Question

Let \( f \) be a continuous function from \( [0, 1] \) to the set of all real numbers. Then which one of the following statements is NOT true?

• A. For any sequence \( \{x_n\}_{n \geq 1} \) in \( [0, 1] \), \( \sum_{n=1}^{\infty} \frac{f(x_n)}{n^2} \) is absolutely convergent
• B. If \( |f(x)| = 1 \) for all \( x \in [0, 1] \), then \( \left| \int_0^1 f(x) dx \right| = 1 \)
• C. If \( \{x_n\}_{n \geq 1} \) is a sequence in \( [0, 1] \) such that \( \{f(x_n)\}_{n \geq 1} \) is convergent, then \( \{x_n\}_{n \geq 1} \) is convergent
• D. If \( f \) is also monotonically increasing, then the image of \( f \) is given by \( [f(0), f(1)] \)

Hint:

- A continuous function on a closed interval maps the interval to another closed interval, but the convergence of \( f(x_n) \) does not guarantee the convergence of \( x_n \).
- Monotonicity helps to determine the image of a function over an interval.

Answer 1

The Correct Option is C

1) Understanding the question:
We are given that \( f \) is continuous, and we are asked to identify which statement is NOT true. Let's examine each option:
2) Analysis of the options:
(A) True: The series \( \sum_{n=1}^{\infty} \frac{f(x_n)}{n^2} \) is absolutely convergent. Since \( n^2 \) grows rapidly, and \( f(x_n) \) is bounded for \( x_n \in [0, 1] \), the series converges absolutely.
(B) True: If \( |f(x)| = 1 \) for all \( x \in [0, 1] \), then the integral \( \int_0^1 f(x) dx \) can be either 1 or -1, depending on the behavior of the function. Therefore, \( \left| \int_0^1 f(x) dx \right| = 1 \).
(C) False: The convergence of the sequence \( \{f(x_n)\}_{n \geq 1} \) does not imply that the sequence \( \{x_n\}_{n \geq 1} \) is convergent. A counterexample can be found where the function \( f \) is continuous but the sequence \( \{x_n\}_{n \geq 1} \) oscillates while still making \( f(x_n) \) converge.
(D) True: If \( f \) is monotonically increasing, then the image of \( f \) on the interval \( [0, 1] \) is given by the interval \( [f(0), f(1)] \). This follows from the intermediate value theorem for continuous functions.
The correct answer is (C).