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Convergence for the norm of a sequence in $ell^{infty}$

$ell^infty$ and $ell^1$When $ell^2$-convergence implies $ell^1$- convergence?Do sequences in $ell^p$ belong to some $ell^q$ for $q<p$?Showing a sequence is in $ell^2$Limit for sequence $a_{m+n}leq a_m+a_n$Show $Tx in ell ^2$ for every $x in ell ^{infty}$Find $|T|_{ell^infty rightarrow ell^2}$Finding the “operatorial” norm of an operatorShowing that $a_n in ell^{infty}$Why ${xinell^infty mid |x|_{ell^infty }leq 1}$ is not completely bounded?

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I still don't know if the statement below is true. It seems to be true but I couldn't find a proof. The statement is the following: for each $n geq 1$, let $x_n = (a_1(n),a_2(n),ldots,a_n(n),0,0,ldots) in ell^{infty}$, where $ell^{infty}$ is defined to be the space of all bounded sequences. If for any $1 leq k leq n$, $a_k(n) rightarrow 0$ as $n rightarrow infty$ then $||x_n||_{infty} = sup (a_1(n),a_2(n),ldots,a_n(n),0,0,ldots) rightarrow 0$ as $n rightarrow infty$? Any help is welcome.

real-analysis functional-analysis

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asked Mar 19 at 16:21

JeromeJerome

1

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add a comment | 

0

$begingroup$

I still don't know if the statement below is true. It seems to be true but I couldn't find a proof. The statement is the following: for each $n geq 1$, let $x_n = (a_1(n),a_2(n),ldots,a_n(n),0,0,ldots) in ell^{infty}$, where $ell^{infty}$ is defined to be the space of all bounded sequences. If for any $1 leq k leq n$, $a_k(n) rightarrow 0$ as $n rightarrow infty$ then $||x_n||_{infty} = sup (a_1(n),a_2(n),ldots,a_n(n),0,0,ldots) rightarrow 0$ as $n rightarrow infty$? Any help is welcome.

real-analysis functional-analysis

share|cite|improve this question

asked Mar 19 at 16:21

JeromeJerome

1

$endgroup$

add a comment | 

$begingroup$

I still don't know if the statement below is true. It seems to be true but I couldn't find a proof. The statement is the following: for each $n geq 1$, let $x_n = (a_1(n),a_2(n),ldots,a_n(n),0,0,ldots) in ell^{infty}$, where $ell^{infty}$ is defined to be the space of all bounded sequences. If for any $1 leq k leq n$, $a_k(n) rightarrow 0$ as $n rightarrow infty$ then $||x_n||_{infty} = sup (a_1(n),a_2(n),ldots,a_n(n),0,0,ldots) rightarrow 0$ as $n rightarrow infty$? Any help is welcome.

real-analysis functional-analysis

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asked Mar 19 at 16:21

JeromeJerome

1

$endgroup$

share|cite|improve this question

asked Mar 19 at 16:21

JeromeJerome

1

share|cite|improve this question

asked Mar 19 at 16:21

JeromeJerome

1

asked Mar 19 at 16:21

JeromeJerome

1

asked Mar 19 at 16:21

JeromeJerome

1

2 Answers
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1

$begingroup$

This is not true. Define
$$a_{k}(n)=begin{cases}1text{ if }0leq nleq 2k\ 0text{ if }n>2kend{cases}.$$
Then $|x_{n}|_{infty}=1$ for all $n$.

share|cite|improve this answer

answered Mar 19 at 16:25

Floris ClaassensFloris Claassens

1,33229

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  $begingroup$
  Also doesn't work. For a fixed $n > 1$, $a_{n-1}(n) = 1$ which does not converge to zero and this is on assumptions.
  $endgroup$
  – Jerome
  Mar 19 at 16:41
  

  
  
  
  


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  $begingroup$
  @Jerome: No, it is not. The convergence is for a fixed $k$.
  $endgroup$
  – tomasz
  Mar 19 at 18:23
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint: Consider $a_k(n):=ndelta_{k,n}$ (this is Kronecker delta).

share|cite|improve this answer

answered Mar 19 at 16:30

tomasztomasz

24k23482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the assumptions, $a_n(n) rightarrow 0$. In your example, $a_n(n) = n delta_{n,n} = n$ does not converge to zero.
  $endgroup$
  – Jerome
  Mar 19 at 16:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jerome: No. in the assumptions, for every $k$, you have $a_k(n)to 0$. In my example, $a_k(n)=0$ for all $n>k$, so in particular, $a_k(n)to 0$.
  $endgroup$
  – tomasz
  Mar 19 at 18:21
  
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

This is not true. Define
$$a_{k}(n)=begin{cases}1text{ if }0leq nleq 2k\ 0text{ if }n>2kend{cases}.$$
Then $|x_{n}|_{infty}=1$ for all $n$.

share|cite|improve this answer

answered Mar 19 at 16:25

Floris ClaassensFloris Claassens

1,33229

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also doesn't work. For a fixed $n > 1$, $a_{n-1}(n) = 1$ which does not converge to zero and this is on assumptions.
  $endgroup$
  – Jerome
  Mar 19 at 16:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jerome: No, it is not. The convergence is for a fixed $k$.
  $endgroup$
  – tomasz
  Mar 19 at 18:23
  

  
  
  
  

add a comment | 

1

$begingroup$

This is not true. Define
$$a_{k}(n)=begin{cases}1text{ if }0leq nleq 2k\ 0text{ if }n>2kend{cases}.$$
Then $|x_{n}|_{infty}=1$ for all $n$.

share|cite|improve this answer

answered Mar 19 at 16:25

Floris ClaassensFloris Claassens

1,33229

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also doesn't work. For a fixed $n > 1$, $a_{n-1}(n) = 1$ which does not converge to zero and this is on assumptions.
  $endgroup$
  – Jerome
  Mar 19 at 16:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jerome: No, it is not. The convergence is for a fixed $k$.
  $endgroup$
  – tomasz
  Mar 19 at 18:23
  

  
  
  
  

add a comment | 

$begingroup$

This is not true. Define
$$a_{k}(n)=begin{cases}1text{ if }0leq nleq 2k\ 0text{ if }n>2kend{cases}.$$
Then $|x_{n}|_{infty}=1$ for all $n$.

share|cite|improve this answer

answered Mar 19 at 16:25

Floris ClaassensFloris Claassens

1,33229

$endgroup$

share|cite|improve this answer

answered Mar 19 at 16:25

Floris ClaassensFloris Claassens

1,33229

answered Mar 19 at 16:25

Floris ClaassensFloris Claassens

1,33229

answered Mar 19 at 16:25

Floris ClaassensFloris Claassens

1,33229

1

$begingroup$

Hint: Consider $a_k(n):=ndelta_{k,n}$ (this is Kronecker delta).

share|cite|improve this answer

answered Mar 19 at 16:30

tomasztomasz

24k23482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the assumptions, $a_n(n) rightarrow 0$. In your example, $a_n(n) = n delta_{n,n} = n$ does not converge to zero.
  $endgroup$
  – Jerome
  Mar 19 at 16:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jerome: No. in the assumptions, for every $k$, you have $a_k(n)to 0$. In my example, $a_k(n)=0$ for all $n>k$, so in particular, $a_k(n)to 0$.
  $endgroup$
  – tomasz
  Mar 19 at 18:21
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint: Consider $a_k(n):=ndelta_{k,n}$ (this is Kronecker delta).

share|cite|improve this answer

answered Mar 19 at 16:30

tomasztomasz

24k23482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the assumptions, $a_n(n) rightarrow 0$. In your example, $a_n(n) = n delta_{n,n} = n$ does not converge to zero.
  $endgroup$
  – Jerome
  Mar 19 at 16:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jerome: No. in the assumptions, for every $k$, you have $a_k(n)to 0$. In my example, $a_k(n)=0$ for all $n>k$, so in particular, $a_k(n)to 0$.
  $endgroup$
  – tomasz
  Mar 19 at 18:21
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Hint: Consider $a_k(n):=ndelta_{k,n}$ (this is Kronecker delta).

share|cite|improve this answer

answered Mar 19 at 16:30

tomasztomasz

24k23482

$endgroup$

share|cite|improve this answer

answered Mar 19 at 16:30

tomasztomasz

24k23482

answered Mar 19 at 16:30

tomasztomasz

24k23482

answered Mar 19 at 16:30

tomasztomasz

24k23482