We now state and proof our main theorem.

**Theorem 3.1** *Let* *K* *be a nonempty*, *closed and convex subset of a real Hilbert space* *H*. *Let* {T}_{i}:K\to K *be asymptotically nonexpansive mappings with sequences* \{{k}_{n,i}\} *for each* i=1,2,\dots ,N. *Assume that* F:={\bigcap}_{i=1}^{N}F({T}_{i}) *is nonempty*. *Let* \{{x}_{n}\} *be a sequence generated by*

\{\begin{array}{c}{x}_{1}\in K,\phantom{\rule{1em}{0ex}}\mathit{\text{chosen arbitrarily}},\hfill \\ {y}_{n}={P}_{K}[(1-{\alpha}_{n}){x}_{n}],\hfill \\ {x}_{n+1}={\beta}_{n,0}{x}_{n}+{\sum}_{i=1}^{N}{\beta}_{n,i}{T}_{i}^{n}{y}_{n},\phantom{\rule{1em}{0ex}}n\ge 1,\hfill \end{array}

(3.1)

*where* {\alpha}_{n}\in (0,1) *such that* {lim}_{n\to \mathrm{\infty}}{\alpha}_{n}=0, {lim}_{n\to \mathrm{\infty}}\frac{({k}_{n,i}^{2}-1)}{{\alpha}_{n}}=0, *for each* i\in \{1,2,\dots ,N\} *and* {\sum}_{n=1}^{\mathrm{\infty}}{\alpha}_{n}=\mathrm{\infty}, \{{\beta}_{n,i}\}\subset [a,b]\subset (0,1) *for* i=1,2,\dots ,N, *satisfying* {\beta}_{n,0}+{\beta}_{n,1}+\cdots +{\beta}_{n,N}=1 *for each* n\ge 1. *Then* \{{x}_{n}\} *converges strongly to the common minimum*-*norm point of* *F*.

*Proof* Let {x}^{\ast}\in {P}_{F}0. Let {k}_{n}:=max\{{k}_{n,i}:i=1,2,\dots ,N\}. Then from (3.1) and asymptotical nonexpansiveness of {T}_{i}, for each i\in \{1,2,\dots ,N\}, we have that

\begin{array}{rcl}\parallel {y}_{n}-{x}^{\ast}\parallel & =& \parallel {P}_{C}[(1-{\alpha}_{n}){x}_{n}]-{P}_{K}{x}^{\ast}\parallel \\ \le & \parallel (1-{\alpha}_{n}){x}_{n}-{x}^{\ast}\parallel \\ =& \parallel {\alpha}_{n}(0-{x}^{\ast})+(1-{\alpha}_{n})({x}_{n}-{x}^{\ast})\parallel \\ \le & {\alpha}_{n}\parallel {x}^{\ast}\parallel +(1-{\alpha}_{n})\parallel {x}_{n}-{x}^{\ast}\parallel ,\end{array}

(3.2)

and

\begin{array}{rcl}\parallel {x}_{n+1}-{x}^{\ast}\parallel & =& \parallel {\beta}_{n,0}{x}_{n}+\sum _{i=1}^{N}{\beta}_{n,i}{T}_{i}^{n}{y}_{n}-{x}^{\ast}\parallel \\ \le & {\beta}_{n,0}\parallel {x}_{n}-{x}^{\ast}\parallel +\sum _{i=1}^{N}{\beta}_{n,i}\parallel {T}_{i}^{n}{y}_{n}-{x}^{\ast}\parallel \\ \le & {\beta}_{n,0}\parallel {x}_{n}-{x}^{\ast}\parallel +(1-{\beta}_{n,0}){k}_{n}\parallel {y}_{n}-{x}^{\ast}\parallel \\ \le & {\beta}_{n,0}\parallel {x}_{n}-{x}^{\ast}\parallel +(1-{\beta}_{n,0}){k}_{n}[{\alpha}_{n}\parallel {x}^{\ast}\parallel +(1-{\alpha}_{n})\parallel {x}_{n}-{x}^{\ast}\parallel ]\\ \le & [{\beta}_{n,0}+(1-{\beta}_{n,0}){k}_{n}(1-{\alpha}_{n})]\parallel {x}_{n}-{x}^{\ast}\parallel +[(1-{\beta}_{n,0}){k}_{n}{\alpha}_{n}]\parallel {x}^{\ast}\parallel \\ \le & {\delta}_{n}\parallel {x}^{\ast}\parallel +[1-(1-\u03f5){\delta}_{n}]\parallel {x}_{n}-{x}^{\ast}\parallel ,\end{array}

(3.3)

where {\delta}_{n}=(1-{\beta}_{n,0}){k}_{n}{\alpha}_{n}, since there exists {N}_{0}>0 such that \frac{({k}_{n}-1)}{{\alpha}_{n}}\le \u03f5{k}_{n} for all n\ge {N}_{0} and for some \u03f5>0 satisfying (1-\u03f5){\delta}_{n}\le 1. Thus, by induction,

\parallel {x}_{n+1}-{x}^{\ast}\parallel \le max\{\parallel {x}_{0}-{x}^{\ast}\parallel ,{(1-\u03f5)}^{-1}\parallel {x}^{\ast}\parallel \},\phantom{\rule{1em}{0ex}}\mathrm{\forall}n\ge {N}_{0},

which implies that \{{x}_{n}\} and hence \{{y}_{n}\} is bounded. Moreover, from (3.2) and Lemma 2.1, we obtain that

\begin{array}{rcl}{\parallel {y}_{n}-{x}^{\ast}\parallel}^{2}& =& {\parallel {P}_{K}[(1-{\alpha}_{n}){x}_{n}]-{P}_{K}{x}^{\ast}\parallel}^{2}\\ \le & {\parallel {\alpha}_{n}(0-{x}^{\ast})+(1-{\alpha}_{n})({x}_{n}-{x}^{\ast})\parallel}^{2}\\ \le & (1-{\alpha}_{n}){\parallel {x}_{n}-{x}^{\ast}\parallel}^{2}-2{\alpha}_{n}\u3008{x}^{\ast},{y}_{n}-{x}^{\ast}\u3009.\end{array}

(3.4)

Furthermore, from (3.1), Lemma 2.2 and asymptotical nonexpansiveness of {T}_{i}, for each i=1,2,\dots ,N, we have that

\begin{array}{rcl}{\parallel {x}_{n+1}-{x}^{\ast}\parallel}^{2}& =& {\parallel {\beta}_{n,0}{x}_{n}+\sum _{i=1}^{N}{\beta}_{n,i}{T}_{i}^{n}{y}_{n}-{x}^{\ast}\parallel}^{2}\\ \le & {\beta}_{n,0}{\parallel {x}_{n}-{x}^{\ast}\parallel}^{2}+\sum _{i=1}^{N}{\beta}_{n,i}{\parallel {T}_{i}^{n}{y}_{n}-{x}^{\ast}\parallel}^{2}\\ -\sum _{i=1}^{N}{\beta}_{n,0}{\beta}_{n,i}{\parallel {x}_{n}-{T}_{i}^{n}{y}_{n}\parallel}^{2}\\ \le & {\beta}_{n,0}{\parallel {x}_{n}-{x}^{\ast}\parallel}^{2}+(1-{\beta}_{n,0}){k}_{n}^{2}{\parallel {y}_{n}-{x}^{\ast}\parallel}^{2}\\ -\sum _{i=1}^{N}{\beta}_{n,0}{\beta}_{n,i}{\parallel {x}_{n}-{T}_{i}^{n}{y}_{n}\parallel}^{2},\end{array}

which implies, using (3.4), that

\begin{array}{rcl}{\parallel {x}_{n+1}-{x}^{\ast}\parallel}^{2}& \le & {\beta}_{n,0}{\parallel {x}_{n}-{x}^{\ast}\parallel}^{2}+(1-{\beta}_{n,0}){k}_{n}^{2}[(1-{\alpha}_{n}){\parallel {x}_{n}-{x}^{\ast}\parallel}^{2}\\ -2{\alpha}_{n}\u3008{x}^{\ast},{y}_{n}-{x}^{\ast}\u3009]-\sum _{i=1}^{N}{\beta}_{n,0}{\beta}_{n,i}{\parallel {x}_{n}-{T}_{i}^{n}{y}_{n}\parallel}^{2}\\ \le & (1-{\theta}_{n}){\parallel {x}_{n}-{x}^{\ast}\parallel}^{2}-2{\theta}_{n}\u3008{x}^{\ast},{y}_{n}-{x}^{\ast}\u3009+({k}_{n}^{2}-1)M\\ -\sum _{i=1}^{N}{\beta}_{n,0}{\beta}_{n,i}{\parallel {x}_{n}-{T}_{i}^{n}{y}_{n}\parallel}^{2}\end{array}

(3.5)

\le (1-{\theta}_{n}){\parallel {x}_{n}-{x}^{\ast}\parallel}^{2}-2{\theta}_{n}\u3008{x}^{\ast},{y}_{n}-{x}^{\ast}\u3009+({k}_{n}^{2}-1)M

(3.6)

for some M>0, where {\theta}_{n}:={\alpha}_{n}(1-{\beta}_{n,0}) for all n\in N.

Now, we consider the following two cases.

Case 1. Suppose that there exists {n}_{0}\in \mathbb{N} such that \{\parallel {x}_{n}-{x}^{\ast}\parallel \} is non-increasing for all n\ge {n}_{0}. In this situation, \{\parallel {x}_{n}-{x}^{\ast}\parallel \} is convergent. Then from (3.5), we have that {\sum}_{i=1}^{N}{\beta}_{n,0}{\beta}_{n,i}{\parallel {x}_{n}-{T}_{i}^{n}{y}_{n}\parallel}^{2}\to 0, which implies that

{x}_{n}-{T}_{i}^{n}{y}_{n}\to 0,\phantom{\rule{1em}{0ex}}\text{as}n\to \mathrm{\infty},

(3.7)

for each i\in \{1,2,\dots ,N\}. Moreover, from (3.1) and (3.7) and the fact that {\alpha}_{n}\to 0, we get that

\parallel {x}_{n+1}-{x}_{n}\parallel ={\beta}_{n,1}\parallel {T}_{1}^{n}{y}_{n}-{x}_{n}\parallel +\cdots +{\beta}_{n,N}\parallel {T}_{N}^{n}{y}_{n}-{x}_{n}\parallel \to 0,

(3.8)

and

\begin{array}{rcl}\parallel {y}_{n}-{x}_{n}\parallel & =& \parallel {P}_{C}[(1-{\alpha}_{n}){x}_{n}]-{P}_{k}{x}_{n}\parallel \\ \le & \parallel -{\alpha}_{n}{x}_{n}\parallel \to 0,\end{array}

(3.9)

as n\to \mathrm{\infty} and hence

\parallel {y}_{n+1}-{y}_{n}\parallel \le \parallel {y}_{n+1}-{x}_{n+1}\parallel +\parallel {x}_{n+1}-{x}_{n}\parallel +\parallel {x}_{n}-{y}_{n}\parallel \to 0,

(3.10)

as n\to \mathrm{\infty}. Furthermore, from (3.7) and (3.9), we get that

\parallel {y}_{n}-{T}_{i}^{n}{y}_{n}\parallel \le \parallel {y}_{n}-{x}_{n}\parallel +\parallel {x}_{n}-{T}_{i}^{n}{y}_{n}\parallel \to 0,\phantom{\rule{1em}{0ex}}\text{as}n\to \mathrm{\infty}.

(3.11)

Therefore, since

\begin{array}{rcl}\parallel {y}_{n}-{T}_{i}{y}_{n}\parallel & \le & \parallel {y}_{n}-{y}_{n+1}\parallel +\parallel {y}_{n+1}-{T}_{i}^{n+1}{y}_{n+1}\parallel +\parallel {T}_{i}^{n+1}{y}_{n+1}-{T}_{i}^{n+1}{y}_{n}\parallel \\ +\parallel {T}_{i}^{n+1}{y}_{n}-{T}_{i}{y}_{n}\parallel ,\\ \le & \parallel {y}_{n}-{y}_{n+1}\parallel +\parallel {y}_{n+1}-{T}_{i}^{n+1}{y}_{n+1}\parallel +{k}_{n+1}\parallel {y}_{n+1}-{y}_{n}\parallel \\ +\parallel {T}_{i}\left({T}_{i}^{n}{y}_{n}\right)-{T}_{i}{y}_{n}\parallel ,\end{array}

(3.12)

we have from (3.10), (3.11), (3.12) and uniform continuity of {T}_{i} that

\parallel {y}_{n}-{T}_{i}{y}_{n}\parallel \to 0,\phantom{\rule{1em}{0ex}}\text{as}n\to \mathrm{\infty},\text{for each}i=1,2,\dots ,N.

(3.13)

Let \{{y}_{{n}_{k}}\} be a subsequence of \{{y}_{n}\} such that

\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008{x}^{\ast},{y}_{n}-{x}^{\ast}\u3009=\underset{k\to \mathrm{\infty}}{lim}\u3008{x}^{\ast},{y}_{{n}_{k}}-{x}^{\ast}\u3009,

and {y}_{{n}_{k}}\rightharpoonup z. Then from (3.9), we have that {x}_{{n}_{k}}\rightharpoonup z. Therefore, by Lemma 2.3, we obtain that

\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008{x}^{\ast},{y}_{n}-{x}^{\ast}\u3009=\underset{k\to \mathrm{\infty}}{lim}\u3008{x}^{\ast},{y}_{{n}_{k}}-{x}^{\ast}\u3009=\u3008{x}^{\ast},z-{x}^{\ast}\u3009\ge 0.

(3.14)

Now, we show that {x}_{n+1}\to {x}^{\ast}, as n\to \mathrm{\infty}. But from (3.13) and Lemma 2.4, we get that z\in F({T}_{i}) for each i\in \{1,2,\dots ,N\} and hence z\in {\bigcap}_{i=1}^{N}F({T}_{i}). Then from (3.6), we get that

\begin{array}{rcl}{\parallel {x}_{n+1}-{x}^{\ast}\parallel}^{2}& \le & (1-{\theta}_{n}){\parallel {x}_{n}-{x}^{\ast}\parallel}^{2}-2{\theta}_{n}\u3008{x}^{\ast},{y}_{n}-{x}^{\ast}\u3009\\ +({k}_{n}^{2}-1)M\end{array}

(3.15)

for some M>0. But note that {\theta}_{n} satisfies {lim}_{n}{\theta}_{n}=0 and {\sum}_{n=1}^{\mathrm{\infty}}{\theta}_{n}=\mathrm{\infty}. Thus, it follows from (3.15) and Lemma 2.5 that \parallel {x}_{n}-{x}^{\ast}\parallel \to 0, as n\to \mathrm{\infty}. Consequently, {x}_{n}\to {x}^{\ast}.

Case 2. Suppose that there exists a subsequence \{{n}_{i}\} of \{n\} such that

\parallel {x}_{{n}_{i}}-{x}^{\ast}\parallel <\parallel {x}_{{n}_{i}+1}-{x}^{\ast}\parallel

for all i\in \mathbb{N}. Then by Lemma 2.6, there exists a nondecreasing sequence \{{m}_{k}\}\subset \mathbb{N} such that {m}_{k}\to \mathrm{\infty}, \parallel {x}_{{m}_{k}}-{x}^{\ast}\parallel \le \parallel {x}_{{m}_{k}+1}-{x}^{\ast}\parallel and \parallel {x}_{k}-{x}^{\ast}\parallel \le \parallel {x}_{{m}_{k}+1}-{x}^{\ast}\parallel for all k\in \mathbb{N}. Then from (3.5) and the fact that {\theta}_{n}\to 0, we have

This implies that {x}_{{m}_{k}}-{T}_{i}^{{m}_{k}}{y}_{{m}_{k}}\to 0, as k\to \mathrm{\infty}. Thus, following the method of Case 1, we obtain that {x}_{{m}_{k}}-{y}_{{m}_{k}}\to 0 and {y}_{{m}_{k}}-{T}_{i}{y}_{{m}_{k}}\to 0 as k\to \mathrm{\infty} for each i=1,2,\dots ,N and hence there exists {z}^{\prime}\in F such that

\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008{x}^{\ast},{y}_{n}-{x}^{\ast}\u3009=\underset{k\to \mathrm{\infty}}{lim}\u3008{x}^{\ast},{y}_{{n}_{k}}-{x}^{\ast}\u3009=\u3008{x}^{\ast},{z}^{\prime}-{x}^{\ast}\u3009\ge 0.

(3.16)

Then from (3.6), we get that

\begin{array}{rcl}{\parallel {x}_{{m}_{k}+1}-{x}^{\ast}\parallel}^{2}& \le & (1-{\theta}_{{m}_{k}}){\parallel {x}_{{m}_{k}}-{x}^{\ast}\parallel}^{2}-2{\theta}_{{m}_{k}}\u3008{x}^{\ast},{y}_{{m}_{k}}-{x}^{\ast}\u3009\\ +({k}_{{m}_{k}}^{2}-1)M.\end{array}

(3.17)

Since \parallel {x}_{{m}_{k}}-{x}^{\ast}\parallel \le \parallel {x}_{{m}_{k}+1}-{x}^{\ast}\parallel, (3.17) implies that

\begin{array}{rcl}{\theta}_{{m}_{k}}{\parallel {x}_{{m}_{k}}-{x}^{\ast}\parallel}^{2}& \le & {\parallel {x}_{{m}_{k}}-{x}^{\ast}\parallel}^{2}-{\parallel {x}_{{m}_{k}+1}-{x}^{\ast}\parallel}^{2}-2{\theta}_{{m}_{k}}\u3008{x}^{\ast},{y}_{{m}_{k}}-{x}^{\ast}\u3009\\ +({k}_{{m}_{k}}^{2}-1)M\\ \le & -2{\theta}_{{m}_{k}}\u3008{x}^{\ast},{y}_{{m}_{k}}-{x}^{\ast}\u3009+({k}_{{m}_{k}}^{2}-1)M.\end{array}

In particular, since {\theta}_{{m}_{k}}>0, we have that

{\parallel {x}_{{m}_{k}}-{x}^{\ast}\parallel}^{2}\le -2\u3008{x}^{\ast},{y}_{{m}_{k}}-{x}^{\ast}\u3009+\frac{({k}_{{m}_{k}}^{2}-1)}{{\theta}_{{m}_{k}}}M.

Thus, from (3.16) and the fact that \frac{({k}_{{m}_{k}}^{2}-1)}{{\theta}_{{m}_{k}}}\to 0, we obtain that \parallel {x}_{{m}_{k}}-{x}^{\ast}\parallel \to 0 as k\to \mathrm{\infty}. This together with (3.17) gives \parallel {x}_{{m}_{k}+1}-{x}^{\ast}\parallel \to 0 as k\to \mathrm{\infty}. But \parallel {x}_{k}-{x}^{\ast}\parallel \le \parallel {x}_{{m}_{k}+1}-{x}^{\ast}\parallel for all k\in \mathbb{N}, thus we obtain that {x}_{k}\to {x}^{\ast}. Therefore, from the above two cases, we can conclude that \{{x}_{n}\} converges strongly to a point {x}^{\ast} of *F* which is the common minimum-norm fixed point of the family \{{T}_{i},i=1,2,\dots ,N\} and the proof is complete. □

If in Theorem 3.1 we assume that N=1, then we get the following corollary.

**Corollary 3.2** *Let* *K* *be a nonempty*, *closed and convex subset of a real Hilbert space* *H*. *Let* T:K\to K *be an asymptotically nonexpansive mapping with a sequence* \{{k}_{n}\}. *Assume that* F(T) *is nonempty*. *Let* \{{x}_{n}\} *be a sequence generated by*

\{\begin{array}{c}{x}_{1}\in C,\phantom{\rule{1em}{0ex}}\mathit{\text{chosen arbitrarily}},\hfill \\ {y}_{n}={P}_{K}[(1-{\alpha}_{n}){x}_{n}],\hfill \\ {x}_{n+1}={\beta}_{n}{x}_{n}+(1-{\beta}_{n}){T}^{n}{y}_{n},\phantom{\rule{1em}{0ex}}n\ge 1,\hfill \end{array}

(3.18)

*where* {\alpha}_{n}\in (0,1) *such that* {lim}_{n\to \mathrm{\infty}}{\alpha}_{n}=0, {lim}_{n\to \mathrm{\infty}}\frac{({k}_{n}^{2}-1)}{{\alpha}_{n}}=0 *and* {\sum}_{n=1}^{\mathrm{\infty}}{\alpha}_{n}=\mathrm{\infty}, \{{\beta}_{n}\}\subset [a,b]\subset (0,1) *for each* n\ge 1. *Then* \{{x}_{n}\} *converges strongly to the minimum*-*norm fixed point of* *T*.

If in Theorem 3.1 we assume that each {T}_{i} is nonexpansive for i=1,2,\dots ,N, then the method of proof of Theorem 3.1 provides the following corollary.

**Corollary 3.3** *Let* *K* *be a nonempty*, *closed and convex subset of a real Hilbert space* *H*. *Let* {T}_{i}:K\to K *be nonexpansive mappings with* F:={\bigcap}_{i=1}^{N}F({T}_{i}) *nonempty*. *Let* \{{x}_{n}\} *be a sequence generated by*

\{\begin{array}{c}{x}_{1}\in K,\phantom{\rule{1em}{0ex}}\mathit{\text{chosen arbitrarily}},\hfill \\ {y}_{n}={P}_{K}[(1-{\alpha}_{n}){x}_{n}],\hfill \\ {x}_{n+1}={\beta}_{n,0}{x}_{n}+{\sum}_{i=1}^{N}{\beta}_{n,i}{T}_{i}{y}_{n},\phantom{\rule{1em}{0ex}}n\ge 1,\phantom{\rule{0.25em}{0ex}}\hfill \end{array}

(3.19)

*where* {\alpha}_{n}\in (0,1) *such that* {lim}_{n\to \mathrm{\infty}}{\alpha}_{n}=0 *and* {\sum}_{n=1}^{\mathrm{\infty}}{\alpha}_{n}=\mathrm{\infty}, \{{\beta}_{n,i}\}\subset [a,b]\subset (0,1), *for* i=1,2,\dots ,N, *satisfying* {\beta}_{n,0}+{\beta}_{n,1}+\cdots +{\beta}_{n,N}=1 *for each* n\ge 1. *Then* \{{x}_{n}\} *converges strongly to the common minimum*-*norm point of* *F*.

If in Corollary 3.3 we assume that N=1, then we have the following corollary.

**Corollary 3.4** *Let* *K* *be a nonempty*, *closed and convex subset of a real Hilbert space* *H*. *Let* T:K\to K *be a nonexpansive mapping with* F(T) *nonempty*. *Let* \{{x}_{n}\} *be a sequence generated by*

\{\begin{array}{c}{x}_{1}\in K,\phantom{\rule{1em}{0ex}}\mathit{\text{chosen arbitrarily}},\hfill \\ {y}_{n}={P}_{K}[(1-{\alpha}_{n}){x}_{n}],\hfill \\ {x}_{n+1}={\beta}_{n}{x}_{n}+(1-{\beta}_{n})T{y}_{n},\phantom{\rule{1em}{0ex}}n\ge 1,\hfill \end{array}

(3.20)

*where* {\alpha}_{n}\in (0,1) *such that* {lim}_{n\to \mathrm{\infty}}{\alpha}_{n}=0 *and* {\sum}_{n=1}^{\mathrm{\infty}}{\alpha}_{n}=\mathrm{\infty}, \{{\beta}_{n}\}\subset [a,b]\subset (0,1) *for each* n\ge 1. *Then* \{{x}_{n}\} *converges strongly to the minimum*-*norm point of* F(T).