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: Analyzing each option:

(A) There exists $\alpha \in (a,c)$ such that $f(\alpha) = \alpha$

Define $g(x) = f(x) - x$. Then:

• $g(a) = f(a) - a = b - a > 0$
• $g(c) = f(c) - c = a - c < 0$

By the Intermediate Value Theorem, since $g$ is continuous and changes sign on $[a,c]$, there exists $\alpha \in (a,c)$ such that $g(\alpha) = 0$, meaning $f(\alpha) = \alpha$.

TRUE

(B) There exists $\beta \in (a,b)$ such that $f(\beta) = \beta$

Consider $g(x) = f(x) - x$ on $[a,b]$:

• $g(a) = b - a > 0$
• $g(b) = c - b > 0$

Both values are positive, so we cannot guarantee a sign change in $(a,b)$. The Intermediate Value Theorem doesn't apply.

NOT NECESSARILY TRUE

(C) There exists $\gamma \in (a,b)$ such that $(f \circ f)(\gamma) = \gamma$

Define $h(x) = f(f(x)) - x$. Then:

• $h(a) = f(f(a)) - a = f(b) - a = c - a > 0$
• $h(b) = f(f(b)) - b = f(c) - b = a - b < 0$

By the Intermediate Value Theorem, since $h$ is continuous and changes sign on $[a,b]$, there exists $\gamma \in (a,b)$ such that $h(\gamma) = 0$, meaning $(f \circ f)(\gamma) = \gamma$.

TRUE

(D) There exists $\delta \in (a,c)$ such that $(f \circ f \circ f)(\delta) = \delta$

Define $k(x) = f(f(f(x))) - x$. Note that:

• $f(f(f(a))) = f(f(b)) = f(c) = a$, so $k(a) = a - a = 0$

This means $\delta = a$ is already a fixed point of $f^3$. However, the question asks for $\delta \in (a,c)$ (open interval).

Let's check other values:

• $k(a) = 0$
• $k(c) = f(f(f(c))) - c = f(f(a)) - c = f(b) - c = c - c = 0$

So both endpoints are fixed points. By continuity, if $k$ changes sign in $(a,c)$, there's another fixed point. Even if $k$ doesn't change sign, since $k(a) = 0$, and by continuity considerations, there must exist $\delta \in (a,c)$ such that $k(\delta) = 0$.

Actually, more rigorously: Consider that $f(a) = b \in (a,c)$, $f(b) = c$, $f(c) = a$. The cycle $a \to b \to c \to a$ ensures that $f^3$ has fixed points throughout the interval.

TRUE 

Answer: (A), (C), and (D) are true