# Convergence of a sequence on the unit sphere of Bahach or Hilbert space

Let $$X$$ be a Banach or Hilbert space and $$A$$ be a bounded linear operator on $$X$$, and fix an element $$x \in X$$. Then I want to know that are there any good ways or theories to deal with the convergence of the following sequence $$\{ y_n\}$$: \begin{align} y_n = \frac{A^nx}{||A^nx||}. \end{align} I'm sorry for an ambiguous question. I welcome any kind of comments.

• If your space $X$ is reflexive (satisfied by Hilbert spaces for example), then the unit ball is compact in the weak topology. In this topology you will thus always find possible limits. – s.harp Dec 5 '18 at 9:46
• Such a weak limit might be zero, which might come unexpected. – daw Dec 5 '18 at 11:07

It does not always converge. As a counter example suppose a finite dimensional space $$X = \mathbb{R}^2$$, and $$$$A = \begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix},\quad x = \begin{bmatrix}1\\1\end{bmatrix}$$$$ Of course $$A$$ is bounded. Then
$$$$y_n = \begin{cases} \displaystyle{\frac{1}{\sqrt{5}}}\begin{bmatrix}2\\-1 \end{bmatrix}, & n=2m+1, \\ \displaystyle{\frac{1}{\sqrt{2}}}\begin{bmatrix}1\\1 \end{bmatrix}, & n=2m. \end{cases}$$$$
In finite dimensional spaces, this is indeed the power iteration that converges to the eigenvector corresponding to the largest eigenvalue (in magnitude) of matrix $$A$$, and the convergence rate is determined by the ratio of first two largest eigenvalues in magnitude, $$|\lambda_1 / \lambda_2|$$.
I assume more than boundedness is needed for convergence of $$y_n$$. Perhaps if $$A$$ is compact, you may show similar argument for rate of convergence based on the singular values.
• I think it is clear that the sequence need not converge, for example even in one dimension you have the linear map $\Bbb R\to \Bbb R$, $x\mapsto -x$ for which the sequence will always alternate between two limit points. I think the interesting question is: When does it have limit points? In finite dimensions it is clear that the sequence always admits limit points, as the unit sphere is compact. – s.harp Dec 12 '18 at 11:42