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Homework Help: Bound Sequences

  1. Oct 31, 2011 #1
    1. The problem statement, all variables and given/known data

    a) Prove that [itex]\ell_\infty \mathbb({R})[/itex] is a subspace of [itex]\ell \mathbb({R})
    [/itex]

    b) Show that [itex]\left \| \right \|_\infty[/itex] is a norm on [itex]\ell_\infty (\mathbb{R})[/itex]

    3. The attempt at a solution

    For a) I guess we have to show that [itex]\vec{x} + \vec{y} \in \ell_\infty \mathbb({R})[/itex] and [itex]\alpha \vec({x}) \in \ell_\infty \mathbb({R})[/itex]

    but I dont know how to proceed....thanks
     
    Last edited by a moderator: Oct 31, 2011
  2. jcsd
  3. Oct 31, 2011 #2

    Mark44

    Staff: Mentor

    Start by assuming that x and y are in [itex]\ell_\infty \mathbb({R})[/itex], and say what it means for a vector to be in that space. Then add the vectors together. Is the sum in that space as well? Same thing for the scalar multiplication part.
     
  4. Oct 31, 2011 #3
    If we let [itex] \ell_\infty \mathbb({R})=\left \{ \vec x=(x_n), \vec y=(y_n) \in \ell_\infty \right \}[/itex]

    If x and y are vectors in [itex]\ell_\infty[/itex] then x+y is also in [itex]\ell_\infty[/itex]
    If [itex]\alpha \in \mathbb{R}[/itex] then [itex]\alpha \vec x[/itex] is also in [itex]\ell_\infty[/itex].......?
     
  5. Oct 31, 2011 #4

    Mark44

    Staff: Mentor

    No, how is [itex] \ell_\infty (R)[/itex] defined?
     
  6. Oct 31, 2011 #5
    This is the only definition I have in my notes with the addition of inserting the vector y

    [itex]\ell_\infty \mathbb({R})=\left \{ \vec x=(x_n), \vec y=(y_n) \in \ell_\infty \mathbb({R}) \right \}[/itex]
     
  7. Oct 31, 2011 #6

    Deveno

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    Science Advisor

    if {xn} is a bounded sequence and {yn} is a bounded sequence,

    is {xn+yn} a bounded sequence?
     
  8. Oct 31, 2011 #7
    Yes, I believe so...
     
  9. Oct 31, 2011 #8
    Mark asked a good question. This is something I dont have in my notes...?
     
  10. Oct 31, 2011 #9

    Deveno

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    (excerpt from wikipedia): If p = ∞, then ℓ is defined to be the space of all bounded sequences (in K, the range of the sequences).

    thus ℓ(R) is the space of all bounded real sequences.
     
  11. Oct 31, 2011 #10

    Mark44

    Staff: Mentor

    Which is what you need to show. Also that {axn} is a bounded sequence.

    Now, how is "bounded sequence" defined?
     
  12. Oct 31, 2011 #11

    Deveno

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    Let [itex]\ell_\infty \mathbb({R})[/itex] be the set of bounded real sequences with k > 0 such that [itex]\left | x_n \right |\le k[/itex]

    this is from another post of yours. it has the information you need.
     
  13. Nov 2, 2011 #12
    Let [itex]\ell_\infty \mathbb({R})[/itex] be the set of bounded real sequences with k > 0 such that [itex]\left | x_n \right |\le k[/itex]

    if [itex]\vec x, \vec y \in \ell_\infty \mathbb({R})[/itex] then there exist [itex]n_1, n_2 \in N[/itex] such that

    [itex]\vec x = (x_1,x_2,...x_{n1},0,0...)[/itex] and [itex]\vec y = (y_1,y_2,...y_{n2},0,0...) \therefore \vec x +\vec y= (x_1+y_1, x_2+y_2....x_n+y_n,0,0...)[/itex] where [itex]k \ge |x_n|, |y_n|[/itex]

    [itex]\alpha \vec x = (\alpha x_1, \alpha x_2,....\alpha x_n,0,0..)[/itex] where

    [itex]\alpha \in \mathbb{R}[/itex]................?
     
  14. Nov 2, 2011 #13

    Mark44

    Staff: Mentor

    If every element xi in the sequence satisfies |xi| <= k, what can you say about αxi?
     
  15. Nov 2, 2011 #14
    I am not sure...it is possible that [itex]|\alpha x_i|>=k[/itex]...?

    I dont see any constraint which states that [itex]\alpha x_i[/itex] cannot be >= to k.....
     
  16. Nov 2, 2011 #15

    Mark44

    Staff: Mentor

    Sure, that's possible, but it's not very relevant.

    Can't you show that |αxi| is <= some other constant?
     
  17. Nov 2, 2011 #16
    Based on the question asked and all the info I have in #12, I dont see any other constant at play....
    I wouldnt understand how or when one would look at some 'other' constant.
     
  18. Nov 2, 2011 #17

    Mark44

    Staff: Mentor

    If |x| < 3, then certainly 2|x| < 6, right?

    On a side note, could you ease up a bit on the LaTeX, especially for symbols that don't actually require it? These threads with lots of LaTeX take a long time to render on my browser. Many of the things that you write can be done using the Quick Symbols that appear on the right after you click Go Advanced.

    Everything below is done without using LaTeX.
    αxi
    πr2
    ∫x2dx
     
  19. Nov 2, 2011 #18
    OK, thanks for the advice on LaTex.

    So yes, that example makes sense....based on that the vector x is closed under multiplication. How about my attempt for addition as above? Hopefully I have a) answered.

    For part b) I have to show that || ||_∞ satifies 4 specific axioms, right?

    Thanks
     
  20. Nov 2, 2011 #19

    Mark44

    Staff: Mentor

    In the previous post, #18, why are you working with finite sequences? Addition in l is term-by-term. All you have to show is that, if x and y are bounded sequences, then x + y is also a bounded sequence.

    No, vectors aren't closed under scalar multiplication - the set that they belong to, l(R) in this case, is closed under scalar multiplication.

    For the b part, verify the properties in the definition of a norm.
     
  21. Nov 2, 2011 #20

    I dont know how to write it any other way...

    x, y ε l_∞(R)
     
  22. Nov 2, 2011 #21
    Ok. Since we know that x and y are bounded sequences then all I need to show is that

    if x_n=(x_1,x_2,x-3...) and y_n=(y_1,y_2....) then x_n+y_n=(x_1+y_1, x_2+y_2.....) Hence the set l∞(R) is closed under addition.......?
     
  23. Nov 2, 2011 #22

    Mark44

    Staff: Mentor

    Yes, that's really all you need to say. Vector addition and scalar multiplication in l(R) are the same as in l(R).

    Here's the link to the wiki page that deveno mentioned yesterday - http://en.wikipedia.org/wiki/Bounded_sequences.
     
  24. Nov 2, 2011 #23


    This attempt is based on a similar example in (R^3, || ||)

    if x_n and y_n are each bound sequences with |x_n| >=0 for n=1,2,3 and similarly for y_n, then

    axiom 1: ||x_n||∞= |x_1|+|x_2|+|x_3|... >=0 THis axiom holds.

    Axiom 2: ||x_n||∞=0 IFF |x_1+|x_2|+|x_3|=0, ie each x_n=0

    Axiom 3: ||ax_n||∞= |ax_1|+|ax_2|+|ax_3|
    = |a|(x_1+x_2+x_3)
    = |a| |x_n|∞


    Axiom 4: ||x_n+y_n||∞= |x_1+y_1| +| x_2+y_2|

    ||x_n+y_n||∞<= |x_1|+|x_2|+|y_1|+|y_2| therefore

    ||x_n+y_n||∞<= ||x_n||∞ + ||y_n||∞

    All 4 axioms hold therefore || ||∞ is a norm on l∞(R)............?
     
  25. Nov 3, 2011 #24
    Have I covered all axioms correctly in this? I could not find any other axiom regarding infinity...?
     
  26. Nov 3, 2011 #25

    Mark44

    Staff: Mentor

    You haven't actually used the definition of the infinity norm (|| ||) anywhere. The norm you seem to be using is the taxicab norm, not the infinity norm. The space here is bounded sequences, so each vector x in the space is an infinite sequence, not just a point in R3.

    See (again) http://en.wikipedia.org/wiki/Bounded_sequences for a description of norms in lp spaces (including l).
    There are other wiki articles on Lp spaces and norms in general.
     
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