# Conditions for the sequence being weakly convergent

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 $0\in H$.

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$F$ cannot be a compact opeator. – Tomás Oct 9 '12 at 11:25
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. – Davide Giraudo Oct 9 '12 at 12:22
@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. – blindman Oct 9 '12 at 23:27
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). – Davide Giraudo Oct 10 '12 at 7:44
@Davide Giraudo: I reduced my question to the case $F$ is an affine mapping. – blindman Oct 11 '12 at 0:19

## 1 Answer

I think the question is too broad to find a simple answer. In any case, let us restrict to linear, bounded $F$. It is possible that $||F|| > 1$ and $u^0$ can be found according to your description, but I think that would be difficult to find criteria in that case (generically, I would imagine that $||F^nx||$ runs off to $+\infty$ so $F^nx$ does not converge weakly by a classical theorem, and the only way to help this is to restrict $F$ to a subspace of which it is a contraction, but then we might as well start with that)

A result you might find relevant is Jacobs–Glicksberg–de Leeuw decomposition (a great book that mentions this theorem is Eisner's Stability of Operators and Operator Semigroups), which says that $H$ splits into a direct sum of $H_r = \operatorname{span}\{x \in H \ : \ F x = \gamma x \text{ for some$\gamma \in \mathbb{C}$with$|\gamma| = 1$} \}$ and $H_s = \{x \in H \ : \ 0 \text{ is a weak accumulation point of } \{F^nx\}_n \}$. This would mean that if $F$ has no eigenvalues on the unit circle (and we probably don't want that in any case), then $F^n x$ has $0$ as an accumulation point, for any choice of $x$.

It is not quite what you want - accumulation point is not yet the limit - but I hope it will be helpful.

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