9.1 Weak topology on normed spaces

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

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• 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

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 $1 < p < \infty$)