Banach spaces

**surjective** · February 23rd 2010, 12:33 AM

Hello,

I have been trying to solve two questions concerning norm-defining and subspace. This I am trying to do with respect to Banach-spaces. Allow me to introduce the problem (it's a bit long so bear with me):

I am considering a sequence $\{w_{k}\}_{k=1}^{\infty}$ of positive real numbers. We define the weighted $\ell^{1}$ -space $\ell_{w}^{1}(\mathbb{N})$ by:

$\ell_{w}^{1}(\mathbb{N}) = \left\lbrace \{x_{k} \}_{k=1}^{\infty} \mid x_{k} \in \mathbb{C}, \sum_{k=1}^{\infty} |x_{k}|w_{k} < \infty \right\rbrace$

Question 1:

I want to show that the expression $\Vert \cdot \Vert$ given by:

$\Vert \{x_{k} \}_{k=1}^{\infty} \Vert = \sum_{k=1}^{\infty} |x_{k}|w_{k}$

defines a norm on $\ell_{w}^{1}(\mathbb{N})$ .

Question 2:

I now consider a special choice:

$w_{k}=2^{k}$

and wan't to show that $\ell_{w}^{1}(\mathbb{N})$ is a subspace of $\ell^{1}(\mathbb{N})$ .

The following is what I have done so far:

Solution to question 1:

The following three condition must be fulfilled in order for a function $\Vert \cdot \Vert$ to define a norm:

a) $\Vert \textbf{v} \Vert \geq 0, \forall \textbf{v} \in V$ and $\Vert \textbf{v} \Vert=0 \Leftrightarrow \textbf{v}=\textbf{0}$

b) $\Vert \alpha \textbf{v}\Vert = |\alpha|\Vert \textbf{v}\Vert, \forall \textbf{v} \in V, \alpha \in \mathbb{C}$

c) $\Vert \textbf{v}+\textbf{w}\Vert \leq \Vert \textbf{v} \Vert + \Vert \textbf{w} \Vert, \forall \textbf{v},\textbf{w} \in V$

Showing a

For the sequences $\{x_{k}\}_{k=1}^{\infty}$ and $\{w_{k}\}_{k=1}^{\infty}$ we have:

$\Vert \{x_{k} \}_{k=1}^{\infty} \Vert \geq 0 \Leftrightarrow \sum_{k=1}^{\infty} |x_{k}|w_{k} \geq 0 \Leftrightarrow |x_{1}|w_{1} + |x_{2}|w_{2} + \ldots + |x_{k}|w_{k} + \cdots \geq 0$

It is clear that the last inequality is satisfied only if $\{x_{k}\}_{k=1}^{\infty},\{w_{k}\}_{k=1}^{\infty} \neq \{0\}_{k=1}^{\infty}$ . Furthermore we see that $\Vert \{x_{k} \}_{k=1}^{\infty} \Vert=0$ only if $\{x_{k} \}_{k=1}^{\infty}=\{w_{k} \}_{k=1}^{\infty}=\{ 0\}_{k=1}^{\infty}$ . The first requirement is hereby fulfilled.

Showing b

For a scalar $\alpha \in \mathbb{C}$ we have that:

$|\alpha| \cdot \Vert \{x_{k} \}_{k=1}^{\infty} \Vert =$
$|\alpha|\cdot \sum_{k=1}^{\infty} |x_{k}|w_{k} =$
$|\alpha|\cdot ( |x_{1}|w_{1} + |x_{2}|w_{2} + \ldots + |x_{k}|w_{k} +\cdots ) =$
$|\alpha||x_{1}|w_{1} + |\alpha||x_{2}|w_{2} + \ldots + |\alpha||x_{k}|w_{k} + \cdots =$
$\sum_{k=1}^{\infty} |\alpha||x_{k}|w_{k} =$
$\Vert \alpha \{x_{k} \}_{k=1}^{\infty} \Vert$

Requirement 2 is hereby met.

Showing c

Having trouble. Assistance needed.

Solution to question 2

Having trouble. Help needed.

Sorry for the long post but I hope that someone can be of assistance regarding showing of c and question 2 . Your help is greatly appreciated.

Thank you very much.

**HallsofIvy** · February 23rd 2010, 04:17 AM

Originally Posted by surjective

Hello,

I have been trying to solve two questions concerning norm-defining and subspace. This I am trying to do with respect to Banach-spaces. Allow me to introduce the problem (it's a bit long so bear with me):

I am considering a sequence $\{w_{k}\}_{k=1}^{\infty}$ of positive real numbers. We define the weighted $\ell^{1}$ -space $\ell_{w}^{1}(\mathbb{N})$ by:

$\ell_{w}^{1}(\mathbb{N}) = \left\lbrace \{x_{k} \}_{k=1}^{\infty} \mid x_{k} \in \mathbb{C}, \sum_{k=1}^{\infty} |x_{k}|w_{k} < \infty \right\rbrace$

Question 1:

I want to show that the expression $\Vert \cdot \Vert$ given by:

$\Vert \{x_{k} \}_{k=1}^{\infty} \Vert = \sum_{k=1}^{\infty} |x_{k}|w_{k}$

defines a norm on $\ell_{w}^{1}(\mathbb{N})$ .

Question 2:

I now consider a special choice:

$w_{k}=2^{k}$

and wan't to show that $\ell_{w}^{1}(\mathbb{N})$ is a subspace of $\ell^{1}(\mathbb{N})$ .

The following is what I have done so far:

Solution to question 1:

The following three condition must be fulfilled in order for a function $\Vert \cdot \Vert$ to define a norm:

a) $\Vert \textbf{v} \Vert \geq 0, \forall \textbf{v} \in V$ and $\Vert \textbf{v} \Vert=0 \Leftrightarrow \textbf{v}=\textbf{0}$

b) $\Vert \alpha \textbf{v}\Vert = |\alpha|\Vert \textbf{v}\Vert, \forall \textbf{v} \in V, \alpha \in \mathbb{C}$

c) $\Vert \textbf{v}+\textbf{w}\Vert \leq \Vert \textbf{v} \Vert + \Vert \textbf{w} \Vert, \forall \textbf{v},\textbf{w} \in V$

Showing a

For the sequences $\{x_{k}\}_{k=1}^{\infty}$ and $\{w_{k}\}_{k=1}^{\infty}$ we have:

$\Vert \{x_{k} \}_{k=1}^{\infty} \Vert \geq 0 \Leftrightarrow \sum_{k=1}^{\infty} |x_{k}|w_{k} \geq 0 \Leftrightarrow |x_{1}|w_{1} + |x_{2}|w_{2} + \ldots + |x_{k}|w_{k} + \cdots \geq 0$

It is clear that the last inequality is satisfied only if $\{x_{k}\}_{k=1}^{\infty},\{w_{k}\}_{k=1}^{\infty} \neq \{0\}_{k=1}^{\infty}$ . Furthermore we see that $\Vert \{x_{k} \}_{k=1}^{\infty} \Vert=0$ only if $\{x_{k} \}_{k=1}^{\infty}=\{w_{k} \}_{k=1}^{\infty}=\{ 0\}_{k=1}^{\infty}$ . The first requirement is hereby fulfilled.

Showing b

For a scalar $\alpha \in \mathbb{C}$ we have that:

$|\alpha| \cdot \Vert \{x_{k} \}_{k=1}^{\infty} \Vert =$
$|\alpha|\cdot \sum_{k=1}^{\infty} |x_{k}|w_{k} =$
$|\alpha|\cdot ( |x_{1}|w_{1} + |x_{2}|w_{2} + \ldots + |x_{k}|w_{k} +\cdots ) =$
$|\alpha||x_{1}|w_{1} + |\alpha||x_{2}|w_{2} + \ldots + |\alpha||x_{k}|w_{k} + \cdots =$
$\sum_{k=1}^{\infty} |\alpha||x_{k}|w_{k} =$
$\Vert \alpha \{x_{k} \}_{k=1}^{\infty} \Vert$

Requirement 2 is hereby met.

Showing c

Having trouble. Assistance needed.

$||u+ v||= \sum_{k=1}^\infty x_k(u_k+ v_k)$
Since that is an infinite sum, it is equal to the limit of its partial sums so look at $\sum_{k=1}^N |x_k(u_k+ v_k)|= \sum_{k=1}^N |x_ku_k+ x_kv_k|\le \sum_{k=1}^N |x_ku_k|+ \sum_{k=1}^N |x_kv_k|$ .

Now take the limit as N goes to infinity.

Solution to question 2

Having trouble. Help needed.

Sorry for the long post but I hope that someone can be of assistance regarding showing of c and question 2 . Your help is greatly appreciated.

Thank you very much.

For every $k\ge 1$ , $1\le 2^k$ so $|w_k|\le 2|^kw_k|$ . Use the "comparison test".

**surjective** · February 23rd 2010, 06:30 AM

Hello,

Thanks for the reply. I am afraid that your considerations are not quite clear to me (i am a slow-learner and need to have things cut out). Would it be possible if you could explain it one more time?

Thank you in advance

**Drexel28** · February 23rd 2010, 07:51 AM

Originally Posted by surjective

Hello,

Thanks for the reply. I am afraid that your considerations are not quite clear to me (i am a slow-learner and need to have things cut out). Would it be possible if you could explain it one more time?

Thank you in advance

HallsOfIvy merely pointed out that if we have to infinite sequences $\{x_n\},\{y_n\}$ in $\mathbb{C}$ that $\|x_n+y_n\|=\sum_{n=1}^{\infty}\left|x_n+y_n\right |$ . But, $\sum_{n=1}^{M}\left|x_n+y_n\right|\leqslant\sum_{n =1}^{M}\left\{|x_n|+|y_n|\right\}=\sum_{n=1}^{M}|x _n|+\sum_{n=1}^{M}|y_n|$ . Taking the limit of both sides of this inequality gives the answer.

**surjective** · February 23rd 2010, 08:27 AM

Hey,

Could you also be kind and elaborate on questions 2 ?

Thanks.

**Jose27** · February 23rd 2010, 10:48 AM

Originally Posted by surjective

Hello,

I have been trying to solve two questions concerning norm-defining and subspace. This I am trying to do with respect to Banach-spaces. Allow me to introduce the problem (it's a bit long so bear with me):

I am considering a sequence $\{w_{k}\}_{k=1}^{\infty}$ of positive real numbers. We define the weighted $\ell^{1}$ -space $\ell_{w}^{1}(\mathbb{N})$ by:

$\ell_{w}^{1}(\mathbb{N}) = \left\lbrace \{x_{k} \}_{k=1}^{\infty} \mid x_{k} \in \mathbb{C}, \sum_{k=1}^{\infty} |x_{k}|w_{k} < \infty \right\rbrace$

Question 1:

I want to show that the expression $\Vert \cdot \Vert$ given by:

$\Vert \{x_{k} \}_{k=1}^{\infty} \Vert = \sum_{k=1}^{\infty} |x_{k}|w_{k}$

defines a norm on $\ell_{w}^{1}(\mathbb{N})$ .

Question 2:

I now consider a special choice:

$w_{k}=2^{k}$

and wan't to show that $\ell_{w}^{1}(\mathbb{N})$ is a subspace of $\ell^{1}(\mathbb{N})$ .

The following is what I have done so far:

Solution to question 1:

The following three condition must be fulfilled in order for a function $\Vert \cdot \Vert$ to define a norm:

a) $\Vert \textbf{v} \Vert \geq 0, \forall \textbf{v} \in V$ and $\Vert \textbf{v} \Vert=0 \Leftrightarrow \textbf{v}=\textbf{0}$

b) $\Vert \alpha \textbf{v}\Vert = |\alpha|\Vert \textbf{v}\Vert, \forall \textbf{v} \in V, \alpha \in \mathbb{C}$

c) $\Vert \textbf{v}+\textbf{w}\Vert \leq \Vert \textbf{v} \Vert + \Vert \textbf{w} \Vert, \forall \textbf{v},\textbf{w} \in V$

Showing a

For the sequences $\{x_{k}\}_{k=1}^{\infty}$ and $\{w_{k}\}_{k=1}^{\infty}$ we have:

$\Vert \{x_{k} \}_{k=1}^{\infty} \Vert \geq 0 \Leftrightarrow \sum_{k=1}^{\infty} |x_{k}|w_{k} \geq 0 \Leftrightarrow |x_{1}|w_{1} + |x_{2}|w_{2} + \ldots + |x_{k}|w_{k} + \cdots \geq 0$

It is clear that the last inequality is satisfied only if $\{x_{k}\}_{k=1}^{\infty},\{w_{k}\}_{k=1}^{\infty} \neq \{0\}_{k=1}^{\infty}$ . Furthermore we see that $\Vert \{x_{k} \}_{k=1}^{\infty} \Vert=0$ only if $\{x_{k} \}_{k=1}^{\infty}=\{w_{k} \}_{k=1}^{\infty}=\{ 0\}_{k=1}^{\infty}$ . The first requirement is hereby fulfilled.

Showing b

For a scalar $\alpha \in \mathbb{C}$ we have that:

$|\alpha| \cdot \Vert \{x_{k} \}_{k=1}^{\infty} \Vert =$
$|\alpha|\cdot \sum_{k=1}^{\infty} |x_{k}|w_{k} =$
$|\alpha|\cdot ( |x_{1}|w_{1} + |x_{2}|w_{2} + \ldots + |x_{k}|w_{k} +\cdots ) =$
$|\alpha||x_{1}|w_{1} + |\alpha||x_{2}|w_{2} + \ldots + |\alpha||x_{k}|w_{k} + \cdots =$
$\sum_{k=1}^{\infty} |\alpha||x_{k}|w_{k} =$
$\Vert \alpha \{x_{k} \}_{k=1}^{\infty} \Vert$

Requirement 2 is hereby met.

Showing c

Having trouble. Assistance needed.

Solution to question 2

Having trouble. Help needed.

Sorry for the long post but I hope that someone can be of assistance regarding showing of c and question 2 . Your help is greatly appreciated.

Thank you very much.

If $w_k= \frac{1}{2^k}$ then clearly if we have $x_k=1$ for all $k$ , $(x_k)\in \ell _w ^1 (\mathbb{N} )$ but $x_k \notin \ell ^1 (\mathbb{N} )$ so $\ell _w ^1 (\mathbb{N} )$ is not a subspace of $\ell ^1 (\mathbb{N} )$

**surjective** · February 23rd 2010, 12:59 PM

Hey,

I have to disappoint you but I still need some clarification on question 2.

The question says $w_{k}=2^{k}$ and we then have to show that $\ell_{w}^{1}(\mathbb{N})$ is indeed a subspace of $\ell^{1}(\mathbb{N})$ . Assistance needed.

Thanks.

**Drexel28** · February 23rd 2010, 02:14 PM

Originally Posted by surjective

Hey,

I have to disappoint you but I still need some clarification on question 2.

The question says $w_{k}=2^{k}$ and we then have to show that $\ell_{w}^{1}(\mathbb{N})$ is indeed a subspace of $\ell^{1}(\mathbb{N})$ . Assistance needed.

Thanks.

You are clearly having a lot of problems. Could you help us out by telling us specifically what you're troubled by?

**surjective** · February 23rd 2010, 03:09 PM

Hey,

Yes I am having a hard time understanding. I will try my best to point out my problem.

When showing that a subset $S$ of a vectorspace $V$ is indeed a subspace of $V$ then the following three condition must hold:

1) $S$ is nonempty
2) $S$ is closed under addition
3) $S$ is closed under multiplication

I would like to use this very formalized way of showing that $\ell_{w}^{1}(\mathbb{N})$ is a subspace of $\ell^{1}(\mathbb{N})$ .

The space $\ell_{w}^{1}(\mathbb{N})$ consists of sequences $\{x_{k} \}_{k=1}^{\infty}$ ( $x_{k}\in \mathbb{C}$ ) for which it must hold that :

$\sum_{k=1}^{\infty} |x_{k}|w_{k} < \infty$

i.e. that the norm of $\{x_{k} \}_{k=1}^{\infty}$ converges to a physical number.

The space $\ell_{w}^{1}(\mathbb{N})$ is obviously nonempty. Then how do I procede. I must try to have a systemized way of showing 2) and 3) otherwise my mind will fail me again

sorry.

Thank you for your patience.

**mabruka** · February 23rd 2010, 03:34 PM

2) $\ell_{w}^{1}(\mathbb{N})$ is closed under addition

Take $x,y \in \ell_{w}^{1}(\mathbb{N})$ .

We must verify that $x+y\in \ell_{w}^{1}(\mathbb{N})$ ,where $x+y=\{x_k+y_k\}_{k=1}^\infty$

For each k we have

$|x_k+y_k|\leq |x_k| + |y_k|$ using triangle inequality, so if we multiply by the corresponding $w_k$ we get

$|x_k+y_k|w_k\leq |x_k|w_k + |y_k|w_k$ for each k

Adding (rigourously ) N terms and then taking the limit we have

$\sum_{k=1}^{\infty} |x_k+y_k|w_k \leq \sum_{k=1}^{\infty}|x_k|w_k + \sum_{k=1}^{\infty} |y_k|w_k < \infty$ because $x,y \in \ell_{w}^{1}(\mathbb{N})$

Hence $x+y \in$ $\ell_{w}^{1}(\mathbb{N})$

Is that ok ?

**mabruka** · February 23rd 2010, 03:45 PM

3) Show it is closed under scalar multiplication

You give it a try

**surjective** · February 23rd 2010, 04:07 PM

Hey,

This is beautiful. It is systematic and orderes. I will give the scalar multiplication a try.

Could you refer me to some online books on functional analysis?

Thanks.

**mabruka** · February 23rd 2010, 05:00 PM

Give Introductory Functional Analysis with Applications, Erwin Kreyszig
a look.

I think this one in particular could be pretty helpful to you.

**Enrique2** · February 24th 2010, 12:27 AM

Originally Posted by surjective

Hello,

Question 2:

I now consider a special choice:

$w_{k}=2^{k}$

and wan't to show that $\ell_{w}^{1}(\mathbb{N})$ is a subspace of $\ell^{1}(\mathbb{N})$ .

Take $(x_n)_n\in \ell_{w}^{1}(\mathbb{N})$ .
Then you can show that $\lim\sup\{(2^{n+1}|x_{n+1}|)/(2^n|x_n |)\}\leq 1$ , otherwise you could find $m>1$ such that
$(2^{n+1}|x_{n+1}|)/(2^n|x_n |)>m$ for infinitely many n, and from this is not dificult to see that
the series $\sum_n2^n|x_n|$ would'nt be convergent. This implies
$\lim\sup_n\{|x_{n+1}|/|x_n|\}\leq 1/2$ , which implies the desired convergence of $\sum_n|x_n|$

By the way, sorry for the estupid argument, the solution of Hallsofivy is much easier and natural!!!
|x_k|\leq 2^k |x_k|, simply!!

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