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












1












$begingroup$


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.






functional-analysis hilbert-spaces dynamical-systems banach-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – s.harp
    Dec 5 '18 at 9:46










  • $begingroup$
    Such a weak limit might be zero, which might come unexpected.
    $endgroup$
    – daw
    Dec 5 '18 at 11:07
















1












$begingroup$


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.






functional-analysis hilbert-spaces dynamical-systems banach-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – s.harp
    Dec 5 '18 at 9:46










  • $begingroup$
    Such a weak limit might be zero, which might come unexpected.
    $endgroup$
    – daw
    Dec 5 '18 at 11:07














1












1








1





$begingroup$


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.






functional-analysis hilbert-spaces dynamical-systems banach-spaces







share|cite|improve this question









$endgroup$




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.






functional-analysis hilbert-spaces dynamical-systems banach-spaces




functional-analysis hilbert-spaces dynamical-systems banach-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 9:38









tnrtnrtnrtnr

61




61












  • $begingroup$
    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.
    $endgroup$
    – s.harp
    Dec 5 '18 at 9:46










  • $begingroup$
    Such a weak limit might be zero, which might come unexpected.
    $endgroup$
    – daw
    Dec 5 '18 at 11:07


















  • $begingroup$
    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.
    $endgroup$
    – s.harp
    Dec 5 '18 at 9:46










  • $begingroup$
    Such a weak limit might be zero, which might come unexpected.
    $endgroup$
    – daw
    Dec 5 '18 at 11:07
















$begingroup$
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.
$endgroup$
– s.harp
Dec 5 '18 at 9:46




$begingroup$
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.
$endgroup$
– s.harp
Dec 5 '18 at 9:46












$begingroup$
Such a weak limit might be zero, which might come unexpected.
$endgroup$
– daw
Dec 5 '18 at 11:07




$begingroup$
Such a weak limit might be zero, which might come unexpected.
$endgroup$
– daw
Dec 5 '18 at 11:07










1 Answer
1






active

oldest

votes


















1












$begingroup$

It does not always converge. As a counter example suppose a finite dimensional space $X = mathbb{R}^2$, and
begin{equation}
A = begin{bmatrix} 1 & 1 \ 0 & -1 end{bmatrix},quad x = begin{bmatrix}1\1end{bmatrix}
end{equation}
Of course $A$ is bounded. Then



begin{equation}
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}
end{equation}



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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You are right @LutzL, I'll edit the answer.
    $endgroup$
    – Sia
    Dec 12 '18 at 9:10










  • $begingroup$
    I think it is clear that the sequence need not converge, for example even in one dimension you have the linear map $Bbb Rto Bbb R$, $xmapsto -x$ 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.
    $endgroup$
    – s.harp
    Dec 12 '18 at 11:42











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

It does not always converge. As a counter example suppose a finite dimensional space $X = mathbb{R}^2$, and
begin{equation}
A = begin{bmatrix} 1 & 1 \ 0 & -1 end{bmatrix},quad x = begin{bmatrix}1\1end{bmatrix}
end{equation}
Of course $A$ is bounded. Then



begin{equation}
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}
end{equation}



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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You are right @LutzL, I'll edit the answer.
    $endgroup$
    – Sia
    Dec 12 '18 at 9:10










  • $begingroup$
    I think it is clear that the sequence need not converge, for example even in one dimension you have the linear map $Bbb Rto Bbb R$, $xmapsto -x$ 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.
    $endgroup$
    – s.harp
    Dec 12 '18 at 11:42
















1












$begingroup$

It does not always converge. As a counter example suppose a finite dimensional space $X = mathbb{R}^2$, and
begin{equation}
A = begin{bmatrix} 1 & 1 \ 0 & -1 end{bmatrix},quad x = begin{bmatrix}1\1end{bmatrix}
end{equation}
Of course $A$ is bounded. Then



begin{equation}
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}
end{equation}



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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You are right @LutzL, I'll edit the answer.
    $endgroup$
    – Sia
    Dec 12 '18 at 9:10










  • $begingroup$
    I think it is clear that the sequence need not converge, for example even in one dimension you have the linear map $Bbb Rto Bbb R$, $xmapsto -x$ 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.
    $endgroup$
    – s.harp
    Dec 12 '18 at 11:42














1












1








1





$begingroup$

It does not always converge. As a counter example suppose a finite dimensional space $X = mathbb{R}^2$, and
begin{equation}
A = begin{bmatrix} 1 & 1 \ 0 & -1 end{bmatrix},quad x = begin{bmatrix}1\1end{bmatrix}
end{equation}
Of course $A$ is bounded. Then



begin{equation}
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}
end{equation}



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.






share|cite|improve this answer











$endgroup$



It does not always converge. As a counter example suppose a finite dimensional space $X = mathbb{R}^2$, and
begin{equation}
A = begin{bmatrix} 1 & 1 \ 0 & -1 end{bmatrix},quad x = begin{bmatrix}1\1end{bmatrix}
end{equation}
Of course $A$ is bounded. Then



begin{equation}
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}
end{equation}



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 12 '18 at 9:15

























answered Dec 12 '18 at 4:46









SiaSia

1064




1064












  • $begingroup$
    You are right @LutzL, I'll edit the answer.
    $endgroup$
    – Sia
    Dec 12 '18 at 9:10










  • $begingroup$
    I think it is clear that the sequence need not converge, for example even in one dimension you have the linear map $Bbb Rto Bbb R$, $xmapsto -x$ 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.
    $endgroup$
    – s.harp
    Dec 12 '18 at 11:42


















  • $begingroup$
    You are right @LutzL, I'll edit the answer.
    $endgroup$
    – Sia
    Dec 12 '18 at 9:10










  • $begingroup$
    I think it is clear that the sequence need not converge, for example even in one dimension you have the linear map $Bbb Rto Bbb R$, $xmapsto -x$ 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.
    $endgroup$
    – s.harp
    Dec 12 '18 at 11:42
















$begingroup$
You are right @LutzL, I'll edit the answer.
$endgroup$
– Sia
Dec 12 '18 at 9:10




$begingroup$
You are right @LutzL, I'll edit the answer.
$endgroup$
– Sia
Dec 12 '18 at 9:10












$begingroup$
I think it is clear that the sequence need not converge, for example even in one dimension you have the linear map $Bbb Rto Bbb R$, $xmapsto -x$ 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.
$endgroup$
– s.harp
Dec 12 '18 at 11:42




$begingroup$
I think it is clear that the sequence need not converge, for example even in one dimension you have the linear map $Bbb Rto Bbb R$, $xmapsto -x$ 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.
$endgroup$
– s.harp
Dec 12 '18 at 11:42


















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