Weak Convergence in Hilbert Spaces
Definition
Let $(H,\langle \cdot \rangle)$ be a Hilbert space, and let $\left\{ x_{n} \right\}$ be a sequence in $H$. For all $y\in H$, if the following equation holds, $\left\{ x_{n} \right\}$ is said to converge weakly and is denoted as $x_{n} \rightharpoonup x$. $$ \langle x_{n}, y \rangle \to \langle x , y \rangle ,\quad \forall y\in H $$
Following the ‘w’ in weak, it can also be denoted as:
$$ x_{n} \overset{\text{w}}{\to} x $$
Or
$$ x_{n} \to x \quad \text{weakly} $$
Explanation
To emphasize that it is not strong convergence, the term “converges” is sometimes written as converges strongly. That is,
$$ \begin{align*} &x_{n} \text{ converges to } x \\ =\ & x_{n} \text{ converges in norm to } x \\ =\ & x_{n} \text{ converges strongly to } x \end{align*} $$
On the other hand, the naming of weak convergence arises because it does not guarantee convergence convergence. Conversely, norm convergence is essentially the same as convergence in metric spaces, so in many cases, norm convergence and convergence are not strictly distinguished. In normed spaces, the distance can be defined as follows.
$$ d(x,y):=\left\| x-y \right\|,\quad x,y\in H $$
Then, for $\left\{ x_{n} \right\}$ satisfying $\lim \limits_{n \to \infty}x_{n}=x$,
$$ \lim \limits_{n \to \infty} d(x_{n},y)=d(x_{n},y) \iff \lim \limits_{n \to \infty} \left\| x_{n}-y \right\| =\left\| x-y \right\| $$
However, it can be understood that this does not hold in the case of inner products. By the Cauchy-Schwarz inequality, the following formula can be obtained.
$$ \left| \left\langle x_{n} , y \right\rangle \right| \le \left\| x_{n} \right\| \left\| y \right\| $$
Therefore,
$$ \lim \limits_{n \to \infty} \left\| x_{n}-x \right\|=0 \begin{array}{c} \implies \\ \ \ \not \!\!\!\!\impliedby \end{array} \lim \limits_{n \to \infty} \left\langle x_{n}-x,y \right\rangle=0,\ \forall y\in H $$
can be understood.
$$ x_{n} \to x \implies x_{n} \rightharpoonup x $$
Proof
Assume $x_{n} \to x$. Then, by the Cauchy-Schwarz inequality,
$$ \begin{align*} \left| \langle x_{n},y \rangle -\langle x,y \rangle \right| &= \left| \langle x_{n}-x, y \rangle \right| \\ & \le \left\| x_{n}-x \right\| \left\| y \right\| \end{align*} $$
Since it is assumed that $\lim \limits_{n\to\infty} \left\| x_{n} -x \right\|=0$,
$$ \lim \limits_{n\to\infty} \left| \langle x_{n},y \rangle -\langle x,y \rangle \right| =\lim \limits_{n\to\infty} \left\| x_{n}-x \right\| \left\| y \right\|=0 $$
■