Weak Convergence in Hilbert Space

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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 $x$

if and only if:

$\innerprod {x_n} y \to \innerprod x y$ 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 $x$

if and only if:

$\innerprod y {x_n} \to \innerprod y x$ 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