Cauchy Convergence in Normed Vector Spaces

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Homework Help Overview

The discussion revolves around the properties of norms in the vector space C[a,b] and the behavior of a specific sequence of functions under these norms. The original poster presents a problem involving the equivalence of norms and the Cauchy convergence of a sequence of functions defined on the interval [-1,1].

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the two norms and the existence of a constant K for the inequalities presented. They discuss the Cauchy sequence properties of the given functions and question the implications of convergence in the context of continuity.

Discussion Status

Some participants have provided insights into the Cauchy sequence properties for the 1-norm, suggesting calculations to demonstrate convergence. Others express uncertainty regarding the behavior of the sequence under the infinity norm and the implications of pointwise limits not being continuous.

Contextual Notes

There is a focus on the definitions of Cauchy convergence and the completeness of C([-1,1]) with respect to the sup-norm. Participants are navigating the constraints of the problem, particularly regarding the continuity of limits and the nature of convergence in normed spaces.

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Homework Statement



Fix a<b in R, and consider the two norms Norm(f)1:=Integralab( Modulus(f)) and Norm(f)Infinity:= sup{Mod(f(x)): a <= x <= b} on the vector space C[a,b]. This question shows that they are not equivalent.
a. Show that there is K in R such that for all f in C[a,b],
Norm(f)1 <= K*Norm(f)Infinity
b. Show that there is no K in R such that for all f in C[a,b],
Norm(f)infinity <= K*Norm(f)1
c. Consider the sequence of functions (fn) in C[-1,1] given by fn(x):=x1/(2n-1)
Is (fn) a cauchy sequence in (C[-1,1], Norm(f)1)? Why?
Is (fn) a cauchy sequence in (C[-1,1], Norm(f)Infinity)? Why?

Homework Equations



Cauchy Convergence Definition:
For all E>0, There exists M(E)>0, in the natural numbers, such that for all m>=n>=M(E), Norm(fm-fn)<E

The Attempt at a Solution



Found the first two parts relatively easy I think, for a taking k to be (b-a) and for b, considering the sequence of functions
fn(x):= { n2-nx for 0<=x<=1/n
0 otherwise
However, for the proof of cauchy sequences I am at a loss, as we cannot use normal convergence in the normed vector space as it converged to something that isn't continuous. So I am trying to use the definition of cauchy convergence, for the first part I have tried splitting up the integral into [0,1] and [-1,0] but have no idea for the second part.
 
Last edited:
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For the 1-norm this is a Cauchy sequence.
You need to prove that [tex]\|f_p-f_q\|_1\rightarrow 0[/tex]as [tex]p,q\rightarrow +\infty[/tex].
So, you'll need to calculate

[tex]\|f_p-f_q\|_1=\int_{-1}^1{|x^{1/(2p-1)}-x^{1/(2q-1)}|}[/tex]

and then prove that this expression goes to 0 as [tex]p,q\rightarrow +\infty[/tex].
 
I somewhat understand the first part yes, however I really cannot see what to do for the second part, as I know it doesn't have cauchy convergence, however, intuitively, the supremum tends to 0 eventually, as given an x, they will both tend to 1 as p,q tends to infinity, and thus the supremum tends to 0 (assuming x isn't 0)
 
C([-1,1]) with the sup-norm is complete, thus all Cauchy convergent sequences converges. But if a sequences converges uniformly, then it converges pointswize. So you need to show that the pointswise limit of your sequence is not continuous.
 

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