**Theorem 3.1** *Let* *C* *be a nonempty*, *closed and convex subset of a real Hilbert space* *H*, *let* {\{{T}_{n}\}}_{n=1}^{\mathrm{\infty}}:C\to C *be a countable family of uniformly closed and uniformly Lipschitz pseudocontractive mappings with Lipschitzian constants* {L}_{n}, *let* L:={sup}_{n\ge 1}{L}_{n}. *Assume that the interior of* F:={\bigcap}_{n=1}^{\mathrm{\infty}}F({T}_{n}) *is nonempty*. *Let* \{{x}_{n}\} *be a sequence generated from an arbitrary* {x}_{0}\in C *by the following algorithm*:

\{\begin{array}{c}{z}_{n}=(1-{\gamma}_{n}){x}_{n}+{\gamma}_{n}{T}_{n}{x}_{n},\hfill \\ {y}_{n}=(1-{\beta}_{n}){x}_{n}+{\beta}_{n}{T}_{n}{z}_{n},\hfill \\ {x}_{n+1}=(1-{\alpha}_{n}){x}_{n}+{\alpha}_{n}{T}_{n}{y}_{n},\hfill \end{array}

(3.1)

*where* \{{\alpha}_{n}\},\{{\beta}_{n}\},\{{\gamma}_{n}\}\subset (0,1) *satisfying the following conditions*: (i) {\alpha}_{n}\le {\beta}_{n}\le {\gamma}_{n}, \mathrm{\forall}n\ge 0; (ii) {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}{\alpha}_{n}=\alpha >0; (iii) {sup}_{n\ge 1}{\gamma}_{n}\le \gamma *with* {\gamma}^{3}{L}^{4}+2{\gamma}^{2}{L}^{3}+{\gamma}^{2}{L}^{2}+\gamma {L}^{2}+2\gamma <1. *Then* \{{x}_{n}\} *converges strongly to* {x}^{\ast}\in F.

*Proof* Suppose that p\in F. Then from (3.1) and Lemma 2.1, we have that

In addition, using (3.1), we have that

\begin{array}{rl}{\parallel {z}_{n}-{T}_{n}{z}_{n}\parallel}^{2}=& {\parallel (1-{\gamma}_{n})({x}_{n}-{T}_{n}{z}_{n})+{\gamma}_{n}({T}_{n}{x}_{n}-{T}_{n}{z}_{n})\parallel}^{2}\\ =& (1-{\gamma}_{n}){\parallel {x}_{n}-{T}_{n}{z}_{n}\parallel}^{2}+{\gamma}_{n}{\parallel {T}_{n}{x}_{n}-{T}_{n}{z}_{n}\parallel}^{2}\\ -{\gamma}_{n}(1-{\gamma}_{n}){\parallel {x}_{n}-{T}_{n}{x}_{n}\parallel}^{2}\\ \le & (1-{\gamma}_{n}){\parallel {x}_{n}-{T}_{n}{z}_{n}\parallel}^{2}+{\gamma}_{n}{L}^{2}{\parallel {x}_{n}-{z}_{n}\parallel}^{2}\\ -{\gamma}_{n}(1-{\gamma}_{n}){\parallel {x}_{n}-{T}_{n}{x}_{n}\parallel}^{2}\\ =& (1-{\gamma}_{n}){\parallel {x}_{n}-{T}_{n}{z}_{n}\parallel}^{2}+{\gamma}_{n}^{3}{L}^{2}{\parallel {x}_{n}-{T}_{n}{x}_{n}\parallel}^{2}\\ -{\gamma}_{n}(1-{\gamma}_{n}){\parallel {x}_{n}-{T}_{n}{x}_{n}\parallel}^{2}\\ =& (1-{\gamma}_{n}){\parallel {x}_{n}-{T}_{n}{z}_{n}\parallel}^{2}+{\gamma}_{n}({\gamma}_{n}^{2}{L}^{2}+{\gamma}_{n}-1){\parallel {x}_{n}-{T}_{n}{x}_{n}\parallel}^{2}.\end{array}

(3.5)

Substituting (3.4) and (3.5) into (3.3), we obtain that

\begin{array}{rl}{\parallel {y}_{n}-p\parallel}^{2}\le & (1-{\beta}_{n}){\parallel {x}_{n}-p\parallel}^{2}+{\beta}_{n}({\parallel {x}_{n}-p\parallel}^{2}+{\gamma}_{n}^{2}{\parallel {x}_{n}-{T}_{n}{x}_{n}\parallel}^{2})\\ +{\beta}_{n}[(1-{\gamma}_{n}){\parallel {x}_{n}-{T}_{n}{z}_{n}\parallel}^{2}+{\gamma}_{n}({\gamma}_{n}^{2}{L}^{2}+{\gamma}_{n}-1){\parallel {x}_{n}-{T}_{n}{x}_{n}\parallel}^{2}]\\ -{\beta}_{n}(1-{\beta}_{n}){\parallel {x}_{n}-{T}_{n}{z}_{n}\parallel}^{2}\\ =& {\parallel {x}_{n}-p\parallel}^{2}+{\beta}_{n}{\gamma}_{n}({\gamma}_{n}^{2}{L}^{2}+2{\gamma}_{n}-1){\parallel {x}_{n}-{T}_{n}{x}_{n}\parallel}^{2}\\ +{\beta}_{n}({\beta}_{n}-{\gamma}_{n}){\parallel {x}_{n}-{T}_{n}{z}_{n}\parallel}^{2}.\end{array}

(3.6)

Since

Then, substituting (3.8) into (3.7), we obtain that

\begin{array}{rl}{\parallel {y}_{n}-{T}_{n}{y}_{n}\parallel}^{2}\le & (1-{\beta}_{n}){\parallel {x}_{n}-{T}_{n}{y}_{n}\parallel}^{2}+{\beta}_{n}{L}^{2}{({\gamma}_{n}-{\beta}_{n}+{\beta}_{n}{\gamma}_{n}L)}^{2}{\parallel {x}_{n}-{T}_{n}{x}_{n}\parallel}^{2}\\ -{\beta}_{n}(1-{\beta}_{n}){\parallel {x}_{n}-{T}_{n}{z}_{n}\parallel}^{2}.\end{array}

(3.9)

Substituting (3.6) and (3.9) into (3.2), we obtain that

\begin{array}{rcl}{\parallel {x}_{n+1}-p\parallel}^{2}& \le & (1-{\alpha}_{n}){\parallel {x}_{n}-p\parallel}^{2}+{\alpha}_{n}[{\parallel {x}_{n}-p\parallel}^{2}\\ +{\beta}_{n}{\gamma}_{n}({\gamma}_{n}^{2}{L}^{2}+2{\gamma}_{n}-1){\parallel {x}_{n}-{T}_{n}{x}_{n}\parallel}^{2}\\ +{\beta}_{n}({\beta}_{n}-{\gamma}_{n}){\parallel {x}_{n}-{T}_{n}{z}_{n}\parallel}^{2}]\\ +{\alpha}_{n}[(1-{\beta}_{n}){\parallel {x}_{n}-{T}_{n}{y}_{n}\parallel}^{2}+{\beta}_{n}{L}^{2}({\gamma}_{n}-{\beta}_{n}\\ +{{\beta}_{n}{\gamma}_{n}L)}^{2}{\parallel {x}_{n}-{T}_{n}{x}_{n}\parallel}^{2}-{\beta}_{n}(1-{\beta}_{n}){\parallel {x}_{n}-{T}_{n}{z}_{n}\parallel}^{2}]\\ -{\alpha}_{n}(1-{\alpha}_{n}){\parallel {x}_{n}-{T}_{n}{y}_{n}\parallel}^{2}\\ =& {\parallel {x}_{n}-p\parallel}^{2}+[{\alpha}_{n}{\beta}_{n}{\gamma}_{n}({\gamma}_{n}^{2}{L}^{2}+2{\gamma}_{n}-1)\\ +{\alpha}_{n}{\beta}_{n}{L}^{2}{({\gamma}_{n}-{\beta}_{n}+{\beta}_{n}{\gamma}_{n}L)}^{2}]{\parallel {x}_{n}-{T}_{n}{x}_{n}\parallel}^{2}\\ +{\alpha}_{n}({\alpha}_{n}-{\beta}_{n}){\parallel {x}_{n}-{T}_{n}{y}_{n}\parallel}^{2}\\ +{\alpha}_{n}{\beta}_{n}(2{\beta}_{n}-{\gamma}_{n}-1){\parallel {x}_{n}-{T}_{n}{z}_{n}\parallel}^{2}.\end{array}

(3.10)

Since from condition (iii), we have that

\begin{array}{c}\gamma ({\gamma}^{2}{L}^{2}+\gamma {L}^{2}+2\gamma -1)+{\gamma}^{2}{L}^{2}{(1+\gamma L)}^{2}<0\\ \Downarrow \\ {\gamma}_{n}({\gamma}_{n}^{2}{L}^{2}+{\gamma}_{n}{L}^{2}+2{\gamma}_{n}-1)+{L}^{2}{({\gamma}_{n}+{\gamma}_{n}^{2}L)}^{2}<0\\ \Downarrow \\ {\alpha}_{n}{\beta}_{n}{\gamma}_{n}({\gamma}_{n}^{2}{L}^{2}+2{\gamma}_{n}-1)+{\alpha}_{n}{\beta}_{n}{L}^{2}{({\gamma}_{n}-{\beta}_{n}+{\beta}_{n}{\gamma}_{n}L)}^{2}<0.\end{array}

Again from condition (i), we have that {\alpha}_{n}-{\beta}_{n}\le 0 and 2{\beta}_{n}-{\gamma}_{n}-1\le 0. So, inequality (3.10) implies that

\begin{array}{rcl}{\parallel {x}_{n+1}-p\parallel}^{2}& \le & {\parallel {x}_{n}-p\parallel}^{2}-[{\alpha}_{n}{\beta}_{n}{\gamma}_{n}(1-{\gamma}_{n}^{2}{L}^{2}-2{\gamma}_{n})\\ -{\alpha}_{n}{\beta}_{n}{L}^{2}{({\gamma}_{n}-{\beta}_{n}+{\beta}_{n}{\gamma}_{n}L)}^{2}]{\parallel {x}_{n}-{T}_{n}{x}_{n}\parallel}^{2}.\end{array}

(3.11)

Then

{\parallel {x}_{n+1}-p\parallel}^{2}\le {\parallel {x}_{n}-p\parallel}^{2}.

(3.12)

It is obviously that {lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}-p\parallel exists, then \{\parallel {x}_{n}-p\parallel \} is bounded. This implies that \{{x}_{n}\}, \{{T}_{n}{x}_{n}\}, \{{z}_{n}\}, \{{T}_{n}{z}_{n}\}, \{{y}_{n}\} and \{{T}_{n}{y}_{n}\} are also bounded.

Furthermore, from (2.3), we have that

\varphi (p,{x}_{n})=\varphi (p,{x}_{n+1})+\varphi ({x}_{n+1},{x}_{n})+2\u3008{x}_{n+1}-p,{x}_{n}-{x}_{n+1}\u3009.

This implies that

\u3008{x}_{n+1}-p,{x}_{n}-{x}_{n+1}\u3009+\frac{1}{2}\varphi ({x}_{n+1},{x}_{n})=\frac{1}{2}(\varphi (p,{x}_{n})-\varphi (p,{x}_{n+1})).

(3.13)

Moreover, since the interior of *F* is nonempty, there exists {p}^{\ast}\in F and r>0 such that {p}^{\ast}+rh\in F whenever \parallel h\parallel \le 1. Thus, from the fact that \varphi (x,y)={\parallel x-y\parallel}^{2}, and (3.12) and (3.13), we get that

\begin{array}{rcl}0& \le & \u3008{x}_{n+1}-({p}^{\ast}+rh),{x}_{n}-{x}_{n+1}\u3009+\frac{1}{2}\varphi ({x}_{n+1},{x}_{n})\\ =& \frac{1}{2}(\varphi ({p}^{\ast}+rh,{x}_{n})-\varphi ({p}^{\ast}+rh,{x}_{n+1})).\end{array}

(3.14)

Then from (3.13) and (3.14), we obtain that

\begin{array}{rcl}r\u3008h,{x}_{n}-{x}_{n+1}\u3009& \le & \u3008{x}_{n+1}-{p}^{\ast},{x}_{n}-{x}_{n+1}\u3009+\frac{1}{2}\varphi ({x}_{n+1},{x}_{n})\\ =& \frac{1}{2}(\varphi ({p}^{\ast},{x}_{n})-\varphi ({p}^{\ast},{x}_{n+1})),\end{array}

and hence

\u3008h,{x}_{n}-{x}_{n+1}\u3009\le \frac{1}{2r}(\varphi ({p}^{\ast},{x}_{n})-\varphi ({p}^{\ast},{x}_{n+1})).

Since *h* with \parallel h\parallel \le 1 is arbitrary, we have

\parallel {x}_{n}-{x}_{n+1}\parallel \le \frac{1}{2r}(\varphi ({p}^{\ast},{x}_{n})-\varphi ({p}^{\ast},{x}_{n+1})).

So, if n>m, then we get that

\begin{array}{rcl}\parallel {x}_{m}-{x}_{n}\parallel & =& \parallel {x}_{m}-{x}_{m+1}+{x}_{m+1}-\cdots -{x}_{n-1}+{x}_{n-1}-{x}_{n}\parallel \\ \le & \sum _{i=m}^{n-1}\parallel {x}_{i}-{x}_{i+1}\parallel \\ \le & \frac{1}{2r}\sum _{i=m}^{n-1}(\varphi ({p}^{\ast},{x}_{i})-\varphi ({p}^{\ast},{x}_{i+1}))\\ =& \frac{1}{2r}(\varphi ({p}^{\ast},{x}_{m})-\varphi ({p}^{\ast},{x}_{n})).\end{array}

But we know that \{\varphi ({p}^{\ast},{x}_{n})\} converges. Therefore, we obtain that \{{x}_{n}\} is a Cauchy sequence. Since *C* is closed subset of *H*, there exists {x}^{\ast}\in C such that

{x}_{n}\to {x}^{\ast}.

(3.15)

Furthermore, from (3.11) and conditions (i), (ii) and (iii), we get that

from which it follows that

\underset{n\to}{lim}\parallel {x}_{n}-{T}_{n}{x}_{n}\parallel =0.

(3.16)

Since {\{{T}_{n}\}}_{n=1}^{\mathrm{\infty}} are uniformly closed, then from (3.15) and (3.16), we obtain that {x}^{\ast}\in {\bigcap}_{n=1}^{\mathrm{\infty}}F({T}_{n})=F. The proof is complete. □

**Theorem 3.2** *Let* *C* *be a nonempty*, *closed and convex subset of a real Hilbert space* *H*, *let* {T}_{n}:C\to C *be a finite family of uniformly closed and uniformly Lipschitz pseudocontractive mappings with Lipschitzian constants* {L}_{n}, n=1,2,\dots ,N. *Assume that the interior of* F:={\bigcap}_{n=1}^{N}F({T}_{n}) *is nonempty*. *Let* \{{x}_{n}\} *be a sequence generated from an arbitrary* {x}_{0}\in C *by the following algorithm*:

\{\begin{array}{c}{z}_{n}=(1-{\gamma}_{n}){x}_{n}+{\gamma}_{n}{T}_{n}{x}_{n},\hfill \\ {y}_{n}=(1-{\beta}_{n}){x}_{n}+{\beta}_{n}{T}_{n}{z}_{n},\hfill \\ {x}_{n+1}=(1-{\alpha}_{n}){x}_{n}+{\alpha}_{n}{T}_{n}{y}_{n},\hfill \end{array}

(3.17)

*where* {T}_{n}:={T}_{n(modN)} *and* \{{\alpha}_{n}\},\{{\beta}_{n}\},\{{\gamma}_{n}\}\subset (0,1) *satisfying the following conditions*: (i) {\alpha}_{n}\le {\beta}_{n}\le {\gamma}_{n}, \mathrm{\forall}n\ge 0; (ii) {lim\hspace{0.17em}inf}_{n\mathrm{\infty}}{\alpha}_{n}=\alpha >0; (iii) {sup}_{n\ge 1}{\gamma}_{n}\le \gamma *with* {\gamma}^{3}{L}^{4}+2{\gamma}^{2}{L}^{3}+{\gamma}^{2}{L}^{2}+\gamma {L}^{2}+2\gamma <1 *for* L:=max\{{L}_{n}:n=1,2,\dots ,N\}. *Then* \{{x}_{n}\} *converges strongly to* {x}^{\ast}\in F.

If in Theorem 3.1, we consider a single Lipschitz pseudocontractive mapping, then we may change the conditions of Theorem 3.1.

**Theorem 3.3** *Let* *C* *be a nonempty*, *closed and convex subset of a real Hilbert space* *H*, *let* T:C\to C *be a Lipschitz pseudocontractive mappings with Lipschitzian constants* *L*. *Assume that the interior of* F(T) *is nonempty*. *Let* \{{x}_{n}\} *be a sequence generated from an arbitrary* {x}_{0}\in C *by the following algorithm*:

\{\begin{array}{c}{z}_{n}=(1-{\gamma}_{n}){x}_{n}+{\gamma}_{n}T{x}_{n},\hfill \\ {y}_{n}=(1-{\beta}_{n}){x}_{n}+{\beta}_{n}T{z}_{n},\hfill \\ {x}_{n+1}=(1-{\alpha}_{n}){x}_{n}+{\alpha}_{n}T{y}_{n},\hfill \end{array}

(3.18)

*where* \{{\alpha}_{n}\},\{{\beta}_{n}\},\{{\gamma}_{n}\}\subset (0,1) *satisfying the following conditions*: (i) {\alpha}_{n}\le {\beta}_{n}\le {\gamma}_{n}, \mathrm{\forall}n\ge 0; (ii) \sum {\alpha}_{n}{\beta}_{n}{\gamma}_{n}=\mathrm{\infty}; (iii) {sup}_{n\ge 1}{\gamma}_{n}\le \gamma *with* {\gamma}^{3}{L}^{4}+2{\gamma}^{2}{L}^{3}+{\gamma}^{2}{L}^{2}+\gamma {L}^{2}+2\gamma <1. *Then* \{{x}_{n}\} *converges strongly to* {x}^{\ast}\in F(T).

*Proof* Following the method of proof of Theorem 3.1, we obtain that {x}_{n}\to {x}^{\ast}\in C.

Furthermore, from (3.10) and conditions (i) and (ii), we obtain (3.11). From (3.11) and conditions (iii) and (iv), we obtain that

from which it follows that

\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}inf}\parallel {x}_{n}-T{x}_{n}\parallel =0,

and hence there exists a subsequence \{{x}_{{n}_{k}}\} of \{{x}_{n}\} such that

\underset{n\to \mathrm{\infty}}{lim}\parallel T{x}_{{n}_{k}}-{x}_{{n}_{k}}\parallel =0.

Thus, {x}_{{n}_{k}}\to {x}^{\ast} and the continuity of *T* imply that {x}^{\ast}=T{x}^{\ast}, and hence {x}^{\ast}\in F(T). □