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Normed Linear Space Part 1

  1. Nov 2, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider the normed linear space [itex]\ell_\infty \mathbb({R})[/itex]. Let [itex]x= \frac{n-1}{n}[/itex], [itex]y=(1/n) and z=2^n[/itex]

    a) Are x,y,z each in [itex]\ell_\infty \mathbb({R})[/itex]

    b) What is x+y? What is [itex]2^{1/2} y?[/itex]

    c) Calculate [itex]||x||_\infty[/itex], [itex]||y||_\infty[/itex], [itex]||x+y||_\infty[/itex] and [itex]||2^{1/2} y ||_\infty[/itex]

    For a) Does that mean are x,y,z subspaces of [itex]\ell_\infty \mathbb({R})[/itex]?
     
  2. jcsd
  3. Nov 2, 2011 #2

    Mark44

    Staff: Mentor

    No, x, y, and z are sequences. The question is asking whether they belong to [itex]\ell_\infty \mathbb({R})[/itex].
     
  4. Nov 2, 2011 #3
    x and y are each an element in l_∞ as they are bounded above

    z is NOT an element as its not bounded above

    In b) what does it mean by 'what is'. Is that asking to evaluate?
     
  5. Nov 2, 2011 #4

    Mark44

    Staff: Mentor

    You could answer b by evaluating x + y, and then saying whether it is in l(R).
     
  6. Nov 2, 2011 #5


    x+y =n implies (n) =(1,2,3,4...) which is not in l∞(R) sicne it is not bounded above.

    2^(1/2)*y= 2^(n+1/2) which is not in l∞(R) either.......?
     
  7. Nov 2, 2011 #6

    Mark44

    Staff: Mentor

    Yes, but here's how to say the first part better. x + y = {n} = {1, 2, 3, ...}
    No, 21/2y = 21/2{1/n}. That means 21/2 times each element of the sequence {1/n}.
     
  8. Nov 2, 2011 #7
    Thanks..that was a mistake on my part. So hence it IS in l∞(R)

    For c) do I attempt to evaluate each?
     
  9. Nov 2, 2011 #8

    Mark44

    Staff: Mentor

    Yes and yes.
    Your textbook should have a definition of the infinity norm - if not, there's one in the link I posted in the other thread.
     
  10. Nov 3, 2011 #9

    I like Serena

    User Avatar
    Homework Helper

    Hmm, I get a different result for x+y... :confused:
     
  11. Nov 3, 2011 #10

    Mark44

    Staff: Mentor


    Let's make that x + y = {1} = {1, 1, 1, ...}
     
  12. Nov 8, 2011 #11
    For x:

    1-1/n =(0,1/2,2/3,3/4...) and is bounded.

    [itex]||1-1/n||_∞[/itex]=.........? I am not sure how to calculate this. If some one can illustrate this one, I can try the others...? I know [itex]||1-1/n||_∞=sup |x_n|[/itex] but dont know how to calculate |x_n|

    Thanks
     
  13. Nov 8, 2011 #12

    Mark44

    Staff: Mentor

    See http://en.wikipedia.org/wiki/Lp_space, in the section titled Lp spaces, starting with "One also defines the ∞-norm as ...".

    I believe that this is the norm to be used in your problem.
     
  14. Nov 8, 2011 #13
    THanks Mark but how to I evalute |x_n|?
     
  15. Nov 8, 2011 #14

    Mark44

    Staff: Mentor

    What is the supremum of the set {0, 1/2, 2/3, 3/4, ..., n/(n + 1), ...}?
     
  16. Nov 8, 2011 #15
    I think the supremum is the smallest real number that is greater than or equal to every number in the set, hence t must be 0...?

    What do i do next?

    Is it something along the lines of

    [itex]|1-1/n|=\sqrt{(1-1/n)^2}=0[/itex] where n=1
     
    Last edited: Nov 8, 2011
  17. Nov 8, 2011 #16

    Mark44

    Staff: Mentor

    Yes.
    No. 0 is smaller than, not larger than every number in the sequence except the first.
     
  18. Nov 8, 2011 #17
    Then it doesnt exist, there is no sup...?
     
  19. Nov 8, 2011 #18

    Mark44

    Staff: Mentor

    Yes! So what is the smallest number that is >= each number in the sequence {0, 1/2, 2/3, 3/4, ..., n/(n+1), ...}
    No, it can't be 0. This number is too small, because it's not larger than 1/2 (for example). 2589 is larger than all the numbers in the sequence, but can you find one that is not so large?
     
  20. Nov 9, 2011 #19
    1 would be greater than or equal to every number in the set....?
     
  21. Nov 9, 2011 #20

    Deveno

    User Avatar
    Science Advisor

    a reasonable guess, can you prove it?
     
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