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Weak convergence (Hilbert space)

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(Redirected from Banach–Saks theorem)

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

Definition

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

{\displaystyle \lim _{n\to \infty }\langle x_{n},y\rangle =\langle x,y\rangle }

for all {\displaystyle y} in {\displaystyle 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.[1]

Properties

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  • 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 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 {\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}\to x} weakly and {\displaystyle \lVert x_{n}\rVert \to \lVert x\rVert }, then {\displaystyle x_{n}\to 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 weak and strong convergence are equivalent.

Example

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The first 3 curves in the sequence fn=sin(nx)
The first three 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 Lp 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

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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 xH, we have

{\displaystyle \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

{\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

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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

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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}(\Omega )}, where {\displaystyle \mu } is the measure on {\displaystyle \Omega } and {\displaystyle {\frac {1}{p}}+{\frac {1}{q}}=1} are conjugate indices.

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.

See also

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References

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  1. ^ "redirect". dept.math.lsa.umich.edu. Retrieved 2024-09-17.
Weak convergence (Hilbert space)
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