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 ${\displaystyle (x_{n})}$ in a Hilbert space H is said to converge weakly to a point x in H if

${\displaystyle \langle x_{n},y\rangle \to \langle x,y\rangle }$

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

${\displaystyle x_{n}\rightharpoonup x}$

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 ${\displaystyle x_{n}}$ 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.
• As a consequence of the principle of uniform boundedness, every weakly convergent sequence is bounded.
• The norm is (sequentially) weakly lower-semicontinuous: if ${\displaystyle x_{n}}$ converges weakly to x, then

${\displaystyle \Vert x\Vert \leq \liminf _{n\to \infty }\Vert x_{n}\Vert ,}$

and this inequality is strict whenever the convergence is not strong. For example, infinite orthonormal sequences converge weakly to zero, as demonstrated below.

• If ${\displaystyle x_{n}}$ converges weakly to ${\displaystyle x}$ and we have the additional assumption that ${\displaystyle \lVert x_{n}\rVert \to \lVert x\rVert }$, then ${\displaystyle x_{n}}$ converges to ${\displaystyle x}$ strongly:

${\displaystyle \langle x-x_{n},x-x_{n}\rangle =\langle x,x\rangle +\langle x_{n},x_{n}\rangle -\langle x_{n},x\rangle -\langle x,x_{n}\rangle \rightarrow 0.}$

• 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 ${\displaystyle f_{n}(x)=\sin(nx)}$ on ${\displaystyle [0,2\pi ]}$. As ${\displaystyle n\rightarrow \infty }$ ${\displaystyle f_{n}}$ converges weakly to ${\displaystyle f=0}$.

The Hilbert space ${\displaystyle L^{2}[0,2\pi ]}$ is the space of the square-integrable functions on the interval ${\displaystyle [0,2\pi ]}$ equipped with the inner product defined by

${\displaystyle \langle f,g\rangle =\int _{0}^{2\pi }f(x)\cdot g(x)\,dx,}$

(see L^p space). The sequence of functions ${\displaystyle f_{1},f_{2},\ldots }$ defined by

${\displaystyle f_{n}(x)=\sin(nx)}$

converges weakly to the zero function in ${\displaystyle L^{2}[0,2\pi ]}$, as the integral

${\displaystyle \int _{0}^{2\pi }\sin(nx)\cdot g(x)\,dx.}$

tends to zero for any square-integrable function ${\displaystyle g}$ on ${\displaystyle [0,2\pi ]}$ when ${\displaystyle n}$ goes to infinity, which is by Riemann–Lebesgue lemma, i.e.

${\displaystyle \langle f_{n},g\rangle \to \langle 0,g\rangle =0.}$

Although ${\displaystyle f_{n}}$ has an increasing number of 0's in ${\displaystyle [0,2\pi ]}$ as ${\displaystyle n}$ goes to infinity, it is of course not equal to the zero function for any ${\displaystyle n}$. Note that ${\displaystyle f_{n}}$ does not converge to 0 in the ${\displaystyle L_{\infty }}$ or ${\displaystyle L_{2}}$ 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 ${\displaystyle e_{n}}$ which was constructed to be orthonormal, that is,

${\displaystyle \langle e_{n},e_{m}\rangle =\delta _{mn}}$

where ${\displaystyle \delta _{mn}}$ 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

${\displaystyle \sum _{n}|\langle e_{n},x\rangle |^{2}\leq \|x\|^{2}}$ (Bessel's inequality)

where equality holds when {e_n} is a Hilbert space basis. Therefore

${\displaystyle |\langle e_{n},x\rangle |^{2}\rightarrow 0}$ (since the series above converges, its corresponding sequence must go to zero)

i.e.

${\displaystyle \langle e_{n},x\rangle \rightarrow 0.}$

Banach–Saks theorem

The Banach–Saks theorem states that every bounded sequence ${\displaystyle x_{n}}$ contains a subsequence ${\displaystyle x_{n_{k}}}$ and a point x such that

${\displaystyle {\frac {1}{N}}\sum _{k=1}^{N}x_{n_{k}}}$

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 ${\displaystyle (x_{n})}$ in a Banach space B is said to converge weakly to a point x in B if

${\displaystyle f(x_{n})\to f(x)}$

for any bounded linear functional ${\displaystyle f}$ defined on ${\displaystyle B}$, that is, for any ${\displaystyle f}$ in the dual space ${\displaystyle B'}$. If ${\displaystyle B}$ is an Lp space on ${\displaystyle \Omega }$, and ${\displaystyle p<\infty }$ then, any such ${\displaystyle f}$ has the form

${\displaystyle f(x)=\int _{\Omega }x\,y\,d\mu }$

For some ${\displaystyle y\in \,L^{q}(B)}$ where ${\displaystyle {\frac {1}{p}}+{\frac {1}{q}}=1}$ and ${\displaystyle \mu }$ is the measure on ${\displaystyle \Omega }$.

In the case where ${\displaystyle B}$ is a Hilbert space, then, by the Riesz representation theorem,

${\displaystyle f(\cdot )=\langle \cdot ,y\rangle }$

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