# Normed Linear Space Part 1

## 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
Mentor

## 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 $\ell_\infty \mathbb({R})$.

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

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
Mentor
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}.

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?

Mark44
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.

I like Serena
Homework Helper
Hmm, I get a different result for x+y...

Mark44
Mentor
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, ...}

## 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 dont know how to calculate |x_n|

Thanks

Mark44
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.

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|?

Mark44
Mentor
What is the supremum of the set {0, 1/2, 2/3, 3/4, ..., n/(n + 1), ...}?

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

Last edited:
Mark44
Mentor
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

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...?

Mark44
Mentor
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?

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
Mentor
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.

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.

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

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
Homework Helper
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?

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
Mentor
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.