Convergence and Continuity of Cauchy Sequences with Fixed Points

  • Thread starter Thread starter manooba
  • Start date Start date
  • Tags Tags
    Cauchy Sequence
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
10 replies · 3K views
manooba
Messages
8
Reaction score
0
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
 
on Phys.org
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
 
Well, you initially begin with
[tex]\left|x_2 - x_1\right|=\left|f(x_1) - f(x_0)\right|\le \lambda\left|x_1 - x_0\right|[/tex]
if you take one more step down the sequence, you can see that
[tex]\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|[/tex]
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?
 
I already gave you the initial push in post #2 after "we have:". You only need to adjust your function inequality a bit now.