Are x, y, and z elements of the normed linear space ℓ∞(ℝ)?

[bugatti79]

Homework Statement

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

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

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

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

For a) Does that mean are x,y,z subspaces of \ell_\infty \mathbb({R})?

 

[Mark44]

Homework Statement

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

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

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

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

For a) Does that mean are x,y,z subspaces of \ell_\infty \mathbb({R})?

No, x, y, and z are sequences. The question is asking whether they belong to \ell_\infty \mathbb({R}).

 

[bugatti79]

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?

 

[Mark44]

You could answer b by evaluating x + y, and then saying whether it is in l_∞(R).

 

[bugatti79]

1. Homework Statement

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

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...?

 

[Mark44]

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, 2^1/2y = 2^1/2{1/n}. That means 2^1/2 times each element of the sequence {1/n}.

 

[bugatti79]

No, 2^1/2y = 2^1/2{1/n}. That means 2^1/2 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?

 

[Mark44]

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.

 

[I like Serena]

Hmm, I get a different result for x+y...

 

[Mark44]

Hmm, I get a different result for x+y...

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, ...}

 

[bugatti79]

Homework Statement

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

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

For x:

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

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

Thanks

 

[Mark44]

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

I believe that this is the norm to be used in your problem.

 

[bugatti79]

See http://en.wikipedia.org/wiki/Lp_space, in the section titled L^p 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|?

 

[Mark44]

What is the supremum of the set {0, 1/2, 2/3, 3/4, ..., n/(n + 1), ...}?

 

[bugatti79]

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

|1-1/n|=\sqrt{(1-1/n)^2}=0 where n=1

 

[Mark44]

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

|1-1/n|=\sqrt{(1-1/n)^2}=0 where n=1

 

[bugatti79]

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...?

 

[Mark44]

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?

 

[bugatti79]

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...?

 

[Deveno]

1 would be greater than or equal to every number in the set...?

a reasonable guess, can you prove it?

 

[Mark44]

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.

 

[bugatti79]

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.

 

[Deveno]

|1 - 1/n| is never 1.

but...perhaps there might be a least upper bound. what are our likely candidates for this office?

 

[I like Serena]

What is:
\lim\limits_{n \to \infty} (1 - {1 \over n})

 

[bugatti79]

What is:
\lim\limits_{n \to \infty} (1 - {1 \over n})

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?

 

[I like Serena]

I don't understand your uncertainty either.

I'm just going to say yes, it is 1.

 

[Deveno]

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:
\lim_{n \to \infty} f(n) = L
mean?

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

 

[bugatti79]

what does:
\lim_{n \to \infty} f(n) = L
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.

 

[Mark44]

what does:
\lim_{n \to \infty} f(n) = L
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.

You didn't answer Deveno's question. He asked, what is sup{f(n)}?

 

[bugatti79]

You didn't answer Deveno's question. He asked, what is sup{f(n)}?

Is it L?

 

[I like Serena]

Is it L?

Would you be willing to change that to:

It is L!​

?

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

 

[bugatti79]

Would you be willing to change that to:

It is L!​

?

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

I don't really get "functional analysis"...but I will battle on anyway :-) Thanks

 

[Mark44]

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||_∞ = \lim_{n \to \infty}(1 - 1/n) = 1.