Strongly convergence and Uniformly convergence in Banach space

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Let $X,Y$ be Banach spaces, $T_n:Xto Y$ are bounded linear operators and $S:Xto Y$ is compact operator. Suppose for all $xin X$, $||T_nx||leq ||Sx||$ and $T_n$ strong convergent to $T$(bounded linear operator) . Then $T_n$ convergent uniformly.

My idea: If $T_n$ not convergent uniformly, there exist $epsilon$ and $x_n$ s.t.,$||x_n||=1$, $||(T_n-T)x_n|| geq epsilon$
Then, there are subsequence s.t., $Sx_n_k$ convergent.
But I don't know the relation $T_n$ to $S$, so I can't prove.

banach-spaces uniform-convergence compact-operators strong-convergence

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edited Jul 25 at 18:57

asked Jul 25 at 16:21

Yan Wen Lie

1017

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up vote
3
down vote

favorite

2

Let $X,Y$ be Banach spaces, $T_n:Xto Y$ are bounded linear operators and $S:Xto Y$ is compact operator. Suppose for all $xin X$, $||T_nx||leq ||Sx||$ and $T_n$ strong convergent to $T$(bounded linear operator) . Then $T_n$ convergent uniformly.

My idea: If $T_n$ not convergent uniformly, there exist $epsilon$ and $x_n$ s.t.,$||x_n||=1$, $||(T_n-T)x_n|| geq epsilon$
Then, there are subsequence s.t., $Sx_n_k$ convergent.
But I don't know the relation $T_n$ to $S$, so I can't prove.

banach-spaces uniform-convergence compact-operators strong-convergence

share|cite|improve this question

edited Jul 25 at 18:57

asked Jul 25 at 16:21

Yan Wen Lie

1017

add a comment | 

up vote
3
down vote

favorite

2

up vote
3
down vote

favorite

2

2

Let $X,Y$ be Banach spaces, $T_n:Xto Y$ are bounded linear operators and $S:Xto Y$ is compact operator. Suppose for all $xin X$, $||T_nx||leq ||Sx||$ and $T_n$ strong convergent to $T$(bounded linear operator) . Then $T_n$ convergent uniformly.

My idea: If $T_n$ not convergent uniformly, there exist $epsilon$ and $x_n$ s.t.,$||x_n||=1$, $||(T_n-T)x_n|| geq epsilon$
Then, there are subsequence s.t., $Sx_n_k$ convergent.
But I don't know the relation $T_n$ to $S$, so I can't prove.

banach-spaces uniform-convergence compact-operators strong-convergence

share|cite|improve this question

edited Jul 25 at 18:57

asked Jul 25 at 16:21

Yan Wen Lie

1017

Let $X,Y$ be Banach spaces, $T_n:Xto Y$ are bounded linear operators and $S:Xto Y$ is compact operator. Suppose for all $xin X$, $||T_nx||leq ||Sx||$ and $T_n$ strong convergent to $T$(bounded linear operator) . Then $T_n$ convergent uniformly.

My idea: If $T_n$ not convergent uniformly, there exist $epsilon$ and $x_n$ s.t.,$||x_n||=1$, $||(T_n-T)x_n|| geq epsilon$
Then, there are subsequence s.t., $Sx_n_k$ convergent.
But I don't know the relation $T_n$ to $S$, so I can't prove.

banach-spaces uniform-convergence compact-operators strong-convergence

share|cite|improve this question

edited Jul 25 at 18:57

asked Jul 25 at 16:21

Yan Wen Lie

1017

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 18:57

edited Jul 25 at 18:57

edited Jul 25 at 18:57

asked Jul 25 at 16:21

Yan Wen Lie

1017

asked Jul 25 at 16:21

Yan Wen Lie

1017

asked Jul 25 at 16:21

Yan Wen Lie

1017

1017

add a comment | 

add a comment | 

1 Answer
1

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up vote
2
down vote



accepted

Firstly, observe that we may assume that $T=0$.

Reason: For any $xin X$, we have $||T_nx||leq||Sx||$. Letting
$nrightarrowinfty$, we have $||Tx||leq||Sx||$. Now
begineqnarray*
||(T_n-T)x|| & leq & ||T_nx||+||Tx||\
& leq & ||Sx||+||Sx||\
& = & ||(2S)x||.
endeqnarray*
$T_n-Tmid ninmathbbN$ is a sequence of bounded linear
map from $X$ into $Y$, with $(T_n-T)xrightarrow0$ for each $xin X$
and that $||(T_n-T)x||leq||(2S)x||$. Note that $2S$ is still
a compact linear map from $X$ into $Y$. Now replace the original
$T_n$ with $T_n-T$.

Let us rephrase the question: Let $X$ and $Y$ be Banach spaces.
Let $T_n:Xrightarrow Y$ be a bounded linear map, $S:Xrightarrow Y$
be a compact linear map. Suppose that $T_nxrightarrow0$ for all
$xin X$ and that $||T_nx||leq||Sx||$ for all $xin X$. Prove that
$||T_n||rightarrow0$.

Proof: Prove by contradiction. Suppose the contrary that $||T_n||notrightarrow0$.
By passing to a subsequence, without loss of generality, we may assume
that there exists $varepsilon_0>0$ such that $||T_n||>varepsilon_0$
for all $n$. Let $B=xin Xmid$. For each $n$, choose
$x_nin B$ such that $||T_nx_n||>varepsilon_0$. Since $overlineS(B)$
is a compact subset in $Y$ and the sequence $Sx_nmid ninmathbbNsubseteqoverlineS(B)$.
The sequence $Sx_nmid ninmathbbN$ has a convergent subsequence.
By passing to a suitable subsequence, without loss of generality,
we may assume that $Sx_nmid ninmathbbN$ is convergent.
Choose $NinmathbbN$ such that $||Sx_n-Sx_m||<fracvarepsilon_04$
whenever $m,ngeq N$. Since $T_nx_Nrightarrow0$ as $nrightarrowinfty$,
there exists $n_0>N$ such that $||T_n_0x_N||<fracvarepsilon_04$.
Note that $||T_n_0(x_n_0-x_N)||leq||S(x_n_0-x_N)||<fracvarepsilon_04$.
It follows that
begineqnarray*
||T_n_0x_n_0|| & leq & ||T_n_0(x_n_0-x_N)||+||T_n_0x_N||\
& < & fracvarepsilon_02
endeqnarray*
which is a contradiction.

share|cite|improve this answer

edited Jul 25 at 21:29

answered Jul 25 at 21:23

Danny Pak-Keung Chan

1,82128

• 
  
  
  

  
  

  
  
  
  
  
  

  
  One point I would like to point out: For a metric space, it is compact if and only if it is sequentially compact. I used this fact implicitly.
  – Danny Pak-Keung Chan
  Jul 26 at 21:13
  
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Firstly, observe that we may assume that $T=0$.

Reason: For any $xin X$, we have $||T_nx||leq||Sx||$. Letting
$nrightarrowinfty$, we have $||Tx||leq||Sx||$. Now
begineqnarray*
||(T_n-T)x|| & leq & ||T_nx||+||Tx||\
& leq & ||Sx||+||Sx||\
& = & ||(2S)x||.
endeqnarray*
$T_n-Tmid ninmathbbN$ is a sequence of bounded linear
map from $X$ into $Y$, with $(T_n-T)xrightarrow0$ for each $xin X$
and that $||(T_n-T)x||leq||(2S)x||$. Note that $2S$ is still
a compact linear map from $X$ into $Y$. Now replace the original
$T_n$ with $T_n-T$.

Let us rephrase the question: Let $X$ and $Y$ be Banach spaces.
Let $T_n:Xrightarrow Y$ be a bounded linear map, $S:Xrightarrow Y$
be a compact linear map. Suppose that $T_nxrightarrow0$ for all
$xin X$ and that $||T_nx||leq||Sx||$ for all $xin X$. Prove that
$||T_n||rightarrow0$.

Proof: Prove by contradiction. Suppose the contrary that $||T_n||notrightarrow0$.
By passing to a subsequence, without loss of generality, we may assume
that there exists $varepsilon_0>0$ such that $||T_n||>varepsilon_0$
for all $n$. Let $B=xin Xmid$. For each $n$, choose
$x_nin B$ such that $||T_nx_n||>varepsilon_0$. Since $overlineS(B)$
is a compact subset in $Y$ and the sequence $Sx_nmid ninmathbbNsubseteqoverlineS(B)$.
The sequence $Sx_nmid ninmathbbN$ has a convergent subsequence.
By passing to a suitable subsequence, without loss of generality,
we may assume that $Sx_nmid ninmathbbN$ is convergent.
Choose $NinmathbbN$ such that $||Sx_n-Sx_m||<fracvarepsilon_04$
whenever $m,ngeq N$. Since $T_nx_Nrightarrow0$ as $nrightarrowinfty$,
there exists $n_0>N$ such that $||T_n_0x_N||<fracvarepsilon_04$.
Note that $||T_n_0(x_n_0-x_N)||leq||S(x_n_0-x_N)||<fracvarepsilon_04$.
It follows that
begineqnarray*
||T_n_0x_n_0|| & leq & ||T_n_0(x_n_0-x_N)||+||T_n_0x_N||\
& < & fracvarepsilon_02
endeqnarray*
which is a contradiction.

share|cite|improve this answer

edited Jul 25 at 21:29

answered Jul 25 at 21:23

Danny Pak-Keung Chan

1,82128

• 
  
  
  

  
  

  
  
  
  
  
  

  
  One point I would like to point out: For a metric space, it is compact if and only if it is sequentially compact. I used this fact implicitly.
  – Danny Pak-Keung Chan
  Jul 26 at 21:13
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Firstly, observe that we may assume that $T=0$.

Reason: For any $xin X$, we have $||T_nx||leq||Sx||$. Letting
$nrightarrowinfty$, we have $||Tx||leq||Sx||$. Now
begineqnarray*
||(T_n-T)x|| & leq & ||T_nx||+||Tx||\
& leq & ||Sx||+||Sx||\
& = & ||(2S)x||.
endeqnarray*
$T_n-Tmid ninmathbbN$ is a sequence of bounded linear
map from $X$ into $Y$, with $(T_n-T)xrightarrow0$ for each $xin X$
and that $||(T_n-T)x||leq||(2S)x||$. Note that $2S$ is still
a compact linear map from $X$ into $Y$. Now replace the original
$T_n$ with $T_n-T$.

Let us rephrase the question: Let $X$ and $Y$ be Banach spaces.
Let $T_n:Xrightarrow Y$ be a bounded linear map, $S:Xrightarrow Y$
be a compact linear map. Suppose that $T_nxrightarrow0$ for all
$xin X$ and that $||T_nx||leq||Sx||$ for all $xin X$. Prove that
$||T_n||rightarrow0$.

Proof: Prove by contradiction. Suppose the contrary that $||T_n||notrightarrow0$.
By passing to a subsequence, without loss of generality, we may assume
that there exists $varepsilon_0>0$ such that $||T_n||>varepsilon_0$
for all $n$. Let $B=xin Xmid$. For each $n$, choose
$x_nin B$ such that $||T_nx_n||>varepsilon_0$. Since $overlineS(B)$
is a compact subset in $Y$ and the sequence $Sx_nmid ninmathbbNsubseteqoverlineS(B)$.
The sequence $Sx_nmid ninmathbbN$ has a convergent subsequence.
By passing to a suitable subsequence, without loss of generality,
we may assume that $Sx_nmid ninmathbbN$ is convergent.
Choose $NinmathbbN$ such that $||Sx_n-Sx_m||<fracvarepsilon_04$
whenever $m,ngeq N$. Since $T_nx_Nrightarrow0$ as $nrightarrowinfty$,
there exists $n_0>N$ such that $||T_n_0x_N||<fracvarepsilon_04$.
Note that $||T_n_0(x_n_0-x_N)||leq||S(x_n_0-x_N)||<fracvarepsilon_04$.
It follows that
begineqnarray*
||T_n_0x_n_0|| & leq & ||T_n_0(x_n_0-x_N)||+||T_n_0x_N||\
& < & fracvarepsilon_02
endeqnarray*
which is a contradiction.

share|cite|improve this answer

edited Jul 25 at 21:29

answered Jul 25 at 21:23

Danny Pak-Keung Chan

1,82128

• 
  
  
  

  
  

  
  
  
  
  
  

  
  One point I would like to point out: For a metric space, it is compact if and only if it is sequentially compact. I used this fact implicitly.
  – Danny Pak-Keung Chan
  Jul 26 at 21:13
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Firstly, observe that we may assume that $T=0$.

Reason: For any $xin X$, we have $||T_nx||leq||Sx||$. Letting
$nrightarrowinfty$, we have $||Tx||leq||Sx||$. Now
begineqnarray*
||(T_n-T)x|| & leq & ||T_nx||+||Tx||\
& leq & ||Sx||+||Sx||\
& = & ||(2S)x||.
endeqnarray*
$T_n-Tmid ninmathbbN$ is a sequence of bounded linear
map from $X$ into $Y$, with $(T_n-T)xrightarrow0$ for each $xin X$
and that $||(T_n-T)x||leq||(2S)x||$. Note that $2S$ is still
a compact linear map from $X$ into $Y$. Now replace the original
$T_n$ with $T_n-T$.

Let us rephrase the question: Let $X$ and $Y$ be Banach spaces.
Let $T_n:Xrightarrow Y$ be a bounded linear map, $S:Xrightarrow Y$
be a compact linear map. Suppose that $T_nxrightarrow0$ for all
$xin X$ and that $||T_nx||leq||Sx||$ for all $xin X$. Prove that
$||T_n||rightarrow0$.

Proof: Prove by contradiction. Suppose the contrary that $||T_n||notrightarrow0$.
By passing to a subsequence, without loss of generality, we may assume
that there exists $varepsilon_0>0$ such that $||T_n||>varepsilon_0$
for all $n$. Let $B=xin Xmid$. For each $n$, choose
$x_nin B$ such that $||T_nx_n||>varepsilon_0$. Since $overlineS(B)$
is a compact subset in $Y$ and the sequence $Sx_nmid ninmathbbNsubseteqoverlineS(B)$.
The sequence $Sx_nmid ninmathbbN$ has a convergent subsequence.
By passing to a suitable subsequence, without loss of generality,
we may assume that $Sx_nmid ninmathbbN$ is convergent.
Choose $NinmathbbN$ such that $||Sx_n-Sx_m||<fracvarepsilon_04$
whenever $m,ngeq N$. Since $T_nx_Nrightarrow0$ as $nrightarrowinfty$,
there exists $n_0>N$ such that $||T_n_0x_N||<fracvarepsilon_04$.
Note that $||T_n_0(x_n_0-x_N)||leq||S(x_n_0-x_N)||<fracvarepsilon_04$.
It follows that
begineqnarray*
||T_n_0x_n_0|| & leq & ||T_n_0(x_n_0-x_N)||+||T_n_0x_N||\
& < & fracvarepsilon_02
endeqnarray*
which is a contradiction.

share|cite|improve this answer

edited Jul 25 at 21:29

answered Jul 25 at 21:23

Danny Pak-Keung Chan

1,82128

Firstly, observe that we may assume that $T=0$.

Reason: For any $xin X$, we have $||T_nx||leq||Sx||$. Letting
$nrightarrowinfty$, we have $||Tx||leq||Sx||$. Now
begineqnarray*
||(T_n-T)x|| & leq & ||T_nx||+||Tx||\
& leq & ||Sx||+||Sx||\
& = & ||(2S)x||.
endeqnarray*
$T_n-Tmid ninmathbbN$ is a sequence of bounded linear
map from $X$ into $Y$, with $(T_n-T)xrightarrow0$ for each $xin X$
and that $||(T_n-T)x||leq||(2S)x||$. Note that $2S$ is still
a compact linear map from $X$ into $Y$. Now replace the original
$T_n$ with $T_n-T$.

Let us rephrase the question: Let $X$ and $Y$ be Banach spaces.
Let $T_n:Xrightarrow Y$ be a bounded linear map, $S:Xrightarrow Y$
be a compact linear map. Suppose that $T_nxrightarrow0$ for all
$xin X$ and that $||T_nx||leq||Sx||$ for all $xin X$. Prove that
$||T_n||rightarrow0$.

Proof: Prove by contradiction. Suppose the contrary that $||T_n||notrightarrow0$.
By passing to a subsequence, without loss of generality, we may assume
that there exists $varepsilon_0>0$ such that $||T_n||>varepsilon_0$
for all $n$. Let $B=xin Xmid$. For each $n$, choose
$x_nin B$ such that $||T_nx_n||>varepsilon_0$. Since $overlineS(B)$
is a compact subset in $Y$ and the sequence $Sx_nmid ninmathbbNsubseteqoverlineS(B)$.
The sequence $Sx_nmid ninmathbbN$ has a convergent subsequence.
By passing to a suitable subsequence, without loss of generality,
we may assume that $Sx_nmid ninmathbbN$ is convergent.
Choose $NinmathbbN$ such that $||Sx_n-Sx_m||<fracvarepsilon_04$
whenever $m,ngeq N$. Since $T_nx_Nrightarrow0$ as $nrightarrowinfty$,
there exists $n_0>N$ such that $||T_n_0x_N||<fracvarepsilon_04$.
Note that $||T_n_0(x_n_0-x_N)||leq||S(x_n_0-x_N)||<fracvarepsilon_04$.
It follows that
begineqnarray*
||T_n_0x_n_0|| & leq & ||T_n_0(x_n_0-x_N)||+||T_n_0x_N||\
& < & fracvarepsilon_02
endeqnarray*
which is a contradiction.

share|cite|improve this answer

edited Jul 25 at 21:29

answered Jul 25 at 21:23

Danny Pak-Keung Chan

1,82128

share|cite|improve this answer

share|cite|improve this answer

edited Jul 25 at 21:29

edited Jul 25 at 21:29

edited Jul 25 at 21:29

answered Jul 25 at 21:23

Danny Pak-Keung Chan

1,82128

answered Jul 25 at 21:23

Danny Pak-Keung Chan

1,82128

answered Jul 25 at 21:23

Danny Pak-Keung Chan

1,82128

1,82128

• 
  
  
  

  
  

  
  
  
  
  
  

  
  One point I would like to point out: For a metric space, it is compact if and only if it is sequentially compact. I used this fact implicitly.
  – Danny Pak-Keung Chan
  Jul 26 at 21:13
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  One point I would like to point out: For a metric space, it is compact if and only if it is sequentially compact. I used this fact implicitly.
  – Danny Pak-Keung Chan
  Jul 26 at 21:13
  
  

  
  
  
  

One point I would like to point out: For a metric space, it is compact if and only if it is sequentially compact. I used this fact implicitly.
– Danny Pak-Keung Chan
Jul 26 at 21:13

One point I would like to point out: For a metric space, it is compact if and only if it is sequentially compact. I used this fact implicitly.
– Danny Pak-Keung Chan
Jul 26 at 21:13

add a comment | 

 

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