(

Functional analysis,

Banach spaces:)

Let B be a Banach space, and let B^{*} be its dual vector space: the set of continuous linear functionals on B.
The *weak topology* on B is the topology whose base is given by the sets

{ x∈B | φ_{1}(x)∈U_{1}, ..., φ_{k}(x)∈U_{k} },

where φ

_{i}∈B

^{*} are functionals on B and U

_{i} are

open sets of

**R** or

**C** (depending where we're working;

it doesn't really matter).

#### What it means:

This topology is defined in terms of convergence of sequences of elements of B. A sequence x_{1},x_{2},...∈B *converges to x in the weak topology* iff for any φ∈B^{*}, the sequence φ(x_{i}) converges to φ(x). The linear functionals can be thought of as providing the values of various coordinates for B; we want convergence in any possible coordinate, instead of convergence in the distance that B's norm defines. For a finite dimensional vector space B, this happens iff the sequence of x_{i}'s converges (in the topology defined by the norm of B).

But in an infinite dimensional space, the weak topology really is weaker. If lim_{i->∞}x_{i} = x then for any continuous φ we also have lim_{i->∞}φ(x_{i}) = φ(x). The converse is not true!

For instance, take B=l

^{2}: the set of all sequences of

real numbers a=(a

_{1},a

_{2},...) for which ||a||

_{2} =

sqrt(∑

_{i}|a

_{i}|

^{2}) < ∞. Then B

^{*}=l

^{2}=B also; we'll write elements of B

^{*} as φ=(f

_{1},f

_{2},...). φ acts on a by

φ(a) = ∑_{i} f_{i}*a_{i}.

Now take the sequence of "

basis vectors" for B, {e

_{i}}

_{i}, where e

_{i} is the sequence with all elements except for the i'th element 0, and the i'th element 1:

e_{1} = (1,0,0,0,...,0,...)

e_{2} = (0,1,0,0,...,0,...)

e_{3} = (0,0,1,0,...,0,...)

...

This sequence does

*not* converge in the norm of l

^{2}: the distance ||e

_{i}-e

_{j}||

_{2}=2 for all i≠j, so it's not a

Cauchy sequence.

But the sequence *does* converge to 0 in the weak topology! For any φ∈B^{*}=l^{2}, since ∑_{i}|f_{i}|^{2}<∞, we must have lim_{i->∞}f_{i} = 0. So

lim_{i->∞}φ(e_{i}) =
lim_{i->∞}f_{i} = 0

In other words, if we look at the k'th coordinates of all the e

_{i}'s, that sequence converges to 0 (indeed, it

*is* zero, starting at i=k+1).

Thus, weak convergence need not imply "strong" convergence (in norm), even in a Hilbert space.