Question

For an integer \( n \), let \( f_n(x) = xe^{-nx }\), where \( x \in [0, 1] \). Let \( S := \{f_n : n \geq 1\} \). Consider the metric space \( (C([0, 1]), d) \), where \[ d(f, g) = \sup_{x \in [0, 1]} |f(x) - g(x)|, \quad f, g \in C([0, 1]). \] Which of the following statement(s) is/are true?}

• A. \( S \) is an equi-continuous family of continuous functions
  (2) (3) (4)
• B. \( S \) is closed in \( (C([0, 1]), d) \)
• C. \( S \) is bounded in \( (C([0, 1]), d) \)
• D. \( S \) is compact in \( (C([0, 1]), d) \)

Hint:

For questions on function families in metric spaces, analyze continuity, equi-continuity, boundedness, and compactness separately.

Answer 1

The Correct Option is A

Step 1: Checking equi-continuity of \( S \). The functions \( f_n(x) = xe^{-nx} \) are continuous for all \( n \geq 1 \), and for each \( \epsilon>0 \), the variation in \( f_n(x) \) can be made arbitrarily small by choosing \( x \) close enough to a fixed point. Hence, \( S \) is equi-continuous.
Step 2: Checking whether \( S \) is closed. The limit of a sequence of functions in \( S \) need not belong to \( S \) (e.g., \( f_n \to 0 \) pointwise as \( n \to \infty \), but \( 0 \notin S \)). Thus, \( S \) is not closed. 
Step 3: Checking boundedness of \( S \). For any \( f_n(x) \in S \), we have \[ |f_n(x)| = |xe^{-nx}| \leq \max_{x \in [0, 1]} |xe^{-nx}| \leq \frac{1}{e}. \] Thus, \( S \) is bounded in \( (C([0, 1]), d) \). 
Step 4: Checking compactness of \( S \). The family \( S \) is not compact because it is not closed (as shown above), violating a necessary condition for compactness in a metric space. 
Step 5: Conclusion. The correct answers are \( {(1), (3)} \).