Let b > 0. Consider the half-open half-closed interval (0, b]. Let f : (0, b] → R be

uniformly continuous on (0, b].

2

(a) Prove that there exists a ∈ (0, b] such that |f (x) − f (y)| < 1 for all x, y ∈ (0, a].

Hence show that there exists K > 0 such that |f (x)| ≤ K for all x ∈ (0, a].

(b) Prove that f is bounded on (0, b].

(c) Is the function g : (0, 1] → R defined by g(x) = 1

x sin( 1

x ) uniformly continuous on

(0, 1]? Justify your answer.

(Hint: Part (ii) would be useful).

To tackle these questions, let's carefully analyze each part and use properties of uniformly continuous functions to derive the desired results.

### (a) Existence of \( a \in (0, b] \)

**Proof:**

1. Since \( f \) is uniformly continuous on \( (0, b] \), for \(\epsilon = 1\), there exists \(\delta > 0\) such that for all \( x, y \in (0, b] \), if \( |x - y| < \delta \), then \( |f(x) - f(y)| < 1 \).

2. Choose \( a = \min\{b, \delta\} \). This ensures that \( a \) lies in \( (0, b] \).

3. Now, for all \( x, y \in (0, a] \) with \( |x - y| < \delta \) (and since \( a \leq \delta \)), we have \( |f(x) - f(y)| < 1 \).

**Conclusion:**
Thus, there exists an \( a \in (0, b] \) such that \( |f(x) - f(y)| < 1 \) for all \( x, y \in (0, a] \).

**Bounding \( f \) on \( (0, a] \):**

1. Choose any \( x_0 \in (0, a] \).

2. Since \( |f(x) - f(x_0)| < 1 \) for all \( x \in (0, a] \), it follows that:
   \[
   |f(x)| \leq |f(x_0)| + 1 \text{ for all } x \in (0, a]
   \]

3. Let \( K = |f(x_0)| + 1 \). Then \( K > 0 \) and \( |f(x)| \leq K \) for all \( x \in (0, a] \).

### (b) Boundedness of \( f \) on \( (0, b] \)

**Proof:**

1. Divide \( (0, b] \) into two parts: \( (0, a] \) and \( [a, b] \).

2. From part (a), \( f \) is bounded on \( (0, a] \) by \( K \).

3. On the closed interval \( [a, b] \), the function \( f \) is uniformly continuous (because \( f \) is uniformly continuous on the whole \( (0, b] \)) and continuous on a compact set, hence bounded.

4. Let \( M = \max_{x \in [a, b]} |f(x)| \).

5. Therefore, \( f \) is bounded on \( (0, b] \) by \( \max(K, M) \).

### (c) Uniform continuity of \( g(x) = \frac{1}{x} \sin\left(\frac{1}{x}\right) \) on \( (0, 1] \)

**Reasoning:**

1. Let's analyze the behavior of \( g(x) \) as \( x \to 0 \).

2. Consider any two points \( x, y \in (0, 1] \).

3. If \( |x - y| < \delta \), where \( \delta \) is small, the behavior of \( g \) depends on \( \frac{1}{x} \) and \( \frac{1}{y} \). For small \( x \), \( \frac{1}{x} \) becomes large, causing the sine function to oscillate rapidly.

4. This rapid oscillation implies that for small changes in \( x \), \( |g(x) - g(y)| \) can be large due to the factor \( \frac{1}{x} \).

5. Hence, there does not exist a single \( \delta \) that works uniformly for all \( \epsilon > 0 \) for small \( x \), implying \( g(x) \) is not uniformly continuous on \( (0, 1] \).

**Conclusion:**
The function \( g(x) = \frac{1}{x} \sin\left(\frac{1}{x}\right) \) is **not** uniformly continuous on \( (0, 1] \).

---

Do you want details or have any questions? 

Here are 8 related questions for further exploration:

1. What are some other examples of functions that are uniformly continuous on a given interval?
2. How does the definition of uniform continuity differ from regular continuity?
3. Can you provide an example of a function that is continuous but not uniformly continuous?
4. How can one prove that a function is not uniformly continuous on a given interval?
5. What is the importance of bounding a function in real analysis?
6. How does one handle uniform continuity on half-open intervals compared to closed intervals?
7. Are there any special properties of uniformly continuous functions on bounded intervals?
8. How does the uniform continuity of a function affect its integrability?

**Tip:** When proving properties about uniformly continuous functions, consider dividing the interval into manageable subintervals, especially when dealing with unbounded or partially bounded domains.

Explore how to prove the existence of points where a function is uniformly continuous and bounded on intervals. Includes step-by-step proofs and analysis using the Uniform Continuity Theorem. Suitable for advanced undergraduate students studying real analysis.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation