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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 $\bar{u}$.

blindman
  • 3,225
  • 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 01 '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 01 '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 01 '13 at 23:55
  • @grew: Dear Sir. Do you have any comments in my question? Thank you again for your construction. – blindman Apr 04 '13 at 07:35
  • If have thought about it a little bit, but didn't found a counterexample. How do you prove the weak convergence of the iterative scheme? – gerw Apr 04 '13 at 07:36
  • Is $F$ assumed to be continuous? Can you include here the proof that the sequence converges weakly to a solution? – timur Apr 05 '13 at 23:23
  • @grew: I included a short proof that the iterative sequence converges weakly to a solution. – blindman Apr 06 '13 at 00:46
  • @timur: I included the short proof of the weak convergence of the iterative sequence. From the Cauchy Schwarz in equality and the following inequality $$ \langle F(u)-F(v), u-v\rangle\geq \gamma|F(u)-F(v)|^2, $$ we imply that $$ \gamma|F(u)-F(v)|^2\leq |F(u)-F(v)||u-v|, $$ or equivalently $$ |F(u)-F(v)|\leq \frac{1}{\gamma}|F(u)-F(v)|. $$ Hence $F$ is continuous. – blindman Apr 06 '13 at 00:48
  • Thanks! So your question is basically if $\bar u = u^*$? – timur Apr 06 '13 at 01:44
  • @timur: Okie. But my problem may have a lot of solutions. – blindman Apr 06 '13 at 01:57
  • I would like to ask more comments and kind help in this question? How can we find the help? – blindman Apr 10 '13 at 04:10

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