Question

Let \( a, b, c \in \mathbb{R} \) such that \( a < b < c. \) Which of the following is/are true for any continuous function \( f : \mathbb{R} \to \mathbb{R} \) satisfying \( f(a) = b, f(b) = c \) and \( f(c) = a? \)

• A. There exists \( \alpha \in (a, c) \) such that \( f(\alpha) = \alpha \)
• B. There exists \( \beta \in (a, b) \) such that \( f(\beta) = \beta \)
• C. There exists \( \gamma \in (a, b) \) such that \( (f \circ f)(\gamma) = \gamma \)
• D. There exists \( \delta \in (a, c) \) such that \( (f \circ f \circ f)(\delta) = \delta \)

Hint:

If a function cyclically permutes three distinct points, the third iterate must have a fixed point due to continuity and the Intermediate Value Theorem.

Answer 1

The Correct Option is A, C, D

Step 1: Apply the Intermediate Value Theorem (IVT).
Given \( f \) is continuous and maps \( a \to b, b \to c, c \to a \). Hence, \( f : [a, c] \to [a, c] \).

Step 2: Define compositions of \( f \).
We have \( f(a) = b, f(b) = c, f(c) = a. \) Compute \( f \circ f \circ f \): \[ (f \circ f \circ f)(a) = f(f(f(a))) = f(f(b)) = f(c) = a. \] Thus, \( f \circ f \circ f(a) = a \) and similarly for \( b \) and \( c \), it permutes cyclically.

Step 3: Use fixed-point theorem.
Since \( f \circ f \circ f \) is continuous and maps the closed interval \([a, c]\) into itself, by the Intermediate Value Theorem, there must exist at least one point \( \delta \in (a, c) \) such that \[ (f \circ f \circ f)(\delta) = \delta. \]

Step 4: Analyze other options.
- (A) and (B): \( f \) need not have a fixed point directly because it cyclically shifts the interval. - (C): \( f \circ f \) may not fix any point since it maps \( a \to c \to a \). Hence, only (D) must be true.

Final Answer: Option (D).