Question

(Alspach) Consider \(K=\left\\{f \in L_{1}[0,1] ; 0 \leq f \leq 2, \int_{0}^{1} f d t=1\right\\}\) and \(T: K \rightarrow K\) defined as $$ T(f)(t)=\left\\{\begin{array}{cl} \min (2 f(2 t), 2) & \text { if } 0 \leq t \leq \frac{1}{2} \\ \max (0,2 f(2 t-1)-2) & \text { if } \frac{1}{2}

Short answer

The operator \(T\) is nonexpansive and does not have any fixed points in the set \(K\).

Step-by-step solution

Understand the Function Space and Mapping

Consider the set \[K = \{ f \,\in\, L_1[0,1] \mid 0 \leq f \leq 2, \int_0^1 f(t) \, dt = 1 \}\].This set contains integrable functions on the interval [0, 1] that are bounded between 0 and 2 and have an integral equal to 1. The operator \( T: K \to K \) is defined piecewise for \( t \in [0, 1] \).

Define the Operator T

The operator \(T(f)(t)\) is given by:\[ T(f)(t) = \left\{ \begin{array}{ll} \min (2 f(2 t), 2) & \text{if } 0 \leq t \leq \frac{1}{2} \max (0, 2 f(2t - 1) - 2) & \text{if } \frac{1}{2} < t \leq 1 \end{array} \right. \]This definition is crucial for analyzing the properties of \(T\).

Show T is a Nonexpansive Map

A map \(T\) is nonexpansive if \[ \| T(f) - T(g) \|_1 \leq \| f - g \|_1 \text{ for all } f, g \in K. \]Consider two functions \(f, g \in K\). For \( 0 \leq t \leq \frac{1}{2} \), \[ |T(f)(t) - T(g)(t)| = |\min(2f(2t), 2) - \min(2g(2t), 2)| \leq 2 |f(2t) - g(2t)|. \]For \( \frac{1}{2} < t \leq 1 \), \[ |T(f)(t) - T(g)(t)| = |\max(0, 2f(2t-1) - 2) - \max(0, 2g(2t-1) - 2)| \leq 2 |f(2t-1) - g(2t-1)|. \]This shows \(T\) does not expand the distance between any two functions in \(K\).

Show T Has No Fixed Point

A fixed point \(f \in K\) would satisfy \(T(f) = f\). Assume \(f\) is such a fixed point. Then,\for \(0 \leq t \leq \frac{1}{2} \), \[ f(t) = \min(2f(2t), 2),\]and for \(\frac{1}{2} < t \leq 1 \), \[ f(t) = \max(0, 2f(2t-1) - 2)\].This implies \(f\) is piecewise linear, but constructing such a function that fits the integral constraint \(\int_0^1 f(t) dt = 1\) and the piecewise conditions simultaneously leads to a contradiction. Therefore, no function in \(K\) can be a fixed point of \(T\).

Key concepts

Weakly compact sets

In the context of functional analysis, a weakly compact set is one where every sequence has a weakly convergent subsequence. This means, roughly, that although the sequence of functions might not converge in the traditional sense (strong convergence), there is a way to observe convergence in a weaker sense.
For the set K described in the exercise, this weak compactness plays an important role since it ensures that despite how functions in K may behave, we can always find some subsequence that converges weakly.
This property is crucial for proving existence results in many areas, especially in Banach space theory and the study of fixed points.
Understanding weak compactness requires familiarity with concepts like weak topology, Banach spaces, and sequential compactness. These elements intertwine to form the fundamental underpinnings of more advanced fixed point theorems.

Fixed point theory

Fixed point theory is an area of mathematical analysis dealing with the existence of points that are invariant under certain mappings. A point \(f\) is called a fixed point of a mapping \(T\) if \(T(f) = f\).
This concept has broad applications, ranging from differential equations to game theory and economics.
In the exercise, proving that \(T\) has no fixed point adds to understanding function behaviors in a constrained set.
The Banach Fixed Point Theorem is a cornerstone of this theory, typically applied in a complete metric space to guarantee a unique fixed point for contraction mappings. Even though \(T\) in the exercise is nonexpansive, methods from fixed point theory are used to analyze it.
Nonexpansive maps, which do not shrink distances but also do not expand them, present more challenging scenarios. Understanding the lack of a fixed point for \(T\) involves proving that no function \(f\) within the set \(K\) can satisfy \(T(f) = f\).

Superreflexive Banach spaces

A Banach space is a complete normed vector space. Superreflexive Banach spaces are a special class where every Banach space that is finitely representable within them is also reflexive. Reflexive spaces are ones where every bounded sequence has a weakly convergent subsequence.
These spaces are crucial in functional analysis because they exhibit more regular behavior.
The exercise hints at the complexity of fixed point theory within superreflexive Banach spaces. It is known that in general reflexive spaces, nonexpansive mappings have fixed points within bounded closed convex subsets. However, proving this specifically for superreflexive spaces remains a challenging open problem.
Superreflexivity implies uniform convexity in finite-dimensional subspaces, providing more control over approximations. The study of such spaces is essential for deepening the understanding of functional analysis and has implications in optimization and theoretical computer science.