(
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 | φ1(x)∈U1, ..., φk(x)∈Uk },
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 x1,x2,...∈B converges to x in the weak topology iff for any φ∈B*, the sequence φ(xi) 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 xi's converges (in the topology defined by the norm of B).
But in an infinite dimensional space, the weak topology really is weaker. If limi->∞xi = x then for any continuous φ we also have limi->∞φ(xi) = 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 fi*ai.
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:
e1 = (1,0,0,0,...,0,...)
e2 = (0,1,0,0,...,0,...)
e3 = (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*=l2, since ∑i|fi|2<∞, we must have limi->∞fi = 0. So
limi->∞φ(ei) =
limi->∞fi = 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.