In this section, we establish the existence of a solution for a nonlinear implicit eigenvalue problem under normalized conditions by using the Browder degree theory for class ({S}_{+}).
Recall that a mapping H:[0,1]\times \overline{\mathrm{\Omega}}\to {X}^{\ast} is of class ({S}_{+}) if the following condition holds:
For any sequence \{{u}_{j}\} in \overline{\mathrm{\Omega}} with {u}_{j}\rightharpoonup {u}_{0} and any sequence \{{t}_{j}\} in [0,1] with {t}_{j}\to {t}_{0} for which
\underset{j\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008H({t}_{j},{u}_{j}),{u}_{j}{u}_{0}\u3009\le 0,
we have {u}_{j}\to {u}_{0} and H({t}_{j},{u}_{j})\rightharpoonup H({t}_{0},{u}_{0}); see [2, 6].
As mentioned in the introduction, Kartsatos and Skrypnik [4] gave the following result provided that property (\mathcal{P}) is fulfilled. In a more concrete situation, we adopt the normalization method considered in [3].
Theorem 2.1 Let Ω be a bounded open set in X with 0\in \mathrm{\Omega}. Let T:D(T)\subset X\to {X}^{\ast} be a maximal monotone operator such that the closure of Ω is included in the interior of D(T) with T(0)=0 and T satisfies condition ({T}_{\mathrm{\infty}}^{(0)}) on \overline{\mathrm{\Omega}}. Assume that C:[0,\mathrm{\infty})\times \overline{\mathrm{\Omega}}\to {X}^{\ast} is demicontinuous, bounded and satisfies condition ({S}_{+}) such that C(0,x)=0 for all x\in \overline{\mathrm{\Omega}} and C(t,x) is continuous in t uniformly with respect to x\in \overline{\mathrm{\Omega}}. Further assume that
(c1) There exists a positive number \mathcal{N} such that the weak sequential closure of the set
G=\{\frac{C(\lambda ,x)}{\parallel C(\lambda ,x)\parallel}:\lambda \ge \mathcal{N},x\in \overline{\mathrm{\Omega}},\parallel Jx+Tx\parallel \le 2M(\lambda )\}
does not contain the origin 0, where
M(\lambda )=sup\{\parallel C(\lambda ,x)\parallel :x\in \overline{\mathrm{\Omega}}\}.
(c2) {lim}_{\lambda \to \mathrm{\infty}}m(\lambda )=\mathrm{\infty}, where m(\lambda )=inf\{\parallel C(\lambda ,x)\parallel :x\in \overline{\mathrm{\Omega}}\}.
Then the following statements hold:

(a)
For each \epsilon >0, there exists a point ({\lambda}_{\epsilon},{x}_{\epsilon}) in (0,\mathrm{\infty})\times \partial \mathrm{\Omega} such that
T{x}_{\epsilon}+C({\lambda}_{\epsilon},{x}_{\epsilon})+\epsilon J{x}_{\epsilon}=0.

(b)
If 0\notin T(\partial \mathrm{\Omega}) and T satisfies condition ({S}_{q}) on ∂ Ω, then the implicit eigenvalue problem
has a solution ({\lambda}_{0},{x}_{0}) in (0,\mathrm{\infty})\times \partial \mathrm{\Omega}.
Proof (a) For our aim, we use the Browder degree {d}_{B} given in [6]. Let ε be any positive number. We first prove that there is a number \mathrm{\Lambda}\in (0,\mathrm{\infty}) such that
{d}_{B}(T+C(\mathrm{\Lambda},\cdot )+\epsilon J,\mathrm{\Omega},0)=0.
(2.1)
Assume the contrary. For a sequence \{{\mathrm{\Lambda}}_{n}\} in (0,\mathrm{\infty}) with {\mathrm{\Lambda}}_{n}\to \mathrm{\infty}, the following occurs:
For each n\in \mathbb{N}, either there exists a point {x}_{n}\in \mathrm{\Omega} such that T{x}_{n}+C({\mathrm{\Lambda}}_{n},{x}_{n})+\epsilon J{x}_{n}=0, in view of {d}_{B}(T+C({\mathrm{\Lambda}}_{n},\cdot )+\epsilon J,\mathrm{\Omega},0)\ne 0, or there exists a point {x}_{n}\in \partial \mathrm{\Omega} such that T{x}_{n}+C({\mathrm{\Lambda}}_{n},{x}_{n})+\epsilon J{x}_{n}=0. Thus, we get a sequence \{{x}_{n}\} in \overline{\mathrm{\Omega}} such that
T{x}_{n}+C({\mathrm{\Lambda}}_{n},{x}_{n})+\epsilon J{x}_{n}=0.
(2.2)
This implies
\parallel J{x}_{n}+T{x}_{n}\parallel \le \parallel (1\epsilon )J{x}_{n}\parallel +\parallel C({\mathrm{\Lambda}}_{n},{x}_{n})\parallel \le 2M({\mathrm{\Lambda}}_{n})
for sufficiently large n. Since the sequence \{{\parallel C({\mathrm{\Lambda}}_{n},{x}_{n})\parallel}^{1}C({\mathrm{\Lambda}}_{n},{x}_{n})\} is bounded in the reflexive Banach space {X}^{\ast}, we may suppose that {\parallel C({\mathrm{\Lambda}}_{n},{x}_{n})\parallel}^{1}C({\mathrm{\Lambda}}_{n},{x}_{n}) converges weakly to some {h}_{0}\in {X}^{\ast}. Using (c1), it is clear that {h}_{0}\ne 0. It follows from (2.2) that
\frac{T{x}_{n}}{\parallel T{x}_{n}\parallel}\rightharpoonup {h}_{0}.
(2.3)
However, \parallel C({\mathrm{\Lambda}}_{n},{x}_{n})\parallel \to \mathrm{\infty} by (c2) implies \parallel T{x}_{n}\parallel \to \mathrm{\infty}, which is a contradiction to condition ({T}_{\mathrm{\infty}}^{(0)}). Hence assertion (2.1) holds.
Next, we consider a mapping H:[0,1]\times \overline{\mathrm{\Omega}}\to {X}^{\ast} given by
H(t,x)=Tx+C(t\mathrm{\Lambda},x)+\epsilon Jx\phantom{\rule{1em}{0ex}}\text{for}(t,x)\in [0,1]\times \overline{\mathrm{\Omega}}.
Then H is of class ({S}_{+}). To prove this, let \{{u}_{j}\} be any sequence in \overline{\mathrm{\Omega}} with {u}_{j}\rightharpoonup {u}_{0} and \{{t}_{j}\} be any sequence in [0,1] with {t}_{j}\to {t}_{0} such that
\underset{j\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008H({t}_{j},{u}_{j}),{u}_{j}{u}_{0}\u3009\le 0.
(2.4)
Since the operators T and J are monotone, it follows from
\u3008H({t}_{j},{u}_{j}),{u}_{j}{u}_{0}\u3009=\u3008T{u}_{j},{u}_{j}{u}_{0}\u3009+\u3008C({t}_{j}\mathrm{\Lambda},{u}_{j}),{u}_{j}{u}_{0}\u3009+\epsilon \u3008J{u}_{j},{u}_{j}{u}_{0}\u3009
(2.5)
that
\u3008H({t}_{j},{u}_{j}),{u}_{j}{u}_{0}\u3009\ge \u3008T{u}_{0},{u}_{j}{u}_{0}\u3009+\u3008C({t}_{j}\mathrm{\Lambda},{u}_{j}),{u}_{j}{u}_{0}\u3009+\epsilon \u3008J{u}_{0},{u}_{j}{u}_{0}\u3009.
(2.6)
By (2.4) and (2.6), we have
\underset{j\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008C({t}_{j}\mathrm{\Lambda},{u}_{j}),{u}_{j}{u}_{0}\u3009\le 0.
(2.7)
There are two cases to consider. If {t}_{0}=0, then C({t}_{j}\mathrm{\Lambda},{u}_{j})\to 0 and so
\underset{j\to \mathrm{\infty}}{lim}\u3008C({t}_{j}\mathrm{\Lambda},{u}_{j}),{u}_{j}{u}_{0}\u3009=0.
(2.8)
Since (2.5) implies
\u3008H({t}_{j},{u}_{j}),{u}_{j}{u}_{0}\u3009\ge \u3008T{u}_{0},{u}_{j}{u}_{0}\u3009+\u3008C({t}_{j}\mathrm{\Lambda},{u}_{j}),{u}_{j}{u}_{0}\u3009+\epsilon \u3008J{u}_{j},{u}_{j}{u}_{0}\u3009,
it follows from (2.4) and (2.8) that
\underset{j\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008J{u}_{j},{u}_{j}{u}_{0}\u3009\le 0.
Since J satisfies condition ({S}_{+}), we obtain
which implies
T{u}_{j}\rightharpoonup T{u}_{0},\phantom{\rule{2em}{0ex}}C({t}_{j}\mathrm{\Lambda},{u}_{j})\rightharpoonup C(0,{u}_{0}),\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}J{u}_{j}\to J{u}_{0}
on observing that T is demicontinuous on \overline{\mathrm{\Omega}}. This means that H({t}_{j},{u}_{j})\rightharpoonup H(0,{u}_{0}). If {t}_{0}>0, we have
\begin{array}{c}\underset{j\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008C({t}_{0}\mathrm{\Lambda},{u}_{j}),{u}_{j}{u}_{0}\u3009\hfill \\ \phantom{\rule{1em}{0ex}}\le \underset{j\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008C({t}_{j}\mathrm{\Lambda},{u}_{j}),{u}_{j}{u}_{0}\u3009+\underset{j\to \mathrm{\infty}}{lim\hspace{0.17em}sup}[\u3008C({t}_{j}\mathrm{\Lambda},{u}_{j})C({t}_{0}\mathrm{\Lambda},{u}_{j}),{u}_{j}{u}_{0}\u3009]\hfill \end{array}
and hence by (2.7),
\underset{j\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008C({t}_{0}\mathrm{\Lambda},{u}_{j}),{u}_{j}{u}_{0}\u3009\le \underset{j\to \mathrm{\infty}}{lim\hspace{0.17em}sup}[\parallel C({t}_{j}\mathrm{\Lambda},{u}_{j})C({t}_{0}\mathrm{\Lambda},{u}_{j})\parallel \parallel {u}_{j}{u}_{0}\parallel ]=0.
Since C satisfies condition ({S}_{+}), we get {u}_{j}\to {u}_{0} from which T{u}_{j}\rightharpoonup T{u}_{0}, C({t}_{j}\mathrm{\Lambda},{u}_{j})\rightharpoonup C({t}_{0}\mathrm{\Lambda},{u}_{0}), and J{u}_{j}\to J{u}_{0}. Consequently, H({t}_{j},{u}_{j})\rightharpoonup H({t}_{0},{u}_{0}). We have just shown that the mapping H is of class ({S}_{+}).
We are now ready to apply the degree theory of Browder [6, 7]. Then we have
{d}_{B}(H(0,\cdot ),\mathrm{\Omega},0)={d}_{B}(T+\epsilon J,\mathrm{\Omega},0)=1.
The last equality is based on Theorem 3 in [6] because the operator T+\epsilon J is strictly monotone and demicontinuous on \overline{\mathrm{\Omega}} and satisfies condition ({S}_{+}). On the other hand, it is shown in (2.1) that
{d}_{B}(H(1,\cdot ),\mathrm{\Omega},0)={d}_{B}(T+C(\mathrm{\Lambda},\cdot )+\epsilon J,\mathrm{\Omega},0)=0.
Hence, in view of Theorem 4 in [6], there exist {t}_{0}\in [0,1] and {x}_{0}\in \partial \mathrm{\Omega} such that
T{x}_{0}+C({t}_{0}\mathrm{\Lambda},{x}_{0})+\epsilon J{x}_{0}=0.
It follows from the injectivity of T+\epsilon J that {t}_{0}>0. Consequently, if we let {\lambda}_{\epsilon}:={t}_{0}\mathrm{\Lambda} and {x}_{\epsilon}:={x}_{0}, then we have {\lambda}_{\epsilon}\in (0,\mathrm{\infty}) and T{x}_{\epsilon}+C({\lambda}_{\epsilon},{x}_{\epsilon})+\epsilon J{x}_{\epsilon}=0.

(b)
Let \{{\epsilon}_{n}\} be a sequence in (0,\mathrm{\infty}) such that {\epsilon}_{n}\to 0. According to statement (a), there exists a sequence \{({\lambda}_{{\epsilon}_{n}},{x}_{{\epsilon}_{n}})\} in (0,\mathrm{\infty})\times \partial \mathrm{\Omega} such that
T{x}_{{\epsilon}_{n}}+C({\lambda}_{{\epsilon}_{n}},{x}_{{\epsilon}_{n}})+{\epsilon}_{n}J{x}_{{\epsilon}_{n}}=0.
If we set {x}_{n}:={x}_{{\epsilon}_{n}} and {\lambda}_{n}:={\lambda}_{{\epsilon}_{n}}, it can be written in the form
T{x}_{n}+C({\lambda}_{n},{x}_{n})+{\epsilon}_{n}J{x}_{n}=0.
(2.9)
Without loss of generality, we may suppose that
{\lambda}_{n}\to {\lambda}_{0},\phantom{\rule{2em}{0ex}}{x}_{n}\rightharpoonup {x}_{0},\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}C({\lambda}_{n},{x}_{n})\rightharpoonup {c}^{\ast},
(2.10)
where {\lambda}_{0}\in [0,\mathrm{\infty}], {x}_{0}\in X, and {c}^{\ast}\in {X}^{\ast}. Note that {\lambda}_{0} belongs to (0,\mathrm{\infty}). In fact, if {\lambda}_{0}=0, then C({\lambda}_{n},{x}_{n})\to 0 implies T{x}_{n}\to 0. Since T satisfies condition ({S}_{q}) on ∂ Ω, we obtain that {x}_{n}\to {x}_{0}\in \partial \mathrm{\Omega} and therefore T{x}_{0}=0, which contradicts the hypothesis that 0\notin T(\partial \mathrm{\Omega}). If {\lambda}_{0}=\mathrm{\infty}, then (c2) implies \parallel C({\lambda}_{n},{x}_{n})\parallel \to \mathrm{\infty} and so \parallel T{x}_{n}\parallel \to \mathrm{\infty}. As in (2.3), a similar argument proves that {\parallel T{x}_{n}\parallel}^{1}T{x}_{n} converges weakly to some nonzero vector, which contradicts condition ({T}_{\mathrm{\infty}}^{(0)}). Thus we have shown that {\lambda}_{0}\in (0,\mathrm{\infty}).
For the next aim, we now show that
\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008C({\lambda}_{n},{x}_{n}),{x}_{n}{x}_{0}\u3009\le 0.
(2.11)
Assume that (2.11) is false. Then there exists a subsequence of \{{x}_{n}\}, denoted again by \{{x}_{n}\}, such that
\underset{n\to \mathrm{\infty}}{lim}\u3008C({\lambda}_{n},{x}_{n}),{x}_{n}{x}_{0}\u3009>0.
Hence we obtain from (2.9) that
\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008T{x}_{n},{x}_{n}{x}_{0}\u3009<0.
Noticing by (2.9) and (2.10) that T{x}_{n}\rightharpoonup {c}^{\ast}, we get
\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008T{x}_{n},{x}_{n}\u3009<\underset{n\to \mathrm{\infty}}{lim}\u3008T{x}_{n},{x}_{0}\u3009=\u3008{c}^{\ast},{x}_{0}\u3009.
(2.12)
For every x\in D(T), we have by the monotonicity of T
\begin{array}{rl}\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}inf}\u3008T{x}_{n},{x}_{n}\u3009& \ge \underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}inf}[\u3008T{x}_{n},x\u3009+\u3008Tx,{x}_{n}x\u3009]\\ =\u3008{c}^{\ast},x\u3009+\u3008Tx,{x}_{0}x\u3009,\end{array}
which implies along with (2.12)
\u3008{c}^{\ast}Tx,{x}_{0}x\u3009>0.
(2.13)
By the maximal monotonicity of T, we have {x}_{0}\in D(T) and T{x}_{0}={c}^{\ast}. Letting x={x}_{0} in (2.13), we get a contradiction. Therefore, (2.11) is true.
Since C({\lambda}_{n},{x}_{n})C({\lambda}_{0},{x}_{n})\to 0, it follows from (2.11) and
\u3008C({\lambda}_{n},{x}_{n}),{x}_{n}{x}_{0}\u3009=\u3008C({\lambda}_{n},{x}_{n})C({\lambda}_{0},{x}_{n}),{x}_{n}{x}_{0}\u3009+\u3008C({\lambda}_{0},{x}_{n}),{x}_{n}{x}_{0}\u3009
that
\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008C({\lambda}_{0},{x}_{n}),{x}_{n}{x}_{0}\u3009\le 0.
Since C satisfies condition ({S}_{+}) and {\lambda}_{0}\in (0,\mathrm{\infty}), we have {x}_{n}\to {x}_{0}\in \partial \mathrm{\Omega}, which implies C({\lambda}_{n},{x}_{n})\rightharpoonup C({\lambda}_{0},{x}_{0}). Hence we obtain from (2.9) that T{x}_{n}\rightharpoonup C({\lambda}_{0},{x}_{0}). By Lemma 1.1, we conclude that T{x}_{0}+C({\lambda}_{0},{x}_{0})=0. This completes the proof. □
Remark 2.2 In the proof of Theorem 2.1, the demicontinuity of T on \overline{\mathrm{\Omega}} is needed to show that H is of class ({S}_{+}). This is guaranteed under an additional condition \overline{\mathrm{\Omega}}\subset intD(T). Actually, local boundedness of T on intD(T) implies the demicontinuity of T; see, e.g., [5].