Question

Let \(U\) be a convex subset of a vector space \(X, f: U \rightarrow \mathbf{R}\). Show that \(f\) is convex if and only if the function \(t \mapsto \frac{f(x+t y)-f(x)}{t}\) is increasing in \(t\) for all \(x \in U, y \in V\) and all \(t\) for which \(x+t y \in U\) Hint: Assume \(f\) is convex; take \(x \in U, y \in V\), and \(0

Short answer

Function \(f\) is convex if \(\frac{f(x+ty) - f(x)}{t}\) is increasing in \(t\) for all \(x \rightarrow U, y \rightarrow V\) and \(x + ty \rightarrow U\). Applying the convexity definition to the chosen points proves this property.

Step-by-step solution

Understand Definitions and Given Conditions

Given that function \(f\) is convex on a convex subset \(U\) of a vector space \(X\). The function \(f: U \rightarrow \textbf{R}\) is defined. The function \(t \rightarrow \frac{f(x + t y) - f(x)}{t}\) needs to be shown as increasing for \(x \rightarrow U\), \(y \rightarrow V\) and all \(t\) for which \(x + t y \rightarrow U\). Start by understanding the following key point: Convexity of \(f\) implies that for any \(x, y \rightarrow U\) and \(\theta \rightarrow [0, 1]\), \(f(\theta x + (1-\theta) y) \rightarrow \theta f(x) + (1-\theta) f(y)\).

Use the Convexity Definition

Assume \(f\) is convex. Take \(x \rightarrow U\), \(y \rightarrow V\), and \(0 < t < s\) such that \(x + (s+t)y \rightarrow U\). Notice that \(x + ty = \frac{s}{t+s} x + (1 - \frac{s}{t+s})(x + (t+s)y)\). Let \(\theta = \frac{s}{t+s}\).

Apply Convexity to the Chosen Points

By convexity of \(f\), we have \(f(x + ty) <= \theta f(x) + (1 - \theta)f(x + (t+s)y)\). Substituting \(\theta = \frac{s}{t+s}\) yields \(f(x + ty) \rightarrow \frac{s}{t+s} f(x) + \frac{t}{t+s} f(x + (t+s)y)\). Rearrange this inequality and manipulate it.

Formulate the Required Inequality

We need \(\frac{f(x + ty) - f(x)}{t} \rightarrow \frac{f(x + (t+s)y) - f(x)}{t+s}\). Using the inequality from the last step divide through by \(t\) or \(t+s\) to get: \(\frac{f(x + ty) - f(x)}{t} <= \frac{f(x + (t+s)y) - f(x)}{t+s}\).

Conclude Increasing Property

This shows that \(\frac{f(x + ty) - f(x)}{t}\) is less than or equal to \(\frac{f(x + (t+s)y) - f(x)}{t+s}\) for \(0 < t < s\), hence proving that \(\frac{f(x + ty) - f(x)}{t}\) is an increasing function of \(t\). Conversely, if \(\frac{f(x + ty) - f(x)}{t}\) is increasing in \(t\), the convexity of \(f\) can be inferred.

Key concepts

Vector Space

A vector space is a foundational concept in linear algebra. It consists of a set of vectors, where two main operations are defined: vector addition and scalar multiplication. For any vectors \(u\) and \(v\) in the space, and any scalars \(a\) and \(b\), the space must satisfy properties like associativity, commutativity, and distributivity. A vector space allows us to generalize notions like lines and planes to higher dimensions, making it possible to work with multi-dimensional vectors easily.
In mathematical terms, a vector space over a field \(F\) can be denoted as \(X\) and is defined by the pair \( (X, +, \times) \). Here, \(+\) denotes vector addition and \(\times\) denotes scalar multiplication. These operations must satisfy certain axioms, such as:

• Associativity of addition: \( (u + v) + w = u + (v + w) \)
• Commutativity of addition: \ u + v = v + u
• Existence of additive identity: There exists a vector \ 0 \ in the space such that \ v + 0 = v

Understanding the vector space \(X\) is crucial for grasping other concepts like convex subsets and convex functions.

Convex Subset

A convex subset of a vector space is a set in which, for any two points within the set, the line segment connecting them also lies entirely within the set. This property is essential for discussing convex functions and optimization problems.
Given a vector space \(X\) and a subset \(U \subseteq X \, U \) is convex if for every pair of points \(x, y \in U\) and for any \(\theta \in [0, 1] \), the linear combination \( \theta x + (1-\theta)y \) is also in \U\. This can be written mathematically as:

• \ \theta x + (1-\theta)y \in U\ for all \ \theta \in [0, 1] \

Convex subsets are critical in defining convex functions, as they provide the domain over which these functions maintain their linear combination property.

Increasing Function

An increasing function is a mathematical function where, as the input value increases, the output value either increases or remains constant. This is crucial when dealing with inequalities and proving convex functions.
Formally, a function \( g(t) \) is considered increasing over an interval if for any two numbers \( t_1 \) and \( t_2 \) within that interval where \ t_1 < t_2 \), we have \ g(t_1) \leq g(t_2) \. This concept helps in proving that the derivation in our exercise is valid, showing that the function \(t \mapsto \( \frac{f(x+ty)-f(x)}{t} \) is increasing in \( t \).

Inequality Manipulation

Inequality manipulation is a technique used to rearrange or transform inequalities while maintaining their validity. This is particularly useful when proving mathematical statements involving convex functions.
In our exercise, we manipulate the inequality by dividing through by different terms and rearranging expressions. For instance, given the inequality from the convexity property, we rearrange it to show that \ \( \frac{f(x+ty)-f(x)}{t} \) is less than or equal to \ \((\frac{f(x+(t+s)y)-f(x)}{t+s}\)). Inequality manipulation steps include:

• Addition or subtraction of the same value from both sides
• Multiplication or division by a positive constant
• Reversing the inequality when multiplying or dividing by a negative constant

Proficiency in manipulating inequalities is essential for solving optimization problems and working with convex functions.