4. Suppose f is defined on R and f satisfies |f(x) − f (y)| ≤ C\x − y|ª for all x, y € R. Here C and a are two given positive constants. Prove that f is uniformly continuous on R.

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Author:Erwin Kreyszig
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### Continuity and Uniform Continuity in Real Analysis

#### Problem Statement

4. Suppose \( f \) is defined on \( \mathbb{R} \) and \( f \) satisfies
\[
|f(x) - f(y)| \leq C|x - y|^\alpha
\]
for all \( x, y \in \mathbb{R} \). Here \( C \) and \( \alpha \) are two given positive constants. Prove that \( f \) is uniformly continuous on \( \mathbb{R} \).

#### Explanation

This problem is asking you to prove that a function \( f \), which is defined on the set of real numbers \( \mathbb{R} \) and satisfies a certain condition, is uniformly continuous on \( \mathbb{R} \). The condition provided is an inequality involving the absolute differences between values of \( f \) at different points.

- \( |f(x) - f(y)| \) represents the absolute difference between \( f(x) \) and \( f(y) \).
- \( C \) and \( \alpha \) are given positive constants.

This inequality suggests that the rate of change of \( f(x) \) is controlled by the term \( C|x - y|^\alpha \). Such conditions are often used to ensure that \( f \) does not change too rapidly, a key property needed for uniform continuity.

#### Detailed Solution Outline

1. **Understanding Uniform Continuity**:
   - A function \( f \) is uniformly continuous on \( \mathbb{R} \) if, for any \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that for all \( x, y \in \mathbb{R} \), \( |x - y| < \delta \) implies \( |f(x) - f(y)| < \epsilon \).

2. **Given Condition**:
   - We are given that \( |f(x) - f(y)| \leq C|x - y|^\alpha \) for some positive constants \( C \) and \( \alpha \).

3. **Proving Uniform Continuity**:
   - To prove that \( f \) is uniformly continuous, we need to find a \( \delta \) for a given \( \epsilon > 0 \).

4. **Steps**:
   - Start
Transcribed Image Text:### Continuity and Uniform Continuity in Real Analysis #### Problem Statement 4. Suppose \( f \) is defined on \( \mathbb{R} \) and \( f \) satisfies \[ |f(x) - f(y)| \leq C|x - y|^\alpha \] for all \( x, y \in \mathbb{R} \). Here \( C \) and \( \alpha \) are two given positive constants. Prove that \( f \) is uniformly continuous on \( \mathbb{R} \). #### Explanation This problem is asking you to prove that a function \( f \), which is defined on the set of real numbers \( \mathbb{R} \) and satisfies a certain condition, is uniformly continuous on \( \mathbb{R} \). The condition provided is an inequality involving the absolute differences between values of \( f \) at different points. - \( |f(x) - f(y)| \) represents the absolute difference between \( f(x) \) and \( f(y) \). - \( C \) and \( \alpha \) are given positive constants. This inequality suggests that the rate of change of \( f(x) \) is controlled by the term \( C|x - y|^\alpha \). Such conditions are often used to ensure that \( f \) does not change too rapidly, a key property needed for uniform continuity. #### Detailed Solution Outline 1. **Understanding Uniform Continuity**: - A function \( f \) is uniformly continuous on \( \mathbb{R} \) if, for any \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that for all \( x, y \in \mathbb{R} \), \( |x - y| < \delta \) implies \( |f(x) - f(y)| < \epsilon \). 2. **Given Condition**: - We are given that \( |f(x) - f(y)| \leq C|x - y|^\alpha \) for some positive constants \( C \) and \( \alpha \). 3. **Proving Uniform Continuity**: - To prove that \( f \) is uniformly continuous, we need to find a \( \delta \) for a given \( \epsilon > 0 \). 4. **Steps**: - Start
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