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

We start with a normed linear space . The dual space of , denoted by , is the collection of all continuous linear functionals, i.e., the set of all mapping satisfying

,

and

when .

**Definition 1**. In , the *strong topology* is the norm topology, i.e., we can talk about an open set of , for example in the following sense: is said to be open if and only if for each , there exists such that .

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

*Proof*. We first recall that a linear functional is said to be bounded if there is a positive number such that for all .

Now we assume is continuous but not bounded; then for any choice of , one has . Clearly, can be replaced by any multiple of ; if we normalize so that

then but . This shows the lack of boundedness implies the lack of continuity.

Now we assume is bounded. For arbitray and , one gets

;

this shows that boundedness implies continuity.

**Definition 2**. In , 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

.

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

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

**Definition 3**. In the dual space of , the *weak* topology* is the crudest topology in which all linear functionals

are continuous.

If 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 is compact in the weak* topology.

We end this topic by the following theorem

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