In this section we prove the strong convergence theorems of the modified Siteration process in a CAT(0) space.
Theorem 1 Let K be a nonempty closed convex subset of a complete CAT(0) space X, T:K\to K be asymptotically quasinonexpansive mapping with F(T)\ne \mathrm{\varnothing} and \{{u}_{n}\} be a nonnegative real sequence with {\sum}_{n=1}^{\mathrm{\infty}}{u}_{n}<\mathrm{\infty}. Suppose that \{{x}_{n}\} is defined by the iteration process (1.7). If
\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}inf}d({x}_{n},F(T))=0\phantom{\rule{1em}{0ex}}\mathit{\text{or}}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}d({x}_{n},F(T))=0,
where d(x,F(T))={inf}_{z\in F(T)}d(x,z), then the sequence \{{x}_{n}\} converges strongly to a fixed point of T.
Proof Let p\in F(T). Since T is an asymptotically quasinonexpansive mapping, there exists a sequence \{{u}_{n}\}\in [0,\mathrm{\infty}) with the property {lim}_{n\to \mathrm{\infty}}{u}_{n}=0 and such that
d({T}^{n}x,p)\le (1+{u}_{n})d(x,p)
for all x\in K and p\in F(T). By combining this inequality and Lemma 1, we get
\begin{array}{rcl}d({y}_{n},p)& =& d((1{b}_{n}){x}_{n}\oplus {b}_{n}{T}^{n}{x}_{n},p)\\ \le & (1{b}_{n})d({x}_{n},p)+{b}_{n}d({T}^{n}{x}_{n},p)\\ \le & (1{b}_{n})d({x}_{n},p)+{b}_{n}(1+{u}_{n})d({x}_{n},p)\\ =& (1+{b}_{n}{u}_{n})d({x}_{n},p).\end{array}
(3.1)
Also,
\begin{array}{rcl}d({x}_{n+1},p)& =& d((1{a}_{n}){T}^{n}{x}_{n}\oplus {a}_{n}{T}^{n}{y}_{n},p)\\ \le & (1{a}_{n})d({T}^{n}{x}_{n},p)+{a}_{n}d({T}^{n}{y}_{n},p)\\ \le & (1{a}_{n})(1+{u}_{n})d({x}_{n},p)+{a}_{n}(1+{u}_{n})d({y}_{n},p)\\ \le & (1{a}_{n})(1+{u}_{n})d({x}_{n},p)+{a}_{n}(1+{u}_{n})(1+{b}_{n}{u}_{n})d({x}_{n},p)\\ \le & (1{a}_{n})(1+{u}_{n})d({x}_{n},p)+{a}_{n}{(1+{u}_{n})}^{2}d({x}_{n},p)\\ =& (1+{u}_{n})(1{a}_{n}+{a}_{n}+{a}_{n}{u}_{n})d({x}_{n},p)\\ \le & (1+{u}_{n})(1+{u}_{n})d({x}_{n},p)\\ =& {(1+{u}_{n})}^{2}d({x}_{n},p).\end{array}
(3.2)
When x\ge 0 and 1+x\le {e}^{x}, we have {(1+x)}^{2}\le {e}^{2x}. Thus,
\begin{array}{rcl}d({x}_{n+m},p)& \le & {(1+{u}_{n+m1})}^{2}d({x}_{n+m1},p)\\ \le & {e}^{2{u}_{n+m1}}d({x}_{n+m1},p)\\ \le & \cdots \\ \le & {e}^{2{\sum}_{k=n}^{n+m1}{u}_{k}}d({x}_{n},p).\end{array}
Let {e}^{2{\sum}_{k=n}^{n+m1}{u}_{k}}=M. Thus, there exits a constant M>0 such that
d({x}_{n+m},p)\le Md({x}_{n},p)
for all n,m\in \mathbb{N} and p\in F(T). By (3.2),
d({x}_{n+1},p)\le {(1+{u}_{n})}^{2}d({x}_{n},p).
This gives
d({x}_{n+1},F(T))\le {(1+{u}_{n})}^{2}d({x}_{n},F(T))=(1+2{u}_{n}+{u}_{n}^{2})d({x}_{n},F(T)).
Since {\sum}_{n=1}^{\mathrm{\infty}}{u}_{n}<\mathrm{\infty}, we have {\sum}_{n=1}^{\mathrm{\infty}}(2{u}_{n}+{u}_{n}^{2})<\mathrm{\infty}. Lemma 2 and {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}d({x}_{n},F(T))=0 or {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty}}d({x}_{n},F(T))=0 gives that
\underset{n\to \mathrm{\infty}}{lim}d({x}_{n},F(T))=0.
(3.3)
Now, we show that \{{x}_{n}\} is a Cauchy sequence in K. Since {lim}_{n\to \mathrm{\infty}}d({x}_{n},F(T))=0, for each \u03f5>0, there exists {n}_{1}\in \mathbb{N} such that
d({x}_{n},F(T))<\frac{\u03f5}{M+1}
for all n>{n}_{1}. Thus, there exists {p}_{1}\in F(T) such that
d({x}_{n},{p}_{1})<\frac{\u03f5}{M+1}\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}n>{n}_{1}
and we obtain that
\begin{array}{rcl}d({x}_{n+m},{x}_{n})& \le & d({x}_{n+m},{p}_{1})+d({p}_{1},{x}_{n})\\ \le & Md({x}_{n},{p}_{1})+d({x}_{n},{p}_{1})\\ =& (M+1)d({x}_{n},{p}_{1})\\ <& (M+1)\frac{\u03f5}{M+1}=\u03f5\end{array}
for all m,n>{n}_{1}. Therefore, \{{x}_{n}\} is a Cauchy sequence in K. Since the set K is complete, the sequence \{{x}_{n}\} must be convergence to a point in K. Let {lim}_{n\to \mathrm{\infty}}{x}_{n}=p\in K. Here after, we show that p is a fixed point. By {lim}_{n\to \mathrm{\infty}}{x}_{n}=p, for all {\u03f5}_{1}>0, there exists {n}_{2}\in \mathbb{N} such that
d({x}_{n},p)<\frac{{\u03f5}_{1}}{2(2+{u}_{1})}
(3.4)
for all n>{n}_{2}. From (3.3), for each {\u03f5}_{1}>0, there exists {n}_{3}\in \mathbb{N} such that
d({x}_{n},F(T))<\frac{{\u03f5}_{1}}{2(4+3{u}_{1})}
for all n>{n}_{3}. In particular, inf\{d({x}_{{n}_{3}},p):p\in F(T)\}<\frac{{\u03f5}_{1}}{2(4+3{u}_{1})}. Thus, there must exist {p}^{\star}\in F(T) such that
d({x}_{{n}_{3}},{p}^{\star})<\frac{{\u03f5}_{1}}{2(4+3{u}_{1})}\phantom{\rule{1em}{0ex}}\text{for all}n{n}_{3}.
(3.5)
From (3.4) and (3.5),
\begin{array}{rcl}d(Tp,p)& \le & d(Tp,{p}^{\star})+d({p}^{\star},T{x}_{{n}_{3}})+d(T{x}_{{n}_{3}},{p}^{\star})+d({p}^{\star},{x}_{{n}_{3}})+d({x}_{{n}_{3}},p)\\ \le & d(Tp,{p}^{\star})+2d(T{x}_{{n}_{3}},{p}^{\star})+d({x}_{{n}_{3}},{p}^{\star})+d({x}_{{n}_{3}},p)\\ \le & (1+{u}_{1})d(p,{p}^{\star})+2(1+{u}_{1})d({x}_{{n}_{3}},{p}^{\star})+d({x}_{{n}_{3}},{p}^{\star})+d({x}_{{n}_{3}},p)\\ \le & (1+{u}_{1})d(p,{x}_{{n}_{3}})+(1+{u}_{1})d({x}_{{n}_{3}},{p}^{\star})+2(1+{u}_{1})d({x}_{{n}_{3}},{p}^{\star})\\ +d({x}_{{n}_{3}},{p}^{\star})+d({x}_{{n}_{3}},p)\\ =& (2+{u}_{1})d({x}_{{n}_{3}},p)+(4+3{u}_{1})d({x}_{{n}_{3}},{p}^{\star})\\ <& (2+{u}_{1})\frac{{\u03f5}_{1}}{2(2+{u}_{1})}+(4+3{u}_{1})\frac{{\u03f5}_{1}}{2(4+3{u}_{1})}={\u03f5}_{1}.\end{array}
Since {\u03f5}_{1} is arbitrary, so d(Tp,p)=0, i.e., Tp=p. Therefore, p\in F(T). This completes the proof. □
Remark 2 Let the hypothesis of Theorem 1 be satisfied and T:K\to K be an asymptotically nonexpansive or quasinonexpansive mapping. From Remark 1, the class of asymptotically quasinonexpansive mappings includes quasinonexpansive mappings and asymptotically nonexpansive mappings. Then the sequence \{{x}_{n}\} converges strongly to a fixed point of T.
Now, we give the following corollaries which have been proved by Theorem 1.
Corollary 1 Under the hypothesis of Theorem 1, T satisfies the following conditions:

(1)
{lim}_{n\to \mathrm{\infty}}d({x}_{n},T{x}_{n})=0.

(2)
If the sequence \{{z}_{n}\} in K satisfies {lim}_{n\to \mathrm{\infty}}d({z}_{n},T{z}_{n})=0, then
\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}inf}d({z}_{n},F(T))=0\phantom{\rule{1em}{0ex}}\mathit{\text{or}}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}d({z}_{n},F(T))=0.
Then the sequence \{{x}_{n}\} converges strongly to a fixed point of T.
Proof It follows from the hypothesis that {lim}_{n\to \mathrm{\infty}}d({x}_{n},T{x}_{n})=0. From (2),
\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}inf}d({x}_{n},F(T))=0\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}d({x}_{n},F(T))=0.
Therefore, the sequence \{{x}_{n}\} must converge to a fixed point of T by Theorem 1. □
Corollary 2 Under the hypothesis of Theorem 1, T satisfies the following conditions:

(1)
{lim}_{n\to \mathrm{\infty}}d({x}_{n},T{x}_{n})=0.

(2)
There exists a function f:[0,\mathrm{\infty})\to [0,\mathrm{\infty}) which is rightcontinuous at 0, f(0)=0 and f(r)>0 for all r>0 such that
d(x,Tx)\ge f\left(d(x,F(T))\right)\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}x\in K,
where d(x,F(T))={inf}_{z\in F(T)}d(x,z).
Then the sequence \{{x}_{n}\} converges strongly to a fixed point of T.
Proof It follows from the hypothesis that
\underset{n\to \mathrm{\infty}}{lim}f(d({x}_{n},F(T))\le \underset{n\to \mathrm{\infty}}{lim}d({x}_{n},T{x}_{n})=0.
That is,
\underset{n\to \mathrm{\infty}}{lim}f(d({x}_{n},F(T))=0.
Since f:[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is rightcontinuous at 0 and f(0)=0, therefore we have
\underset{n\to \mathrm{\infty}}{lim}d({x}_{n},F(T))=0.
Thus, {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}d({x}_{n},F(T))={lim\hspace{0.17em}sup}_{n\to \mathrm{\infty}}d({x}_{n},F(T))=0. By Theorem 1, the sequence \{{x}_{n}\} converges strongly to q, a fixed point of T. This completes the proof. □
Finally, we give the following theorem which has a different hypothesis from Theorem 1.
Theorem 2 Let K be a nonempty closed convex subset of a complete CAT(0) space X, T:K\to K be an asymptotically quasinonexpansive mapping with F(T)\ne \mathrm{\varnothing} and \{{u}_{n}\} be a nonnegative real sequence with {\sum}_{n=1}^{\mathrm{\infty}}{u}_{n}<\mathrm{\infty}. Suppose that \{{x}_{n}\} is defined by the iteration process (1.7). If T is semicompact and {lim}_{n\to \mathrm{\infty}}d({x}_{n},T{x}_{n})=0, then the sequence \{{x}_{n}\} converges strongly to a fixed point of T.
Proof From the hypothesis, we have {lim}_{n\to \mathrm{\infty}}d({x}_{n},T{x}_{n})=0. Also, since T is semicompact, there exists a subsequence \{{x}_{{n}_{k}}\} of \{{x}_{n}\} such that {x}_{{n}_{k}}\to p\in K. Hence,
d(p,Tp)=\underset{n\to \mathrm{\infty}}{lim}d({x}_{{n}_{k}},T{x}_{{n}_{k}})=0.
Thus, p\in F(T). By (3.2),
\begin{array}{rcl}d({x}_{n+1},p)& \le & {(1+{u}_{n})}^{2}d({x}_{n},p)\\ =& (1+2{u}_{n}+{u}_{n}^{2})d({x}_{n},p).\end{array}
Since {\sum}_{n=1}^{\mathrm{\infty}}{u}_{n}<\mathrm{\infty}, we have {\sum}_{n=1}^{\mathrm{\infty}}(2{u}_{n}+{u}_{n}^{2})<\mathrm{\infty}. By Lemma 2, {lim}_{n\to \mathrm{\infty}}d({x}_{n},p) exists and {x}_{{n}_{k}}\to p\in F(T) gives that {x}_{n}\to p\in F(T). This completes the proof. □