• If a sequence converges strongly (that is, if it converges in norm), then it converges weakly as well. • Since every closed and bounded set is weakly
relatively compact (its closure in the weak topology is compact), every
bounded sequence x_n in a Hilbert space
H contains a weakly convergent subsequence. Note that closed and bounded sets are not in general weakly compact in Hilbert spaces (consider the set consisting of an
orthonormal basis in an infinite-dimensional Hilbert space which is closed and bounded but not weakly compact since it doesn't contain 0). However, bounded and weakly closed sets are weakly compact so as a consequence every convex bounded closed set is weakly compact. • As a consequence of the
principle of uniform boundedness, every weakly convergent sequence is bounded. • The norm is (sequentially) weakly
lower-semicontinuous: if x_n converges weakly to
x, then ::\Vert x\Vert \le \liminf_{n\to\infty} \Vert x_n \Vert, :and this inequality is strict whenever the convergence is not strong. For example, infinite orthonormal sequences converge weakly to zero, as demonstrated below. • If x_n \to x weakly and \lVert x_n \rVert \to \lVert x \rVert, then x_n \to x strongly: ::\langle x - x_n, x - x_n \rangle = \langle x, x \rangle + \langle x_n, x_n \rangle - \langle x_n, x \rangle - \langle x, x_n \rangle \rightarrow 0. • If the Hilbert space is finite-dimensional, i.e. a
Euclidean space, then weak and strong convergence are equivalent.
Example Image:Sinfrequency.jpg|alt=The first 3 curves in the sequence fn=sin(nx)|thumb|350px|The first three functions in the sequence f_n(x) = \sin(n x) on [0, 2 \pi]. As n \rightarrow \infty f_n converges weakly to f =0. The Hilbert space L^2[0, 2\pi] is the space of the
square-integrable functions on the interval [0, 2\pi] equipped with the inner product defined by :\langle f,g \rangle = \int_0^{2\pi} f(x)\cdot g(x)\,dx, (see
Lp space). The sequence of functions f_1, f_2, \ldots defined by :f_n(x) = \sin(n x) converges weakly to the zero function in L^2[0, 2\pi], as the integral :\int_0^{2\pi} \sin(n x)\cdot g(x)\,dx. tends to zero for any square-integrable function g on [0, 2\pi] when n goes to infinity, which is by
Riemann–Lebesgue lemma, i.e. :\langle f_n,g \rangle \to \langle 0,g \rangle = 0. Although f_n has an increasing number of 0's in [0,2 \pi] as n goes to infinity, it is of course not equal to the zero function for any n. Note that f_n does not converge to 0 in the L_\infty or L_2 norms. This dissimilarity is one of the reasons why this type of convergence is considered to be "weak."
Weak convergence of orthonormal sequences Consider a sequence e_n which was constructed to be orthonormal, that is, :\langle e_n, e_m \rangle = \delta_{mn} where \delta_{mn} equals one if
m =
n and zero otherwise. We claim that if the sequence is infinite, then it converges weakly to zero. A simple proof is as follows. For
x ∈
H, we have : \sum_n | \langle e_n, x \rangle |^2 \leq \| x \|^2 (
Bessel's inequality) where equality holds when {
en} is a Hilbert space basis. Therefore : | \langle e_n, x \rangle |^2 \rightarrow 0 (since the series above converges, its corresponding sequence must go to zero) i.e. : \langle e_n, x \rangle \rightarrow 0 . ==Banach–Saks theorem==