Qn : Does a continuous function always have a fixed point in [0, 1]?

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SUMMARY

The discussion centers on the existence of a fixed point for continuous functions f and g mapping from the interval [0, 1] onto itself. It is established that if f and g are continuous, then there exists an x0 in [0, 1] such that f(g(x0)) = g(f(x0)). The proof involves defining a continuous function h(x) = f(g(x)) - g(f(x)), demonstrating that h(a) ≤ 0 and h(b) ≥ 0 for specific points a and b in [0, 1]. This guarantees a fixed point by the Intermediate Value Theorem.

PREREQUISITES
  • Understanding of continuous functions
  • Familiarity with the Intermediate Value Theorem
  • Knowledge of function composition
  • Basic concepts of fixed points in mathematical analysis
NEXT STEPS
  • Study the Intermediate Value Theorem in detail
  • Explore fixed point theorems, such as Brouwer's Fixed Point Theorem
  • Learn about continuous mappings and their properties
  • Investigate examples of continuous functions on closed intervals
USEFUL FOR

Mathematicians, students of calculus, and anyone interested in the properties of continuous functions and fixed point theory will benefit from this discussion.

ice_kid
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I hope someone can help me wif this qnestion.

Qn : Let f ; g be continuous functions from [0, 1] onto [0, 1]. Prove that there is
x0 ∈ [0, 1] such that f (g(x0)) = g(f (x0)).

Thanks in advance.
 
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ice_kid said:
I hope someone can help me wif this qnestion.

Qn : Let f ; g be continuous functions from [0, 1] onto [0, 1]. Prove that there is
x0 ∈ [0, 1] such that f (g(x0)) = g(f (x0)).

Thanks in advance.

Since both functions are onto [0,1], there are points a and b in [0,1] such that f(g(a)) = 0 and f(g(b)) = 1.

Then we must have
[tex]f(g(a)) - g(f(a)) \leq 0[/tex]
and
[tex]f(g(b)) - g(f(b)) \geq 0[/tex].

The function h defined by [tex]h(x) = f(g(x)) - g(f(x))[/tex] is continuous. So...

Maybe you can take it from here.
 

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