In this section, we prove some fixed point theorems for expanding mappings without continuity in the following theorems.

**Theorem 2.1** *Let* (X,d) *be a complete cone metric space*. *Suppose the mapping* f:X\to X *is onto and satisfies*

d(fx,fy)\ge {a}_{1}d(x,y)+{a}_{2}d(x,fx)+{a}_{3}d(y,fy)+{a}_{4}d(x,fy)+{a}_{5}d(y,fx)

(2.1)

*for all* x,y\in X, *where* {a}_{i} (i=1,2,3,4,5) *satisfies* {a}_{1}+{a}_{2}+{a}_{3}>1 *and* {a}_{3}\le 1+{a}_{4}. *Then* *f* *has a fixed point*.

*Proof* Since *f* is an onto mapping, for each {x}_{0}\in X, there exists f{x}_{1}={x}_{0}. Continuing this process, we can define \{{x}_{n}\} by {x}_{n}=f{x}_{n+1}, n=0,1,2,\dots . Without loss of generality, we suppose that {x}_{n-1}\ne {x}_{n} for all n\ge 1. According to (2.1), we have

\begin{array}{rcl}d({x}_{n},{x}_{n-1})& =& d(f{x}_{n+1},f{x}_{n})\\ \ge & {a}_{1}d({x}_{n+1},{x}_{n})+{a}_{2}d({x}_{n+1},f{x}_{n+1})+{a}_{3}d({x}_{n},f{x}_{n})\\ +{a}_{4}d({x}_{n+1},f{x}_{n})+{a}_{5}d({x}_{n},f{x}_{n+1})\\ =& {a}_{1}d({x}_{n+1},{x}_{n})+{a}_{2}d({x}_{n+1},{x}_{n})+{a}_{3}d({x}_{n},{x}_{n-1})+{a}_{4}d({x}_{n+1},{x}_{n-1})+{a}_{5}d({x}_{n},{x}_{n}).\end{array}

By d({x}_{n+1},{x}_{n-1})\ge d({x}_{n+1},{x}_{n})-d({x}_{n-1},{x}_{n}), the above inequality implies that

d({x}_{n+1},{x}_{n})\le \frac{1-{a}_{3}+{a}_{4}}{{a}_{1}+{a}_{2}+{a}_{4}}d({x}_{n},{x}_{n-1}).

Let h=\frac{1-{a}_{3}+{a}_{4}}{{a}_{1}+{a}_{2}+{a}_{4}}. By {a}_{3}\le 1+{a}_{4} and {a}_{1}+{a}_{2}+{a}_{3}>1, we know {a}_{1}+{a}_{2}+{a}_{4}>1-{a}_{3}+{a}_{4}\ge 0 and h\in [0,1). Hence, we get

d({x}_{n+1},{x}_{n})\le hd({x}_{n},{x}_{n-1}).

So, by the triangle inequality, for any n>m, we see

\begin{array}{rl}d({x}_{n},{x}_{m})& \le d({x}_{n},{x}_{n-1})+d({x}_{n-1},{x}_{n-2})+\cdots +d({x}_{m+1},{x}_{m})\\ \le ({h}^{n-1}+{h}^{n-2}+\cdots +{h}^{m})d({x}_{1},{x}_{0})\\ \le \frac{{h}^{m}}{1-h}d({x}_{1},{x}_{0}).\end{array}

Thus, as h\in [0,1), we can choose a natural number {N}_{0} such that \frac{{h}^{m}}{1-h}d({x}_{1},{x}_{0})\ll c for each c\gg \theta and m>{N}_{0}. Hence, we see

d({x}_{n},{x}_{m})\ll c\phantom{\rule{1em}{0ex}}\text{for all}nm{N}_{0}.

Therefore, \{{x}_{n}\} is a Cauchy sequence in (X,d).

Since *X* is complete, there exists q\in X such that f{x}_{n+1}={x}_{n}\to q as n\to \mathrm{\infty}. Consequently, we can find a p\in X such that fp=q. Now, we show that p=q. Substituting x=p, y={x}_{n+1} in (2.1), we get

\begin{array}{rl}d(q,{x}_{n})& =d(fp,f{x}_{n+1})\\ \ge {a}_{1}d(p,{x}_{n+1})+{a}_{2}d(p,fp)+{a}_{3}d({x}_{n+1},f{x}_{n+1})+{a}_{4}d(p,f{x}_{n+1})+{a}_{5}d({x}_{n+1},fp).\end{array}

For the second and fourth term on the right-hand side, we have d(p,q)\ge d(p,{x}_{n+1})-d(q,{x}_{n+1}) and d(p,{x}_{n})\ge d(p,{x}_{n+1})-d({x}_{n},{x}_{n+1}). For the left-hand side, d(q,{x}_{n})\le d(q,{x}_{n+1})+d({x}_{n+1},{x}_{n}). It follows that

({a}_{1}+{a}_{2}+{a}_{4})d(p,{x}_{n+1})\le (1+{a}_{2}-{a}_{5})d(q,{x}_{n+1})+(1-{a}_{3}+{a}_{4})d({x}_{n+1},{x}_{n}).

Now, we have

\begin{array}{rl}d(p,{x}_{n+1})& \le \frac{1+{a}_{2}-{a}_{5}}{{a}_{1}+{a}_{2}+{a}_{4}}d(q,{x}_{n+1})+\frac{1-{a}_{3}+{a}_{4}}{{a}_{1}+{a}_{2}+{a}_{4}}d({x}_{n+1},{x}_{n})\\ \le \frac{1+{a}_{2}-{a}_{5}}{{a}_{1}+{a}_{2}+{a}_{4}}d(q,{x}_{n+1})+d({x}_{n+1},{x}_{n}).\end{array}

If 1+{a}_{2}-{a}_{5}>0 for each c\gg \theta, we can choose a natural number {N}_{1} such that d({x}_{n+1},{x}_{n})\ll \frac{c}{2} and d(q,{x}_{n+1})\ll \frac{({a}_{1}+{a}_{2}+{a}_{4})c}{2(1+{a}_{2}-{a}_{5})} for n\ge {N}_{1}. Thus, we obtain

d(p,{x}_{n+1})\ll \frac{c}{2}+\frac{c}{2}=c.

If 1+{a}_{2}-{a}_{5}\le 0 for n\ge {N}_{1},

d(p,{x}_{n+1})\le d({x}_{n+1},{x}_{n})\ll \frac{c}{2}\ll c.

Therefore, {x}_{n+1}\to p. From Lemma 1.3, we see p=q. The conclusion is true. □

Taking some particular value of {a}_{i} (i=1,2,3,4,5) in Theorem 2.1, we obtain several new results in the following.

**Corollary 2.2** *Let* (X,d) *be a complete cone metric space*. *Suppose the mapping* f:X\to X *is onto and satisfies*

d(fx,fy)\ge kd(x,y)+ld(x,fx)+pd(y,fy)

*for all* x,y\in X, *where* p\le 1 *and* k+l+p>1. *Then* *f* *has a fixed point*.

**Corollary 2.3** *Let* (X,d) *be a complete cone metric space*. *Suppose the mapping* f:X\to X *is onto and satisfies*

d(fx,fy)\ge kd(x,y)+ld(fx,y)

*for all* x,y\in X, *where* *k*, *l* *are constants and* k>1. *Then* *f* *has a fixed point*.

**Remark 2.4** Obviously, in our theorem and its corollaries above, we delete the continuity of the mappings which is essential in the results of [9]. Moreover, in Corollary 2.2 we delete k\ge -1, l>1, which is essential in Theorem 2.6 in [9]. In Corollary 2.3 we delete l\ge 0, which is essential in Theorem 2.5 in [9]. Theorem 2.3 in [9] is a special case of Theorem 2.1 with {a}_{1}={a}_{4}={a}_{5}=0, {a}_{2}={a}_{3}=K, and *f* is continuous.

Now, we introduce some common fixed point theorems for two expanding mappings which satisfy generalized expansive conditions without continuity of the mappings.

**Theorem 2.5** *Let* (X,d) *be a complete cone metric space*. *Suppose mappings* f,g:X\to X *are onto and satisfy*

d(fx,gy)\ge {a}_{1}d(x,y)+{a}_{2}d(x,fx)+{a}_{3}d(y,gy)+{a}_{4}d(x,gy)+{a}_{5}d(y,fx)

(2.2)

*for all* x,y\in X, *where* {a}_{i} (i=1,2,3,4,5) *satisfies* {a}_{1}+{a}_{2}+{a}_{3}>1 *and* {a}_{2}\le 1+{a}_{5}, {a}_{3}\le 1+{a}_{4}. *Then* *f* *and* *g* *have a common fixed point*.

*Proof* Suppose {x}_{0} is an arbitrary point in *X*. Since *f*, *g* are onto, there exist {x}_{1},{x}_{2}\in X such that {x}_{0}=g{x}_{1}, {x}_{1}=f{x}_{2}. Continuing this process, we can define \{{x}_{n}\} by {x}_{2n}=g{x}_{2n+1}, {x}_{2n+1}=f{x}_{2n+2}, n=0,1,2,\dots . By (2.2), we have

\begin{array}{rcl}d({x}_{2n+1},{x}_{2n})& =& d(f{x}_{2n+2},g{x}_{2n+1})\\ \ge & {a}_{1}d({x}_{2n+2},{x}_{2n+1})+{a}_{2}d({x}_{2n+2},f{x}_{2n+2})+{a}_{3}d({x}_{2n+1},g{x}_{2n+1})\\ +{a}_{4}d({x}_{2n+2},g{x}_{2n+1})+{a}_{5}d({x}_{2n+1},f{x}_{2n+2})\\ =& {a}_{1}d({x}_{2n+2},{x}_{2n+1})+{a}_{2}d({x}_{2n+2},{x}_{2n+1})+{a}_{3}d({x}_{2n+1},{x}_{2n})\\ +{a}_{4}d({x}_{2n+2},{x}_{2n})+{a}_{5}d({x}_{2n+1},{x}_{2n+1}).\end{array}

Since d({x}_{2n+2},{x}_{2n})\ge d({x}_{2n+2},{x}_{2n+1})-d({x}_{2n},{x}_{2n+1}), the above inequality implies that

(1-{a}_{3}+{a}_{4})d({x}_{2n},{x}_{2n+1})\ge ({a}_{1}+{a}_{2}+{a}_{4})d({x}_{2n+1},{x}_{2n+2}).

Similarly, it can be shown that

\begin{array}{rcl}d({x}_{2n-1},{x}_{2n})& =& d(f{x}_{2n},g{x}_{2n+1})\\ \ge & {a}_{1}d({x}_{2n},{x}_{2n+1})+{a}_{2}d({x}_{2n},f{x}_{2n})+{a}_{3}d({x}_{2n+1},g{x}_{2n+1})\\ +{a}_{4}d({x}_{2n},g{x}_{2n+1})+{a}_{5}d({x}_{2n+1},f{x}_{2n})\\ =& {a}_{1}d({x}_{2n},{x}_{2n+1})+{a}_{2}d({x}_{2n},{x}_{2n-1})+{a}_{3}d({x}_{2n+1},{x}_{2n})\\ +{a}_{4}d({x}_{2n},{x}_{2n})+{a}_{5}d({x}_{2n+1},{x}_{2n-1}),\end{array}

which also implies that

(1-{a}_{2}+{a}_{5})d({x}_{2n-1},{x}_{2n})\ge ({a}_{1}+{a}_{3}+{a}_{5})d({x}_{2n},{x}_{2n+1}).

Let M=\frac{1-{a}_{3}+{a}_{4}}{{a}_{1}+{a}_{2}+{a}_{4}}, N=\frac{1-{a}_{2}+{a}_{5}}{{a}_{1}+{a}_{3}+{a}_{5}}. From {a}_{1}+{a}_{2}+{a}_{3}>1 and {a}_{2}\le 1+{a}_{5}, {a}_{3}\le 1+{a}_{4}, we see {a}_{1}+{a}_{2}+{a}_{4}>1-{a}_{3}+{a}_{4}\ge 0 and {a}_{1}+{a}_{3}+{a}_{5}>1-{a}_{2}+{a}_{5}\ge 0. Thus, h=MN\in [0,1). Now, by induction we have

\begin{array}{rl}d({x}_{2n+2},{x}_{2n+1})& \le Md({x}_{2n+1},{x}_{2n})\le MNd({x}_{2n},{x}_{2n-1})\\ \le {M}^{2}Nd({x}_{2n-1},{x}_{2n-2})\le \cdots \le M{h}^{n}d({x}_{1},{x}_{0})\end{array}

and

d({x}_{2n+1},{x}_{2n})\le Nd({x}_{2n},{x}_{2n-1})\le \cdots \le {h}^{n}d({x}_{1},{x}_{0}).

Hence, for any n>m, we deduce

\begin{array}{rl}d({x}_{2n+1},{x}_{2m+1})& \le d({x}_{2n+1},{x}_{2n})+d({x}_{2n},{x}_{2n-1})+\cdots +d({x}_{2m+2},{x}_{2m+1})\\ \le (\sum _{i=m+1}^{n}{h}^{i}+M\sum _{i=m}^{n-1}{h}^{i})d({x}_{1},{x}_{0})\\ \le (\frac{{h}^{m+1}}{1-h}+\frac{M{h}^{m}}{1-h})d({x}_{1},{x}_{0})\\ =(N+1)\frac{M{h}^{m}}{1-h}d({x}_{1},{x}_{0}).\end{array}

In an analogous way, we gain

and

d({x}_{2n+1},{x}_{2m})\le (M+1)\frac{{h}^{m}}{1-h}d({x}_{1},{x}_{0}).

Thus, for n>m>0,

d({x}_{n},{x}_{m})\le max\{(N+1)\frac{M{h}^{m}}{1-h},(M+1)\frac{{h}^{m}}{1-h}\}d({x}_{1},{x}_{0})={\lambda}_{m}d({x}_{1},{x}_{0}),

where {\lambda}_{m}\to 0 as m\to \mathrm{\infty}.

For each c\gg \theta, choose \delta >0 such that c-x\in intP, where \parallel x\parallel <\delta, *i.e.*, x\ll c. For this *δ*, we can choose a natural number {N}_{2} such that \parallel {\lambda}_{m}d({x}_{1},{x}_{0})\parallel <\delta for m>{N}_{2}. Thus, we get

d({x}_{n},{x}_{m})\le {\lambda}_{m}d({x}_{1},{x}_{0})\ll c\phantom{\rule{1em}{0ex}}\text{for all}nm{N}_{2}.

Therefore, \{{x}_{n}\} is a Cauchy sequence in (X,d).

As *X* is complete, there exists q\in X such that {x}_{n}\to q as n\to \mathrm{\infty}. It is equivalent to {x}_{2n}=g{x}_{2n+1}\to q, {x}_{2n+1}=f{x}_{2n+2}\to q as n\to \mathrm{\infty}. Since *f*, *g* are onto, there exist u,p\in X such that fu=gp=q. Now, we show that u=p=q. By (2.2), we have

\begin{array}{rcl}d(f{x}_{2n+2},gp)& \ge & {a}_{1}d({x}_{2n+2},p)+{a}_{2}d({x}_{2n+2},f{x}_{2n+2})+{a}_{3}d(p,gp)\\ +{a}_{4}d({x}_{2n+2},gp)+{a}_{5}d(p,f{x}_{2n+2}),\end{array}

that is,

d({x}_{2n+1},q)\ge {a}_{1}d({x}_{2n+2},p)+{a}_{2}d({x}_{2n+2},{x}_{2n+1})+{a}_{3}d(p,q)+{a}_{4}d({x}_{2n+2},q)+{a}_{5}d(p,{x}_{2n+1}).

From the fact that d(p,q)\ge d(p,{x}_{2n+2})-d(q,{x}_{2n+2}), d(p,{x}_{2n+1})\ge d(p,{x}_{2n+2})-d({x}_{2n+1},{x}_{2n+2}) and d({x}_{2n+1},q)\le d({x}_{2n+1},{x}_{2n+2})+d({x}_{2n+2},q), we get

({a}_{1}+{a}_{3}+{a}_{5})d(p,{x}_{2n+2})\le (1+{a}_{3}-{a}_{4})d({x}_{2n+2},q)+(1-{a}_{2}+{a}_{5})d({x}_{2n+1},{x}_{2n+2}).

Now, we have

\begin{array}{rl}d(p,{x}_{2n+2})& \le \frac{1+{a}_{3}-{a}_{4}}{{a}_{1}+{a}_{3}+{a}_{5}}d({x}_{2n+2},q)+\frac{1-{a}_{2}+{a}_{5}}{{a}_{1}+{a}_{3}+{a}_{5}}d({x}_{2n+1},{x}_{2n+2})\\ \le \frac{2}{{a}_{1}+{a}_{3}+{a}_{5}}d({x}_{2n+2},q)+d({x}_{2n+1},{x}_{2n+2}).\end{array}

For each c\gg \theta, we can choose a natural number {N}_{3} such that d({x}_{2n+1},{x}_{2n+2})\ll \frac{c}{2} and d({x}_{2n+2},q)\ll \frac{({a}_{1}+{a}_{3}+{a}_{5})c}{4} for n\ge {N}_{3}. Hence, we obtain d(p,{x}_{2n+2})\ll \frac{c}{2}+\frac{c}{2}=c, *i.e.*, {x}_{2n+2}\to p. By Lemma 1.3, we know p=q, gq=q. Similarly, we also have

\begin{array}{rcl}d(fu,g{x}_{2n+1})& \ge & {a}_{1}d(u,{x}_{2n+1})+{a}_{2}d(u,fu)+{a}_{3}d({x}_{2n+1},g{x}_{2n+1})\\ +{a}_{4}d(u,g{x}_{2n+1})+{a}_{5}d({x}_{2n+1},fu).\end{array}

As in the previous proof, it is not difficult to get q=u, *i.e.*, fq=q. Therefore, fq=gq=q. □

**Corollary 2.6** *Let* (X,d) *be a complete cone metric space*. *Suppose mappings* f,g:X\to X *are onto and satisfy*

d(fx,gy)\ge \alpha d(x,y)+\beta [d(x,fx)+d(y,gy)]+\gamma [d(x,gy)+d(y,fx)]

*for all* x,y\in X, *where* \beta \le 1+\gamma *and* \alpha +2\beta >1. *Then* *f* *and* *g* *have a common fixed point*.

**Corollary 2.7** *Let* (X,d) *be a complete cone metric space*. *Suppose mappings* f,g:X\to X *are onto and satisfy*

*for all* x,y\in X, *where* k>1 *is a constant*. *Then* *f* *and* *g* *have a unique common fixed point*.

**Corollary 2.8** *Let* (X,d) *be a complete cone metric space*. *Suppose the mapping* f:X\to X *is onto and satisfies*

d({f}^{p}x,{f}^{q}y)\ge kd(x,y)

*for all* x,y\in X, *where* *p*, *q* *are positive integers and* k>1 *is a constant*. *Then* *f* *has a unique fixed point*.

*Proof* Let f={f}^{p}, g={f}^{q}. Since *f* is an onto mapping, f={f}^{p}, g={f}^{q} are onto mappings, the conditions of Corollary 2.7 are satisfied. □

**Remark 2.9** In Corollary 2.8, we obtain Corollary 2.2 in [9] when we take p=q.

Now, we present the following examples. In Example 1, we gain a fixed point for one expanding mapping of the situation when Corollary 2.2 can be applied, while the results in [9] cannot. In Example 2, we obtain the common fixed point for two expanding mappings in a cone metric space.

**Example 1** Let X=[1,+\mathrm{\infty}), E={C}_{\mathbb{R}}^{2}([0,1]) with \parallel x\parallel ={\parallel x\parallel}_{\mathrm{\infty}}+{\parallel {x}^{\prime}\parallel}_{\mathrm{\infty}} and P=\{x\in E:x(t)\ge 0,t\in [0,1]\} (this cone is not normal). Define d:X\times X\to E by d(x,y)=|x-y|\phi, where \phi :[0,1]\to \mathbb{R} such that \phi (t)={e}^{t}. Consider the mapping

f(x)=\{\begin{array}{cc}2{x}^{2}-x,\hfill & 1\le x<2;\hfill \\ \frac{5}{2}x,\hfill & x\ge 2,\hfill \end{array}

which implies that *f* is onto in *X*. Taking k=2, l=\frac{1}{4}, p=-\frac{1}{4}, for 1\le x<2, all the conditions of Corollary 2.2 are fulfilled. Indeed, since 0<\frac{1}{2}x+\frac{1}{2}y+\frac{3}{2}<2x+2y-1, we have

\begin{array}{rl}d(fx,fy)& =|2{x}^{2}-x-(2{y}^{2}-y)|{e}^{t}\\ =|(x-y)(2x+2y-1)|{e}^{t}\\ \ge |x-y|(\frac{1}{2}x+\frac{1}{2}y+\frac{3}{2}){e}^{t}\\ =2|x-y|{e}^{t}+\frac{1}{2}|(x-y)(x+y-1)|{e}^{t}\\ \ge 2|x-y|{e}^{t}+\frac{1}{4}|2x-2{x}^{2}|{e}^{t}-\frac{1}{4}|2y-2{y}^{2}|{e}^{t}.\end{array}

For x\ge 2, since f(x) is increasing in *x*, we have

d(fx,fy)=|\frac{5}{2}x-\frac{5}{2}y|{e}^{t}\ge 2|x-y|{e}^{t}+\frac{1}{4}|x-\frac{5}{2}x|{e}^{t}-\frac{1}{4}|y-\frac{5}{2}y|{e}^{t}.

Therefore, we can apply Corollary 2.2 and conclude that *f* has a (unique) fixed point 0 in *X*. Since *f* is not continuous in *X* and l<1, Theorem 2.6 in [9] is not applicable. Hence, our theorems have improved and generalized the main results in [9].

**Example 2** Let X=\{1,2,3\} and d:X\times X\to {\mathbb{R}}^{2} be defined by d(x,y)=(0,0) for x=y and

d(2,3)=d(3,2)=(0,0),\phantom{\rule{2em}{0ex}}d(2,1)=d(1,2)=(1,1),\phantom{\rule{2em}{0ex}}d(1,3)=d(3,1)=(1,1).

Then (X,d) is a complete cone metric space. Further, define mappings f,g:X\to X as follows:

f(x)=\{\begin{array}{cc}1,\hfill & x=1,\hfill \\ 3,\hfill & x=2,\hfill \\ 2,\hfill & x=3;\hfill \end{array}\phantom{\rule{2em}{0ex}}g(x)=\{\begin{array}{cc}1,\hfill & x=1,\hfill \\ 2,\hfill & x=2,\hfill \\ 3,\hfill & x=3,\hfill \end{array}

which implies that *f*, *g* are onto in *X*. Note that

d(fx,gy)\ge {a}_{1}d(x,y)+{a}_{2}d(x,fx)+{a}_{3}d(y,gy)+{a}_{4}d(x,gy)+{a}_{5}d(y,fx)

for all x,y\in X by taking {a}_{1}=-\frac{1}{7}, {a}_{2}=-\frac{2}{7}, {a}_{3}=\frac{11}{7}, {a}_{4}=\frac{5}{7}, {a}_{5}=\frac{3}{7}. Thus, all the conditions of Theorem 2.5 are fulfilled. Then *f* and *g* have a unique common fixed point 1 in *X*.

**Remark 2.10** Obviously, in the above two examples, we obtain the (common) fixed point which essentially needs the structure of a cone metric and not an ordinary metric on *X*. Then the results in a metric space in [10] cannot be applied to these examples.