In mathematics, weak convergence in a Hilbert space is the convergence of a sequence of points in the weak topology.
Definition
A sequence of points <math>(x_n)</math> in a Hilbert space <math>H</math> is said to converge weakly to a point <math>x</math> in <math>H</math> if
:<math>\lim_{n\to\infty}\langle x_n,y \rangle = \langle x,y \rangle</math>
for all <math>y</math> in <math>H</math>. Here, <math>\langle \cdot, \cdot \rangle</math> is understood to be the inner product on the Hilbert space. The notation
:<math>x_n \rightharpoonup x</math>
is sometimes used to denote this kind of convergence.
Properties
- 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 <math>x_n</math> 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 <math>x_n</math> converges weakly to x, then
::<math>\Vert x\Vert \le \liminf_{n\to\infty} \Vert x_n \Vert, </math>
:and this inequality is strict whenever the convergence is not strong. For example, infinite orthonormal sequences converge weakly to zero, as demonstrated below.
- If <math>x_n \to x</math> weakly and <math>\lVert x_n \rVert \to \lVert x \rVert</math>, then <math> x_n \to x</math> strongly:
::<math>\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.</math>
- 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 <math>f_n(x) = \sin(n x)</math> on <math>[0, 2 \pi]</math>. As <math>n \rightarrow \infty</math> <math>f_n</math> converges weakly to <math>f =0</math>.]]
The Hilbert space <math>L^2[0, 2\pi]</math> is the space of the square-integrable functions on the interval <math>[0, 2\pi]</math> equipped with the inner product defined by
:<math>\langle f,g \rangle = \int_0^{2\pi} f(x)\cdot g(x)\,dx,</math>
(see L<sup>p</sup> space). The sequence of functions <math>f_1, f_2, \ldots</math> defined by
:<math>f_n(x) = \sin(n x)</math>
converges weakly to the zero function in <math>L^2[0, 2\pi]</math>, as the integral
:<math>\int_0^{2\pi} \sin(n x)\cdot g(x)\,dx.</math>
tends to zero for any square-integrable function <math>g</math> on <math>[0, 2\pi]</math> when <math>n</math> goes to infinity, which is by Riemann–Lebesgue lemma, i.e.
:<math>\langle f_n,g \rangle \to \langle 0,g \rangle = 0.</math>
Although <math>f_n</math> has an increasing number of 0's in <math>[0,2 \pi]</math> as <math>n</math> goes to infinity, it is of course not equal to the zero function for any <math>n</math>. Note that <math>f_n</math> does not converge to 0 in the <math>L_\infty</math> or <math>L_2</math> 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 <math>e_n</math> which was constructed to be orthonormal, that is,
:<math>\langle e_n, e_m \rangle = \delta_{mn}</math>
where <math>\delta_{mn}</math> 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
:<math> \sum_n | \langle e_n, x \rangle |^2 \leq \| x \|^2</math> (Bessel's inequality)
where equality holds when {e<sub>n</sub>} is a Hilbert space basis. Therefore
:<math> | \langle e_n, x \rangle |^2 \rightarrow 0</math> (since the series above converges, its corresponding sequence must go to zero)
i.e.
:<math> \langle e_n, x \rangle \rightarrow 0 .</math>
Banach–Saks theorem
The Banach–Saks theorem states that every bounded sequence <math>x_n</math> contains a subsequence <math>x_{n_k}</math> and a point x such that
:<math>\frac{1}{N}\sum_{k=1}^N x_{n_k}</math>
converges strongly to x as N goes to infinity.
Generalizations
The definition of weak convergence can be extended to Banach spaces. A sequence of points <math>(x_n)</math> in a Banach space B is said to converge weakly to a point x in B if
<math display=block>f(x_n) \to f(x)</math>
for any bounded linear functional <math>f</math> defined on <math>B</math>, that is, for any <math>f</math> in the dual space <math>B'</math>. If <math>B</math> is an Lp space on <math>\Omega</math> and <math>p<+\infty</math>, then any such <math>f</math> has the form
<math display=block>f(x) = \int_{\Omega} x\,y\,d\mu</math>
for some <math>y\in\,L^q(\Omega)</math>, where <math>\mu</math> is the measure on <math>\Omega</math> and <math>\frac{1}{p}+\frac{1}{q}=1</math> are conjugate indices.
In the case where <math>B</math> is a Hilbert space, then, by the Riesz representation theorem,
<math display=block>f(\cdot) = \langle \cdot,y \rangle</math>
for some <math>y</math> in <math>B</math>, so one obtains the Hilbert space definition of weak convergence.
See also
- Operator topologies – topologies on the set of operators on a Hilbert space
