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Formalizing converging norm

Given k unbounded sequences, must there exist a sequence such that the product of all k sequences with it is converging, diverging?Norm, adjoint operator and compactness os some operatorsIs the norm on $ell^infty$ induced by an inner product?Boundedness of two sequencesWhy define norm in $L_p$ in that way?Why do sequences need to converge with respect to a norm?Completeness of $ell^2$ done rightNeed help on proving triangle inequality for norm.Can we define an inner product such that its induced norm is $ell^1$-norm or others?Any examples of unbounded linear operators between $ell^infty$ and $ell^infty$? $ell^p$ and $ell^p$? $ell^p$ and $ell^infty$?

0

$begingroup$

Let $x in ell^2$. Define $(u_N)$ as the sequence of vectors $u_N in U$ such that $u_n = x_n$ if $n le N$, $0$ otherwise.

Then $lVert u_N rVert$ is an increasing function which converges to finite $lVert x rVert$.

The statement seems obvious, but how can I make this claim more rigorous? Saying the norm of $u_N$ increases doesn't seem convincing.

I suppose it depends on how norm is defined. If defined as an infinite sum which is defined as the limit of a finite summation, then the result follows by definition ...

real-analysis lp-spaces

share|cite|improve this question

edited Mar 22 at 10:18

qwr

asked Mar 22 at 9:12

qwrqwr

6,68242755

$endgroup$

add a comment | 

0

$begingroup$

Let $x in ell^2$. Define $(u_N)$ as the sequence of vectors $u_N in U$ such that $u_n = x_n$ if $n le N$, $0$ otherwise.

Then $lVert u_N rVert$ is an increasing function which converges to finite $lVert x rVert$.

The statement seems obvious, but how can I make this claim more rigorous? Saying the norm of $u_N$ increases doesn't seem convincing.

I suppose it depends on how norm is defined. If defined as an infinite sum which is defined as the limit of a finite summation, then the result follows by definition ...

real-analysis lp-spaces

share|cite|improve this question

edited Mar 22 at 10:18

qwr

asked Mar 22 at 9:12

qwrqwr

6,68242755

$endgroup$

add a comment | 

$begingroup$

Let $x in ell^2$. Define $(u_N)$ as the sequence of vectors $u_N in U$ such that $u_n = x_n$ if $n le N$, $0$ otherwise.

Then $lVert u_N rVert$ is an increasing function which converges to finite $lVert x rVert$.

The statement seems obvious, but how can I make this claim more rigorous? Saying the norm of $u_N$ increases doesn't seem convincing.

I suppose it depends on how norm is defined. If defined as an infinite sum which is defined as the limit of a finite summation, then the result follows by definition ...

real-analysis lp-spaces

share|cite|improve this question

edited Mar 22 at 10:18

qwr

asked Mar 22 at 9:12

qwrqwr

6,68242755

$endgroup$

share|cite|improve this question

edited Mar 22 at 10:18

qwr

asked Mar 22 at 9:12

qwrqwr

6,68242755

share|cite|improve this question

edited Mar 22 at 10:18

qwr

asked Mar 22 at 9:12

qwrqwr

6,68242755

asked Mar 22 at 9:12

qwrqwr

6,68242755

asked Mar 22 at 9:12

qwrqwr

6,68242755

3 Answers
3

active

oldest

votes

1

$begingroup$

$sqrt ^2$ increases to $sqrt sumlimits_k=1^infty $. This is just deinition of the infinite sum $sumlimits_k=1^infty |x_k|^2$.

share|cite|improve this answer

edited Mar 22 at 9:41

answered Mar 22 at 9:14

Kavi Rama MurthyKavi Rama Murthy

72.9k53170

$endgroup$

add a comment | 

0

$begingroup$

$newcommandnorm[1]lVert #1 rVert$What I wanted to show is actually both $normu_N to normx$ and $normx - u_N to 0$.

Using the definition of norm as a finite value such that
$$normx^2 = sum_n=1^infty x_n^2 = lim_N to infty sum_n=1^N x_n^2 = lim_N to infty normu_N^2$$

we see clearly $lVert u_N rVert$ is an increasing function and $normu_N to normx$. Therefore
$$normx - u_N^2 = sum_n=N+1^infty x_n^2 = sum_n=1^infty x_n^2 - sum_n=1^N x_n^2 = normx^2 - normu_N^2 to 0$$

share|cite|improve this answer

answered Mar 29 at 19:43

qwrqwr

6,68242755

$endgroup$

add a comment | 

-1

$begingroup$

$$|x|^2geqsum_n=1^N+1x_n^2=|u_N+1|^2=x_1^2+cdots + x_N+1^2geq x_1^2+cdots + x_N^2=|u_N|^2$$

share|cite|improve this answer

answered Mar 22 at 9:14

uniquesolutionuniquesolution

9,4971823

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ok but this doesn't show norm of $u_N$ actually converges to norm of $x$
  $endgroup$
  – qwr
  Mar 22 at 9:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes it does, because a bounded nondecreasing sequence converges to its supremum. I leave it to you to show that the supremum is the correct limit.
  $endgroup$
  – uniquesolution
  Mar 22 at 9:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ok, so I need to show the supremum is norm of $x$ ?
  $endgroup$
  – qwr
  Mar 22 at 9:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you need to show that.
  $endgroup$
  – uniquesolution
  Mar 22 at 9:20
  

  
  
  
  

add a comment | 

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1

$begingroup$

$sqrt ^2$ increases to $sqrt sumlimits_k=1^infty $. This is just deinition of the infinite sum $sumlimits_k=1^infty |x_k|^2$.

share|cite|improve this answer

edited Mar 22 at 9:41

answered Mar 22 at 9:14

Kavi Rama MurthyKavi Rama Murthy

72.9k53170

$endgroup$

add a comment | 

1

$begingroup$

$sqrt ^2$ increases to $sqrt sumlimits_k=1^infty $. This is just deinition of the infinite sum $sumlimits_k=1^infty |x_k|^2$.

share|cite|improve this answer

edited Mar 22 at 9:41

answered Mar 22 at 9:14

Kavi Rama MurthyKavi Rama Murthy

72.9k53170

$endgroup$

add a comment | 

$begingroup$

$sqrt ^2$ increases to $sqrt sumlimits_k=1^infty $. This is just deinition of the infinite sum $sumlimits_k=1^infty |x_k|^2$.

share|cite|improve this answer

edited Mar 22 at 9:41

answered Mar 22 at 9:14

Kavi Rama MurthyKavi Rama Murthy

72.9k53170

$endgroup$

share|cite|improve this answer

edited Mar 22 at 9:41

answered Mar 22 at 9:14

Kavi Rama MurthyKavi Rama Murthy

72.9k53170

answered Mar 22 at 9:14

Kavi Rama MurthyKavi Rama Murthy

72.9k53170

answered Mar 22 at 9:14

Kavi Rama MurthyKavi Rama Murthy

72.9k53170

0

$begingroup$

$newcommandnorm[1]lVert #1 rVert$What I wanted to show is actually both $normu_N to normx$ and $normx - u_N to 0$.

Using the definition of norm as a finite value such that
$$normx^2 = sum_n=1^infty x_n^2 = lim_N to infty sum_n=1^N x_n^2 = lim_N to infty normu_N^2$$

we see clearly $lVert u_N rVert$ is an increasing function and $normu_N to normx$. Therefore
$$normx - u_N^2 = sum_n=N+1^infty x_n^2 = sum_n=1^infty x_n^2 - sum_n=1^N x_n^2 = normx^2 - normu_N^2 to 0$$

share|cite|improve this answer

answered Mar 29 at 19:43

qwrqwr

6,68242755

$endgroup$

add a comment | 

0

$begingroup$

$newcommandnorm[1]lVert #1 rVert$What I wanted to show is actually both $normu_N to normx$ and $normx - u_N to 0$.

Using the definition of norm as a finite value such that
$$normx^2 = sum_n=1^infty x_n^2 = lim_N to infty sum_n=1^N x_n^2 = lim_N to infty normu_N^2$$

we see clearly $lVert u_N rVert$ is an increasing function and $normu_N to normx$. Therefore
$$normx - u_N^2 = sum_n=N+1^infty x_n^2 = sum_n=1^infty x_n^2 - sum_n=1^N x_n^2 = normx^2 - normu_N^2 to 0$$

share|cite|improve this answer

answered Mar 29 at 19:43

qwrqwr

6,68242755

$endgroup$

add a comment | 

$begingroup$

$newcommandnorm[1]lVert #1 rVert$What I wanted to show is actually both $normu_N to normx$ and $normx - u_N to 0$.

Using the definition of norm as a finite value such that
$$normx^2 = sum_n=1^infty x_n^2 = lim_N to infty sum_n=1^N x_n^2 = lim_N to infty normu_N^2$$

we see clearly $lVert u_N rVert$ is an increasing function and $normu_N to normx$. Therefore
$$normx - u_N^2 = sum_n=N+1^infty x_n^2 = sum_n=1^infty x_n^2 - sum_n=1^N x_n^2 = normx^2 - normu_N^2 to 0$$

share|cite|improve this answer

answered Mar 29 at 19:43

qwrqwr

6,68242755

$endgroup$

share|cite|improve this answer

answered Mar 29 at 19:43

qwrqwr

6,68242755

answered Mar 29 at 19:43

qwrqwr

6,68242755

answered Mar 29 at 19:43

qwrqwr

6,68242755

-1

$begingroup$

$$|x|^2geqsum_n=1^N+1x_n^2=|u_N+1|^2=x_1^2+cdots + x_N+1^2geq x_1^2+cdots + x_N^2=|u_N|^2$$

share|cite|improve this answer

answered Mar 22 at 9:14

uniquesolutionuniquesolution

9,4971823

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ok but this doesn't show norm of $u_N$ actually converges to norm of $x$
  $endgroup$
  – qwr
  Mar 22 at 9:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes it does, because a bounded nondecreasing sequence converges to its supremum. I leave it to you to show that the supremum is the correct limit.
  $endgroup$
  – uniquesolution
  Mar 22 at 9:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ok, so I need to show the supremum is norm of $x$ ?
  $endgroup$
  – qwr
  Mar 22 at 9:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you need to show that.
  $endgroup$
  – uniquesolution
  Mar 22 at 9:20
  

  
  
  
  

add a comment | 

-1

$begingroup$

$$|x|^2geqsum_n=1^N+1x_n^2=|u_N+1|^2=x_1^2+cdots + x_N+1^2geq x_1^2+cdots + x_N^2=|u_N|^2$$

share|cite|improve this answer

answered Mar 22 at 9:14

uniquesolutionuniquesolution

9,4971823

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ok but this doesn't show norm of $u_N$ actually converges to norm of $x$
  $endgroup$
  – qwr
  Mar 22 at 9:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes it does, because a bounded nondecreasing sequence converges to its supremum. I leave it to you to show that the supremum is the correct limit.
  $endgroup$
  – uniquesolution
  Mar 22 at 9:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ok, so I need to show the supremum is norm of $x$ ?
  $endgroup$
  – qwr
  Mar 22 at 9:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you need to show that.
  $endgroup$
  – uniquesolution
  Mar 22 at 9:20
  

  
  
  
  

add a comment | 

$begingroup$

$$|x|^2geqsum_n=1^N+1x_n^2=|u_N+1|^2=x_1^2+cdots + x_N+1^2geq x_1^2+cdots + x_N^2=|u_N|^2$$

share|cite|improve this answer

answered Mar 22 at 9:14

uniquesolutionuniquesolution

9,4971823

$endgroup$

share|cite|improve this answer

answered Mar 22 at 9:14

uniquesolutionuniquesolution

9,4971823

answered Mar 22 at 9:14

uniquesolutionuniquesolution

9,4971823

answered Mar 22 at 9:14

uniquesolutionuniquesolution

9,4971823