Convergence and Continuity of Cauchy Sequences with Fixed Points

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SUMMARY

The discussion centers on the proof of continuity for a function f: [a,b] → [a,b] that satisfies the condition |f(x) - f(y)| ≤ λ|x - y| with 0 < λ < 1. Participants confirm that the sequence defined by X_n+1 = f(X_n) converges and its limit L is a fixed point of f, meaning f(L) = L. The continuity proof involves demonstrating that for any arbitrary point c in [a, b] and ε > 0, the function's properties ensure |f(x) - f(c)| can be bounded appropriately, leading to the conclusion that the sequence is Cauchy.

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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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