Convergence and Continuity of Cauchy Sequences with Fixed Points

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    Cauchy Sequence
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Homework Help Overview

The discussion revolves around the continuity of a function defined on a closed interval and the convergence of a sequence generated by iterating this function. The function satisfies a contraction mapping condition, which is central to the problem.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the implications of the contraction mapping condition for proving continuity. There are attempts to establish the convergence of the sequence defined by iterating the function, with some participants questioning how to demonstrate that the sequence is Cauchy.

Discussion Status

Some participants provide initial reasoning and guidance on continuity and convergence, while others express uncertainty and seek further clarification on specific steps. Multiple interpretations of the problem are being explored, particularly regarding the application of the function's properties.

Contextual Notes

There is a noted emphasis on making individual efforts in solving the homework problem, with reminders that assistance will follow once attempts are made. Participants are also navigating through the definitions and properties relevant to the function and sequence in question.

manooba
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Let f : [a,b] → [a,b] satisfy

|f(x)-f(y)| ≤ λ|x-y|

where 0<λ<1. Prove f is continuous. Choose any Xo ε [a,b] and for n ≥ 1 define X_n+1 = f(Xn). Prove that the sequence (Xn) is convergent and that its limit L is a 'fixed point' of f, namely f(L)=L
 
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The continuity part seems pretty straightforward, unless I'm mistaken. Let c be an arbitraty point in [a, b]. Let ε > 0 be given. Let δ = ε. Them for all x such that |x - c| < δ, we have: |f(x) - f(c)| <= ...
 
radou said:
The continuity part seems pretty straightforward, unless I'm mistaken. Let c be an arbitraty point in [a, b]. Let ε > 0 be given. Let δ = ε. Them for all x such that |x - c| < δ, we have: |f(x) - f(c)| <= ...

reckon you can help me further please i am really struggling
 
manooba said:
reckon you can help me further please i am really struggling

No. You must make your own efforts on homework problems. We are not here to solve the questions for you. We will help you as soon as you make an attempt.
 
Well, you initially begin with
\left|x_2 - x_1\right|=\left|f(x_1) - f(x_0)\right|\le \lambda\left|x_1 - x_0\right|
if you take one more step down the sequence, you can see that
\left|x_3 - x_2\right|=\left|f(x_2) - f(x_1)\right|\le \lambda\left|x_2 - x_1\right|\le \lambda^2\left|x_1 - x_0\right|
I don't think it's difficult to see how this generalizes.

Can you see how this implies the sequence is Cauchy? What do we know about Cauchy sequences? Finally, what do we know about the limit of a sequence in a continuous function?
 
And for the first part, can you continue where I wrote "..."? What does your function satisfy, by definition?
 
radou said:
And for the first part, can you continue where I wrote "..."? What does your function satisfy, by definition?

sorry the answers no :/
satisfy's |f(x)-f(y)| ≤ λ|x-y) where 0<λ<1
 
Last edited:
manooba said:
satisfy's |f(x)-f(y)| ≤ λ|x-y) where 0<λ<1

OK. Now, we're looking at all x such that |x - c| < δ, right? So, apply your function property to these x's...
 
radou said:
OK. Now, we're looking at all x such that |x - c| < δ, right? So, apply your function property to these x's...

ok i get that, so what do i plug into the x's?
 
  • #10
I already gave you the initial push in post #2 after "we have:". You only need to adjust your function inequality a bit now.
 
  • #11
lknkjn
 

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