Often in functional analysis one needs local algebraic linearity. Thus, one of the interactions of the algebraic and topological structure of a topological vector is manifested in the important properties of the class of convex functions. So far, we have allowed the convex functions defined on the convex subsets of a vector space to be real valued. We will extend the definition of convexity to the valued functions in a Banach lattice.

Definition 1.1 Let $E$ be a Banach lattice. A function $f:C\to E$ on a convex set C in a vector space $X$ is:

1) order convex (denoted by o-convex) if for all $x,y\in C$ and all $0\le \alpha \le 1$, $f\left(\alpha x+\left(1-\alpha \right)y\right)\le \alpha f\left(x\right)+\left(1-\alpha \right)f\left(y\right)$.

2) strictly o-convex if for all $x,y\in C$ with $x\ne y$ and all $0<\alpha <1$, $f\left(\alpha x+\left(1-\alpha \right)y\right)<\alpha f\left(x\right)+\left(1-\alpha \right)f\left(y\right)$.

3) o-concave (respectively, strictly o-concave) if $-f$ is an o-convex (respectively, strictly o-concave) function.

It is easy to realize that, $f$ is o-convex if and only if,

$f\left(\underset{k=1}{\overset{n}{{\displaystyle \sum }}}{\alpha }_k{x}_k\right)\le \underset{k=1}{\overset{n}{{\displaystyle \sum }}}{\alpha }_{k}f\left(x_k\right)$

for every convex combination $\underset{k=1}{\overset{n}{{\displaystyle \sum }}}{\alpha }_k{x}_k$.

Example 1.1 Here are some familiar examples of o-convex mappings.

• Obviously, any convex real function is o-convex.

• Let $E$ be a Banach lattice. The absolute value $x\mapsto \left|x\right|$ is an o-convex mappings from $E$ to $E$.

• Let $A$ be a commutative unital real Banach algebra. The set of all multiplicative linear functionals on $A$ is denoted by ${\Delta }_A$. It is well known that ${\Delta }_A$, endowed with the Gelfand topology, is compact and the Gelfand representation $\varphi$ of $A$ into $C\left({\Delta }_A\right)$ is an homomorphism ( [1] , Theorem 13). Thus $\varphi$ is o-convex.

Proposition 1.1 A function $f:C\to E$ on a convex subset of a vector space into a Banach lattice $E$ is o-convex if and only if its epigraph, $\text{epi}\left(f\right)=\left\{\left(x,\chi \right)\in C\times E:\chi \ge f\left(x\right)\right\}$, is convex. Similarly, $f$ is o-concave if and only if its hypograph is concave.

Proof. We prove the first part of this proposition. The remaining assertion is identical. Suppose that $f$ is o-convex, then for $\left(x_1,{\chi }_1\right),\left(x_2,{\chi }_2\right)\in \text{epi}\left(f\right)$ and $\alpha \in \left[0,1\right]$ we have

$\begin{array}{c}{\alpha }_1{\chi }_1+\left(1-\alpha \right){\chi }_2\ge {\alpha }_{1}f\left(x_1\right)+\left(1-\alpha \right)f\left(x_2\right)\\ \ge f\left(\alpha x_1+\left(1-\alpha \right)x_2\right)\end{array}$

So, $\left({\alpha }_1{\chi }_1+\left(1-\alpha \right){\chi }_2,{\alpha }_1{x}_1+\left(1-\alpha \right)x_2\right)\in \text{epi}\left(f\right)$. The “only if” part stems from the fact that $\left(x_1,f\left(x_1\right)\right)\in \text{epi}\left(f\right)$ and $\left(x_2,f\left(x_2\right)\right)\in \text{epi}\left(f\right)$. ◼

Proposition 1.2 The collection of o-convex functions on a fixed convex set $C$ into a Banach lattice $E$ has the following properties:

1) Sums and nonnegative scalar multiples of o-convex functions are o-convex.

2) The (finite) pointwise order limit of a net of o-convex functions is o-convex.

3) The (finite) pointwise supremum of a family of o-convex functions is o-convex.

Proof. The first statement is trivial. For the second assertion, consider a net $\left\{f_i\right\}$ of o-convex functions (finite) pointwise order convergent to $f$, that is, for any finite part $F$ of $C$, there is a net $\left\{{\chi }_i\right\}$ (with the same directed set) satisfying ${\chi }_i\downarrow 0$ and $\left|f_i\left(z\right)-f\left(z\right)\right|\le {\chi }_i$ for each $i$ and every $z\in F$. Let $x,y\in C$ and $\alpha \in \left[0,1\right]$. For $F=\left\{x,y,\alpha x+\left(1-\alpha \right)y\right\}$ we have:

$\begin{array}{c}f\left(\alpha x+\left(1-\alpha \right)y\right)\le {\chi }_i+f_i\left(\alpha x+\left(1-\alpha \right)y\right)\\ \le {\chi }_i+\alpha f_i\left(x\right)+\left(1-\alpha \right)f_i\left(y\right)\\ \le {\chi }_i+\alpha \left[f\left(x\right)+f_i\left(x\right)-f\left(x\right)\right]\\ +\left(1-\alpha \right)\left[f\left(y\right)+f_i\left(y\right)-f\left(y\right)\right]\\ \le {\chi }_i+\alpha f\left(x\right)+\left(1-\alpha \right)f\left(y\right)+{\chi }_i\\ \le \left[2{\chi }_i+\alpha f\left(x\right)+\left(1-\alpha \right)f\left(y\right)\right]\downarrow \alpha f\left(x\right)+\left(1-\alpha \right)f\left(y\right)\end{array}$

So, $f$ is o-convex.

Now, let $f_1,f_2,\cdots ,f_n$ be o-convex functions on a convex set $C$ into a Banach lattice $E$. For all $x\in C$, we define $f\left(x\right)=\underset{1\le k\le n}{{\displaystyle V}}{f}_k\left(x\right)$. It is easy to see that:

$\begin{array}{c}f\left(\alpha x+\left(1-\alpha \right)y\right)=\underset{1\le k\le n}{{\displaystyle V}}{f}_k\left(\alpha x+\left(1-\alpha \right)y\right)\\ \le \underset{1\le k\le n}{{\displaystyle V}}\left[\alpha f_k\left(x\right)+\left(1-\alpha \right)f_k\left(y\right)\right]\\ \le \underset{1\le k\le n}{{\displaystyle V}}\left[\alpha f_k\left(x\right)\right]+\underset{1\le k\le n}{{\displaystyle V}}\left[\left(1-\alpha \right)f_k\left(y\right)\right]\\ \le \alpha \underset{1\le k\le n}{{\displaystyle V}}{f}_k\left(x\right)+\left(1-\alpha \right)\underset{1\le k\le n}{{\displaystyle V}}{f}_k\left(y\right)\\ \le \alpha f\left(x\right)+\left(1-\alpha \right)f\left(y\right).\end{array}$

So, $f$ is o-convex, what completes the proof. ◼

Proposition 1.3 Let $f:C\to E$ be an o-convex function, $x\in C$ and $x\mp z\in C$. Then for all $\alpha \in \left[0,1\right]$,

$\left|f\left(x+\alpha z\right)-f\left(x\right)\right|\le \alpha \left(\left[f\left(x+z\right)-f\left(x\right)\right]\vee \left[f\left(x-z\right)-f\left(x\right)\right]\right).$

Proof. $f\left(x+\alpha z\right)\le \left(1-\alpha \right)f\left(x\right)+\alpha f\left(x+z\right)$ because the hypothesis and the equality: $x+\alpha z=\left(1-\alpha \right)x+\alpha \left(x+z\right)$. Rearranging terms yields

$f\left(x+\alpha z\right)-f\left(x\right)\le \alpha \left[f\left(x+z\right)-f\left(x\right)\right]$(1.1)

$\le \alpha \left(\left[f\left(x+z\right)-f\left(x\right)\right]\vee \left[f\left(x-z\right)-f\left(x\right)\right]\right)$(1.2)

Replacing $z$ by $-z$ in (1.1) gives

$f\left(x-\alpha z\right)-f\left(x\right)\le \alpha \left[f\left(x-z\right)-f\left(x\right)\right]$(1.3)

Since $x=\frac{1}{2}\left(x+\alpha z\right)+\frac{1}{2}\left(x-\alpha z\right)$, we have $f\left(x\right)\le \frac{1}{2}f\left(x+\alpha z\right)+\frac{1}{2}f\left(x-\alpha z\right)$. Multiplying by two and rearranging terms we obtain

$f\left(x\right)-f\left(x+\alpha z\right)\le f\left(x-\alpha z\right)-f\left(x\right)$(1.4)

(1.3) implies

$\begin{array}{c}f\left(x\right)-f\left(x+\alpha z\right)\le f\left(x-\alpha z\right)-f\left(x\right)\\ \le \alpha \left[f\left(x-z\right)-f\left(x\right)\right]\\ \le \alpha \left(\left[f\left(x+z\right)-f\left(x\right)\right]\vee \left[f\left(x-z\right)-f\left(x\right)\right]\right)\end{array}$(1.5)

With definition of the absolute value in mind, (1.2) in conjunction with (1.5) yields the conclusion of the proposition. ◼

Recall that a subset $A$ of a Riesz space $X$ is order bounded, from above if there is a vector $u$ (called an upper bound of $A$) that dominates each element of $A$, that is, satisfying $a\le u$ for each $a\in A$. Sets order bounded from below are defined similarly. A box or an order interval, is any set of the form

$\left[a,b\right]=\left\{x\in X:a\le x\le b\right\}$

Definition 1.2 A mapping $f:X\to E$ between Riesz spaces is o-bounded above (respectively, o-bounded) on a subset $V$ of $X$, if $f\left(V\right)$ is order bounded from above (respectively, if $f\left(V\right)\subset \left[a,b\right]$ for some box $\left[a,b\right]$ of $E$).

We would have liked an o-convex function $f:X\to E$ to be order continuous, but this is not true even in the trivial case when $E=ℝ$. Indeed, let $X=C\left[0,1\right]$, we emphasize: There is no nonzero $\sigma$-order continuous linear functional on the Riesz space $X$ (see for example ( [2] , p. 329)). However, for the topological continuity we have the following, which generalizes a similar result well known for the convex (real) functions.

Theorem 1.1 If an o-convex function $f:C\to E$ is o-bounded above in a neighborhood of an interior point $x\in C$, then $f$ is continuous at $x$.

Proof. We may assume that for some $x\in C$ there exist an open ball $V$ of radius $\eta$ at 0 and some $\chi \in E$ satisfying $x+V\subset C$ and $f\left(y\right)\le f\left(x\right)+\chi$ for each $y\in x+V$. Fix $\epsilon >0$ and choose some $0<\alpha <1$ so that $\alpha \Vert \chi \Vert <\epsilon$. From Proposition1.3, it follows that for each $y\in x+\alpha V$ we have $\left|f\left(y\right)-f\left(x\right)\right|\le \alpha \chi$. Now, the norm of $E$ is lattice, then $\Vert f\left(y\right)-f\left(x\right)\Vert \le \alpha \Vert \chi \Vert <\epsilon$. ◼

Remark 1.1

Provided that the interior $Int\left(E_{+}\right)$ of the cone $E_{+}$ is non-empty, semicontinuity can be generalized to vector functions as follows (For more details on the impact of a cone’s properties on the Riesz space it generates, the reader is referred to [3] ).

Definition 1.3 A mapping $f:X\to E$ from a topological space $X$ into a Banach lattice $E$ is:

$♦$ Lower o-semicontinuous if for each $c\in E$ the set

$\left\{x\in X:f\left(x\right)-c\in Int\left(E_{+}\right)\right\}$

is open.

$♦$ Upper o-semicontinuous if for each $c\in E$ the set

$\left\{x\in X:c-f\left(x\right)\in Int\left(E_{+}\right)\right\}$

is open.

Obviously, a mapping $f$ is lower o-semicontinuous if and only $-f$ is upper o-semicontinuous, and vice versa.

The classic example of a lower (resp. upper) o-semicontinuous mapping is given by the lower (resp. upper) semicontinuous real functions. Now assume that $E$ is a Banach lattice with an order unit $e$. It is well known that the principal ideal $E_e$ generated by $e$ coincides with $E$ which when provided with the norm ${\Vert x\Vert }_{\infty }=\text{inf}\left\{\lambda >0:\left|x\right|\le \lambda e\right\}$ becomes an AM-space with unit. Let $f:ℝ\to ℝ$ be a continuous real function. Then the mapping $\tilde{f}:E\to E$ defined by $\tilde{f}\left(x\right)=f\left({\Vert x\Vert }_{\infty }\right)e$ is a lower and upper o-semicontinuous mapping.

The $E$-valued mapping on a Banach lattice is a useful device, but it needs to be handled with care. For example, the complement of the set $\left\{x\in X:f\left(x\right)\ge c\right\}$ in $X$ is not at all the set $\left\{x\in X:f\left(x\right)<c\right\}$. However, the following lemma reduces this difficulty by reducing us to functions with real values.

Proposition 1.4 If $E_{+}$ (respectively, ${E^{\prime }}_{+}$) is the positive cone of a Banach lattice $E$ (respectively, of $E\text{'}$), then $x\notin E_{+}$ if and only if it exists $\phi \in {E^{\prime }}_{+}$ such that $\phi \left(x\right)<0$.

Proof. The definition of the positive cone ${E^{\prime }}_{+}$ gives a sense of lemma. Conversely, if $x\notin E_{+}$, since $E_{+}$ is closed and convex, it follows from the Hahn-Banach theorem that there is a $\phi \in E^{\prime }$ with $\phi \left(x\right)<\phi \left(y\right)$ for all $y\in E_{+}$. Thus $\phi \left(x\right)<0=f\left(0\right)$ and $\phi \left(x\right)<\phi \left(ny\right)$ for all non-negative integer number $n$. So $\phi \left(x\right)<0\le \phi \left(y\right)$ for all $y\in E_{+}$. ◼

As a first application of the above definitions, we have the following result.

Theorem 1.2 For an o-convex mapping $f:C\to E$ on an open convex subset $C\subset X$, the following are equivalent.

(a) $f$ is continuous on $C$.

(b) $f$ is upper o-semicontinuous.

(c) $f$ is o-bounded above on a neighborhood of each point in $C$.

(d) $f$ is o-bounded above on a neighborhood of some point in $C$.

(e) $f$ is continuous at some point in $C$.

Proof. (a) $\Rightarrow$(b) is obvious.

(b) $\Rightarrow$(c); Assume that $f$ is upper o-semicontinuous. Let $x\in C$ and $a\in Int\left(E_{+}\right)-\left\{0\right\}$. Then the set $\left\{y\in E:f\left(x\right)+a-f\left(y\right)\in Int\left(E_{+}\right)\right\}$ is an open neighborhood of $x$ on which $f$ is o-bounded above.

(c) $\Rightarrow$(d) Obvious.

(d) $\Rightarrow$(e) This is Theorem 1.1.

(e) $\Rightarrow$(a) Suppose $f$ is continuous at the point $x$, and let $y$ be any other point in $C$. Since $C$ is open and convex, therefore $C$ does not contain extreme points. This implies that there exist $z\in C$ and $0<\lambda <1$ such that $y=\lambda x+\left(1-\lambda \right)z$. Fix $\epsilon >0$ and choose some circled neighborhood $V$ of zero

so that $\Vert f\left(x\right)-f\left(x+v\right)\Vert <\frac{\epsilon }{\lambda }$ for all $v\in V$. We claim that, $\Vert f\left(y\right)-f\left(y+v\right)\Vert <\epsilon$ for all $v\in \lambda V$. Indeed, let $v\in V$, Then $y+\lambda v=\lambda \left(x+v\right)+\left(1-\lambda \right)z\in C$ and the o-convexity of $f$ implies

$\begin{array}{c}f\left(y+\lambda v\right)=f\left(\lambda \left(x+v\right)+\left(1-\lambda \right)z\right)\\ \le \lambda f\left(x+v\right)+\left(1-\lambda \right)f\left(z\right)\end{array}$

and

$\begin{array}{c}f\left(y\right)=f\left(\lambda x+\left(1-\lambda \right)z\right)\\ \le \lambda f\left(x\right)+\left(1-\lambda \right)f\left(z\right)\end{array}$

Thus

$\begin{array}{c}f\left(y+\lambda v\right)-f\left(y\right)\le \lambda \left(f\left(\left(x+v\right)-f\left(x\right)\right)\right)\\ \le \lambda \left|f\left(\left(x+v\right)-f\left(x\right)\right)\right|\end{array}$(1.6)

and

$\begin{array}{c}f\left(y\right)-f\left(y+\lambda v\right)\le \lambda \left(f\left(x\right)-f\left(\left(x+v\right)\right)\right)\\ \le \lambda \left|f\left(\left(x+v\right)-f\left(x\right)\right)\right|\end{array}$(1.7)

This shows that

$\left|f\left(y\right)-f\left(y+\lambda v\right)\right|\le \lambda \left|f\left(\left(x+v\right)-f\left(x\right)\right)\right|$(1.8)

Then

$\Vert f\left(y\right)-f\left(y+\lambda v\right)\Vert \le \lambda \Vert f\left(\left(x+v\right)-f\left(x\right)\right)\Vert <\epsilon$

So, $f$ is continuous at $y$. ◼

Lipchitzian and contractive real functions have important properties that we want to extend to infinite dimensional analysis. For this purpose, we adopt the following definition.

Definition 2.1 A mapping f from a subset $B$ of a normed space $\left(X,\Vert \Vert \right)$ to a Banach lattice $E$ is order Lipschitzian on $B$ if there exists $e\in E_{+}$ such that for every $y,z\in B$

$\left|f\left(y\right)-f\left(z\right)\right|\le \Vert y-z\Vert e$

If moreover $\Vert e\Vert <1$ then $f$ is called an order contraction.

The following gives examples of order Lipschitzian functions.

Theorem 2.1 Let $f:C\to E$ be a positive o-convex function from a convex subset $C\subset X$ into a Banach lattice $E$. If $f$ is continuous at some interior point $x$ of $C$, then $f$ is order Lipschitzian on a neighborhood of $x$.

Proof. Since $f$ is continuous at $x$, it follows from Theorem 1.2 that there exists $e\in E_{+}$ and $\delta >0$ satisfying $B_{2\delta }\left(x\right)\subset C$ and $f\left(y\right)\le e$. So, $w,z\in B_{2\delta }\left(x\right)$ implies $0\le f\left(w\right)\le e$ and $-e\le -f\left(z\right)\le 0$. By addition, we achieve $\left|f\left(w\right)-f\left(z\right)\right|\le e$, for all $w,z\in B_{2\delta }$. Let $y,z\in B_{2\delta }\left(x\right)$ and $\alpha =\Vert y-z\Vert$. Then

$w=y+\frac{\delta }{\alpha }\left(y+z\right)$ belongs to $B_{2\delta }$ and we have $y=\frac{\alpha }{\alpha +\delta }w+\frac{\delta }{\alpha +\delta }z$. Therefore

$f\left(y\right)\le \frac{\alpha }{\alpha +\delta }f\left(w\right)+\frac{\delta }{\alpha +\delta }f\left(z\right)$

Subtracting $f\left(z\right)$ from each side gives

$\begin{array}{c}f\left(y\right)-f\left(z\right)\le \frac{\alpha }{\alpha +\delta }\left[f\left(w\right)-f\left(z\right)\right]\\ \le \frac{\alpha }{\alpha +\delta }e\\ \le \alpha e\end{array}$

Switching the roles of $y$ and $z$ allows us to conclude

$\left|f\left(y\right)-f\left(z\right)\right|\le \Vert y-z\Vert e$

◼

A net $\left\{x_{\alpha }\right\}$ in a Riesz space $E$ is order convergent to some $x\in E$, written $\left\{x_{\alpha }\right\}\stackrel{o}{\to }x$, if there is a net $\left\{q_{\alpha }\right\}$ (with the same directed set) satisfying $\left\{q_{\alpha }\right\}\downarrow 0$ and $\left|x_{\alpha }-x\right|\le q_{\alpha }$ for each $\alpha$. A function $f:E\to F$ between two Riesz spaces is order uniformly continuous if $\left\{y_{\alpha }-z_{\alpha }\right\}\stackrel{o}{\to }0$ in $E$ implies $\left\{f\left(y_{\alpha }\right)-f\left(z_{\alpha }\right)\right\}\stackrel{o}{\to }0$ in $F$.

Proposition 2.1 If $f;X\to E$ is order Lipschitz continuous and $\Vert .\Vert$ is an order continuous lattice norm on $X$, then $f$ is order uniformly continuous.

Proof. Obvious. ◼

Now we will generalize, to convex order applications, one of the important themes of the analysis, namely the extreme points of a convex functions on a compact convex set. Let $C$ be a convex subset of a vector space $X$. Recall that an extreme subset of $C$, is a nonempty subset $F$ of $C$ with the property that if $x$ belongs to $F$, it cannot be written as a convex combination of points of $C$ outside $F$. A point $x$ is an extreme point of $C$ if the singleton $\left\{x\right\}$ is an extreme set.

Proposition 2.2 If $f:C\to E$ is o-convex and attains a maximum at some point, then the set of maximizers is an extreme set.

Proof. Suppose $f$ achieves a maximum on $C$; that is, $f$ satisfies the identity $\text{sup}\left\{f\left(x\right):x\in C\right\}=f\left(e\right)$ for some $e\in C$. Put $ℳ=\left\{x\in C:f\left(x\right)=f\left(e\right)\right\}$. Suppose that $x=\alpha y+\left(1-\alpha \right)z\in ℳ$, $0<\alpha <1$ and $y,z\in C$. If $y\notin ℳ$ then $f\left(y\right)<f\left(e\right)$, so

$\begin{array}{c}f\left(e\right)=f\left(x\right)=f\left(\alpha y+\left(1-\alpha \right)z\right)\\ \le \alpha f\left(y\right)+\left(1-\alpha \right)f\left(z\right)\\ <\alpha f\left(e\right)+\left(1-\alpha \right)f\left(e\right)=f\left(e\right)\end{array}$

a contradiction. Hence $y,z\in ℳ$, so $ℳ$ is an extreme subset of $C$. ◼

Recall that the order $\le$ of a Banach lattice $E$ is continuous if $\le$ is a closed subset of $E\times E$. Let us say that $\le$ is upper semicontinuous if $\left\{x\in E:y\le x\right\}$ is closed for each $y$.

Theorem 2.2 Let $K$ be compact and $f:K\to E$ continuous, with $E$ a Banach lattice having continuous order. If $f\left(K\right)$ is closed under suprema (condition C), then $f$ attains a minimum on $K$, and the minimizer set is compact.

Proof. Let $K$ be a compact of a normed vector space $X$ and let $f:K\to E$ be a continuous mapping from $K$ to a Banach lattice $E$. For each $c\in f\left(K\right)$, put $F_c=\left\{x\in K:f\left(x\right)\ge c\right\}$. It follows from the continuity of $f$ and of the order that the nonempty set $F_c$ is closed ( $c=f\left(x\right)$ implies $x\in F_c$). Moreover, the family $ℱ=\left\{F_c:c\in f\left(K\right)\right\}$ has the finite intersection property. In deed, let $F_{c_1},F_{c_2},\cdots ,F_{c_n}$ be a finite family in $ℱ$. Since $F\left(K\right)$ satisfies condition $\left(C\right)$ so, $c_0=\underset{i=1}{\overset{i=n}{{\displaystyle V}}}{c}_i\in f\left(K\right)$. For all $x\in F_{c_0}$ and $1\le i\le n$ we have

$0\le f\left(x\right)-c_0\le f\left(x\right)-c_i$

so, $x\in {\displaystyle \underset{1\le i\le n}{\cap }{F}_{c_i}}$ and $F_{c_0}\subset {\displaystyle \underset{1\le i\le n}{\cap }{F}_{c_i}}$. Since $K$ is compact, ( [2] , Theorem.2.31)

implies that the set of minimizers ${\displaystyle \underset{c\in f\left(K\right)}{\cap }{F}_c}$ is compact and nonempty. ◼

We realize that, in its real context, the assumptions $\left(C\right)$ and $\left(C^{\prime }\right)$ in Theorem 2.2 are ensured from the fact that the order in $ℝ$ is total.

A complete lattice is a lattice in which every nonempty subset that is order bounded from above has a supremum. (Equivalently, if every nonempty subset that is bounded from below has an infimum).

Now consider a vector form of the Contraction Mapping Theorem.

Theorem 2.3 Let $f:B\to B$ be an order contraction on a closed subset $B$ of a Banach lattice $E$, with contraction modulus $\Vert e\Vert <1$. Then f has a unique fixed point $x$, and for any $x_0\in B$, the iterates $x_{n+1}=f\left(x_n\right)$ converges to $x$ with $\Vert x_n-x\Vert \le {\Vert e\Vert }^n\Vert x_0-x\Vert$.

Proof. Let $e\in E_{+}$ such that $\Vert e\Vert <1$ and $\left|f\left(y\right)-f\left(z\right)\right|\le \Vert y-z\Vert e$, for all $y,z\in B$. If $f\left(x\right)=x$ and $f\left(y\right)=y$ then $\left|x-y\right|=\left|f\left(x\right)-f\left(x\right)\right|\le \Vert x-y\Vert e$. Since the norm of $E$ is lattice, we have $\Vert x-y\Vert \le \Vert x-y\Vert \Vert e\Vert$ and hence $\Vert x-y\Vert =0$. Thus $f$ can have at most one fixed point.

Now, if $x_0$ is chosen in $B$ then the formula $x_{n+1}=f\left(x_n\right)$, $n=0,1,2,\cdots$ defines inductively the sequence $\left(x_n\right)$ which satisfies: $\left|x_{n+1}-x_n\right|\le \Vert x_n-x_{n-1}\Vert e$, for every $n\ge 1$. The lattice property verified by the norm of $E$ implies that $\Vert x_{n+1}-x_n\Vert \le \Vert x_n-x_{n-1}\Vert \Vert e\Vert$ and by induction, we see that for all $n⩾1$, $\Vert x_{n+1}-x_n\Vert \le \Vert x_1-x_0\Vert {\Vert e\Vert }^n$. Hence, for $n>m$ the triangle inequality yields

$\begin{array}{c}\Vert x_m-x_n\Vert \le \underset{k=m+1}{\overset{n}{{\displaystyle \sum }}}\Vert x_k-x_{k-1}\Vert \\ \le \Vert x_1-x_0\Vert \underset{k=m+1}{\overset{n}{{\displaystyle \sum }}}{\Vert e\Vert }^k\\ \le \Vert x_1-x_0\Vert \frac{{\Vert e\Vert }^m}{1-\Vert e\Vert }\end{array}$

This implies that $\left(x_n\right)$ is a Cauchy sequence. Since $B$ is closed in the complete space $E$ then, $\left(x_n\right)\to x\in B$. Obviously, $f$ is continuous, and: $x=\underset{n\to \infty }{\lim }{x}_{n+1}=\underset{n\to \infty }{\lim }f\left(x_n\right)=f\left(x\right)$, so $x$ is the fixed point of $f$.