Mixing
Infinite dimensional Banach spaces and sequences
$begingroup$
Let $X$ be an infinite dimensional Banach space.
$a)$ Prove that there is a sequence $x=(x_n)_{n geq 1}$ in $X$ such that $||x_n||=1$ for all $n$ satisfying
begin{equation*}
mathrm{dist}(x_{n+1},mathrm{Span} {x_1,...,x_n })geq 1, ngeq 1
end{equation*}
$b)$ Based on $a)$, show that the closed unit ball in $X$ cannot be compact.
My attempt:
I want to prove it by induction: if $n=0$, then I can take any $0 neq x in X$ and normalize it. Thus $mathrm{dist}(x,mathrm{Span} { 0 })= ||x||=1$ and the inequality is true. Suppose now it's true for $n-1$. Then for $x_1,...,x_n$ I can choose $x_{n+1} notin mathrm{Span}{x_1,...,x_n}$ (since $X$ is infinite dimensional) such that $||x_{n+1}||=1$. Thus $mathrm{dist}(x_{n+1},mathrm{Span} {x_1,...,x_n })=mathrm{inf}_{x in mathrm{Span}{x_1,...,x_n} }d(x_{n+1},x) geq 1$ since $d(x_{n+1},x)=d(x,x_{n+1}) geq mathrm{dist}(x, <x_{n+1}>) geq 1$ ($<x_{n+1}>$ is $1$ dimensional and I can apply the induction hypothesis). In particular I get that $||x_n-x_m|| geq 1$ for $n neq m$
For part $b)$ suppose the closed unit ball is compact, i.e. every sequence in it has a convergent subsequence. But this is not true for my sequence $(x_n)_{n geq1}$ since $||x_n-x_m|| geq 1$ for $n neq m$. Am I right?
sequences-and-series functional-analysis convergence banach-spaces compactness
asked Jan 9 at 10:15
user289143user289143
910313
$endgroup$
add a comment |
$begingroup$
Let $X$ be an infinite dimensional Banach space.
$a)$ Prove that there is a sequence $x=(x_n)_{n geq 1}$ in $X$ such that $||x_n||=1$ for all $n$ satisfying
begin{equation*}
mathrm{dist}(x_{n+1},mathrm{Span} {x_1,...,x_n })geq 1, ngeq 1
end{equation*}
$b)$ Based on $a)$, show that the closed unit ball in $X$ cannot be compact.
My attempt:
I want to prove it by induction: if $n=0$, then I can take any $0 neq x in X$ and normalize it. Thus $mathrm{dist}(x,mathrm{Span} { 0 })= ||x||=1$ and the inequality is true. Suppose now it's true for $n-1$. Then for $x_1,...,x_n$ I can choose $x_{n+1} notin mathrm{Span}{x_1,...,x_n}$ (since $X$ is infinite dimensional) such that $||x_{n+1}||=1$. Thus $mathrm{dist}(x_{n+1},mathrm{Span} {x_1,...,x_n })=mathrm{inf}_{x in mathrm{Span}{x_1,...,x_n} }d(x_{n+1},x) geq 1$ since $d(x_{n+1},x)=d(x,x_{n+1}) geq mathrm{dist}(x, <x_{n+1}>) geq 1$ ($<x_{n+1}>$ is $1$ dimensional and I can apply the induction hypothesis). In particular I get that $||x_n-x_m|| geq 1$ for $n neq m$
For part $b)$ suppose the closed unit ball is compact, i.e. every sequence in it has a convergent subsequence. But this is not true for my sequence $(x_n)_{n geq1}$ since $||x_n-x_m|| geq 1$ for $n neq m$. Am I right?
sequences-and-series functional-analysis convergence banach-spaces compactness
asked Jan 9 at 10:15
user289143user289143
910313
$endgroup$
$begingroup$
Your argument for $(a)$ is flawed. In fact it implies that you can choose $x_{n+1}$ to be any unitary element that is not in $operatorname{span}left{x_1,dots,x_nright}$, which is clearly false. Take for instance $n=1$, $x_1=(1,0)$ in $mathbb{R}^2$ and $x_2=(sqrt{1-varepsilon^2},varepsilon)$ for small $varepsilon$. Then $x_2$ satisfies your conditions but clearly $d(x_2,leftlangle x_1rightrangle)ll 1$.
$endgroup$
– Lorenzo Quarisa
Jan 9 at 10:31
2
$begingroup$
One usually uses Riesz's lemma to prove a) with begin{equation*} mathrm{dist}(x_{n+1},mathrm{Span} {x_1,...,x_n })geq alpha, ngeq 1 end{equation*} for any (fixed) $0<alpha<1$. See this. Riesz's lemma has a slight refinment though, see this. I think you might get your result using this result.
$endgroup$
– David Mitra
Jan 9 at 10:48
add a comment |
$begingroup$
Let $X$ be an infinite dimensional Banach space.
$a)$ Prove that there is a sequence $x=(x_n)_{n geq 1}$ in $X$ such that $||x_n||=1$ for all $n$ satisfying
begin{equation*}
mathrm{dist}(x_{n+1},mathrm{Span} {x_1,...,x_n })geq 1, ngeq 1
end{equation*}
$b)$ Based on $a)$, show that the closed unit ball in $X$ cannot be compact.
My attempt:
I want to prove it by induction: if $n=0$, then I can take any $0 neq x in X$ and normalize it. Thus $mathrm{dist}(x,mathrm{Span} { 0 })= ||x||=1$ and the inequality is true. Suppose now it's true for $n-1$. Then for $x_1,...,x_n$ I can choose $x_{n+1} notin mathrm{Span}{x_1,...,x_n}$ (since $X$ is infinite dimensional) such that $||x_{n+1}||=1$. Thus $mathrm{dist}(x_{n+1},mathrm{Span} {x_1,...,x_n })=mathrm{inf}_{x in mathrm{Span}{x_1,...,x_n} }d(x_{n+1},x) geq 1$ since $d(x_{n+1},x)=d(x,x_{n+1}) geq mathrm{dist}(x, <x_{n+1}>) geq 1$ ($<x_{n+1}>$ is $1$ dimensional and I can apply the induction hypothesis). In particular I get that $||x_n-x_m|| geq 1$ for $n neq m$
For part $b)$ suppose the closed unit ball is compact, i.e. every sequence in it has a convergent subsequence. But this is not true for my sequence $(x_n)_{n geq1}$ since $||x_n-x_m|| geq 1$ for $n neq m$. Am I right?
sequences-and-series functional-analysis convergence banach-spaces compactness
asked Jan 9 at 10:15
user289143user289143
910313
$endgroup$
Let $X$ be an infinite dimensional Banach space.
$a)$ Prove that there is a sequence $x=(x_n)_{n geq 1}$ in $X$ such that $||x_n||=1$ for all $n$ satisfying
begin{equation*}
mathrm{dist}(x_{n+1},mathrm{Span} {x_1,...,x_n })geq 1, ngeq 1
end{equation*}
$b)$ Based on $a)$, show that the closed unit ball in $X$ cannot be compact.
My attempt:
I want to prove it by induction: if $n=0$, then I can take any $0 neq x in X$ and normalize it. Thus $mathrm{dist}(x,mathrm{Span} { 0 })= ||x||=1$ and the inequality is true. Suppose now it's true for $n-1$. Then for $x_1,...,x_n$ I can choose $x_{n+1} notin mathrm{Span}{x_1,...,x_n}$ (since $X$ is infinite dimensional) such that $||x_{n+1}||=1$. Thus $mathrm{dist}(x_{n+1},mathrm{Span} {x_1,...,x_n })=mathrm{inf}_{x in mathrm{Span}{x_1,...,x_n} }d(x_{n+1},x) geq 1$ since $d(x_{n+1},x)=d(x,x_{n+1}) geq mathrm{dist}(x, <x_{n+1}>) geq 1$ ($<x_{n+1}>$ is $1$ dimensional and I can apply the induction hypothesis). In particular I get that $||x_n-x_m|| geq 1$ for $n neq m$
For part $b)$ suppose the closed unit ball is compact, i.e. every sequence in it has a convergent subsequence. But this is not true for my sequence $(x_n)_{n geq1}$ since $||x_n-x_m|| geq 1$ for $n neq m$. Am I right?
sequences-and-series functional-analysis convergence banach-spaces compactness
sequences-and-series functional-analysis convergence banach-spaces compactness
asked Jan 9 at 10:15
user289143user289143
910313
asked Jan 9 at 10:15
user289143user289143
910313
asked Jan 9 at 10:15
user289143user289143
910313
asked Jan 9 at 10:15
user289143user289143
910313
asked Jan 9 at 10:15
user289143user289143
910313
910313
$begingroup$
Your argument for $(a)$ is flawed. In fact it implies that you can choose $x_{n+1}$ to be any unitary element that is not in $operatorname{span}left{x_1,dots,x_nright}$, which is clearly false. Take for instance $n=1$, $x_1=(1,0)$ in $mathbb{R}^2$ and $x_2=(sqrt{1-varepsilon^2},varepsilon)$ for small $varepsilon$. Then $x_2$ satisfies your conditions but clearly $d(x_2,leftlangle x_1rightrangle)ll 1$.
$endgroup$
– Lorenzo Quarisa
Jan 9 at 10:31
2
$begingroup$
One usually uses Riesz's lemma to prove a) with begin{equation*} mathrm{dist}(x_{n+1},mathrm{Span} {x_1,...,x_n })geq alpha, ngeq 1 end{equation*} for any (fixed) $0<alpha<1$. See this. Riesz's lemma has a slight refinment though, see this. I think you might get your result using this result.
$endgroup$
– David Mitra
Jan 9 at 10:48
add a comment |
$begingroup$
Your argument for $(a)$ is flawed. In fact it implies that you can choose $x_{n+1}$ to be any unitary element that is not in $operatorname{span}left{x_1,dots,x_nright}$, which is clearly false. Take for instance $n=1$, $x_1=(1,0)$ in $mathbb{R}^2$ and $x_2=(sqrt{1-varepsilon^2},varepsilon)$ for small $varepsilon$. Then $x_2$ satisfies your conditions but clearly $d(x_2,leftlangle x_1rightrangle)ll 1$.
$endgroup$
– Lorenzo Quarisa
Jan 9 at 10:31
2
$begingroup$
One usually uses Riesz's lemma to prove a) with begin{equation*} mathrm{dist}(x_{n+1},mathrm{Span} {x_1,...,x_n })geq alpha, ngeq 1 end{equation*} for any (fixed) $0<alpha<1$. See this. Riesz's lemma has a slight refinment though, see this. I think you might get your result using this result.
$endgroup$
– David Mitra
Jan 9 at 10:48
$begingroup$
Your argument for $(a)$ is flawed. In fact it implies that you can choose $x_{n+1}$ to be any unitary element that is not in $operatorname{span}left{x_1,dots,x_nright}$, which is clearly false. Take for instance $n=1$, $x_1=(1,0)$ in $mathbb{R}^2$ and $x_2=(sqrt{1-varepsilon^2},varepsilon)$ for small $varepsilon$. Then $x_2$ satisfies your conditions but clearly $d(x_2,leftlangle x_1rightrangle)ll 1$.
$endgroup$
– Lorenzo Quarisa
Jan 9 at 10:31
$begingroup$
Your argument for $(a)$ is flawed. In fact it implies that you can choose $x_{n+1}$ to be any unitary element that is not in $operatorname{span}left{x_1,dots,x_nright}$, which is clearly false. Take for instance $n=1$, $x_1=(1,0)$ in $mathbb{R}^2$ and $x_2=(sqrt{1-varepsilon^2},varepsilon)$ for small $varepsilon$. Then $x_2$ satisfies your conditions but clearly $d(x_2,leftlangle x_1rightrangle)ll 1$.
$endgroup$
– Lorenzo Quarisa
Jan 9 at 10:31
2
2
$begingroup$
One usually uses Riesz's lemma to prove a) with begin{equation*} mathrm{dist}(x_{n+1},mathrm{Span} {x_1,...,x_n })geq alpha, ngeq 1 end{equation*} for any (fixed) $0<alpha<1$. See this. Riesz's lemma has a slight refinment though, see this. I think you might get your result using this result.
$endgroup$
– David Mitra
Jan 9 at 10:48
$begingroup$
One usually uses Riesz's lemma to prove a) with begin{equation*} mathrm{dist}(x_{n+1},mathrm{Span} {x_1,...,x_n })geq alpha, ngeq 1 end{equation*} for any (fixed) $0<alpha<1$. See this. Riesz's lemma has a slight refinment though, see this. I think you might get your result using this result.
$endgroup$
– David Mitra
Jan 9 at 10:48
add a comment |
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$begingroup$
Your argument for $(a)$ is flawed. In fact it implies that you can choose $x_{n+1}$ to be any unitary element that is not in $operatorname{span}left{x_1,dots,x_nright}$, which is clearly false. Take for instance $n=1$, $x_1=(1,0)$ in $mathbb{R}^2$ and $x_2=(sqrt{1-varepsilon^2},varepsilon)$ for small $varepsilon$. Then $x_2$ satisfies your conditions but clearly $d(x_2,leftlangle x_1rightrangle)ll 1$.
$endgroup$
– Lorenzo Quarisa
Jan 9 at 10:31
2
$begingroup$
One usually uses Riesz's lemma to prove a) with begin{equation*} mathrm{dist}(x_{n+1},mathrm{Span} {x_1,...,x_n })geq alpha, ngeq 1 end{equation*} for any (fixed) $0<alpha<1$. See this. Riesz's lemma has a slight refinment though, see this. I think you might get your result using this result.
$endgroup$
– David Mitra
Jan 9 at 10:48