Question

Let \(C\) be a convex closed set in a reflexive Banach space whose norm is locally uniformly convex. To every \(x \in X\) assign \(p(x)\), the closest point of \(C\) to \(x\). Show that \(p\) is continuous.

Short answer

The projection map \(p\), which assigns the closest point in a convex closed subset to any point in a reflexive Banach space, is continuous due to the locally uniformly convex norm.

Step-by-step solution

- Understand the Problem

We need to show that the map that assigns to each point in a reflexive Banach space its closest point in a convex closed subset is continuous. The Banach space has a norm that is locally uniformly convex.

- Define the Projection Map

Let the map be defined as follows: \[ p: X \to C \quad \text{such that} \quad p(x) = \text{argmin}_{y \in C} \| x - y \| \]This map assigns to each \(x \in X\), the point \(p(x) \in C\) which is closest to \(x\).

- Use Properties of Reflexive Banach Spaces

In a reflexive Banach space, the closed convex subset \(C\) ensures the existence and uniqueness of the projection \(p(x)\). This is guaranteed by the Hilbert space projection theorem.

- Show the Continuity of the Projection Map

The map \(p\) is continuous if for all sequences \(x_n \to x \) in \(X\), \(p(x_n) \to p(x)\) in \(C\). Since \(X\) is reflexive and its norm is locally uniformly convex, small changes in \(x_n\) cause small changes in \(p(x_n)\).

- Use Locally Uniform Convexity

Local uniform convexity of the norm in \(X\) allows us to state that: for any \(\epsilon > 0\), there exists \(\delta > 0\) such that if \(\| x_n - x \| < \delta\), then \(\| p(x_n) - p(x) \| < \epsilon\).

- Conclusion

By the definition of local uniform convexity and properties of reflexive spaces, \(\| p(x_n) - p(x) \| < \epsilon\) implies that \(p \) is continuous.

Key concepts

Convex Closed Set

A convex closed set is a fundamental concept in optimization and functional analysis. In simple terms, a set is convex if, for any two points within the set, the line segment connecting them is also within the set. Mathematically, a set \(C\) is convex if for any \(x, y \in C\) and any \(\lambda \in [0, 1]\), we have \(\lambda x + (1 - \lambda ) y \in C\).
Additionally, a set is closed if it contains all its limit points. This means that any point that can be approached by a sequence of points within the set, also belongs to the set.
These properties are crucial in analysis and optimization because many problems can be simplified or solved under these conditions. In our exercise, the closest-point projection is well-defined because the set is convex and closed.

Reflexive Banach Space

A Banach space is a complete normed vector space, meaning it is a vector space equipped with a norm, and every Cauchy sequence in this space converges to a limit within the space. Reflexivity adds another layer: a Banach space is called reflexive if every bounded linear functional attains its supremum on the closed unit ball.
In more intuitive terms, a Banach space \(X\) is reflexive if the natural map from \(X\) to its double dual \(X^{**}\) is surjective, meaning every element of \(X^{**}\) corresponds to an element of \(X\).
This property is significant in our exercise because it ensures the existence and uniqueness of the projection of any point onto the convex closed set \(C\). This result follows from the Hahn-Banach theorem and the geometry of reflexive spaces.

Locally Uniformly Convex Norm

A norm \( \| \, \cdot \, \| \) on a Banach space is locally uniformly convex if, for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that for all \( x, y \) in the unit sphere, if \( \| x - y \| \geq \epsilon \), then \( \| \frac{x+y}{2} \| \leq 1 - \delta \).
This concept implies that the norm exhibits strong convexity properties on every bounded subset. It means that small perturbations in input yield controlled changes in output, which is essential for continuity.
In our exercise, the local uniform convexity of the norm in the reflexive Banach space plays a critical role. It ensures that if a sequence of points \(x_n\) converges to a point \(x\), then the sequence of their projections \(p(x_n)\) also converges to \(p(x)\), aiding in proving the continuity of the projection map.

Continuous Map

A map \(f: X \to Y\) is continuous if small changes in input \(x \in X\) result in small changes in output \(f(x) \in Y\). Formally, \(f\) is continuous if for every sequence \(x_n\) that converges to \(x\), the sequence \(f(x_n)\) converges to \(f(x)\).
In the context of our exercise, we need to show that the projection map \(p\), which assigns to each \(x \in X\) its closest point in the convex closed set \(C\), is continuous. This is validated by the properties of reflexive Banach spaces and the local uniform convexity of the norm.
Through this exercise, we see the power of functional analysis and convexity principles in ensuring useful properties like continuity, which are critical in optimization and many areas of applied mathematics.