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Math Help - Banach spaces

  1. #1
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    Banach spaces

    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<br />
\sum_{k=1}^{\infty} |x_{k}|w_{k} \geq 0 \Leftrightarrow<br />
|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.
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  2. #2
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    Quote Originally Posted by surjective View Post
    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<br />
\sum_{k=1}^{\infty} |x_{k}|w_{k} \geq 0 \Leftrightarrow<br />
|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".
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  3. #3
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    Banach spaces

    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
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by surjective View Post
    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.
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  5. #5
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    Banach Spaces

    Hey,

    Could you also be kind and elaborate on questions 2 ?

    Thanks.
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  6. #6
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    Quote Originally Posted by surjective View Post
    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<br />
\sum_{k=1}^{\infty} |x_{k}|w_{k} \geq 0 \Leftrightarrow<br />
|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} )
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  7. #7
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    Banach spaces

    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.
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by surjective View Post
    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?
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  9. #9
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    Banach Spaces

    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.
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  10. #10
    Member mabruka's Avatar
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    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 ?
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  11. #11
    Member mabruka's Avatar
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    3) Show it is closed under scalar multiplication


    You give it a try
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  12. #12
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    Banach spaces

    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.
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  13. #13
    Member mabruka's Avatar
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    Give Introductory Functional Analysis with Applications, Erwin Kreyszig
    a look.

    I think this one in particular could be pretty helpful to you.
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  14. #14
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    Quote Originally Posted by surjective View Post
    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!!
    Last edited by Enrique2; February 24th 2010 at 03:20 AM. Reason: just noticed better solution in post 2
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