Contraction Mapping Theorem: Proving Continuity and Convergence of a Sequence

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SUMMARY

The discussion focuses on the Contraction Mapping Theorem, specifically proving the continuity and convergence of a sequence defined by a function f on the real numbers R. The theorem states that if there exists a constant c (0 < c < 1) such that |f(x) - f(y)| < c|x - y|, then f is continuous on R. The sequence defined by y_(n+1) = f(y_n) is shown to be a Cauchy sequence, leading to the conclusion that the limit y is a unique fixed point of f. Furthermore, it is demonstrated that for any arbitrary point x in R, the sequence (x, f(x), f(f(x)), ...) converges to this fixed point y.

PREREQUISITES
  • Understanding of the Contraction Mapping Theorem
  • Familiarity with Cauchy sequences
  • Knowledge of continuity in real analysis
  • Proficiency in mathematical notation and terminology
NEXT STEPS
  • Study the implications of the Banach Fixed-Point Theorem
  • Explore examples of contraction mappings in various functions
  • Learn about the relationship between Cauchy sequences and completeness in metric spaces
  • Investigate the uniqueness of fixed points in different mathematical contexts
USEFUL FOR

Mathematicians, students of real analysis, and anyone interested in understanding fixed-point theorems and their applications in various fields such as numerical methods and optimization.

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Homework Statement



Let f be a function defined on all of R and assume there is a constant c such that 0<c<1 and |f(x)-f(y)<c|c-y|
a) Show f is continuous on all of R
b)Pick some point y1 in R and construct the sequence (y1,f(y1),f(f(y1)),...)
In general if y_(n+1)=f(yn) show that the resulting sequence yn is a Cauchy sequence. hence we may let y=limyn
c)Prove that y is a fixed point of f and that is unique in this regard.
d) Finally prove that if x is any arbitrary point in R then the sequence (x,f(x),f(f(x)),...) converges to y defined in (b).

Homework Equations



a) want to show if |x-c|<delta then |f(x)-f(c)|<epsilon
b) A sequence is Cauchy if |an-am|<epsilon
c)want to show f(y)=y


The Attempt at a Solution

 
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Take a). Your "want to show if |x-c|<delta then |f(x)-f(c)|<epsilon" is not precise enough. There will always be some epsilon. In mathematics the terms and the order of terms "For any", "there exists", "such that" are of crucial importance.
 

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