Proving Subspace & Norm on $\ell_\infty (\mathbb{R})$

[bugatti79]

Homework Statement

a) Prove that \ell_\infty \mathbb({R}) is a subspace of \ell \mathbb({R})<br />

b) Show that \left \| \right \|_\infty is a norm on \ell_\infty (\mathbb{R})

The Attempt at a Solution

For a) I guess we have to show that \vec{x} + \vec{y} \in \ell_\infty \mathbb({R}) and \alpha \vec({x}) \in \ell_\infty \mathbb({R})

but I don't know how to proceed...thanks

 

[Mark44]

Homework Statement

a) Prove that \ell_\infty \mathbb({R}) is a subspace of \ell \mathbb({R})<br />

b) Show that \left \| \right \|_\infty is a norm on \ell_\infty (\mathbb{R})

The Attempt at a Solution

For a) I guess we have to show that \vec{x} + \vec{y} \in \ell_\infty \mathbb({R}) and \alpha \vec{x} \in \ell_\infty \mathbb({R})

but I don't know how to proceed...thanks

Start by assuming that x and y are in \ell_\infty \mathbb({R}), and say what it means for a vector to be in that space. Then add the vectors together. Is the sum in that space as well? Same thing for the scalar multiplication part.

 

[bugatti79]

Start by assuming that x and y are in \ell_\infty \mathbb({R}), and say what it means for a vector to be in that space. Then add the vectors together. Is the sum in that space as well? Same thing for the scalar multiplication part.

If we let \ell_\infty \mathbb({R})=\left \{ \vec x=(x_n), \vec y=(y_n) \in \ell_\infty \right \}

If x and y are vectors in \ell_\infty then x+y is also in \ell_\infty
If \alpha \in \mathbb{R} then \alpha \vec x is also in \ell_\infty...?

 

[Mark44]

If we let \ell_\infty \mathbb({R})=\left \{ \vec x=(x_n), \vec y=(y_n) \in \ell_\infty \right \}

No, how is \ell_\infty (R) defined?

If x and y are vectors in \ell_\infty then x+y is also in \ell_\infty
If \alpha \in \mathbb{R} then \alpha \vec x is also in \ell_\infty...?

 

[bugatti79]

No, how is \ell_\infty (R) defined?

This is the only definition I have in my notes with the addition of inserting the vector y

\ell_\infty \mathbb({R})=\left \{ \vec x=(x_n), \vec y=(y_n) \in \ell_\infty \mathbb({R}) \right \}

 

[Deveno]

if {x_n} is a bounded sequence and {y_n} is a bounded sequence,

is {x_n+y_n} a bounded sequence?

 

[bugatti79]

if {x_n} is a bounded sequence and {y_n} is a bounded sequence,

is {x_n+y_n} a bounded sequence?

Yes, I believe so...

 

[bugatti79]

No, how is \ell_\infty (R) defined?

Mark asked a good question. This is something I don't have in my notes...?

 

[Deveno]

(excerpt from wikipedia): If p = ∞, then ℓ_∞ is defined to be the space of all bounded sequences (in K, the range of the sequences).

thus ℓ_∞(R) is the space of all bounded real sequences.

 

[Mark44]

if {x_n} is a bounded sequence and {y_n} is a bounded sequence,

is {x_n+y_n} a bounded sequence?

Yes, I believe so...

Which is what you need to show. Also that {ax_n} is a bounded sequence.

Now, how is "bounded sequence" defined?

 

[Deveno]

Let \ell_\infty \mathbb({R}) be the set of bounded real sequences with k > 0 such that \left | x_n \right |\le k

this is from another post of yours. it has the information you need.

 

[bugatti79]

Which is what you need to show. Also that {ax_n} is a bounded sequence.

Now, how is "bounded sequence" defined?

Let \ell_\infty \mathbb({R}) be the set of bounded real sequences with k > 0 such that \left | x_n \right |\le k

if \vec x, \vec y \in \ell_\infty \mathbb({R}) then there exist n_1, n_2 \in N such that

\vec x = (x_1,x_2,...x_{n1},0,0...) and \vec y = (y_1,y_2,...y_{n2},0,0...) \therefore \vec x +\vec y= (x_1+y_1, x_2+y_2...x_n+y_n,0,0...) where k \ge |x_n|, |y_n|

\alpha \vec x = (\alpha x_1, \alpha x_2,...\alpha x_n,0,0..) where

\alpha \in \mathbb{R}...?

 

[Mark44]

Let \ell_\infty \mathbb({R}) be the set of bounded real sequences with k > 0 such that \left | x_n \right |\le k

if \vec x, \vec y \in \ell_\infty \mathbb({R}) then there exist n_1, n_2 \in N such that

\vec x = (x_1,x_2,...x_{n1},0,0...) and \vec y = (y_1,y_2,...y_{n2},0,0...) \therefore \vec x +\vec y= (x_1+y_1, x_2+y_2...x_n+y_n,0,0...) where k \ge |x_n|, |y_n|

\alpha \vec x = (\alpha x_1, \alpha x_2,...\alpha x_n,0,0..) where

\alpha \in \mathbb{R}...?

If every element x_i in the sequence satisfies |x_i| <= k, what can you say about αx_i?

 

[bugatti79]

I am not sure...it is possible that |\alpha x_i|&gt;=k...?

I don't see any constraint which states that \alpha x_i cannot be >= to k...

 

[Mark44]

I am not sure...it is possible that |\alpha x_i|&gt;=k...?

Sure, that's possible, but it's not very relevant.

Can't you show that |αx_i| is <= some other constant?

I don't see any constraint which states that \alpha x_i cannot be >= to k...

 

[bugatti79]

Sure, that's possible, but it's not very relevant.

Can't you show that |αx_i| is <= some other constant?

Based on the question asked and all the info I have in #12, I don't see any other constant at play...
I wouldn't understand how or when one would look at some 'other' constant.

 

[Mark44]

If |x| < 3, then certainly 2|x| < 6, right?

On a side note, could you ease up a bit on the LaTeX, especially for symbols that don't actually require it? These threads with lots of LaTeX take a long time to render on my browser. Many of the things that you write can be done using the Quick Symbols that appear on the right after you click Go Advanced.

Everything below is done without using LaTeX.
αx_i
πr²
∫x²dx

 

[bugatti79]

Let \ell_\infty \mathbb({R}) be the set of bounded real sequences with k > 0 such that \left | x_n \right |\le k

if \vec x, \vec y \in \ell_\infty \mathbb({R}) then there exist n_1, n_2 \in N such that

\vec x = (x_1,x_2,...x_{n1},0,0...) and \vec y = (y_1,y_2,...y_{n2},0,0...) \therefore \vec x +\vec y= (x_1+y_1, x_2+y_2...x_n+y_n,0,0...) where k \ge |x_n|, |y_n|

OK, thanks for the advice on LaTex.

So yes, that example makes sense...based on that the vector x is closed under multiplication. How about my attempt for addition as above? Hopefully I have a) answered.

For part b) I have to show that || ||_∞ satifies 4 specific axioms, right?

Thanks

 

[Mark44]

In the previous post, #18, why are you working with finite sequences? Addition in l_∞ is term-by-term. All you have to show is that, if x and y are bounded sequences, then x + y is also a bounded sequence.

...based on that the vector x is closed under multiplication...

No, vectors aren't closed under scalar multiplication - the set that they belong to, l_∞(R) in this case, is closed under scalar multiplication.

For the b part, verify the properties in the definition of a norm.

 

[bugatti79]

In the previous post, #18, why are you working with finite sequences? Addition in l_∞ is term-by-term. All you have to show is that, if x and y are bounded sequences, then x + y is also a bounded sequence.

I don't know how to write it any other way...

x, y ε l_∞(R)

 

[bugatti79]

In the previous post, #18, why are you working with finite sequences? Addition in l_∞ is term-by-term. All you have to show is that, if x and y are bounded sequences, then x + y is also a bounded sequence.

Ok. Since we know that x and y are bounded sequences then all I need to show is that

if x_n=(x_1,x_2,x-3...) and y_n=(y_1,y_2...) then x_n+y_n=(x_1+y_1, x_2+y_2...) Hence the set l∞(R) is closed under addition...?

 

[Mark44]

Yes, that's really all you need to say. Vector addition and scalar multiplication in l_∞(R) are the same as in l(R).

Here's the link to the wiki page that deveno mentioned yesterday - http://en.wikipedia.org/wiki/Bounded_sequences.

 

[bugatti79]

1. Homework Statement

b) Show that \left \| \right \|_\infty is a norm on \ell_\infty (\mathbb{R})

This attempt is based on a similar example in (R^3, || ||)

if x_n and y_n are each bound sequences with |x_n| >=0 for n=1,2,3 and similarly for y_n, then

axiom 1: ||x_n||∞= |x_1|+|x_2|+|x_3|... >=0 THis axiom holds.

Axiom 2: ||x_n||∞=0 IFF |x_1+|x_2|+|x_3|=0, ie each x_n=0

Axiom 3: ||ax_n||∞= |ax_1|+|ax_2|+|ax_3|
= |a|(x_1+x_2+x_3)
= |a| |x_n|∞


Axiom 4: ||x_n+y_n||∞= |x_1+y_1| +| x_2+y_2|

||x_n+y_n||∞<= |x_1|+|x_2|+|y_1|+|y_2| therefore

||x_n+y_n||∞<= ||x_n||∞ + ||y_n||∞

All 4 axioms hold therefore || ||∞ is a norm on l∞(R)...?

 

[bugatti79]

Have I covered all axioms correctly in this? I could not find any other axiom regarding infinity...?

 

[Mark44]

This attempt is based on a similar example in (R^3, || ||)

if x_n and y_n are each bound sequences with |x_n| >=0 for n=1,2,3 and similarly for y_n, then

axiom 1: ||x_n||∞= |x_1|+|x_2|+|x_3|... >=0 THis axiom holds.

Axiom 2: ||x_n||∞=0 IFF |x_1+|x_2|+|x_3|=0, ie each x_n=0

Axiom 3: ||ax_n||∞= |ax_1|+|ax_2|+|ax_3|
= |a|(x_1+x_2+x_3)
= |a| |x_n|∞


Axiom 4: ||x_n+y_n||∞= |x_1+y_1| +| x_2+y_2|

||x_n+y_n||∞<= |x_1|+|x_2|+|y_1|+|y_2| therefore

||x_n+y_n||∞<= ||x_n||∞ + ||y_n||∞

All 4 axioms hold therefore || ||∞ is a norm on l∞(R)...?

You haven't actually used the definition of the infinity norm (|| ||_∞) anywhere. The norm you seem to be using is the taxicab norm, not the infinity norm. The space here is bounded sequences, so each vector x in the space is an infinite sequence, not just a point in R³.

See (again) http://en.wikipedia.org/wiki/Bounded_sequences for a description of norms in l^p spaces (including l^∞).
There are other wiki articles on L^p spaces and norms in general.

 

[Deveno]

the "infinity norm" is the supremum (least upper bound) of the absolute value of the individual terms. since our sequences are bounded, the set of absolute values of the terms forms a set of real numbers, which is bounded above (since otherwise the sequence would either be unbounded above, or below, and we aren't considering such sequences).

a set of real numbers bounded from above, has a supremum (this is an intrinsic property of real numbers).

the usual norm, for FINITE sequences of a given length, is what you would call a "2-norm", that is we consider the norm to be the square root of the sum of the squares of the terms. this corresponds to our usual notion of distance in a finite-dimensional euclidean space. but this doesn't work for an infinite number of terms, because the terms may not "decrease fast enough" (those that do are rather special, forming an l₂-space).

the reason sup is used, instead of max, is because we might have a sequence like:

{(2ⁿ- 1)/2ⁿ} which is:

(1/2, 3/4, 7/8, 15/16, 31/32, 63/64,...)

it's pretty clear this sequence is bounded below by 1/2, and bounded above by 1, but there is no "maximum" term.

 

[bugatti79]

You haven't actually used the definition of the infinity norm (|| ||_∞) anywhere. The norm you seem to be using is the taxicab norm, not the infinity norm. The space here is bounded sequences, so each vector x in the space is an infinite sequence, not just a point in R³.

See (again) http://en.wikipedia.org/wiki/Bounded_sequences for a description of norms in l^p spaces (including l^∞).
There are other wiki articles on L^p spaces and norms in general.

I don't know how to prove the 4 axioms for the infinity norm, there is no example in my notes...any suggestions?

Thanks