In this section, we shall establish the existence and uniqueness of coupled fixed point for singlevalued mixed monotone operators, and then we extend the study to the multivalued case.
To verify the main results, we need the following lemma.
Lemma 4.1
If
\((X,\\cdot\)\)
is a Banach space, P
is a normal and reproducing cone in
X, then

(1)
\(X \times X\)
is a Banach space with the norm
\(\(x,y)\=\max \{\x\,\y\\}\).

(2)
\(P\times(P)\)
is a normal and reproducing cone in
\(X \times X\).
Proof
We only prove assertion (2). Noting that a partial order ≲ of \(X\times X\) can be induced by \(P\times(P)\), we have
$$(x_{1},y_{1})\lesssim (x_{2},y_{2})\quad \Longleftrightarrow\quad x_{2}x_{1}\in P, y_{2}y_{1}\in(P)\quad\Longleftrightarrow\quad x_{1}\leq x_{2}, y_{2}\leq y_{1}. $$

(i)
If \((0,0)\lesssim(x_{1},y_{1})\lesssim (x_{2},y_{2})\), then \(0\leq x_{1}\leq x_{2}\), \(0\geq y_{1}\geq y_{2}\), or equivalently, \(0\leqy_{1}\leqy_{2}\). Since P is normal,
$$\x_{1}\\leq K\x_{2}\,\qquad \y_{1}\\leq K \y_{2}\, $$
where K is the normal constant of P.
Hence, we know
$$\bigl\ (x_{1},y_{1})\bigr\ =\max\bigl\{ \x_{1}\, \y_{1}\\bigr\} \leq K\max\bigl\{ \x_{2}\,\y_{2}\\bigr\} =K\bigl\ (x_{2},y_{2})\bigr\ , $$
which implies \(P\times(P)\) is normal with normal constant K.

(ii)
For arbitrary \((x,y)\in X \times X\), since P is reproducing, there exist \(u_{1},v_{1},u_{2},v_{2}\in P\), such that \(x=u_{1}v_{1}\), \(y=u_{2}v_{2}\), or equivalently, \(y=(v_{2})(v_{1})\). Then
$$(x,y)=(u_{1}, v_{2})(u_{2}, v_{1})\in P\times(P)P\times(P), $$
which implies \(P\times(P)\) is a reproducing cone in \(X \times X\).
□
Theorem 4.2
Suppose
\(T:X\times X\rightarrow X\)
is a mixed monotone operator. Assume there exist two linear operators
\(L,S:X\rightarrow X\)
with
\(\L\+\S\ <1\), \(L(P)\subset P\), \(S(P)\subset P\)
such that, for any
\(x_{1}, x_{2}, y_{1}, y_{2}\in X\), \(x_{1}\leq x_{2}\), \(y_{2}\leq y_{1}\)
$$T(x_{2},y_{2}) T(x_{1},y_{1})\leq L(x_{2}x_{1})+S(y_{1}y_{2}). $$
Then
T
has a unique coupled fixed point
\((\overline{x},\overline{y})\)
in
\(X\times X\). Moreover, for every
\((x,y)\in X\times X\),
$$\lim_{n\rightarrow\infty}T^{n}(x,y) = \overline{x},\qquad\lim _{n\rightarrow \infty}T^{n}(y,x) = \overline{y}. $$
Proof
Define a new mapping \(F:X\times X\rightarrow X\times X\) as follows, for \((x,y)\in X\times X\):
$$F(x,y)=\bigl(T(x,y),T(y,x)\bigr). $$
Then T has a coupled fixed point if and only if F has a fixed point.
When T is mixed monotone, then \(\forall x_{1},x_{2},y_{1},y_{2}\in X\),
$$x_{1}\leq x_{2}, y_{2}\leq y_{1}\quad \Rightarrow\quad T(x_{1},y_{1})\leq T(x_{2},y_{2}), T(y_{1},x_{1})\geq T(y_{2},x_{2}), $$
which means
$$(x_{1},y_{1})\lesssim (x_{2},y_{2})\quad \Rightarrow\quad F(x_{1},y_{1})\lesssim F(x_{2},y_{2}) $$
i.e.
F is increasing.
Furthermore, by assumption, for any \(x_{1}, x_{2}, y_{1}, y_{2}\in X\), \(x_{1}\leq x_{2}\), \(y_{2}\leq y_{1}\)
$$T(x_{2},y_{2}) T(x_{1},y_{1})\leq L(x_{2}x_{1})+S(y_{1}y_{2})=L(x_{2}x_{1})S(y_{2}y_{1}), $$
then
$$T(y_{1},x_{1})T(y_{2},x_{2})\leq L(y_{1}y_{2})+S(x_{2}x_{1}), $$
or equivalently,
$$T(y_{2},x_{2})T(y_{1},x_{1}) \geqL(y_{1}y_{2})S(x_{2}x_{1})=S(x_{2}x_{1})+L(y_{2}y_{1}), $$
which implies
$$F(x_{2},y_{2})F(x_{1},y_{1})\lesssim \Phi\bigl[(x_{2},y_{2})(x_{1},y_{1}) \bigr],\quad \forall (x_{1},y_{1})\lesssim (x_{2},y_{2}), $$
where \(\Phi:X\times X\rightarrow X\times X\) is defined as
$$\Phi(x,y)=\bigl(L(x)S(y),S(x)+L(y)\bigr). $$
Note that

(1)
Φ is a linear operator and \(\Phi(P\times(P))\subset P\times(P)\).
In fact, since L, S is linear, Φ is a linear operator on \(X\times X\).
If \((x,y)\in P\times(P)\), i.e.
\(x\geq0\), \(y\leq0\), due to the positivity of L, S, we have \(L(x)\geq0\), \(L(y)\leq0\), \(S(x)\geq0\), \(S(y)\leq0\), then
$$L(x)S(y)\geq0,\qquad S(x)+L(y)\leq0 $$
i.e.
$$\Phi(x,y)\in P\times(P). $$

(2)
Φ is bounded and its operator norm \(\\Phi\\leq\L\ +\S\<1\).
In fact,
$$\begin{aligned} \\Phi\&= \sup_{(x,y)\in X\times X, \(x,y)\=1}\bigl\ \Phi(x,y )\bigr\ \\ &= \sup_{(x,y)\in X\times X, \(x,y)\=1}\max\bigl\{ \bigl\ L(x)S(y)\bigr\ ,\bigl\ L(y)S(x)\bigr\ \bigr\} \\ &\leq\sup_{(x,y)\in X\times X, \(x,y)\=1}\max\bigl\{ \bigl\ L(x)\bigr\ +\bigl\ S(y)\bigr\ ,\bigl\ L(y)\bigr\ +\bigl\ S(x)\bigr\ \bigr\} \\ &\leq \sup_{(x,y)\in X\times X, \max \{\x\,\y\ \}=1}\max\bigl\{ \L\ \x\+\S\\y\,\L\\y\+\S\ \x\ \bigr\} \\ &\leq \L\+\S\. \end{aligned}$$
Then, by [4] Theorem 3.2, F has a unique fixed point \((\overline {x},\overline{y})\in X\times X\). Moreover, for every \((x,y)\in X\times X\),
$$\lim_{n\rightarrow\infty}F^{n}(x,y) = (\overline{x},\overline{y}). $$
Hence, T has a unique coupled fixed point \((\overline{x},\overline {y})\) in \(X\times X\). Moreover, for every \((x,y)\in X\times X\),
$$\lim_{n\rightarrow\infty}T^{n}(x,y) = \overline{x},\qquad\lim _{n\rightarrow \infty}T^{n}(y,x) = \overline{y}. $$
This ends the proof. □
Remark 4.3
If, in addition, the positive linear operators L, S are commutative, i.e.
\(LS=SL\), then the requirement \(\L\+\S\<1\) can be relaxed to \(r(L)+r(S)<1\).
In this case, by mathematical induction, we have
$$\Phi^{2n}(x,y)=\bigl(\bigl(L^{2}+S^{2} \bigr)^{n}(x),\bigl(L^{2}+S^{2}\bigr)^{n}(y) \bigr),\quad \forall(x,y)\in X\times X, $$
then
$$\begin{aligned} \bigl\ \Phi^{2n}\bigr\ &= \sup_{(x,y)\in X\times X, \(x,y)\=1}\bigl\ \Phi(x,y)\bigr\ \\ &= \sup_{(x,y)\in X\times X, \(x,y)\=1}\max\bigl\{ \bigl\ \bigl(L^{2}+S^{2} \bigr)^{n}(x)\bigr\ ,\bigl\ \bigl(L^{2}+S^{2} \bigr)^{n}(y)\bigr\ \bigr\} \\ &\leq \bigl\ \bigl(L^{2}+S^{2}\bigr)^{n}\bigr\ . \end{aligned}$$
Hence, by Gelfand’s formula
$$\begin{aligned} r(\Phi)&= \lim_{n\rightarrow\infty} {\bigl\ (\Phi)^{n}\bigr\ }^{\frac{1}{n}} \\ &= \lim_{n\rightarrow\infty} {\bigl\ (\Phi)^{2n}\bigr\ }^{\frac{1}{2n}} \\ &\leq \lim_{n\rightarrow\infty} {\bigl\ \bigl(L^{2}+S^{2} \bigr)^{n}\bigr\ }^{\frac{1}{2n}} \\ &=\sqrt{r\bigl(L^{2}+S^{2}\bigr)} \\ &\leq \sqrt{{r(L)}^{2}+{r(S)}^{2}} \\ &\leq r(L)+r(S) \\ &< 1. \end{aligned}$$
Then, by [4] Theorem 3.2, F has a unique fixed point \((\overline {x},\overline{y})\in X\times X\).
Now, we extend the study to the multivalued case.
In the same manner as the proof of Theorem 3.1 and Theorem 4.2, we can verify the following coupled fixed point theorem for multivalued mixed monotone operators.
Theorem 4.4
Suppose
\(T:X\times X\rightarrow2^{X}\setminus\{\phi\}\)
is a multivalued mixed monotone operator. Assume the following conditions are satisfied:

(1)
For any
\((x,y)\in X\times X\), \(T(x,y)\)
is a nonempty and closed subset of
X.

(2)
There exist two linear operators
\(L,S:X\rightarrow X\), \(\L\ +\S\<1\), \(L(P)\subset P\), \(S(P)\subset P\)
such that, for any
\(x_{1}, x_{2}, y_{1}, y_{2}\in X\), \(x_{1}\leq x_{2}\), \(y_{2}\leq y_{1}\)
imply:

(i)
for any
\(u\in T(x_{1},y_{1})\), there exists
\(v\in T(x_{2},y_{2})\)
$$0\leq vu \leq L(x_{2}x_{1})+S(y_{1}y_{2}), $$

(ii)
for any
\(v\in T(x_{2},y_{2})\), there exists
\(u\in T(x_{1},y_{1})\)
$$0\leq vu \leq L(x_{2}x_{1})+S(y_{1}y_{2}). $$
Then
T
has a coupled fixed point in
\(X\times X\).