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

In summary: 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.Yes, that looks good. Just be careful with your notation - instead of using ellipses (the dot dot dot), it's better to use summation notation to indicate that there are more terms in the sequence. So instead of x1 + x2 + ... + xn + 0 + 0 + ..., you would write Σi=1n xi + Σi=n+1∞ 0. Does that make sense?For part b, you want to show that || ||_
  • #1
bugatti79
794
1

Homework Statement



a) Prove that [itex]\ell_\infty \mathbb({R})[/itex] is a subspace of [itex]\ell \mathbb({R})
[/itex]

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

The Attempt at a Solution



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

but I don't know how to proceed...thanks
 
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  • #2
bugatti79 said:

Homework Statement



a) Prove that [itex]\ell_\infty \mathbb({R})[/itex] is a subspace of [itex]\ell \mathbb({R})
[/itex]

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

The Attempt at a Solution



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

but I don't know how to proceed...thanks
Start by assuming that x and y are in [itex]\ell_\infty \mathbb({R})[/itex], 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.
 
  • #3
Mark44 said:
Start by assuming that x and y are in [itex]\ell_\infty \mathbb({R})[/itex], 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 [itex] \ell_\infty \mathbb({R})=\left \{ \vec x=(x_n), \vec y=(y_n) \in \ell_\infty \right \}[/itex]

If x and y are vectors in [itex]\ell_\infty[/itex] then x+y is also in [itex]\ell_\infty[/itex]
If [itex]\alpha \in \mathbb{R}[/itex] then [itex]\alpha \vec x[/itex] is also in [itex]\ell_\infty[/itex]...?
 
  • #4
bugatti79 said:
If we let [itex] \ell_\infty \mathbb({R})=\left \{ \vec x=(x_n), \vec y=(y_n) \in \ell_\infty \right \}[/itex]
No, how is [itex] \ell_\infty (R)[/itex] defined?
bugatti79 said:
If x and y are vectors in [itex]\ell_\infty[/itex] then x+y is also in [itex]\ell_\infty[/itex]
If [itex]\alpha \in \mathbb{R}[/itex] then [itex]\alpha \vec x[/itex] is also in [itex]\ell_\infty[/itex]...?
 
  • #5
Mark44 said:
No, how is [itex] \ell_\infty (R)[/itex] defined?

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

[itex]\ell_\infty \mathbb({R})=\left \{ \vec x=(x_n), \vec y=(y_n) \in \ell_\infty \mathbb({R}) \right \}[/itex]
 
  • #6
if {xn} is a bounded sequence and {yn} is a bounded sequence,

is {xn+yn} a bounded sequence?
 
  • #7
Deveno said:
if {xn} is a bounded sequence and {yn} is a bounded sequence,

is {xn+yn} a bounded sequence?

Yes, I believe so...
 
  • #8
Mark44 said:
No, how is [itex] \ell_\infty (R)[/itex] defined?

Mark asked a good question. This is something I don't have in my notes...?
 
  • #9
(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.
 
  • #10
Deveno said:
if {xn} is a bounded sequence and {yn} is a bounded sequence,

is {xn+yn} a bounded sequence?

bugatti79 said:
Yes, I believe so...

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

Now, how is "bounded sequence" defined?
 
  • #11
Let [itex]\ell_\infty \mathbb({R})[/itex] be the set of bounded real sequences with k > 0 such that [itex]\left | x_n \right |\le k[/itex]

this is from another post of yours. it has the information you need.
 
  • #12
Mark44 said:
Which is what you need to show. Also that {axn} is a bounded sequence.

Now, how is "bounded sequence" defined?

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

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

[itex]\vec x = (x_1,x_2,...x_{n1},0,0...)[/itex] and [itex]\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...)[/itex] where [itex]k \ge |x_n|, |y_n|[/itex]

[itex]\alpha \vec x = (\alpha x_1, \alpha x_2,...\alpha x_n,0,0..)[/itex] where

[itex]\alpha \in \mathbb{R}[/itex]...?
 
  • #13
bugatti79 said:
Let [itex]\ell_\infty \mathbb({R})[/itex] be the set of bounded real sequences with k > 0 such that [itex]\left | x_n \right |\le k[/itex]

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

[itex]\vec x = (x_1,x_2,...x_{n1},0,0...)[/itex] and [itex]\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...)[/itex] where [itex]k \ge |x_n|, |y_n|[/itex]

[itex]\alpha \vec x = (\alpha x_1, \alpha x_2,...\alpha x_n,0,0..)[/itex] where

[itex]\alpha \in \mathbb{R}[/itex]...?

If every element xi in the sequence satisfies |xi| <= k, what can you say about αxi?
 
  • #14
I am not sure...it is possible that [itex]|\alpha x_i|>=k[/itex]...?

I don't see any constraint which states that [itex]\alpha x_i[/itex] cannot be >= to k...
 
  • #15
bugatti79 said:
I am not sure...it is possible that [itex]|\alpha x_i|>=k[/itex]...?
Sure, that's possible, but it's not very relevant.

Can't you show that |αxi| is <= some other constant?
bugatti79 said:
I don't see any constraint which states that [itex]\alpha x_i[/itex] cannot be >= to k...
 
  • #16
Mark44 said:
Sure, that's possible, but it's not very relevant.

Can't you show that |αxi| 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.
 
  • #17
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.
αxi
πr2
∫x2dx
 
  • #18
bugatti79 said:
Let [itex]\ell_\infty \mathbb({R})[/itex] be the set of bounded real sequences with k > 0 such that [itex]\left | x_n \right |\le k[/itex]

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

[itex]\vec x = (x_1,x_2,...x_{n1},0,0...)[/itex] and [itex]\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...)[/itex] where [itex]k \ge |x_n|, |y_n|[/itex]

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
 
  • #19
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.

bugatti79 said:
...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.
 
  • #20
Mark44 said:
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)
 
  • #21
Mark44 said:
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...?
 
  • #23
bugatti79 said:
1. Homework Statement

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



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)...?
 
  • #24
Have I covered all axioms correctly in this? I could not find any other axiom regarding infinity...?
 
  • #25
bugatti79 said:
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 lp spaces (including l).
There are other wiki articles on Lp spaces and norms in general.
 
  • #26
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.
 
  • #27
Mark44 said:
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 lp spaces (including l).
There are other wiki articles on Lp 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
 

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

What is a subspace?

A subspace is a subset of a vector space that satisfies all the properties of a vector space, including closure under addition and scalar multiplication.

How do you prove that a subset is a subspace?

To prove that a subset is a subspace, you must show that it satisfies all the properties of a vector space, such as closure under addition and scalar multiplication.

What is the definition of a norm?

A norm is a mathematical function that assigns a positive length or size to every vector in a vector space, satisfying certain properties such as positivity, homogeneity, and the triangle inequality.

How do you prove that a norm is valid?

To prove that a norm is valid, you must show that it satisfies the properties of a norm, such as positivity, homogeneity, and the triangle inequality.

What is the $\ell_\infty (\mathbb{R})$ space?

The $\ell_\infty (\mathbb{R})$ space is the set of all bounded sequences of real numbers, with the norm defined as the maximum absolute value of the elements in the sequence. It is a normed vector space that satisfies all the properties of a vector space and a norm.

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