Appendix 3
Uniform, Strong and Weak Convergence
Sequences of operators may be shown to converge with respect to different topologies than the one induced by the operator norm. The same can be said for sequences of elements in Banach or Hilbert spaces.
To take a concrete example, consider the following case. Let (un)n∈ℕ be an arbitrary Hilbert basis in a Hilbert space 
. For all n ∈ ℕ, we define the linear operator:
From the geometric characterization of projection operators (see Theorem 6.32), we know that An is the orthogonal projector on the vector subspace of generated by u1, . . . , un: Sn = span(u1, . . . , un).
Since any x ∈ may be written as 
, it would seem that the sequence of projectors (An)n∈ℕ converges toward id when n → +∞.
Nevertheless, since Sn ⊂ Sn′ ∀n < n′, we know by Theorem 6.35 that An′ – An is the projector onto 
, thanks to the orthogonality of the vectors (un)n∈ℕ. As we have seen, all orthogonal projectors onto non-trivial subspaces have a unitary norm, that is, ‖An′ – An‖ = 1 ∀n < n′, and thus the sequence (An)n∈ℕ is not a Cauchy sequence in 
() with respect to the operator norm; thus, it cannot be convergent because () is complete and so convergent and Cachy sequences coincide.
The sequence (Anx)n∈ℕ in , however, converges to x for all x ∈ , by the fact that (un)n∈ℕ is a Hilbert basis. This highlights the need to define an alternative form of convergence in order to assign a precise meaning to the intuitive notion that the sequence (An)n∈ℕ converges to id.
Similar examples are encountered in Banach and Hilbert spaces; for this reason, we have organized our presentation of alternative forms of convergence into separate sections for different spaces.
A3.1. Strong and weak convergence in Banach spaces
Let (V, ‖‖) be a Banach space. By definition, a sequence (xn)n∈ℕ ⊂ V converges toward x ∈ V if 
. A different type of convergence can be defined in V by using the continuous linear functionals of its dual space V*.
DEFINITION A3.1 (Weak convergence in a Banach space).– Let V be a Banach space. The sequence (xn)n∈ℕ ⊂ V converges weakly toward x ∈ V if, for all ϕ ∈ V*:
where the convergence in this case is that of sequences of scalars in 
. x is the weak limit of the sequence (xn)n∈ℕ and we write 
, with w for weak.
We note that, for all ϕ ∈ V* and x ∈ V, ‖ϕ(x) ‖ ≼ ‖ ϕ ‖ x ‖, thus, if 
, then:
that is, “standard” convergence implies weak convergence. For this reason, “standard” convergence in a Banach space is also referred to as strong convergence.
Counter-examples show that the inverse is not generally true. Thus, in a Banach space, the topology defined by weak convergence has fewer opens than the topology defined by strong convergence.
A3.2. Strong and weak convergence in a Hilbert space
A Hilbert space is also a Banach space, thus the definition of strong and weak convergence given above also applies to Hilbert spaces. Nevertheless, by the Riesz representation theorem, we know that the action of any continuous linear functional on can be identified with a scalar product. For this reason, an equivalent definition, which is more explicit for the purposes of calculation, can be used for weak convergence in a Hilbert space.
DEFINITION A3.2 (weak convergence in a Hilbert space).– Let be a Hilbert space. The sequence (xn)n∈ℕ ⊂ converges weakly toward x ∈ if, for all y ∈ :
As in the case of Banach spaces, x is said to be the weak limit of the sequence (xn)n∈ℕ and we write 
A very simple counter-example can be used to show that weak convergence does not generally imply strong convergence in a Hilbert space.
Take any y ∈ and xn = un ∀n ∈ ℕ, where (un)n∈ℕ is an arbitrary orthonormal system in . By Bessel’s inequality 
, so the series is convergent and thus its general term tends toward 0.
Since is complete, any series which is absolutely convergent is convergent, hence 
is convergent and then 
; however, this holds if and only if 
for all y ∈ .
Hence, any orthonormal system (un)n∈ℕ in a Hilbert space is weakly convergent toward 0.
However, we know that the distance between any two elements of an orthonormal system is 
, thus (un)n∈ℕ does not verify the Cauchy condition, and therefore it cannot be strongly convergent.
A3.3. Uniform, strong and weak convergence in the Banach algebra ()
In the Banach algebra ((), ‖ ‖), where is any Hilbert space and ‖ ‖ is the operator norm, three different convergences can be defined for a sequence of operators (An)n∈ℕ ⊂ ().
DEFINITION A3.3.– We shall use u, s and w to denote uniform, strong and weak. Let (An)n∈ℕ ⊂ () be a sequence of bounded linear operators on the Hilbert space , and take A ∈ ().
– Uniform convergence (standard convergence, in operator norm):
– Strong convergence:
– Weak convergence:
As we saw at the beginning of this appendix, for any Hilbert basis (un)n∈ℕ, the sequence:
does not converge uniformly id. However, it converges strongly towards the identity operator, since, by the continuity of the norm, we have:
having used the fact that id(x) is not dependent on n and the generalized Fourier expansion on the Hilbert basis (un)n∈ℕ.
It is possible to show that, in (), uniform convergence implies strong convergence, which itself implies weak convergence. On the other hand, as we see from the example shown above, strong convergence does not imply uniform convergence. Other counter-examples can be used to show that weak convergence in () does not imply strong convergence.