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Prove that if two functions \(f, g\) with values in a normed vèctor space are uniformly continuous on a set \(B,\) so also are \(f \pm g\) and \(a f\) for a fixed scalar \(a\). For real functions, prove this also for \(f \vee g\) and \(f \wedge g\) defined by $$(f \vee g)(x)=\max (f(x), g(x))$$ and $$ (f \wedge g)(x)=\min (f(x), g(x)) $$

Short Answer

Expert verified
If \( f \) and \( g \) are uniformly continuous, then so are \( f \pm g \) and \( a f \), and for real \( f, g \), \( f \vee g \) and \( f \wedge g \) are also uniformly continuous.

Step by step solution

01

Understand Uniform Continuity

A function \( f : B \to V \) is uniformly continuous on \( B \) if for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that for all \( x, y \in B \), if \( \|x-y\| < \delta \) then \( \|f(x)-f(y)\| < \epsilon \). This definition applies similarly to \( g \), \( f+g \), \( f-g \), and \( a f \).
02

Prove Uniform Continuity for Sum and Difference

Given that \( f \) and \( g \) are uniformly continuous on \( B \), we can choose \( \delta \) for a given \( \epsilon \) such that if \( \|x-y\| < \delta \), then both \( \|f(x)-f(y)\| < \epsilon/2 \) and \( \|g(x)-g(y)\| < \epsilon/2 \). For \( f \pm g \), \( \|(f\pm g)(x) - (f\pm g)(y)\| \leq \|f(x) - f(y)\| + \|g(x) - g(y)\| < \epsilon \). Thus, \( f \pm g \) is uniformly continuous on \( B \).
03

Prove Uniform Continuity for Scalar Multiplication

Consider a fixed scalar \( a \). \( f \) is uniformly continuous, so for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that \( \|f(x) - f(y)\| < \epsilon / |a| \) for all \( x, y \in B \). Thus, \( \|af(x) - af(y)\| = |a| \cdot \|f(x) - f(y)\| < \epsilon \), ensuring that \( a f \) is uniformly continuous on \( B \).
04

Prove Uniform Continuity for Real-Valued Functions f ∨ g

For real-valued functions, define \((f \vee g)(x) = \max(f(x), g(x))\). Given \( f \) and \( g \) are uniformly continuous, for any \( \epsilon > 0 \), there exists \( \delta > 0 \) such that for \( |x-y| < \delta \), both \( |f(x)-f(y)| < \epsilon \) and \( |g(x)-g(y)| < \epsilon \). Consequently, \( |(f \vee g)(x) - (f \vee g)(y)| \leq \max(|f(x)-f(y)|, |g(x)-g(y)|) < \epsilon \), proving \( f \vee g \) is uniformly continuous.
05

Prove Uniform Continuity for Real-Valued Functions f ∧ g

For real-valued functions \((f \wedge g)(x) = \min(f(x), g(x))\), we use similar reasoning. For \( |x-y| < \delta \), \( |(f \wedge g)(x) - (f \wedge g)(y)| \leq \max(|f(x)-f(y)|, |g(x)-g(y)|) < \epsilon \), proving \( f \wedge g \) is uniformly continuous.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normed Vector Space
In mathematics, a normed vector space is an essential concept in the study of various types of functions and their properties. Think of a normed vector space as a vector space that includes a norm, which is a function that assigns a non-negative length or size to each vector in the space. This norm needs to satisfy certain properties to ensure consistency:
  • It must be non-negative: For any vector \( v \), \( \|v\| \geq 0 \) and \( \|v\| = 0 \) only if \( v \) is the zero vector.
  • It must be scalable with scalar multiplication: For any scalar \( a \) and vector \( v \), \( \|a v\| = |a| \cdot \|v\| \).
  • It must satisfy the triangle inequality: For any vectors \( u \) and \( v \), \( \|u + v\| \leq \|u\| + \|v\| \).
This framework allows for the measurement of vector magnitudes and the distance between vectors, which is crucial in analyzing continuous functions.
Scalar Multiplication
Scalar multiplication is a basic yet powerful operation in mathematics, particularly in vector spaces. It involves multiplying a vector by a scalar, which is essentially a real or complex number. Here’s how it works:
  • Given a vector \( v \) and a scalar \( a \), the product \( a v \) results in a new vector whose direction is the same as \( v \) if \( a > 0 \) and opposite if \( a < 0 \).
  • The magnitude of the new vector \( a v \) is \( |a| \) times the magnitude of \( v \).
Scalar multiplication is used widely in the definition of functions within vector spaces. It helps in stretching or compressing vectors, which is crucial for understanding the behavior of functions like \( a f \). In such cases, if \( f \) is a function and uniformly continuous, \( a f \) is also uniform due to the properties of scalar multiplication and continuity relationships.
Real-Valued Functions
Real-valued functions have their outputs as real numbers—this makes them crucial in various mathematical analyses, including those concerning continuity and limits. A function \( f: B \rightarrow \mathbb{R} \) is real-valued if it associates every element of the set \( B \) with a real number.

Uniform continuity is a vital property when analyzing real-valued functions. It ensures that the function's output doesn't wildly change with infinitesimal inputs. This stability is vital in both theoretical and applied mathematics since it implies predictability and control over function behavior throughout its domain.

Real-valued functions can be combined in various ways, such as through maximum and minimum, sum, and difference, each retaining continuity under specific conditions. A key aspect in the study of real-valued functions is their limits and how their properties, like continuity, behave under operations.

Maximum and Minimum of Functions
The concepts of maximum and minimum play a significant role in analyzing functions, especially in determining their uniform continuity and behavior over a set. For two functions \( f(x) \) and \( g(x) \), we define their maximum and minimum via the functions:
  • Maximum function: \( (f \vee g)(x) = \max(f(x), g(x)) \)
  • Minimum function: \( (f \wedge g)(x) = \min(f(x), g(x)) \)
These derivatives ensure that for any input \( x \), we take the greater value for the maximum function and the lesser for the minimum function.
In the context of uniform continuity, these operations are significant. If each individual function \( f \) and \( g \) is uniformly continuous, the maximum and minimum functions are also uniformly continuous. This is because the change in the maximum or minimum is controlled by the larger change of either \( f \) or \( g \), hence maintaining the bound required for uniform continuity. This makes applying these operations safe without losing the desired properties of the original functions.

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