Show subspace of normed vector is closed under sup norm.

In summary, the conversation discusses how to show that x=(x^0,x^1,x^2,x^3,...) is in c_0, using the definition of the limit of a sequence and the triangle inequality. The sup norm is also mentioned in relation to showing that x_n is Cauchy. There is a clarification about what x represents, whether it is a sequence or just one sequence.
  • #1
Greger
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http://imageshack.us/a/img141/4963/92113198.jpg [Broken]

hey,

I'm having some trouble with this question,

For part a) I know that in order for c_0 to be closed every sequence in c_0 must converge to a limit in c_0 but I am having trouble actually showing that formally with the use of the norm given.

Say x_n is a sequence in c_0 which converges to x then for any ε>0 there exists an N such that whenever n>N |x_n - x| < ε.

So that's just the definition for the limit of a sequence, but I'm having trouble how to use this to show that the limit x is in c_0 for every x_n in c_0.

Would anyone be able to help me out?

Thanks in advance
 
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  • #2
So, you wish to show that [itex]x=(x^0,x^1,x^2,x^3,...)\in c_0[/itex]. Take [itex]\varepsilon>0[/itex]. You'll need to find an N>0 such that for all n>N holds that [itex]|x^n|<\varepsilon[/itex].

So, what you need to do is make [itex]|x^n|[/itex] smaller than something that is smaller than [itex]\varepsilon[/itex]. What could you make [itex]|x^n|[/itex] smaller as? Hint: use the triangle inequality to introduce the [itex]x_k[/itex].
 
  • #3
Oh so using the trick where you add 0,

|x_n|=|x_n - x_k + x_k|≤ |x_n - x_k|+|x_k|

Then for x_k in c_0,

For ε>0 there is an N > 0 such that for all n>N, |x_k| <ε/4

So

|x_n - x_k|+|x_k| < |x_n - x_k|+ε/4

Then |x_n - x_k| will definitely be smaller then ε so

|x_n - x_k|+ε/4 < ε


I am having trouble seeing where the sup norm comes in,

Have I missed something?
 
  • #4
Does that look ok?

That shows that x_n is Cauchy, since you have

||x_n - x_k|| < 3e/4 < e

for n,k > N = max(N_n, N_K)
 
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  • #5
Do you mean by [tex]x= \{ x_1,x_2,..\} [/tex] that x_1 is a sequence of itself?

So for example [tex]x_1 (n) = 0.5^n, n =1, 2, 3,...[/tex]
[tex]x_2(n) = 0.4^n, n =1, 2, 3,...[/tex]

Or is x just ONE sequence?
 

1. What is "show subspace of normed vector is closed under sup norm" referring to?

This refers to the concept of a subspace of a normed vector space being closed under the supremum norm, also known as the "sup norm".

2. What is a subspace?

A subspace is a subset of a vector space that is itself a vector space, meaning it satisfies the same properties as the original vector space.

3. What is a normed vector space?

A normed vector space is a vector space that has a norm function defined on it, which measures the length or size of a vector.

4. What is the supremum norm?

The supremum norm is a type of norm defined on a vector space, which measures the maximum distance between two points in the space.

5. Why is it important for a subspace to be closed under the supremum norm?

This property ensures that the subspace retains its structure and properties, making it easier to perform calculations and proofs within the subspace. It also allows for the use of the supremum norm as a tool for analysis and optimization within the subspace.

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