Question:
Let \(C[0, 1] = \{ f : [0, 1] \to \mathbb{R} : f \text{ is continuous}\}\).
Consider the metric space \((C[0,1], d_\infty)\), where
\[
d_\infty(f, g) = \sup\{ |f(x) - g(x)| : x \in [0, 1] \} \text{ for } f, g \in C[0,1].
\]
Let \(f_0(x) = 0\) for all \(x \in [0,1]\) and
\[
X = \{f \in (C[0, 1], d_\infty) : d_\infty(f_0, f) \ge \frac{1}{2}\}.
\]
Let \(f_1, f_2 \in C[0, 1]\) be defined by \(f_1(x) = x\) and \(f_2(x) = 1-x\) for all \(x \in [0,1]\).
Consider the following statements:
P: \(f_1\) is in the interior of X.
Q: \(f_2\) is in the interior of X.
Which of the following statements is correct?
Let \(C[0, 1] = \{ f : [0, 1] \to \mathbb{R} : f \text{ is continuous}\}\).
Consider the metric space \((C[0,1], d_\infty)\), where \[ d_\infty(f, g) = \sup\{ |f(x) - g(x)| : x \in [0, 1] \} \text{ for } f, g \in C[0,1]. \] Let \(f_0(x) = 0\) for all \(x \in [0,1]\) and \[ X = \{f \in (C[0, 1], d_\infty) : d_\infty(f_0, f) \ge \frac{1}{2}\}. \] Let \(f_1, f_2 \in C[0, 1]\) be defined by \(f_1(x) = x\) and \(f_2(x) = 1-x\) for all \(x \in [0,1]\).
Consider the following statements:
P: \(f_1\) is in the interior of X.
Q: \(f_2\) is in the interior of X.
Which of the following statements is correct?
Consider the metric space \((C[0,1], d_\infty)\), where \[ d_\infty(f, g) = \sup\{ |f(x) - g(x)| : x \in [0, 1] \} \text{ for } f, g \in C[0,1]. \] Let \(f_0(x) = 0\) for all \(x \in [0,1]\) and \[ X = \{f \in (C[0, 1], d_\infty) : d_\infty(f_0, f) \ge \frac{1}{2}\}. \] Let \(f_1, f_2 \in C[0, 1]\) be defined by \(f_1(x) = x\) and \(f_2(x) = 1-x\) for all \(x \in [0,1]\).
Consider the following statements:
P: \(f_1\) is in the interior of X.
Q: \(f_2\) is in the interior of X.
Which of the following statements is correct?
Show Hint
The interior of a set like \(\{x : ||x|| \ge c\}\) is \(\{x : ||x||>c\}\). A point \(x_0\) is in the interior if \(||x_0||>c\). The boundary is \(\{x : ||x|| = c\}\). In this problem, \(||f_1||_\infty = 1\) and \(||f_2||_\infty = 1\), and since \(1>1/2\), both are in the interior of \(X\).
Updated On: Sep 5, 2025
- P is TRUE and Q is FALSE
- P is FALSE and Q is TRUE
Both P and Q are FALSE
Both P and Q are TRUE
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
The problem is set in the space of continuous functions on \([0,1]\) with the supremum norm, \(||f||_\infty = d_\infty(f_0, f)\). The set \(X\) consists of all functions whose norm is at least \(1/2\). A function \(f\) is in the interior of \(X\) if there exists an open ball centered at \(f\) that is entirely contained within \(X\).
Step 2: Key Formula or Approach:
A point \(f\) is in the interior of a set \(X\) if there exists an \(\epsilon>0\) such that the open ball \(B(f, \epsilon) = \{g : ||f-g||_\infty<\epsilon\}\) is a subset of \(X\).
For any \(g \in B(f, \epsilon)\), we must show that \(g \in X\), which means we must show \(||g||_\infty \ge 1/2\). The reverse triangle inequality, \(||g||_\infty \ge |||f||_\infty - ||f-g||_\infty|\), is useful here.
Step 3: Detailed Calculation:
The set \(X\) can be written as \(X = \{f \in C[0,1] : ||f||_\infty \ge 1/2\}\).
Analysis of Statement P:
Consider \(f_1(x) = x\). 1. Check if \(f_1 \in X\). We calculate its norm: \(||f_1||_\infty = \sup_{x \in [0,1]} |x| = 1\). Since \(1 \ge 1/2\), \(f_1\) is in \(X\).
2. Check if \(f_1\) is in the interior of \(X\). We need to find an \(\epsilon>0\) such that for any \(g\) with \(||f_1 - g||_\infty<\epsilon\), we have \(||g||_\infty \ge 1/2\).
Using the reverse triangle inequality: \[ ||g||_\infty = ||f_1 - (f_1 - g)||_\infty \ge ||f_1||_\infty - ||f_1 - g||_\infty \] Since \(||f_1-g||_\infty<\epsilon\), we have \(||g||_\infty>||f_1||_\infty - \epsilon = 1 - \epsilon\).
We want to guarantee that \(||g||_\infty \ge 1/2\). We can achieve this if we set \(1 - \epsilon \ge 1/2\), which means \(\epsilon \le 1/2\). Let's choose \(\epsilon = 1/4\). Then for any \(g \in B(f_1, 1/4)\), it follows that \(||g||_\infty>1 - 1/4 = 3/4\). Since \(3/4 \ge 1/2\), every \(g\) in this ball is in \(X\).
So, \(B(f_1, 1/4) \subseteq X\). This proves that \(f_1\) is an interior point of \(X\). Thus, P is TRUE.
Analysis of Statement Q:
Consider \(f_2(x) = 1-x\). 1. Check if \(f_2 \in X\). We calculate its norm: \(||f_2||_\infty = \sup_{x \in [0,1]} |1-x| = 1\). Since \(1 \ge 1/2\), \(f_2\) is in \(X\). 2. Check if \(f_2\) is in the interior of \(X\). The argument is identical to the one for \(f_1\). Let \(g\) be any function such that \(||f_2 - g||_\infty<\epsilon\). \[ ||g||_\infty \ge ||f_2||_\infty - ||f_2 - g||_\infty>1 - \epsilon \] Again, we can choose \(\epsilon = 1/4\). Then for any \(g \in B(f_2, 1/4)\), we have \(||g||_\infty>1 - 1/4 = 3/4 \ge 1/2\). So, \(B(f_2, 1/4) \subseteq X\). This proves that \(f_2\) is an interior point of \(X\). Thus, Q is TRUE.
Step 4: Final Answer:
Both P and Q are TRUE.
Step 5: Why This is Correct:
Both functions \(f_1\) and \(f_2\) have a norm of 1. The set X consists of functions with a norm of at least 1/2. Because the norms of \(f_1\) and \(f_2\) are strictly greater than 1/2, they are "safely" inside X, not on its boundary. This allows for a small ball of radius \(\epsilon\) around them that is still fully contained in X, making them interior points.
The problem is set in the space of continuous functions on \([0,1]\) with the supremum norm, \(||f||_\infty = d_\infty(f_0, f)\). The set \(X\) consists of all functions whose norm is at least \(1/2\). A function \(f\) is in the interior of \(X\) if there exists an open ball centered at \(f\) that is entirely contained within \(X\).
Step 2: Key Formula or Approach:
A point \(f\) is in the interior of a set \(X\) if there exists an \(\epsilon>0\) such that the open ball \(B(f, \epsilon) = \{g : ||f-g||_\infty<\epsilon\}\) is a subset of \(X\).
For any \(g \in B(f, \epsilon)\), we must show that \(g \in X\), which means we must show \(||g||_\infty \ge 1/2\). The reverse triangle inequality, \(||g||_\infty \ge |||f||_\infty - ||f-g||_\infty|\), is useful here.
Step 3: Detailed Calculation:
The set \(X\) can be written as \(X = \{f \in C[0,1] : ||f||_\infty \ge 1/2\}\).
Analysis of Statement P:
Consider \(f_1(x) = x\). 1. Check if \(f_1 \in X\). We calculate its norm: \(||f_1||_\infty = \sup_{x \in [0,1]} |x| = 1\). Since \(1 \ge 1/2\), \(f_1\) is in \(X\).
2. Check if \(f_1\) is in the interior of \(X\). We need to find an \(\epsilon>0\) such that for any \(g\) with \(||f_1 - g||_\infty<\epsilon\), we have \(||g||_\infty \ge 1/2\).
Using the reverse triangle inequality: \[ ||g||_\infty = ||f_1 - (f_1 - g)||_\infty \ge ||f_1||_\infty - ||f_1 - g||_\infty \] Since \(||f_1-g||_\infty<\epsilon\), we have \(||g||_\infty>||f_1||_\infty - \epsilon = 1 - \epsilon\).
We want to guarantee that \(||g||_\infty \ge 1/2\). We can achieve this if we set \(1 - \epsilon \ge 1/2\), which means \(\epsilon \le 1/2\). Let's choose \(\epsilon = 1/4\). Then for any \(g \in B(f_1, 1/4)\), it follows that \(||g||_\infty>1 - 1/4 = 3/4\). Since \(3/4 \ge 1/2\), every \(g\) in this ball is in \(X\).
So, \(B(f_1, 1/4) \subseteq X\). This proves that \(f_1\) is an interior point of \(X\). Thus, P is TRUE.
Analysis of Statement Q:
Consider \(f_2(x) = 1-x\). 1. Check if \(f_2 \in X\). We calculate its norm: \(||f_2||_\infty = \sup_{x \in [0,1]} |1-x| = 1\). Since \(1 \ge 1/2\), \(f_2\) is in \(X\). 2. Check if \(f_2\) is in the interior of \(X\). The argument is identical to the one for \(f_1\). Let \(g\) be any function such that \(||f_2 - g||_\infty<\epsilon\). \[ ||g||_\infty \ge ||f_2||_\infty - ||f_2 - g||_\infty>1 - \epsilon \] Again, we can choose \(\epsilon = 1/4\). Then for any \(g \in B(f_2, 1/4)\), we have \(||g||_\infty>1 - 1/4 = 3/4 \ge 1/2\). So, \(B(f_2, 1/4) \subseteq X\). This proves that \(f_2\) is an interior point of \(X\). Thus, Q is TRUE.
Step 4: Final Answer:
Both P and Q are TRUE.
Step 5: Why This is Correct:
Both functions \(f_1\) and \(f_2\) have a norm of 1. The set X consists of functions with a norm of at least 1/2. Because the norms of \(f_1\) and \(f_2\) are strictly greater than 1/2, they are "safely" inside X, not on its boundary. This allows for a small ball of radius \(\epsilon\) around them that is still fully contained in X, making them interior points.
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