# On the weak and strong convergence of an iterative sequence

I have some difficulties in the following problem.

I would like to thank for all kind help and construction.

Let $H$ be an infinite dimensional real Hilbert space and $F: H\rightarrow H$ be a mapping. Suppose that there exists $\gamma>0$ such that $$\langle F(u)-F(v), u-v\rangle\geq \gamma\|F(u)-F(v)\|^2\quad \forall u,v \in H.$$ Let $u_0\in H$ and $\lambda\in (0, 2\gamma)$ and $\{u_n\}$ be a sequence given by $$u_{n+1}=u_n-\lambda F(u_n) \quad \forall n\in \mathbb{N}.$$ I would like to construct the above mapping $F$ such that the equation $F(u)=0$ has a solution and the iterative sequence $\{u_n\}$ converges weakly to some solution of the latter equation, but not strongly.

Note. We can prove that the iterative sequence $\{u_n\}$ converges weakly to some solution of the equation $F(u)=0$ provided that the latter equation has a solution.

Let $u_*$ be a solution of $F(u)=0$. We observe that \begin{eqnarray*} \|u_{n+1}-u_*\|^2&=&\|u_n-\lambda F(u_n)-u_*+\lambda F(u_*)\|^2\\ &=& \|u_n-u_*\|^2-2\lambda\langle u_n-u_*, F(u_n)-F(u_*)\rangle+\lambda^2\|F(u_n)-F(u_*)\|^2\\ &\leq& \|u_n-u_*\|^2-2\lambda\gamma\|F(u_n)-F(u_*)\|^2+\lambda^2\|F(u_n)-F(u_*)\|^2\\ &=&\|u_n-u_*\|^2-\lambda(2\gamma-\lambda)\|F(u_n)\|^2\\ &=&\|u_n-u_*\|^2-\frac{2\gamma-\lambda}{\lambda}\|u_n-u_{n+1}\|^2. \end{eqnarray*} From these inequalities we deduce that

• $\{\|u_n-u_*\|\}$ is monotonically decreasing and so it is convergent.

• $\{\|u_n-u_{n+1}\|\}$ and $\{\|F(u_n)\|\}$ tend to $0$.

• $\{u_n\}$ is bounded and so it has a subsequence $\{u_{n_k}\}$ converges weakly to $\bar{u}$.

• $\bar{u}$ is a solution of $F(u)=0$, and so $\{u_n\}$ converges weakly to $\bar{u}$.

• There is a problem with your question: your iterative method does not converge (weakly). Take $H = \mathbb{R}^2$, $F = \mathrm{diag}(\Lambda_1, \Lambda_2)$, for some $\Lambda_2 > \Lambda_1$. Then, your iteration is linear and described by the matrix $\mathrm{diag}(1-\lambda\,\Lambda_1, 1-\lambda\,\Lambda_2)$. For convergence, you need $\lambda \in (0, 2/\Lambda_2)$ and not $(0, 2\,\Lambda_1)$. – gerw Mar 31 '13 at 14:43
• Dear gerw. Thank you for your comments. Remember that the mapping $F$ must satisfy the condition $$\langle F(u)-F(v), u-v\rangle\geq \gamma\|F(u)-F(v)\|^2\quad \forall u,v \in H$$ for some $\gamma>0$. The mapping $F$ in your counterexample seem not to satisfy our condition. – blindman Mar 31 '13 at 23:50
• It satisfies this condition, but for $\gamma = 1/\Lambda_2$ (not $\gamma = \Lambda_1$) and then everything is fine. – gerw Apr 1 '13 at 17:44
• @gerw: Thank you again for your interesting comments. Here we need $F$ must satisfy $$\langle F(u)-F(v), u-v\rangle\geq \gamma\|F(u)-F(v)\|^2\quad \forall u,v \in H$$ but not $$\langle F(u)-F(v), u-v\rangle\geq \gamma\|u-v\|^2\quad \forall u,v \in H$$ – blindman Apr 1 '13 at 23:42
• @grew: In your counterexample, in order to have $$\langle F(u)-F(v), u-v\rangle\geq \gamma\|F(u)-F(v)\|^2\quad \forall u,v \in H=\mathbb{R}^2$$ we must have (choose $u=(x_1, y_1), v=(x_2, y_2)$) $$\lambda_1(x_1-x_2)^2+\lambda_2(y_1-y_2)^2\geq \gamma(\lambda_1^2(x_1-x_2)^2+\lambda_2^2(y_1-y_2)^2).$$ From this we deduce that $\lambda_1, \lambda_2\in (0,\frac{1}{\gamma}]$. Therefore $(0, 2\gamma)\subset (0, \frac{2}{\lambda_2})\subset (0,\frac{2}{\lambda_1})$. – blindman Apr 1 '13 at 23:55