Normed Linear Space Part 1

In summary, The question is about the normed linear space \ell_\infty \mathbb({R}) and the elements x, y, and z in this space. It is asked whether x, y, z are each in \ell_\infty \mathbb({R}) and how to calculate their infinity norms. It is also asked to evaluate x+y and 2^{1/2}y. In order to calculate the infinity norm, the supremum of the set needs to be found. For x, the supremum is the smallest number that is greater than or equal to every number in the sequence {0, 1/2, 2/3, 3/4, ..., n/(n+
  • #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]?
 
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  • #2
bugatti79 said:

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
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
You could answer b by evaluating x + y, and then saying whether it is in l(R).
 
  • #5
bugatti79 said:
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
bugatti79 said:
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, ...}
bugatti79 said:
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
Mark44 said:
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
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
Hmm, I get a different result for x+y... :confused:
 
  • #10
I like Serena said:
Hmm, I get a different result for x+y... :confused:

Mark44 said:
Yes, but here's how to say the first part better. x + y = {n} = {1, 2, 3, ...}


Let's make that x + y = {1} = {1, 1, 1, ...}
 
  • #11
bugatti79 said:

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 don't know how to calculate |x_n|

Thanks
 
  • #13
Mark44 said:
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.

THanks Mark but how to I evalute |x_n|?
 
  • #14
What is the supremum of the set {0, 1/2, 2/3, 3/4, ..., n/(n + 1), ...}?
 
  • #15
Mark44 said:
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
 
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  • #16
bugatti79 said:
I think the supremum is the smallest real number that is greater than or equal to every number in the set,
Yes.
bugatti79 said:
hence t must be 0...?
No. 0 is smaller than, not larger than every number in the sequence except the first.
bugatti79 said:
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
Mark44 said:
Yes.No. 0 is smaller than, not larger than every number in the sequence except the first.
Then it doesn't exist, there is no sup...?
 
  • #18
bugatti79 said:
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), ...}
bugatti79 said:
, 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
Mark44 said:
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
bugatti79 said:
1 would be greater than or equal to every number in the set...?

a reasonable guess, can you prove it?
 
  • #21
bugatti79 said:
1 would be greater than or equal to every number in the set...?
Yes. Why are you uncertain?

Deveno said:
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
Mark44 said:
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 don't know to prove its 1.
 
  • #23
|1 - 1/n| is never 1.

but...perhaps there might be a least upper bound. what are our likely candidates for this office?
 
  • #24
What is:
[tex]\lim\limits_{n \to \infty} (1 - {1 \over n})[/tex]
 
  • #25
I like Serena said:
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?
 
  • #26
I don't understand your uncertainty either.

I'm just going to say yes, it is 1.
 
  • #27
bugatti79 said:
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?

what does:
[tex]\lim_{n \to \infty} f(n) = L[/tex]
mean?

if you have such a limit, and, you know that f(n) < L for all n, what must sup{f(n)} be?
 
  • #28
Deveno said:
what does:
[tex]\lim_{n \to \infty} f(n) = L[/tex]
mean?

if you have such a limit, and, you know that f(n) < L for all n, what must sup{f(n)} be?

That L is a finte real number and in this case it is a 1.
 
  • #29
Deveno said:
what does:
[tex]\lim_{n \to \infty} f(n) = L[/tex]
mean?

if you have such a limit, and, you know that f(n) < L for all n, what must sup{f(n)} be?

bugatti79 said:
That L is a finte real number and in this case it is a 1.
You didn't answer Deveno's question. He asked, what is sup{f(n)}?
 
  • #30
Mark44 said:
You didn't answer Deveno's question. He asked, what is sup{f(n)}?

Is it L?
 
  • #31
bugatti79 said:
Is it L?

Would you be willing to change that to:

It is L!​

?

(Or else please clarify what is still puzzling you. :confused:)
 
  • #32
I like Serena said:
Would you be willing to change that to:

It is L!​

?

(Or else please clarify what is still puzzling you. :confused:)

I don't really get "functional analysis"...but I will battle on anyway :-) Thanks
 
  • #33
So to go back to what you were trying to figure out much earlier in this thread, the infinity norm of x, where x = {1 - 1/n}, ||x|| = [itex]\lim_{n \to \infty}(1 - 1/n)[/itex] = 1.
 

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