Question:
Let
\[
f_n(x) = \frac{x^2}{x^2 + (1 - nx)^2}, \quad x \in [0, 1], \, n = 1, 2, 3, \dots.
\]
Then, which of the following statements is TRUE?
Let
\[
f_n(x) = \frac{x^2}{x^2 + (1 - nx)^2}, \quad x \in [0, 1], \, n = 1, 2, 3, \dots.
\]
Then, which of the following statements is TRUE?
Show Hint
To check equicontinuity, observe how the function behaves as the parameter (in this case, \( n \)) increases. If the function becomes increasingly sensitive to small changes in \( x \), it is likely not equicontinuous.
Updated On: Dec 4, 2025
- \( \{f_n\} \) is not equicontinuous on \( [0, 1] \)
- \( \{f_n\} \) is uniformly convergent on \( [0, 1] \)
- \( \{f_n\} \) is equicontinuous on \( [0, 1] \)
- \( \{f_n\} \) is uniformly bounded and has a subsequence converging uniformly on \( [0, 1] \)
Show Solution
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The Correct Option is A
Solution and Explanation
We are given the sequence of functions \( f_n(x) \) defined by:
\[
f_n(x) = \frac{x^2}{x^2 + (1 - nx)^2}.
\]
To determine which statement is true, we analyze the behavior of the sequence \( \{f_n\} \) on the interval \( [0, 1] \).
Option (A) \( \{f_n\} \) is not equicontinuous on \( [0, 1] \):
Equicontinuity requires that for every \( \epsilon>0 \), there exists a \( \delta>0 \) such that for all \( n \) and all \( x, y \in [0, 1] \), \( |x - y|<\delta \) implies \( |f_n(x) - f_n(y)|<\epsilon \).
For \( f_n(x) \), as \( n \to \infty \), the function becomes increasingly sensitive to changes in \( x \) near \( x = \frac{1}{n} \). This means that the functions \( f_n(x) \) do not exhibit the uniform behavior required for equicontinuity. Specifically, as \( n \) increases, the function \( f_n(x) \) oscillates more near \( x = 0 \), leading to the lack of equicontinuity.
Thus, option (A) is correct: \( \{f_n\} \) is not equicontinuous on \( [0, 1] \).
Option (B) \( \{f_n\} \) is uniformly convergent on \( [0, 1] \):
For uniform convergence, the difference \( |f_n(x) - f(x)| \) must be small for all \( x \in [0, 1] \) and for all \( n \). The sequence \( f_n(x) \) does not converge uniformly to a function on \( [0, 1] \) because the behavior near \( x = 0 \) prevents uniform convergence. Therefore, option (B) is incorrect.
Option (C) \( \{f_n\} \) is equicontinuous on \( [0, 1] \):
As shown in option (A), the sequence \( \{f_n\} \) is not equicontinuous, so option (C) is incorrect.
Option (D) \( \{f_n\} \) is uniformly bounded and has a subsequence converging uniformly on \( [0, 1] \):
The sequence \( \{f_n\} \) is not uniformly bounded due to the behavior of \( f_n(x) \) as \( n \to \infty \), especially near \( x = 0 \). This rules out option (D).
Thus, the correct answer is (A).
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