5
$\begingroup$

Let $H=\ell_2$ be the Hilbert space of the square-summable sequences where $$ \langle x,y\rangle=\sum_{i=1}^{\infty}x_iy_i, \quad \|x\|=\sqrt{\langle x,x\rangle}. $$ Let $F: H\rightarrow H$ be an affine mapping, i.e. $$ F[\lambda u+(1-\lambda)v]=\lambda F(u)+(1-\lambda)F(v), \quad \forall u,v\in H, \lambda\in \mathbb{R}. $$ Let $\{u^k\}$ be a sequence given by $$ u^{k+1}=F(u^k) \quad k\in\mathbb{N}, $$ where $u^0$ is an any point in $H$. Find the conditions on $F$ and $u^0$ such that $\{u^k\}$ is weakly convergent but not strongly convergent.

Example. If $u^0=(1,0,0,\ldots,0,\ldots)$ and $F(u)$ is given by $$ F(u)=(0,u_1, u_2, \ldots, u_n, \ldots) \quad \forall u=(u_1,u_2,\ldots, u_n, \ldots)\in H. $$ The sequence $\{u^k\}$ generated by the formula $u^{k+1}=F(u^k)$ is given by $$ u^0=(1,0,\ldots, 0,\ldots), \quad u^1=(0,1,0,\ldots, 0, \ldots), \ldots, u^n=(0,0,\ldots, 1, 0, \ldots),\ldots $$ is weakly convergent but not strongly convergent to $0\in H$.

  • 1
    $F$ cannot be a compact opeator.2012-10-09
  • 1
    Is the mapping $F$ arbitrary? In this case, the question is strange, for example we can choose, once $x_0$ is fixed, $F$ who map $x_0$ to $e_1$ (the first vector of the classical Hilbert basis), $e_1$ to $e_2$ etc... (if $x_0$ is not one of these vectors) and anything you want for the non concerned vectors.2012-10-09
  • 0
    @Davide Giraudo: Dear Sir. Thank you for your comments. I would like to find some sufficent conditions on the mapping $F$ and $u^0$ such that $\{u^k\}$ is weakly convergent but not strongly convergent.2012-10-09
  • 0
    The problem is that if $F$ is non-linear, we allow a lot of things. At least, we can restrict us to linear $F$ (as linearity is quite compatible with weak convergence).2012-10-10
  • 0
    @Davide Giraudo: I reduced my question to the case $F$ is an affine mapping.2012-10-11
  • 0
    Maybe it would help if you told us what do you need such criteria for. It seems a rather strange question, especially since it's very broad – it sounds kind of like “under what criteria is sum of two squares not prime, but not divisible by 5”. Though the latter is probably easier, I know little of number theory. ;)2013-07-05

1 Answers 1