🧐Functional Analysis Unit 9 Review

The weak topology in normed spaces is a crucial concept in functional analysis. It's a less restrictive topology than the norm topology, allowing for more convergent sequences and compact sets. This makes it a powerful tool for studying linear functionals and operators.

Understanding the weak topology helps in analyzing convergence of sequences and series in infinite-dimensional spaces. It's particularly useful in applications to differential equations, optimization, and quantum mechanics, where weak convergence often occurs naturally.

Weak Topology on Normed Spaces

Weak topology in normed spaces

Defines the coarsest topology on a normed space $X$ that ensures continuity of all continuous linear functionals on $X$
A subset $U \subset X$ is weakly open if for every $x \in U$, there exist continuous linear functionals $f_1, \ldots, f_n \in X^*$ and $\epsilon > 0$ such that the set ${y \in X : |f_i(y) - f_i(x)| < \epsilon, i = 1, \ldots, n}$ is contained in $U$
The basis for the weak topology on $X$ consists of sets of the form ${x \in X : |f_i(x) - a_i| < \epsilon, i = 1, \ldots, n}$, where $f_1, \ldots, f_n \in X^*$, $a_1, \ldots, a_n \in \mathbb{R}$, and $\epsilon > 0$
- These basis sets are called weak neighborhoods ($\ell^p$ spaces, $L^p$ spaces)

Comparison of weak vs norm topologies

To prove the weak topology is coarser than the norm topology, show every weakly open set is open in the norm topology
Let $U$ be a weakly open set in $X$ and $x \in U$
- There exist continuous linear functionals $f_1, \ldots, f_n \in X^*$ and $\epsilon > 0$ such that ${y \in X : |f_i(y) - f_i(x)| < \epsilon, i = 1, \ldots, n} \subset U$
By continuity of linear functionals in the norm topology, there exists $\delta > 0$ such that $|y - x| < \delta$ implies $|f_i(y) - f_i(x)| < \epsilon$ for all $i = 1, \ldots, n$
- This means the open ball $B(x, \delta) = {y \in X : |y - x| < \delta}$ is contained in $U$ ($C[0,1]$, $\ell^1$)
Thus, every weakly open set is open in the norm topology, proving the weak topology is coarser than the norm topology

Weak topology in normed spaces, An Introduction to Fuzzy Topological Spaces

Hausdorff property of weak topology

To show the weak topology is Hausdorff, prove for any two distinct points $x, y \in X$, there exist disjoint weakly open sets $U$ and $V$ such that $x \in U$ and $y \in V$
By the Hahn-Banach theorem, there exists a continuous linear functional $f \in X^*$ such that $f(x) \neq f(y)$
- Let $a = f(x)$ and $b = f(y)$, and choose $\epsilon > 0$ such that $(a - \epsilon, a + \epsilon) \cap (b - \epsilon, b + \epsilon) = \emptyset$
Define weakly open sets $U = {z \in X : |f(z) - a| < \epsilon}$ and $V = {z \in X : |f(z) - b| < \epsilon}$
- $x \in U$, $y \in V$, and $U \cap V = \emptyset$ ($\ell^2$, $L^2$)
Therefore, the weak topology is Hausdorff

Characteristics of weakly closed sets

A subset $C \subset X$ is weakly closed if and only if for every sequence $(x_n)$ in $C$ that converges weakly to $x \in X$, we have $x \in C$
- Equivalently, $C$ is weakly closed if and only if it is closed in the weak topology ($c_0$, $\ell^\infty$)
A subset $K \subset X$ is weakly compact if and only if every sequence in $K$ has a subsequence that converges weakly to a point in $K$
- By the Eberlein-Šmulian theorem, $K$ is weakly compact if and only if it is weakly sequentially compact
In a reflexive Banach space, a subset is weakly compact if and only if it is bounded and weakly closed
- This follows from the Banach-Alaoglu theorem and the fact that the weak topology on a reflexive space coincides with the weak* topology on its dual ($L^p$ spaces with $1 < p < \infty$)

🧐Functional Analysis Unit 9 Review