Normed Linear Space Part 1

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Homework Statement



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]?
 

Answers and Replies

  • #2
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Homework Statement



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]?
No, x, y, and z are sequences. The question is asking whether they belong to [itex]\ell_\infty \mathbb({R})[/itex].
 
  • #3
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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?
 
  • #4
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You could answer b by evaluating x + y, and then saying whether it is in l(R).
 
  • #5
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1. Homework Statement

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



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.......?
 
  • #6
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x+y =n implies (n) =(1,2,3,4...) which is not in l∞(R) sicne it is not bounded above.
Yes, but here's how to say the first part better. x + y = {n} = {1, 2, 3, ...}
2^(1/2)*y= 2^(n+1/2) which is not in l∞(R) either.......?
No, 21/2y = 21/2{1/n}. That means 21/2 times each element of the sequence {1/n}.
 
  • #7
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No, 21/2y = 21/2{1/n}. That means 21/2 times each element of the sequence {1/n}.
Thanks..that was a mistake on my part. So hence it IS in l∞(R)

For c) do I attempt to evaluate each?
 
  • #8
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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.
 
  • #9
I like Serena
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Hmm, I get a different result for x+y... :confused:
 
  • #11
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Homework Statement



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]

c) Calculate [itex]||x||_\infty[/itex], [itex]||y||_\infty[/itex], [itex]||x+y||_\infty[/itex] and [itex]||2^{1/2} y ||_\infty[/itex]
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
 
  • #14
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What is the supremum of the set {0, 1/2, 2/3, 3/4, ..., n/(n + 1), ...}?
 
  • #15
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What is the supremum of the set {0, 1/2, 2/3, 3/4, ..., n/(n + 1), ...}?
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:
  • #16
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I think the supremum is the smallest real number that is greater than or equal to every number in the set,
Yes.
hence t must be 0...?
No. 0 is smaller than, not larger than every number in the sequence except the first.
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
 
  • #17
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Yes.No. 0 is smaller than, not larger than every number in the sequence except the first.
Then it doesnt exist, there is no sup...?
 
  • #18
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I think the supremum is the smallest real number that is greater than or equal to every number in the set
Yes! So what is the smallest number that is >= each number in the sequence {0, 1/2, 2/3, 3/4, ..., n/(n+1), ...}
, hence t must be 0...?
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?
 
  • #19
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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?
1 would be greater than or equal to every number in the set....?
 
  • #20
Deveno
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1 would be greater than or equal to every number in the set....?
a reasonable guess, can you prove it?
 
  • #21
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1 would be greater than or equal to every number in the set....?
Yes. Why are you uncertain?

a reasonable guess, can you prove it?
Although not required in the problem, Deveno's question is reasonable, given the uncertainty in the answer above.
 
  • #22
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Yes. Why are you uncertain?

Although not required in the problem, Deveno's question is reasonable, given the uncertainty in the answer above.
All I know is that |1- 1/n| < 1 but I dont know to prove its 1.
 
  • #23
Deveno
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|1 - 1/n| is never 1.

but...perhaps there might be a least upper bound. what are our likely candidates for this office?
 
  • #24
I like Serena
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What is:
[tex]\lim\limits_{n \to \infty} (1 - {1 \over n})[/tex]
 
  • #25
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What is:
[tex]\lim\limits_{n \to \infty} (1 - {1 \over n})[/tex]
IF n tends to infinity then the modulus tends to 1 but never 1...so does that mean that
||1- 1/n||_∞=sup |1-1/n|=1?
 

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