C(J, E)$ by $$ (T_yx)(t)=(Tx)(t)+y(t). $$ Then $T_y$ is expansive with constant $h\geq 2$ and onto since $f$ maps $E$ onto $E$. By Lemma \ref{lem2}, we know there exists $x^*\in C(J, E)$ such that $T_yx^*=x^*$, that is $ (I-T)x^*=y$. Hence $S(K)\subset (I-T)(E)$. This completes Step 3. \noindent\textbf{Step 4:} Prove that the condition (v) of Theorem \ref{thm2.2} is satisfied. For each $x\in \mathbb{F}(E, K; T, S)$, then by the definition of $\mathbb{F}$ and Lemma \ref{lem2.3} there exists $y\in K$ such that \begin{equation}\label{9999} x=(I-T)^{-1}Sy. \end{equation} Hence, for $t, s\in J$, we obtain from Lemma \ref{lem2.3}, \eqref{9999} and \eqref{4.3} that $$ \|x(t)-x(s)\|\leq |b(t)-b(s)|, $$ which illustrates that $\mathbb{F}(E, K; T, S)$ is equicontinuous in $C(J, E)$. Let $\{x_n\}$ be a sequence in $\mathbb{F}$. Then $\{x_n\}$ is equicontinuous in $C(J, E)$ and there exists $\{y_n\}$ in $K$ with $x_n=Tx_n+Sy_n$. Thus, one has from \eqref{4.2} and \eqref{4.4} that $$ \|x_n(t)\|\leq \frac{1}{h-1}\|(Sy_n)(t)\|\leq b(t), \forall t\in J. $$ It follows that, for each $t\in J$, the set $\{x_n(t)\}$ is relatively weakly compact in $E$. The above discussion tells us that $\{x_n: n\in \mathbb{N}\}$ is relatively weakly compact. The Eberlein-\v{S}mulian theorem implies that $\mathbb{F}$ is relatively weakly compact. This achieves Step 4 \noindent\textbf{Step 5:} Prove that $T$ fulfils the condition (iv) of Theorem \ref{2.2}. By the second part of (H2) and the fact that $\mathbb{F}$ is relatively weakly compact we obtain that $T(\mathbb{F})$ is bounded. Again by the second part of (H2) and the fact that $\mathbb{F}$ is equicontinuous, one can readily deduce that $T(\mathbb{F})$ is also equicontinuous. Now, let $\{x_n\}\subset \mathbb{F}$ with $x_n\rightharpoonup x$ in $C(J, E)$ for some $x\in K$. It follows from (H1) that $(Tx_n)(t)\rightharpoonup (Tx)(t)$. Since $\{Tx_n: n\in \mathbb{N}\}$ is equicontinuous in $C(J, E)$, as before, we conclude that $Tx_n\rightharpoonup Tx$ in $C(J, E)$. The Step 5 is proved. Now, invoking Theorem \ref{thm2.2} we obtain that there is $x^*\in K$ with $Tx^*+Sx^*=x^*$; i.e., $x^*$ is a solution to \eqref{4.1}. This accomplishes the proof. \end{proof} \begin{remark}\label{rem4.1}\rm It is clearly seen that the following locally ``Lipscitizan'' type condition fulfills the second part of (H2): For each bounded subset $U$ of $E$, there exists a continuous function $\psi_U: \mathbb{R}^{+}\to \mathbb{R}^{+}$ with $\psi_U(0)= 0$, such that $\|f(x)-f(y)\|\leq \psi_U(\|x-y\|)$ for all $x, y \in U$. Although the proof of Theorem \ref{thm4.1} is analogous to that of \cite[Theorem 5.1]{Barroso2}, it clarifies some vague points made in \cite{Barroso2}. Moreover, it can be easily known that Theorem \ref{thm4.1} does not contain the corresponding result of \cite[Theorem 5.1]{Barroso2}, vice versa. Therefore, Theorem \ref{thm4.1} and \cite[Theorem 5.1]{Barroso2} are complementary. \end{remark} We conclude this artcile by presenting a class of maps which fulfil the assumptions (H1) and (H2) in Theorem \ref{thm4.1}. Assume that $f: \mathbb{R}^n\to \mathbb{R}^n$ is continuous and is coercive, i.e., $f$ satisfies the inequality $$ (f(x)-f(y), x-y)\geq \alpha(\|x-y\|)\|x-y\|, \quad \forall x, y\in\mathbb{R}^n, $$ where $\alpha(0)=0, \alpha(t)>0$ for all $t>0$; $\lim_{t\to \infty}\alpha(t)=\infty$. Then it is well known that $f$ is surjective. Particularly, if $\alpha(t)=ht$, $h>0$, then $f: \mathbb{R}^n\to \mathbb{R}^n$ is a homeomorphism. Specifically, let us consider the function $f: \mathbb{R}\to \mathbb{R}$ defined by $f(x)= x^k+hx+x_0$, where $h\geq 2$, $k$ is a positive odd number, and $x_0$ is a given constant. Then $f$ satisfies the assumptions (H1) and (H2) in Theorem \ref{thm4.1} for $E=\mathbb{R}$. \subsection*{Acknowledgments} The authors are very grateful to the anonymous referee who carefully read the manuscript, pointed out a misinformation and several vague points, and gave us valuable comments, and suggestions for improving the exposition. 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