Chapter 8: Problem 17
Let \(X\) be an infinite-dimensional separable Banach space. Show that there is a continuous convex function \(f\) on \(X\) such that \(f\) is unbounded on \(B_{X}\)
Short Answer
Expert verified
The norm function \( f(x) = orm{x} \) is continuous, convex, and unbounded on the unit ball in an infinite-dimensional Banach space.
Step by step solution
01
- Understand the problem
Work with the definition: An infinite-dimensional separable Banach space is a space that is both complete and has a countable dense subset. The goal is to find a continuous convex function that is unbounded on the unit ball of this space.
02
- Define the candidate function
Consider the norm function on the Banach space defined as \[ f(x) = orm{x} \] Note that the function is continuous and convex since it is derived from the Banach space norm.
03
- Prove function is continuous
To show continuity, recall that the norm function is a continuous function on the Banach space by the definition of the norm.
04
- Prove function is convex
To prove convexity, for any two points \[ x, y \text{ in } X \text{ and any } t \text{ in } [0, 1], \] the function \[ orm{tx + (1-t)y} \text{ satisfies the convexity inequality} \] \[ orm{tx + (1-t)y} \leq torm{x} + (1-t)orm{y}. \]
05
- Show unboundedness on the unit ball
Consider the unit ball \[B_X = orm{x} \text{ such that } orm{x} \leq 1 \text{ for all } x \text{ in } X.\] In an infinite-dimensional Banach space, there exist sequences of elements with increasing norms ensuring unboundedness.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Separable Banach space
In functional analysis, a Banach space is a vector space with a norm, where any Cauchy sequence converges within the space. A separable Banach space goes a step further: it contains a countable dense subset. This means we can find a countable set of elements such that any element in the Banach space can be arbitrarily approximated by elements from this countable set.
Separable Banach spaces are important because they are 'small' in a sense; they can be described using a countable set despite being infinite-dimensional.
For our problem, we examine an infinite-dimensional separable Banach space. This setting allows us to consider the infinite aspects while ensuring manageability due to the separability.
Separable Banach spaces are important because they are 'small' in a sense; they can be described using a countable set despite being infinite-dimensional.
For our problem, we examine an infinite-dimensional separable Banach space. This setting allows us to consider the infinite aspects while ensuring manageability due to the separability.
Convex function
A function \(f: X \to \real\) is called convex if, for all \(x, y \text{ in } X\) and for all \(t \text{ in [0,1]}\), the following inequality holds:
For our exercise, we consider the norm function
\f(x) = orm{x}
which can be proven to be convex as it adheres to the convexity inequality. This characteristic is crucial in demonstrating specific properties related to the unit ball in a Banach space.
- \(f(tx + (1-t)y) \text{ tends to be less than or equal to } tf(x) + (1-t)f(y)\).
For our exercise, we consider the norm function
\f(x) = orm{x}
which can be proven to be convex as it adheres to the convexity inequality. This characteristic is crucial in demonstrating specific properties related to the unit ball in a Banach space.
Continuous function
A function is continuous if small changes in its input result in small changes in its output. Mathematically, \(f \text{ is continuous at a point } a\) if:
\f(\text{ if } \forall\text{ ε > 0, there exists a } δ > 0 \text{ such that for all } x, \forall x \text{ satisfying } 0 < |x - a| < δ, \text{ it holds that } |f(x) - f(a)| < ε).\text{
For the norm function in our problem, continuity follows directly from the properties of the norm in the Banach space. The norm is a fundamental built-in feature ensuring this function smoothly varies without drastic jumps.
\f(\text{ if } \forall\text{ ε > 0, there exists a } δ > 0 \text{ such that for all } x, \forall x \text{ satisfying } 0 < |x - a| < δ, \text{ it holds that } |f(x) - f(a)| < ε).\text{
For the norm function in our problem, continuity follows directly from the properties of the norm in the Banach space. The norm is a fundamental built-in feature ensuring this function smoothly varies without drastic jumps.
Unit ball
The unit ball in a Banach space \(X\), denoted \(B_X\), is the set of all points in \(X\) with a norm less than or equal to one. Formally:
\f(B_X = { x \text{ in } X | orm{x} \text{ less than or equal to } 1).
It represents the 'ball' in an abstract vector space centered at the origin, encapsulating all elements with norm (distance in this context) not exceeding one.
For our problem, we're interested in properties of functions defined on \(B_X\). Notably, an unbounded function on this set cannot be restricted in its growth, leading to values of arbitrary large magnitude inside \(B_X\).
\f(B_X = { x \text{ in } X | orm{x} \text{ less than or equal to } 1).
It represents the 'ball' in an abstract vector space centered at the origin, encapsulating all elements with norm (distance in this context) not exceeding one.
For our problem, we're interested in properties of functions defined on \(B_X\). Notably, an unbounded function on this set cannot be restricted in its growth, leading to values of arbitrary large magnitude inside \(B_X\).
Norm function
The norm function in a Banach space assigns a non-negative length or size to each vector in the space. For a vector \(x\) in \(X\), the norm is typically denoted as ||x||. The properties of the norm include:
In our case, the norm function itself serves as the candidate function \(f\) proving to be continuous and convex. This makes analyzing its behavior on the unit ball straightforward, demonstrating its unboundedness in infinite-dimensional scenarios.
- Non-negativity: ||x|| \(\text{greater than or equal to}\) 0;
- Definiteness: ||x|| = 0 if and only if x = 0;
- Scalability: ||λx|| = |λ| ||x|| for any scalar λ;
- Triangle Inequality: ||x + y|| \(\text{less than or equal to}\) ||x|| + ||y||.
In our case, the norm function itself serves as the candidate function \(f\) proving to be continuous and convex. This makes analyzing its behavior on the unit ball straightforward, demonstrating its unboundedness in infinite-dimensional scenarios.