# Weak convergence (Hilbert space)

In mathematics, weak convergence in a Hilbert space is convergence of a sequence of points in the weak topology.

## Definition

A sequence of points $(x_n)$ in a Hilbert space H is said to converge weakly to a point x in H if

$\langle x_n,y \rangle \to \langle x,y \rangle$

for all y in H. Here, $\langle \cdot, \cdot \rangle$ is understood to be the inner product on the Hilbert space. The notation

$x_n \rightharpoonup x$

is sometimes used to denote this kind of convergence.

### Weak topology

Weak convergence is in contrast to strong convergence or convergence in the norm, which is defined by

$\Vert x_n -x \Vert \to 0$

where $\Vert x \Vert = \sqrt {\langle x,x \rangle}$ is the norm of x.

The notion of weak convergence defines a topology on H and this is called the weak topology on H. In other words, the weak topology is the topology generated by the bounded functionals on H. It follows from Schwarz inequality that the weak topology is weaker than the norm topology. Therefore convergence in norm implies weak convergence while the converse is not true in general. However, if for each $y, \langle x_n, y \rangle \to \langle x, y \rangle$ and $\| x_n \| \to \|x\|$, then we have $\| x_n - x \| \to 0$ as $n \to \infty.$

On the level of operators, a bounded operator T is also continuous in the weak topology: If xnx weakly, then for all y

$\langle Tx_n, y \rangle = \langle x_n, T^* y \rangle \rightarrow \langle x, T^*y \rangle = \langle Tx, y \rangle.$

## Properties

• If a sequence converges strongly, 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 infinitely 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.
• 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 $x_n$ converges weakly to $x$ and we have the additional assumption that $\lVert x_n \rVert \to \lVert x \rVert$, then $x_n$ converges 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 the concepts of weak convergence and strong convergence are the same.

### Example

The first 3 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, i.e.

$\langle f_n,g \rangle \to \langle 0,g \rangle = 0.$

While $f_n$ will be equal to zero more frequently as $n$ goes to infinity, it is not very similar to the zero function at all. 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 xH, 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

$| \langle e_n, x \rangle |^2 \rightarrow 0$

i.e.

$\langle e_n, x \rangle \rightarrow 0 .$

## Banach-Saks theorem

The Banach-Saks theorem states that every bounded sequence $x_n$ contains a subsequence $x_{n_k}$ and a point x such that

$\frac{1}{N}\sum_{k=1}^N x_{n_k}$

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 $(x_n)$ in a Banach space B is said to converge weakly to a point x in B if

$f(x_n) \to f(x)$

for any bounded linear functional $f$ defined on $B$, that is, for any $f$ in the dual space $B'.$ If $B$ is a Hilbert space, then, by the Riesz representation theorem, any such $f$ has the form

$f(\cdot)=\langle \cdot,y \rangle$

for some $y$ in $B$, so one obtains the Hilbert space definition of weak convergence.