Weak Convergence in Hilbert Space
Theorem
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\HH$.
Let $x \in X$.
Then:
- $\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$
- $\innerprod {x_n} y \to \innerprod x y$ for each $y \in \HH$.
Corollary
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\HH$.
Let $x \in X$.
Then:
- $\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$
- $\innerprod y {x_n} \to \innerprod y x$ for each $y \in \HH$.
Proof
Let $\struct {\HH^\ast, \norm \cdot_{\HH^\ast} }$ be the normed dual space of $\HH$.
Necessary Condition
Suppose that:
- $\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$
For each $y \in \HH$ define the function $f_y : \HH \to \R$ by:
- $\map {f_y} x = \innerprod x y$
for each $x \in \HH$.
From Inner Product with Vector is Bounded Linear Functional, it follows that $f_y$ is a bounded linear functional.
So, by the definition of the normed dual:
- $f_y \in \HH^\ast$
By hypothesis we have:
- $\map {f_y} {x_n} \to \map {f_y} x$
That is:
- $\innerprod {x_n} y \to \innerprod x y$
$\Box$
Sufficient Condition
Suppose that:
- $\innerprod {x_n} y \to \innerprod x y$ for each $y \in \HH$.
Let:
- $f \in \HH^\ast$
Then from the Riesz Representation Theorem (Hilbert Spaces), there exists $z \in \HH$ such that:
- $\map f x = \innerprod x z$
for each $x \in \HH$.
By hypothesis, we have:
- $\innerprod {x_n} z \to \innerprod x z$
so:
- $\map f {x_n} \to \map f x$
Since $f \in \HH^\ast$ was arbitrary, we have:
- $\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$.
$\blacksquare$
Sources
- 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $27.1$: Weak Convergence