# Ngô Quốc Anh

## October 28, 2009

### The weak and weak* topologies: A few words

Filed under: Giải tích 8 (MA5206), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 2:48

The weak and weak* topologies are the weakest in which certain linear functionals are continuous.

We start with a normed linear space $X$. The dual space of $X$, denoted by $X'$, is the collection of all continuous linear functionals, i.e., the set of all mapping $\ell : X \to \mathbb R$ satisfying $\ell(ax)=a \ell (x)$, $\ell(x+y)=\ell(x)+\ell(y)$

and $\displaystyle\lim_{n \to \infty} \ell(x_n) = \ell(x)$ when $\displaystyle\lim_{n \to \infty} \|x_n - x\|=0$.

Definition 1. In $X$, the strong topology is the norm topology, i.e., we can talk about an open set of $X$, for example $U$ in the following sense: $U \subset X$ is said to be open if and only if for each $x_0 \in U$, there exists $\varepsilon>0$ such that $\{ x \in X: \|x-x_0\|<\varepsilon\} \subset U$.

Claim 1. Bounded linear functionals are continuous in the strong topology.

Proof. We first recall that a linear functional $\ell$ is said to be bounded if there is a positive number $c$ such that $|\ell (x)| \leq c\|x\|$ for all $x \in X$.

Now we assume $\ell$ is continuous but not bounded; then for any choice of $c=n$, one has $\ell(x_n) > n \|x_n\|$. Clearly, $x_n$ can be replaced by any multiple of $x_n$; if we normalize $x_n$ so that $\displaystyle \|x_n\|=\frac{1}{\sqrt{n}}$

then $x_n \to 0$ but $\ell (x_n) \to \infty$. This shows the lack of boundedness implies the lack of continuity.

Now we assume $\ell$ is bounded. For arbitray $x_n$ and $x$, one gets $|\ell(x_n)-\ell (x)| = |\ell (x_n-x)| \leq c\|x_n-x\|$;

this shows that boundedness implies continuity.

Definition 2. In $X$, the weak topology is the weakest topology in which all bounded linear functionals are continuous.

The open sets in the weak topology are unions of finite intersections of sets of the form $\{ x : a< \ell(x) < b\}$.

Clearly, in an infinite-dimensional space the intersection of a finite number of sets of the above form is unbounded. This shows that every set that is open in the weak topology is unbounded. In particular, the balls $\{ x : \|x\|

opens in the strong topology, are not open is the weak topology.

Definition 3. In $X'$ the dual space of $X$, the weak* topology is the crudest topology in which all linear functionals $x: X' \to \mathbb R, x(\ell) := \ell(x)$

are continuous.

If $X'$ is nonreflexive, the weak* topology is genuinely coarser than the weak topology, as will be clear from the following theorem due to Alaoglu

Theorem (Alaoglu). The closed unit ball in $X'$ is compact in the weak* topology.

We end this topic by the following theorem

Theorem. The closed unit ball in $X$ is compact in the weak topology if and only if $X$ is reflexive.

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