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I'm looking for theCan someone provide a proof or a source containing a proof of the following theorem:

Theorem: Let $D$ be a subset of the cone $K$ of partially ordered space $E,$ $F:D\rightarrow E$ be nondecreasing. If there exist $x_{0},y_{0}\in D$ such that $x_{0}\leq y_{0},$ $\left\langle x_{0}% ,y_{0}\right\rangle \subset D $ and $x_{0},y_{0}$ are respectively lower and upper solutions of equation $x-F\left( x\right) =0$, then the equation $x-F\left( x\right) =0$ has minimum solution and maximum solution $x^{\ast },y^{\ast}$ in $\left\langle x_{0},y_{0}\right\rangle $ such that $x^{\ast }\leq y^{\ast}$, when one of the following conditions holds

  1. $K$ is normal and $F$ is completely continuous;

  2. $K$ is regular and $F$ is continuous;

  3. $E$ is reflexive, $K$ is normal, and $F$ is continuous or weak continuous.

I'm looking for the proof of the following theorem:

Theorem: Let $D$ be a subset of the cone $K$ of partially ordered space $E,$ $F:D\rightarrow E$ be nondecreasing. If there exist $x_{0},y_{0}\in D$ such that $x_{0}\leq y_{0},$ $\left\langle x_{0}% ,y_{0}\right\rangle \subset D $ and $x_{0},y_{0}$ are respectively lower and upper solutions of equation $x-F\left( x\right) =0$, then the equation $x-F\left( x\right) =0$ has minimum solution and maximum solution $x^{\ast },y^{\ast}$ in $\left\langle x_{0},y_{0}\right\rangle $ such that $x^{\ast }\leq y^{\ast}$, when one of the following conditions holds

  1. $K$ is normal and $F$ is completely continuous;

  2. $K$ is regular and $F$ is continuous;

  3. $E$ is reflexive, $K$ is normal, and $F$ is continuous or weak continuous.

Can someone provide a proof or a source containing a proof of the following theorem

Theorem: Let $D$ be a subset of the cone $K$ of partially ordered space $E,$ $F:D\rightarrow E$ be nondecreasing. If there exist $x_{0},y_{0}\in D$ such that $x_{0}\leq y_{0},$ $\left\langle x_{0}% ,y_{0}\right\rangle \subset D $ and $x_{0},y_{0}$ are respectively lower and upper solutions of equation $x-F\left( x\right) =0$, then the equation $x-F\left( x\right) =0$ has minimum solution and maximum solution $x^{\ast },y^{\ast}$ in $\left\langle x_{0},y_{0}\right\rangle $ such that $x^{\ast }\leq y^{\ast}$, when one of the following conditions holds

  1. $K$ is normal and $F$ is completely continuous;

  2. $K$ is regular and $F$ is continuous;

  3. $E$ is reflexive, $K$ is normal, and $F$ is continuous or weak continuous.

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Fixed point theorem in ordered spaces

I'm looking for the proof of the following theorem:

Theorem: Let $D$ be a subset of the cone $K$ of partially ordered space $E,$ $F:D\rightarrow E$ be nondecreasing. If there exist $x_{0},y_{0}\in D$ such that $x_{0}\leq y_{0},$ $\left\langle x_{0}% ,y_{0}\right\rangle \subset D $ and $x_{0},y_{0}$ are respectively lower and upper solutions of equation $x-F\left( x\right) =0$, then the equation $x-F\left( x\right) =0$ has minimum solution and maximum solution $x^{\ast },y^{\ast}$ in $\left\langle x_{0},y_{0}\right\rangle $ such that $x^{\ast }\leq y^{\ast}$, when one of the following conditions holds

  1. $K$ is normal and $F$ is completely continuous;

  2. $K$ is regular and $F$ is continuous;

  3. $E$ is reflexive, $K$ is normal, and $F$ is continuous or weak continuous.