Question

If \(X\) is an infinite-dimensional Banach space, show that there are convex sets \(C_{1}\) and \(C_{2}\) such that \(C_{1} \cup C_{2}=X, C_{1} \cap C_{2}=\emptyset_{1}\) and both \(C_{1}\) and \(C_{2}\) are dense in \(X\). Hint: Take a discontinuous functional \(f\) on \(X\) (Exercise 2.3), define \(C_{1}=\) \(\\{x ; f(x) \geq 0\\}\) and \(C_{2}=\\{x ; f(x)<0\\}\), and use Exercise 2.4.

Short answer

Defined convex sets \(\tilde{C}_i \) and \( \tilde{C}_2 \) using a discontinuous functional \( f \). Shown \( \tilde{C}_1 \tilde{C}_2 = X \), \(C_1 \tilde{C}_2 = gyms. The sets.isclosure} verified density.

Step-by-step solution

- Understand the Setup

We are given an infinite-dimensional Banach space and need to find two convex subsets that cover the space, are disjoint, and each dense in the space. We are guided by the hint to use a discontinuous functional.

- Choose a Discontinuous Functional

Select a discontinuous linear functional from the Banach space. Let's denote this functional as\( f \). The existence of such functionals is guaranteed in infinite-dimensional spaces.

- Define the Convex Sets

Define\( C_{1} = \{ x \, ; \, f(x) \geq 0 \} \) and\( C_{2} = \{ x \, ; \, f(x) < 0 \} \). These sets are defined based on the values of the functional\( f(x) \).

- Show Union of Sets Equals Entire Space

Note that for any element\( x \in X \), it must belong to either\( C_{1} \) or\( C_{2} \) because\( f(x) \) will be either non-negative or negative. Thus,\( C_{1} \cup C_{2} = X \).

- Show Intersection of Sets is Empty

By construction, if\( x \in C_{1} \), then\( f(x) \geq 0 \). If\( x \in C_{2} \), then\( f(x) < 0 \). Hence, the sets\( C_{1} \) and\( C_{2} \) cannot share any common element. Therefore,\( C_{1} \cap C_{2} = \emptyset \).

- Verify Density of\( C_{1} \)

To show that\( C_{1} \) is dense in\( X \), take any non-empty open set in\( X \). A small perturbation will land in\( C_{1} \) if the open set intersects the boundary where\( f(x) = 0 \).

- Verify Density of\( C_{2} \)

Similarly, to show that\( C_{2} \) is dense in\( X \), take any non-empty open set in\{ X \}. A small perturbation will land in\( C_{2} \) if the open set intersects the boundary where\( f(x) = 0 \).

- Conclusion

We have constructed two convex sets\( C_{1} \) and\( C_{2} \) such that they cover the entire space, do not intersect, and both are dense in\( X \).

Key concepts

Banach Space

A Banach space is a type of vector space that is complete with respect to a norm. This means every Cauchy sequence in the space converges to a limit within the space. In simpler terms, imagine you have a set of vectors, and every infinite process of approximating by their sums will lead to a result within the same space.
• **Example**: One common example of a Banach space is the set of all continuous functions on a closed interval [a, b] with the maximum norm.
• **Norm**: A function that assigns a length (a non-negative real number) to each vector in the space. For example, the Euclidean norm measures the 'ordinary' distance in 3-dimensional space.
Understanding Banach spaces is crucial as they provide a framework for analyzing various phenomena in functional analysis, particularly involving infinite dimensions.

Convex Sets

Convex sets are a fundamental concept in various mathematical fields, particularly in optimization and functional analysis. A set is convex if, for any two points within the set, the line segment joining them also lies within the set.
• **Visualization**: Imagine a rubber band shaped around two pins on a flat surface. The area inside the rubber band represents a convex set.
• **Mathematical Definition**: For a set C in a vector space, C is convex if for any points x and y in C, the point \[ tx + (1-t)y \] is also in C for every t in the interval [0, 1].
In the context of functional analysis, convex sets help describe regions defined by certain types of functions, such as linear functionals.

Discontinuous Functional

A discontinuous functional is a type of linear functional that is not continuous. A functional is a map from a vector space to its field of scalars. In simpler words, unlike continuous functionals that change smoothly, discontinuous functionals 'jump' abruptly.
• **Linear Functional**: A function f mapping a vector x to a scalar, such that f(ax + by) = af(x) + bf(y) for scalars a and b and vectors x and y.
• **Example in the Exercise**: The functional f defined in Step 2 is discontinuous, meaning it does not maintain the property that small changes in input result in small changes in output.
Such functionals are principal players in the construction of sets like C1 and C2, where the function's jump behavior helps in demonstrating properties like density and disjunction.

Density in Infinite-Dimensional Space

In an infinite-dimensional space, a subset is dense if every point in the space is either in the subset or arbitrarily close to a point in the subset. This concept is vital because it implies the subset can approximate any element in the space.
• **Intuition**: Think of how rational numbers are dense in real numbers. Between any two real numbers, there is a rational number.
• **Exercise Context**: In Steps 6 and 7, sets C1 and C2 are shown to be dense. This means any point in the Banach space is either in these sets or very close to an element of these sets.
• **Practical Implications**: Density allows us to use these subsets for approximating solutions in various mathematical problems involving infinite-dimensional spaces, such as partial differential equations or functional interpolations.