设$ H $为Hilbert空间,用$ \left<{\cdot, \cdot}\right> $表示其内积, $ C $为$ H $中非空闭凸子集.设$ T $为$ C $上非线性自映射, $ F(T) $表示$ T $的不动点集.

映射$ f: C\rightarrow C $称为严格压缩的,若存在常数$ \alpha\in(0, 1) $使得

(1.1) $ \begin{align} \left\Vert{f(x)-f(y)}\right\Vert\leq\alpha\left\Vert{x-y}\right\Vert, \ \forall \ x, y\in C. \end{align} $

映射$ T:C\rightarrow H $称为非扩张的,若

(1.2) $ \begin{align} \left\Vert{Tx-Ty}\right\Vert\leq\left\Vert{x-y}\right\Vert, \forall\, \, x, y\in C. \end{align} $

映射$ A:C\rightarrow H $称为单调的,若

$ \left<{Ax-Ay, x-y}\right>\geq0, \, \, \forall\, x, y\in C. $

映射$ A:C\rightarrow H $称为$ \alpha $ -逆强单调的,若存在常数$ \alpha $使得

(1.3) $ \begin{align} \left<{Ax-Ay, x-y}\right>\geq\alpha\left\Vert{Ax-Ay}\right\Vert^2, \, \, \forall\, x, y\in C. \end{align} $

变分不等式是解决一些纯理论和应用科学问题的一种重要工具,例如均衡问题,优化问题等.许多研究者已经提出了一些数值方法来解决变分不等式问题和相关的优化问题,见文献[1-14].

设$ A:C\rightarrow H $为非线性算子,经典的变分不等式就是找$ x^* $满足

(1.4) $ \begin{align} \left<{Ax^*, x-x^*}\right>\geq0, \ \forall\ x\in C. \end{align} $

用$ \rm{VI}(A, C) $记不等式$ (1.4) $的解集. Iiduka, Takahashi和Toyoda^[4]引进了一个迭代算法并证明了由此迭代算法构建的序列弱收敛到上述变分不等式问题的一个元.

进一步, Ceng等^[6]研究了下述一般的变分不等式: $ A, B:C\rightarrow H $为非线性算子,找$ (x^*, y^*)\in C\times C $满足

(1.5) $ \begin{align} \left\{ \begin{array}{ll} \left<{x^*-(I-\lambda A)y^*, x-x^*}\right>\geq0, \, \, &\forall\, x\in C, \\ \left<{y^*-(I-\mu B)x^*, x-y^*}\right>\geq0, \, \, &\forall\, x\in C.\\ \end{array} \right. \end{align} $

他们引入迭代算法并找到了变分不等式(1.5)的解集与一个非扩张映射的不动点集的公共元,这里$ \lambda, \mu $为正实数.

我们考虑更一般的变分不等式:找$ (x^*, y^*)\in C\times C $满足

(1.6) $ \begin{align} \left\{ \begin{array}{ll} \left<{x^*-(I-\lambda A)(ax^*+(1-a)y^*), x-x^*}\right>\geq0, \, \, &\forall\, x\in C, \\ \left<{y^*-(I-\mu B)x^*, x-y^*}\right>\geq0, \, \, &\forall\, x\in C, \\ \end{array} \right. \end{align} $

这里$ 0\leq a<1 $, $ \lambda, \mu $为正实数.如果$ a = 0 $,则(1.6)式变为$ (1.5) $式,因此(1.5)式作为(1.6)式特殊情形.

本文用修改的超梯度方法,我们研究了一个新的粘性隐式算法并证明了由此算法构建的序列强收敛到变分不等式问题(1.6)的解集和非扩张映射的不动点集的公共元.

任取$ x\in H $,众所周知, $ C $中存在唯一点,记为$ P_Cx $,满足

(2.1) $\left\Vert{x-P_Cx}\right\Vert\leq\left\Vert{x-y}\right\Vert, \, \, \forall\, \, y\in C.$

$ P $称为$ H $到$ C $上的投影算子.易知$ P_C $为$ H $到$ C $上的非扩张映射且满足

(2.2) $\left<{x-y, P_Cx-P_Cy}\right>\geq\left\Vert{P_Cx-P_Cy}\right\Vert^2, \, \forall\, x, y\in H.$

进一步, $ P_Cx $满足如下性质: $ P_Cx\in C $且

(2.3) $\left<{x-P_Cx, y-P_Cx}\right>\leq0,$

(2.4) $\left\Vert{x-y}\right\Vert^2\geq\left\Vert{x-P_Cx}\right\Vert^2+\left\Vert{y-P_Cx}\right\Vert^2, \forall\, x\in H, \, y\in C.$

也有

(2.5) $ \begin{align} \left\Vert{x-y}\right\Vert^2 = \left\Vert{x}\right\Vert^2-\left\Vert{y}\right\Vert^2-2\left<{x-y, y}\right>. \end{align} $

为了证明主要结果,需要如下引理.

引理2.1^[11]  设$ \left\{{a_n}\right\} $为非负实数列满足

$ a_{n+1}\leq(1-\alpha_n)a_n+\delta_n, \, \, n\geq0, $

其中$ \left\{{\alpha_n}\right\} $为$ (0, 1) $中序列且$ \left\{{\delta_n}\right\} $为实数列满足

(i) $ \sum\limits_{n = 0}^\infty \alpha_n = \infty $,

(ii) $ \limsup\limits_{n\rightarrow\infty}\frac{\delta_n}{\alpha_n}\leq0 $或$ \sum\limits_{n = 1}^\infty\left\vert{\delta_n}\right\vert<\infty $,

则$ \lim\limits_{n\rightarrow\infty}a_n = 0 $.

引理2.2^[12]  设$ \left\{{x_n}\right\} $和$ \left\{{z_n}\right\} $为Banach空间$ E $中有界序列, $ \left\{{\beta_n}\right\} $为$ [0, 1] $中序列满足: $ 0<\liminf\limits_{n\rightarrow\infty}\beta_n\leq\limsup\limits_{n\rightarrow\infty}\beta_n<1 $.设$ x_{n+1} = \beta_nx_n+(1-\beta_n)z_n $, $ n\geq0 $满足$ \limsup\limits_{n\rightarrow\infty}(\left\Vert{z_{n+1}-z_n}\right\Vert-\left\Vert{x_{n+1}-x_n}\right\Vert)\leq0 $.则$ \lim\limits_{n\rightarrow\infty}\left\Vert{z_n-x_n}\right\Vert = 0 $.

引理2.3^[13]  设$ C $为实Hilbert空间$ H $中非空闭凸子集, $ S $为$ C $上非扩张自映射且满足$ F(T)\neq\emptyset $.则$ I-S $是半闭的,即任给$ C $中序列$ \left\{{x_n}\right\} $且$ \left\{{x_n}\right\} $弱收敛到$ x\in C $,序列$ \left\{{(I-S)x_n}\right\} $强收敛到$ y $,则$ (I-S)x = y $,这里$ I $为$ H $上恒等算子.

通过简单的证明,可以得到下面结果.

引理2.4  设$ C $为实Hilbert空间$ H $中非空闭凸子集, $ A, B:C\rightarrow H $为两个非线性算子.任给$ \lambda, \mu>0 $, $ a\in[0, 1] $.则下述结论等价.

(i) $ (x^*, y^*)\in C\times C $为问题(1.6)的解;

(ii) $ x^* $为映射$ G $的不动点,即$ x^*\in F(G) $,这里$ G:C\rightarrow C $定义为

$ G(x) = P_C(I-\lambda A)(aI+(1-a)P_C(I-\mu B))(x), \ \forall\ x\in C, $

这里$ y^* = P_C(I-\mu B)x^* $.

定理3.1  设$ C $为实Hilbert空间$ H $中非空闭凸子集, $ A, B:C\rightarrow H $分别为$ d_1 $ -逆强单调的与$ d_2 $ -逆强单调的, $ T:C\rightarrow C $为非扩张映射满足$ \Omega = F(T)\cap F(G)\neq\emptyset $,这里$ G $的定义见引理2.4.设$ f $为$ C $上严格压缩自映射,其参数$ \delta\in(0, 1) $.任取$ x_{1}\in C $.定义序列$ \left\{{x_{n}}\right\} $为

(3.1) $ \begin{equation} \left\{ \begin{array}{l} { } v_n = \frac{1}{2}(x_n+z_n), \\ u_n = P_C(I-\mu B)v_n, \\ z_n = P_C(I-\lambda A)(ax_n+(1-a)u_n), \\ y_n = \alpha_nf(x_n)+(1-\alpha_n)z_n, \\ x_{n+1} = \beta_nx_n+(1-\beta_n)Ty_n, \end{array} \right. \end{equation} $

这里$ \lambda\in (0, 2d_1), \mu\in(0, 2d_2) $, $ 0\leq a<1 $.设$ \{\alpha_{n}\} $, $ \{\beta_{n}\} $为$ (0, 1] $中序列满足:

(i) $ \lim\limits_{n\rightarrow\infty}\alpha_{n} = 0 $, $ \sum\limits_{n = 1}^{\infty}\alpha_{n} = \infty $,

(ii) $ 0<\liminf\limits_{n\rightarrow\infty}\beta_n\leq\limsup\limits_{n\rightarrow\infty}\beta_n<1 $,

则$ \left\{{x_n}\right\} $强收敛到$ q\in \Omega $,这里$ q = P_\Omega f(q) $,也满足下述变分不等式

$ \left<{f(q)-q, p-q}\right>\leq0, \, \, \forall\, \, p\in \Omega. $

证  首先证明$ \{x_{n}\} $有界.任取$ x, y\in C $,观察

(3.2) $ \begin{eqnarray} \left\Vert{(I-\lambda A)x-(I-\lambda A)y}\right\Vert^2 & = &\left\Vert{x-y}\right\Vert^2-2\lambda\left<{Ax-Ay, x-y}\right>+\lambda^2\left\Vert{Ax-Ay}\right\Vert^2 \\ &\leq&\left\Vert{x-y}\right\Vert^2-2\lambda d_1\left\Vert{Ax-Ay}\right\Vert^2+\lambda^2\left\Vert{Ax-Ay}\right\Vert^2 \\ & = &\left\Vert{x-y}\right\Vert^2+\lambda(\lambda-2d_1)\left\Vert{Ax-Ay}\right\Vert^2 \\ &\leq&\left\Vert{x-y}\right\Vert^2. \end{eqnarray} $

根据$ \lambda\in(0, 2d_1) $知$ I-\lambda A $是非扩张的.同理可以证明$ I-\mu B $也为非扩张的,从而可证$ G $也是非扩张的.设$ x^*\in \Omega $, $ y^* = P_C(I-\lambda A)x^* $.则

$ \begin{eqnarray*} \left\Vert{z_n-x^*}\right\Vert& = &\left\Vert{Gv_n-Gx^*}\right\Vert \leq \left\Vert{v_n-x^*}\right\Vert\\ & = &\left\Vert{\frac{1}{2}(x_n-x^*)+\frac{1}{2}(z_n-x^*)}\right\Vert\\ &\leq&\frac{1}{2}\left\Vert{x_n-x^*}\right\Vert+\frac{1}{2}\left\Vert{z_n-x^*}\right\Vert, \end{eqnarray*} $

于是

(3.3) $ \begin{equation} \left\Vert{z_n-x^*}\right\Vert\leq\left\Vert{x_n-x^*}\right\Vert. \end{equation} $

由(3.1)式和数学归纳法得

$ \begin{eqnarray*} \left\Vert{x_{n+1}-x^*}\right\Vert & = &\left\Vert{\beta_n(x_n-x^*)+(1-\beta_n)(Ty_n-x^*)}\right\Vert\\ &\leq&\beta_n\left\Vert{x_n-x^*}\right\Vert+(1-\beta_n)\left\Vert{Ty_n-x^*}\right\Vert\\ &\leq&\beta_n\left\Vert{x_n-x^*}\right\Vert+(1-\beta_n)\left\Vert{y_n-x^*}\right\Vert\\ & = &\beta_n\left\Vert{x_n-x^*}\right\Vert+(1-\beta_n)\left\Vert{\alpha_n(f(x_n)-x^*)+(1-\alpha_n)(z_n-x^*)}\right\Vert\\ &\leq&\beta_n\left\Vert{x_n-x^*}\right\Vert+(1-\beta_n)\alpha_n\left\Vert{f(x_n)-f(x^*)}\right\Vert+(1-\beta_n)\alpha_n\left\Vert{f(x^*)-x^*}\right\Vert\\ &&+(1-\beta_n)(1-\alpha_n)\left\Vert{z_n-x^*}\right\Vert\\ &\leq&[\beta_n+(1-\beta_n)\alpha_n\delta+(1-\beta_n)(1-\alpha_n)]\left\Vert{x_n-x^*}\right\Vert+(1-\beta_n)\alpha_n\left\Vert{f(x^*)-x^*}\right\Vert\\ & = &[1-(1-\beta_n)\alpha_n(1-\delta)]\left\Vert{x_n-x^*}\right\Vert+(1-\beta_n)\alpha_n(1-\delta)\frac{\left\Vert{f(x^*)-x^*}\right\Vert}{1-\delta}\\ &\leq&\max\{\left\Vert{x_1-x^*}\right\Vert, \frac{\left\Vert{f(x^*)-x^*}\right\Vert}{1-\delta}\}. \end{eqnarray*} $

因此$ \{x_{n}\} $有界.

下面证明$ \lim\limits_{n\rightarrow\infty}\left\Vert{x_{n+1}-x_n}\right\Vert = 0 $.观察

$ \begin{eqnarray*} \left\Vert{z_{n+1}-z_n}\right\Vert& = &\left\Vert{Gv_{n+1}-Gv_n}\right\Vert \leq\left\Vert{z_{n+1}-z_n}\right\Vert\\ & = &\left\Vert{\frac{1}{2}(x_{n+1}-x_n)+\frac{1}{2}(z_{n+1}-z_n)}\right\Vert\\ &\leq&\frac{1}{2}\left\Vert{x_{n+1}-x_n}\right\Vert+\frac{1}{2}\left\Vert{z_{n+1}-z_n}\right\Vert, \end{eqnarray*} $

从而

(3.4) $ \begin{equation} \left\Vert{z_{n+1}-z_n}\right\Vert\leq\left\Vert{x_{n+1}-x_n}\right\Vert. \end{equation} $

由(3.4)式得

(3.5) $ \begin{eqnarray} \left\Vert{y_{n+1}-y_n}\right\Vert & = &\left\Vert{\alpha_{n+1}f(x_{n+1})+(1-\alpha_{n+1})z_{n+1}-\alpha_nf(x_n)-(1-\alpha_n)z_n}\right\Vert \\ & = &\left\Vert{\alpha_{n+1}(f(x_{n+1})-z_{n+1})+\alpha_n(z_n-f(x_n))+z_{n+1}-z_n}\right\Vert \\ &\leq&\alpha_{n+1}\left\Vert{f(x_{n+1})-z_{n+1}}\right\Vert+\alpha_n\left\Vert{z_n-f(x_n)}\right\Vert+\left\Vert{z_{n+1}-z_n}\right\Vert \\ &\leq&\alpha_{n+1}\left\Vert{f(x_{n+1})-z_{n+1}}\right\Vert+\alpha_n\left\Vert{z_n-f(x_n)}\right\Vert+\left\Vert{x_{n+1}-x_n}\right\Vert, \end{eqnarray} $

于是

$ \begin{eqnarray*} \left\Vert{Ty_{n+1}-Ty_n}\right\Vert &\leq&\left\Vert{y_{n+1}-y_n}\right\Vert\\ &\leq&\alpha_{n+1}\left\Vert{f(x_{n+1})-z_{n+1}}\right\Vert+\alpha_n\left\Vert{z_n-f(x_n)}\right\Vert+\left\Vert{x_{n+1}-x_n}\right\Vert. \end{eqnarray*} $

由条件(i), (ii)知

$ \limsup\limits_{n\rightarrow\infty}(\left\Vert{Ty_{n+1}-Ty_n}\right\Vert-\left\Vert{x_{n+1}-x_n}\right\Vert)\leq0. $

根据引理2.2得

$ \lim\limits_{n\rightarrow\infty}\left\Vert{Ty_n-x_n}\right\Vert = 0. $

于是

(3.6) $ \begin{equation} \lim\limits_{n\rightarrow\infty}\left\Vert{x_{n+1}-x_n}\right\Vert = \lim\limits_{n\rightarrow\infty}(1-\beta_n)\left\Vert{Ty_n-x_n}\right\Vert = 0. \end{equation} $

下面证明$ \lim\limits_{n\rightarrow\infty}\left\Vert{x_n-Gx_n}\right\Vert = 0, \ \lim\limits_{n\rightarrow\infty}\left\Vert{x_n-Tx_n}\right\Vert = 0 $.

由(3.1)式知

(3.7) $ \begin{eqnarray} \left\Vert{u_n-y^*}\right\Vert^2& = &\left\Vert{P_C(I-\mu B)v_n-P_C(I-\mu B)x^*}\right\Vert^2 \\ &\leq&\left\Vert{(I-\mu B)v_n-(I-\mu B)x^*}\right\Vert^2 \\ &\leq&\left\Vert{v_n-x^*}\right\Vert^2-\mu(2d_2-\mu)\left\Vert{Bv_n-Bx^*}\right\Vert^2 \\ & = &\left\Vert{\frac{1}{2}(x_n-x^*)+\frac{1}{2}(z_n-x^*)}\right\Vert^2-\mu(2d_2-\mu)\left\Vert{Bv_n-Bx^*}\right\Vert^2 \\ &\leq&\frac{1}{2}\left\Vert{x_n-x^*}\right\Vert^2+\frac{1}{2}\left\Vert{z_n-x^*}\right\Vert^2-\mu(2d_2-\mu)\left\Vert{Bv_n-Bx^*}\right\Vert^2. \end{eqnarray} $

同理可得

(3.8) $ \begin{eqnarray} \left\Vert{z_n-x^*}\right\Vert^2 & = &\left\Vert{P_C(I-\lambda A)(ax_n+(1-a)u_n)-P_C(I- \lambda A)(ax^*+(1-a)y^*)}\right\Vert^2 \\ &\leq&\left\Vert{(I-\lambda A)(ax_n+(1-a)u_n)-(I- \lambda A)(ax^*+(1-a)y^*)}\right\Vert^2 \\ &\leq&\left\Vert{(ax_n+(1-a)u_n)-(ax^*+(1-a)y^*)}\right\Vert^2 \\ &&-\lambda(2d_1-\lambda)\left\Vert{A(ax_n+(1-a)u_n)-A(ax^*+(1-a)y^*)}\right\Vert^2 \\ &\leq& a\left\Vert{x_n-x^*}\right\Vert^2+(1-a)\left\Vert{u_n-y^*}\right\Vert^2 \\ &&-\lambda(2d_1-\lambda)\left\Vert{A(ax_n+(1-a)u_n)-A(ax^*+(1-a)y^*)}\right\Vert^2. \end{eqnarray} $

(3.7)式代入(3.8)式,得

$ \begin{eqnarray*} \left\Vert{z_n-x^*}\right\Vert^2 &\leq& a\left\Vert{x_n-x^*}\right\Vert^2\\ &&+(1-a)\bigg[\frac{1}{2}\left\Vert{x_n-x^*}\right\Vert^2+\frac{1}{2}\left\Vert{z_n-x^*}\right\Vert^2 -\mu(2d_2-\mu)\left\Vert{Bv_n-Bx^*}\right\Vert^2\bigg]\\ &&-\lambda(2d_1-\lambda)\left\Vert{A(ax_n+(1-a)u_n)-A(ax^*+(1-a)y^*)}\right\Vert^2\\ & = &\frac{1+a}{2}\left\Vert{x_n-x^*}\right\Vert^2+\frac{1-a}{2}\left\Vert{z_n-x^*}\right\Vert^2-(1-a)\mu(2d_2-\mu)\left\Vert{Bv_n-Bx^*}\right\Vert^2\\ &&-\lambda(2d_1-\lambda)\left\Vert{A(ax_n+(1-a)u_n)-A(ax^*+(1-a)y^*)}\right\Vert^2. \end{eqnarray*} $

于是

(3.9) $ \begin{eqnarray} \left\Vert{z_n-x^*}\right\Vert^2&\leq&\left\Vert{x_n-x^*}\right\Vert^2-\frac{1-a}{1+a}2\mu(2d_2-\mu)\left\Vert{Bv_n-Bx^*}\right\Vert^2 \\ &&-\frac{1}{1+a}2\lambda(2d_1-\lambda)\left\Vert{A(ax_n+(1-a)u_n)-A(ax^*+(1-a)y^*)}\right\Vert^2. \end{eqnarray} $

根据(3.1)式知

(3.10) $ \begin{eqnarray} \left\Vert{x_{n+1}-x^*}\right\Vert^2& = &\left\Vert{\beta_n(x_n-x^*)+(1-\beta_n)(Ty_n-x^*)}\right\Vert^2 \\ &\leq&\beta_n\left\Vert{x_n-x^*}\right\Vert^2+(1-\beta_n)\left\Vert{Ty_n-x^*}\right\Vert^2 \\ &\leq&\beta_n\left\Vert{x_n-x^*}\right\Vert^2+(1-\beta_n)\left\Vert{y_n-x^*}\right\Vert^2 \\ & = &\beta_n\left\Vert{x_n-x^*}\right\Vert^2+(1-\beta_n)\left\Vert{\alpha_n(f(x_n)-x^*)+(1-\alpha_n)(z_n-x^*)}\right\Vert^2 \\ &\leq&\beta_n\left\Vert{x_n-x^*}\right\Vert^2+(1-\beta_n)\alpha_n\left\Vert{f(x_n)-x^*}\right\Vert^2+(1-\beta_n)(1-\alpha_n)\left\Vert{z_n-x^*}\right\Vert^2 \\ &\leq&\alpha_n\left\Vert{f(x_n)-x^*}\right\Vert^2+\beta_n\left\Vert{x_n-x^*}\right\Vert^2+(1-\beta_n)\left\Vert{z_n-x^*}\right\Vert^2 \\ &\leq&\alpha_nM_1+\beta_n\left\Vert{x_n-x^*}\right\Vert^2+(1-\beta_n)\left\Vert{z_n-x^*}\right\Vert^2, \end{eqnarray} $

这里$ M_1 = \sup_{n\geq1}\left\Vert{f(x_n)-x^*}\right\Vert^2 $.

将(3.9)式代入(3.10)式得

$ \begin{eqnarray*} \left\Vert{x_{n+1}-x^*}\right\Vert^2 &\leq&\alpha_nM_1+\beta_n\left\Vert{x_n-x^*}\right\Vert^2\\ &&+(1-\beta_n)\bigg[\left\Vert{x_n-x^*}\right\Vert^2-\frac{1-a}{1+a}2\mu(2d_2-\mu)\left\Vert{Bv_n-Bx^*}\right\Vert^2\\ &&-\frac{1}{1+a}2\lambda(2d_1-\lambda)\left\Vert{A(ax_n+(1-a)u_n)-A(ax^*+(1-a)y^*)}\right\Vert^2\bigg]\\ & = &\alpha_nM_1+\left\Vert{x_n-x^*}\right\Vert^2-\frac{1-a}{1+a}2(1-\beta_n)\mu(2d_2-\mu)\left\Vert{Bv_n-Bx^*}\right\Vert^2\\ &&-\frac{1}{1+a}2(1-\beta_n)\lambda(2d_1-\lambda)\left\Vert{A(ax_n+(1-a)u_n)-A(ax^*+(1-a)y^*)}\right\Vert^2, \end{eqnarray*} $

于是

$ \begin{eqnarray*} &&\frac{1-a}{1+a}2(1-\beta_n)\mu(2d_2-\mu)\left\Vert{Bv_n-Bx^*}\right\Vert^2\\ &&+\frac{1}{1+a}2(1-\beta_n)\lambda(2d_1-\lambda)\left\Vert{A(ax_n+(1-a)u_n)-A(ax^*+(1-a)y^*)}\right\Vert^2\\ &\leq&\alpha_nM_1+\left\Vert{x_n-x^*}\right\Vert^2-\left\Vert{x_{n+1}-x^*}\right\Vert^2\\ &\leq&\alpha_nM_1+\left\Vert{x_n-x_{n+1}}\right\Vert(\left\Vert{x_n-x^*}\right\Vert+\left\Vert{x_{n+1}-x^*}\right\Vert). \end{eqnarray*} $

因为$ \lambda\in(0, 2d_1), \ \mu\in(0, 2d_2) $, $ 0\leq a<1 $, $ { }\lim\limits_{n\rightarrow\infty}\alpha_{n} = 0 $和(3.6)式,有

(3.11) $ \begin{eqnarray} \lim\limits_{n\rightarrow\infty}\left\Vert{Bv_n-Bx^*}\right\Vert = 0, \, \lim\limits_{n\rightarrow\infty} \left\Vert{A(ax_n+(1-a)u_n)-A(ax^*+(1-a)y^*)}\right\Vert = 0. \end{eqnarray} $

另外一方面,由(2.2)和(2.5)式得

$ \begin{eqnarray*} \left\Vert{u_n-y^*}\right\Vert^2 & = &\left\Vert{P_C(I-\mu B)v_n-P_C(I-\mu B)x^*}\right\Vert^2\\ &\leq&\left<{(I-\mu B)v_n-(I-\mu B)x^*, u_n-y^*}\right>\\ & = &\left<{v_n-x^*, u_n-y^*}\right>-\mu\left<{Bv_n-Bx^*, u_n-y^*}\right>\\ & = &\frac{1}{2}[\left\Vert{v_n-x^*}\right\Vert^2+\left\Vert{u_n-y^*}\right\Vert^2-\left\Vert{v_n-u_n-(x^*-y^*)}\right\Vert^2]\\ &&+\mu\left<{Bx^*-Bv_n, u_n-y^*}\right>, \end{eqnarray*} $

从而

(3.12) $ \begin{eqnarray} \left\Vert{u_n-y^*}\right\Vert^2 &\leq&\left\Vert{v_n-x^*}\right\Vert^2-\left\Vert{v_n-u_n-(x^*-y^*)}\right\Vert^2+2\mu\left\Vert{Bx^*-Bv_n}\right\Vert\left\Vert{u_n-y^*}\right\Vert \\ & = &\left\Vert{\frac{1}{2}(x_n-x^*)+\frac{1}{2}(z_n-x^*)}\right\Vert^2-\left\Vert{v_n-u_n-(x^*-y^*)}\right\Vert^2 \\ &&+2\mu\left\Vert{Bx^*-Bv_n}\right\Vert\left\Vert{u_n-y^*}\right\Vert \\ &\leq&\frac{1}{2}\left\Vert{x_n-x^*}\right\Vert^2+\frac{1}{2}\left\Vert{z_n-x^*}\right\Vert^2-\left\Vert{v_n-u_n-(x^*-y^*)}\right\Vert^2 \\ &&+2\mu\left\Vert{Bx^*-Bv_n}\right\Vert\left\Vert{u_n-y^*}\right\Vert. \end{eqnarray} $

再次由(2.2)和(2.5)式知

$ \begin{eqnarray*} \left\Vert{z_n-x^*}\right\Vert^2 & = &\left\Vert{P_C(I-\lambda A)(ax_n+(1-a)u_n)-P_C(I- \lambda A)(ax^*+(1-a)y^*)}\right\Vert^2\\ &\leq&\left<{(I-\lambda A)(ax_n+(1-a)u_n)-(I- \lambda A)(ax^*+(1-a)y^*), z_n-x^*}\right>\\ & = &\left<{(ax_n+(1-a)u_n)-(ax^*+(1-a)y^*), z_n-x^*}\right>\\ &&+\lambda\left<{A(ax^*+(1-a)y^*)-A(ax_n+(1-a)u_n), z_n-x^*}\right>\\ & = &\left<{a(x_n-x^*)+(1-a)(u_n-y^*), z_n-x^*}\right>\\ &&+\lambda\left<{A(ax^*+(1-a)y^*)-A(ax_n+(1-a)u_n), z_n-x^*}\right>\\ & = &a\left<{x_n-x^*, z_n-x^*}\right>+(1-a)\left<{u_n-y^*, z_n-x^*}\right>\\ &&+\lambda\left<{A(ax^*+(1-a)y^*)-A(ax_n+(1-a)u_n), z_n-x^*}\right>\\ & = &\frac{a}{2}[\left\Vert{x_n-x^*}\right\Vert^2+\left\Vert{z_n-x^*}\right\Vert^2-\left\Vert{x_n-z_n}\right\Vert^2]\\ &&+\frac{1-a}{2}[\left\Vert{u_n-y^*}\right\Vert^2+\left\Vert{z_n-x^*}\right\Vert^2 -\left\Vert{u_n-z_n+x^*-y^*}\right\Vert^2]\\ &&+\lambda\left\Vert{A(ax^*+(1-a)y^*)-A(ax_n+(1-a)u_n)}\right\Vert\left\Vert{z_n-x^*}\right\Vert, \end{eqnarray*} $

于是

(3.13) $ \begin{eqnarray} \left\Vert{z_n-x^*}\right\Vert^2 &\leq& a\left\Vert{x_n-x^*}\right\Vert^2+(1-a)\left\Vert{u_n-y^*}\right\Vert^2-a\left\Vert{x_n-z_n}\right\Vert^2 \\ &&-(1-a)\left\Vert{u_n-z_n+x^*-y^*}\right\Vert^2 \\ &&+2\lambda\left\Vert{A(ax^*+(1-a)y^*)-A(ax_n+(1-a)u_n)}\right\Vert\left\Vert{z_n-x^*}\right\Vert \\ &\leq& a\left\Vert{x_n-x^*}\right\Vert^2+(1-a)\left\Vert{u_n-y^*}\right\Vert^2-(1-a)\left\Vert{u_n-z_n+x^*-y^*}\right\Vert^2 \\ &&+2\lambda\left\Vert{A(ax^*+(1-a)y^*)-A(ax_n+(1-a)u_n)}\right\Vert\left\Vert{z_n-x^*}\right\Vert. \end{eqnarray} $

将(3.12)式代入(3.13)式,得

$ \begin{eqnarray*} \left\Vert{z_n-x^*}\right\Vert^2 &\leq& a\left\Vert{x_n-x^*}\right\Vert^2+(1-a)\bigg[\frac{1}{2}\left\Vert{x_n-x^*}\right\Vert^2+\frac{1}{2}\left\Vert{z_n-x^*}\right\Vert^2\\ &&-\left\Vert{v_n-u_n-(x^*-y^*)}\right\Vert^2+2\mu\left\Vert{Bx^*-Bv_n}\right\Vert\left\Vert{u_n-y^*}\right\Vert\bigg]\\ &&-(1-a)\left\Vert{u_n-z_n+x^*-y^*}\right\Vert^2\\ &&+2\lambda\left\Vert{A(ax^*+(1-a)y^*)-A(ax_n+(1-a)u_n)}\right\Vert\left\Vert{z_n-x^*}\right\Vert\\ & = &\frac{1+a}{2}\left\Vert{x_n-x^*}\right\Vert^2+\frac{1-a}{2}\left\Vert{z_n-x^*}\right\Vert^2-(1-a)\left\Vert{v_n-u_n-(x^*-y^*)}\right\Vert^2\\ &&+2(1-a)\mu\left\Vert{Bx^*-Bv_n}\right\Vert\left\Vert{u_n-y^*}\right\Vert-(1-a)\left\Vert{u_n-z_n+x^*-y^*}\right\Vert^2\\ &&+2\lambda\left\Vert{A(ax^*+(1-a)y^*)-A(ax_n+(1-a)u_n)}\right\Vert\left\Vert{z_n-x^*}\right\Vert. \end{eqnarray*} $

于是

(3.14) $ \begin{eqnarray} \left\Vert{z_n-x^*}\right\Vert^2 &\leq&\left\Vert{x_n-x^*}\right\Vert^2-\frac{2(1-a)}{1+a}\left\Vert{v_n-u_n-(x^*-y^*)}\right\Vert^2 \\ &&+\frac{1-a}{1+a}4\mu\left\Vert{Bx^*-Bv_n}\right\Vert\left\Vert{u_n-y^*}\right\Vert-\frac{2(1-a)}{1+a}\left\Vert{u_n-z_n+x^*-y^*}\right\Vert^2 \\ &&+\frac{4\lambda}{1+a}\left\Vert{A(ax^*+(1-a)y^*)-A(ax_n+(1-a)u_n)}\right\Vert\left\Vert{z_n-x^*}\right\Vert. \end{eqnarray} $

将(3.14)式代入(3.10)式,有

$ \begin{eqnarray*} \left\Vert{x_{n+1}-x^*}\right\Vert^2 &\leq&\alpha_nM_1+\beta_n\left\Vert{x_n-x^*}\right\Vert^2\\ &&+(1-\beta_n)\bigg[\left\Vert{x_n-x^*}\right\Vert^2 -\frac{2(1-a)}{1+a}\left\Vert{v_n-u_n-(x^*-y^*)}\right\Vert^2 \\ &&+\frac{1-a}{1+a}4\mu\left\Vert{Bx^*-Bv_n}\right\Vert\left\Vert{u_n-y^*}\right\Vert -\frac{2(1-a)}{1+a}\left\Vert{u_n-z_n+x^*-y^*}\right\Vert^2 \\ &&+\frac{4\lambda}{1+a}\left\Vert{A(ax^*+(1-a)y^*)-A(ax_n+(1-a)u_n)}\right\Vert\left\Vert{z_n-x^*}\right\Vert\bigg]\\ & = &\alpha_nM_1+\left\Vert{x_n-x^*}\right\Vert^2-\frac{2(1-a)}{1+a}(1-\beta_n)\left\Vert{v_n-u_n-(x^*-y^*)}\right\Vert^2\\ &&+\frac{1-a}{1+a}4(1-\beta_n)\mu\left\Vert{Bx^*-Bv_n}\right\Vert\left\Vert{u_n-y^*}\right\Vert\\ &&-\frac{2(1-a)}{1+a}(1-\beta_n)\left\Vert{u_n-z_n+x^*-y^*}\right\Vert^2\\ &&+\frac{4\lambda}{1+a}(1-\beta_n)\left\Vert{A(ax^*+(1-a)y^*)-A(ax_n+(1-a)u_n)}\right\Vert\left\Vert{z_n-x^*}\right\Vert, \end{eqnarray*} $

从而

$ \begin{eqnarray*} &&\frac{2(1-a)}{1+a}(1-\beta_n)\left\Vert{v_n-u_n-(x^*-y^*)}\right\Vert^2+\frac{2(1-a)}{1+a}(1-\beta_n)\left\Vert{u_n-z_n+x^*-y^*}\right\Vert^2\\ &\leq&\alpha_nM_1+\left\Vert{x_n-x^*}\right\Vert^2-\left\Vert{x_{n+1}-x^*}\right\Vert^2+\frac{1-a}{1+a}4(1-\beta_n)\mu\left\Vert{Bx^*-Bv_n}\right\Vert\left\Vert{u_n-y^*}\right\Vert\\ &&+\frac{4\lambda}{1+a}(1-\beta_n)\left\Vert{A(ax^*+(1-a)y^*)-A(ax_n+(1-a)u_n)}\right\Vert\left\Vert{z_n-x^*}\right\Vert\\ &\leq&\alpha_nM_1+\left\Vert{x_n-x_{n+1}}\right\Vert(\left\Vert{x_n-x^*}\right\Vert+\left\Vert{x_{n+1}-x^*}\right\Vert)\\ &&+\frac{1-a}{1+a}4(1-\beta_n)\mu\left\Vert{Bx^*-Bv_n}\right\Vert\left\Vert{u_n-y^*}\right\Vert\\ &&+\frac{4\lambda}{1+a}(1-\beta_n)\left\Vert{A(ax^*+(1-a)y^*)-A(ax_n+(1-a)u_n)}\right\Vert\left\Vert{z_n-x^*}\right\Vert. \end{eqnarray*} $

因为$ \lim\limits_{n\rightarrow\infty}\alpha_{n} = 0 $, $ 0\leq a<1 $,条件(ii), (3.6)式和(3.11)式,有

(3.15) $ \begin{eqnarray} \lim\limits_{n\rightarrow\infty}\left\Vert{v_n-u_n-(x^*-y^*)}\right\Vert = 0, \, \lim\limits_{n\rightarrow\infty}\left\Vert{u_n-z_n+x^*-y^*}\right\Vert = 0. \end{eqnarray} $

从而

(3.16) $ \begin{eqnarray} \left\Vert{v_n-z_n}\right\Vert \leq\left\Vert{v_n-u_n-(x^*-y^*)}\right\Vert+\left\Vert{u_n-z_n+x^*-y^*}\right\Vert \rightarrow0, \ \mbox{当}\ n\rightarrow\infty. \end{eqnarray} $

注意到

$ \left\Vert{x_n-z_n}\right\Vert\leq\left\Vert{x_n-v_n}\right\Vert+\left\Vert{v_n-z_n}\right\Vert = \frac{1}{2}\left\Vert{x_n-z_n}\right\Vert+\left\Vert{v_n-z_n}\right\Vert, $

于是

(3.17) $ \begin{eqnarray} \left\Vert{x_n-z_n}\right\Vert\leq2\left\Vert{v_n-z_n}\right\Vert \rightarrow0, \ \mbox{当}\ n\rightarrow\infty. \end{eqnarray} $

由(3.1)式得

$ \left\Vert{x_{n+1}-x_n}\right\Vert = (1-\beta_n)\left\Vert{Ty_n-x_n}\right\Vert, $

于是

$ \left\Vert{Ty_n-x_n}\right\Vert\leq\frac{1}{1-\beta_n}\left\Vert{x_{n+1}-x_n}\right\Vert. $

由条件(ii)和(3.6)式得

(3.18) $ \begin{eqnarray} \lim\limits_{n\rightarrow\infty}\left\Vert{Ty_n-x_n}\right\Vert = 0. \end{eqnarray} $

观察

(3.19) $ \begin{eqnarray} \left\Vert{y_n-z_n}\right\Vert = \alpha_n\left\Vert{f(x_n)-z_n}\right\Vert \rightarrow0, \ \mbox{当}\ n\rightarrow\infty. \end{eqnarray} $

于是

(3.20) $ \begin{eqnarray} \left\Vert{Tx_n-x_n}\right\Vert&\leq&\left\Vert{Tx_n-Tz_n}\right\Vert+\left\Vert{Tz_n-Ty_n}\right\Vert+\left\Vert{Ty_n-x_n}\right\Vert \\ &\leq&\left\Vert{x_n-z_n}\right\Vert+\left\Vert{z_n-y_n}\right\Vert+\left\Vert{Ty_n-x_n}\right\Vert \\ &\rightarrow&0, \ \ \mbox{当}\ n\rightarrow\infty. \end{eqnarray} $

由(3.17)式和(3.19)式知

(3.21) $ \begin{eqnarray} \left\Vert{x_n-y_n}\right\Vert\leq\left\Vert{x_n-z_n}\right\Vert+\left\Vert{z_n-y_n}\right\Vert \rightarrow0, \ \mbox{当}\ n\rightarrow\infty. \end{eqnarray} $

下面证明

(3.22) $ \begin{eqnarray} \limsup\limits_{n\rightarrow\infty}\left<{f(q)-q, x_{n}-q}\right>\leq0, \end{eqnarray} $

这里$ q = P_{\Omega}f(q) $.事实上,存在$ \{x_{n}\} $的子列$ \{x_{n_i}\} $满足

$ \limsup\limits_{n\rightarrow\infty}\left<{f(q)-q, x_{n}-q}\right> = \lim\limits_{i\rightarrow\infty}\left<{f(q)-q, x_{n_i}-q}\right>. $

易知$ P_{\Omega}f $为压缩映射.由Banach压缩映射原理知$ P_{\Omega}f $存在唯一不动点.记为$ q\in C $,即, $ q = P_{\Omega}f(q) $.因为$ \left\{{x_n}\right\} $为$ C $中有界列,不失一般性,假设$ x_{n_i}\rightharpoonup z\in C $.根据引理2.3和(3.17)式,有$ z\in F(G) $.根据引理2.3和(3.20)式,有$ z\in F(T) $.则$ z\in\Omega $.因此

$ \limsup\limits_{n\rightarrow\infty}\left<{f(q)-q, x_{n}-q}\right> = \lim\limits_{i\rightarrow\infty}\left<{f(q)-q, x_{n_i}-q}\right> = \left<{f(q)-q, z-q}\right>\leq0, $

于是(3.22)式成立.由(3.21)式得

$ \limsup\limits_{n\rightarrow\infty}\left<{f(q)-q, y_n-q}\right>\leq0. $

观察

$ \begin{eqnarray*} \left\Vert{y_n-q}\right\Vert^2 & = &\alpha_n\left<{f(x_n)-q, y_n-q}\right>+(1-\alpha_n)\left<{z_n-q, y_n-q}\right>\\ & = &\alpha_n\left<{f(x_n)-f(q), y_n-q}\right>+\alpha_n\left<{f(q)-q, y_n-q}\right>+(1-\alpha_n)\left<{z_n-q, y_n-q}\right>\\ &\leq&\alpha_n\delta\left\Vert{x_n-q}\right\Vert\left\Vert{y_n-q}\right\Vert+(1-\alpha_n)\left\Vert{z_n-q}\right\Vert\left\Vert{y_n-q}\right\Vert+\alpha_n\left<{f(q)-q, y_n-q}\right>\\ &\leq&\alpha_n\delta\left\Vert{x_n-q}\right\Vert\left\Vert{y_n-q}\right\Vert+(1-\alpha_n)\left\Vert{x_n-q}\right\Vert\left\Vert{y_n-q}\right\Vert+\alpha_n\left<{f(q)-q, y_n-q}\right>\\ & = &[1-\alpha_n(1-\delta)]\left\Vert{x_n-q}\right\Vert\left\Vert{y_n-q}\right\Vert+\alpha_n\left<{f(q)-q, y_n-q}\right>\\ &\leq&\frac{1-\alpha_n(1-\delta)}{2}(\left\Vert{x_n-q}\right\Vert^2+\left\Vert{y_n-q}\right\Vert^2)+\alpha_n\left<{f(q)-q, y_n-q}\right>\\ &\leq&\frac{1-\alpha_n(1-\delta)}{2}\left\Vert{x_n-q}\right\Vert^2+\frac{1}{2}\left\Vert{y_n-q}\right\Vert^2+\alpha_n\left<{f(q)-q, y_n-q}\right>, \end{eqnarray*} $

于是

(3.23) $ \begin{eqnarray} \left\Vert{y_n-q}\right\Vert^2\leq[1-\alpha_n(1-\delta)]\left\Vert{x_n-q}\right\Vert^2+\alpha_n(1-\delta)\frac{2\left<{f(q)-q, y_n-q}\right>}{1-\delta}. \end{eqnarray} $

从而

(3.24) $ \begin{eqnarray} \left\Vert{x_{n+1}-q}\right\Vert^2& = &\left\Vert{\beta_n(x_n-q)+(1-\beta_n)(Ty_n-q)}\right\Vert^2 \\ &\leq&\beta_n\left\Vert{x_n-q}\right\Vert^2+(1-\beta_n)\left\Vert{Ty_n-q}\right\Vert^2 \\ &\leq&\beta_n\left\Vert{x_n-q}\right\Vert^2+(1-\beta_n)\left\Vert{y_n-q}\right\Vert^2 \\ &\leq&\beta_n\left\Vert{x_n-q}\right\Vert^2 \\ &&+(1-\beta_n) \bigg[(1-\alpha_n(1-\delta))\left\Vert{x_n-q}\right\Vert^2 +\alpha_n(1-\delta)\frac{2\left<{f(q)-q, y_n-q}\right>}{1-\delta}\bigg] \\ & = &[1-(1-\beta_n)\alpha_n(1-\delta)]\left\Vert{x_n-q}\right\Vert^2+(1-\beta_n)\alpha_n(1-\delta)\frac{2\left<{f(q)-q, y_n-q}\right>}{1-\delta}. \\ \end{eqnarray} $

由引理2.1得$ \left\{{x_n}\right\} $强收敛到$ q $.证毕.

根据定理3.1,容易得到下述结果.

定理3.2  设$ C $为实Hilbert空间$ H $中非空闭凸子集, $ A, B:C\rightarrow H $分别为$ d_1 $ -逆强单调的与$ d_2 $ -逆强单调的, $ T:C\rightarrow C $为非扩张映射满足$ \Omega = F(T)\cap F(G)\neq\emptyset $,这里$ G $的定义见引理2.4.设$ f $为$ C $上严格压缩自映射,其参数$ \delta\in(0, 1) $.任取$ x_{1}\in C $.定义序列$ \left\{{x_{n}}\right\} $为

(3.25) $ \begin{align} \left\{ \begin{array}{l} { } v_n = \frac{1}{2}(x_n+z_n), \\ u_n = P_C(I-\mu B)v_n, \\ z_n = P_C(I-\lambda A)u_n, \\ y_n = \alpha_nf(x_n)+(1-\alpha_n)z_n, \\ x_{n+1} = \beta_nx_n+(1-\beta_n)Ty_n, \end{array} \right. \end{align} $

这里$ \lambda\in (0, 2d_1), \mu\in(0, 2d_2) $.设$ \{\alpha_{n}\} $, $ \{\beta_{n}\} $为$ (0, 1] $中序列满足:

(i) $ { }\lim\limits_{n\rightarrow\infty}\alpha_{n} = 0 $, $ \sum\limits_{n = 1}^{\infty}\alpha_{n} = \infty $,

(ii) $ 0<\liminf\limits_{n\rightarrow\infty}\beta_n\leq\limsup\limits_{n\rightarrow\infty}\beta_n<1 $,

则$ \left\{{x_n}\right\} $强收敛到$ q\in \Omega $,这里$ q = P_\Omega f(q) $,也满足下述变分不等式

$ \left<{f(q)-q, p-q}\right>\leq0, \, \, \forall\, \, p\in \Omega. $

设$ \phi:C\times C\rightarrow {{\Bbb R}} $为二元函数,这里$ {{\Bbb R}} $为实数集.关于$ \phi $的均衡问题就是找点$ x\in C $使得

(4.1) $ \phi(x, y)\geq0, \, \forall\, y\in C. $

不等式$ (4.1) $的解集记为$ EP(\phi) $.均衡问题包括变分不等式问题,不动点问题,优化问题作为特殊情形(参见文献[14]).

为了解决均衡问题,假设$ \phi $满足下述条件(参见文献[14]):

(A1) $ \phi(x, x) = 0, \forall\ x\in C $;

(A2) $ \phi $是单调的,即$ \phi(x, y)+\phi(y, x)\leq0, \forall \ x, y\in C $;

(A3) $ \phi $是上半连续的,即,任给$ x, y, z\in C $,有

$ \limsup\limits_{t\rightarrow0^+}\phi(tz+(1-t)x, y)\leq\phi(x, y); $

(A4)任给$ x\in C $, $ \phi(x, .) $是凸的和弱下半连续的.

引理4.1^[14]  设$ C $为Hilbert空间$ H $中非空闭凸子集, $ \phi:C\times C\rightarrow {{\Bbb R}} $为二元函数且满足条件(A1)–(A4).设$ r>0 $, $ x\in H $.则存在$ z\in C $满足

$ \phi(z, y)+\frac{1}{r}\left<{y-z, z-x}\right>\geq0, \, \, \forall\, \, y\in C. $

引理4.2^[15]  设$ \phi:C\times C\rightarrow {{\Bbb R}} $满足条件(A1)–(A4).给定$ r>0 $和$ x\in H $,定义$ T_r:H\rightarrow C $如下:

$ T_r(x) = \left\{{z\in C:\phi(z, y)+\frac{1}{r}\left<{y-z, z-x}\right>\geq0\, \, \forall\, y\in C, z\in H}\right\}, $

则

(1) $ T_r $是单值的;

(2) $ T_r $为固定非扩张的,即,任给$ x, y\in H $,有$ \left\Vert{T_rx-T_ry}\right\Vert^2\leq\left<{T_rx-T_ry, x-y}\right> $.

从而$ \left\Vert{T_rx-T_ry}\right\Vert\leq\left\Vert{x-y}\right\Vert, \, \forall x, y\in H $,于是, $ T_r $是非扩张的;

(3) $ F(T_r) = EP(\phi), \, \, \forall r>0 $;

(4) $ EP(\phi) $是闭凸的.

我们可得下面结果.

定理4.1  设$ C $为实Hilbert空间$ H $中非空闭凸子集, $ A, B:C\rightarrow H $分别为$ d_1 $ -逆强单调的与$ d_2 $ -逆强单调的,设$ \phi:C\times C\rightarrow {{\Bbb R}} $满足条件(A1)–(A4).假设$ \Omega = F(T)\cap F(G)\neq\emptyset $,这里$ G $的定义见引理2.4.设$ f $为$ C $上严格压缩自映射,其参数$ \delta\in(0, 1) $.任取$ x_{1}\in C $.定义序列$ \left\{{x_{n}}\right\} $为

(4.2) $ \begin{align} \left\{ \begin{array}{l} { } v_n = \frac{1}{2}(x_n+z_n), \\ u_n = P_C(I-\mu B)v_n, \\ z_n = P_C(I-\lambda A)(ax_n+(1-a)u_n), \\ y_n = \alpha_nf(x_n)+(1-\alpha_n)z_n, \\ x_{n+1} = \beta_nx_n+(1-\beta_n)T_ry_n, \end{array} \right. \end{align} $

这里$ r>0 $, $ \lambda\in (0, 2d_1), \mu\in(0, 2d_2) $, $ 0\leq a<1 $.设$ \{\alpha_{n}\} $, $ \{\beta_{n}\} $为$ (0, 1] $中序列满足:

(i) $ { }\lim\limits_{n\rightarrow\infty}\alpha_{n} = 0 $, $ \sum\limits_{n = 1}^{\infty}\alpha_{n} = \infty $,

(ii) $ 0<\liminf\limits_{n\rightarrow\infty}\beta_n\leq\limsup\limits_{n\rightarrow\infty}\beta_n<1 $,

则$ \left\{{x_n}\right\} $强收敛到$ q\in \Omega $,这里$ q = P_\Omega f(q) $,也满足下述变分不等式

$ \left<{f(q)-q, p-q}\right>\leq0, \, \, \forall\, \, p\in \Omega. $