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 $y \in \HH$.

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:


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

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 $x$.