Lecture 14 - Weak Convergence

#math

In general closed bounded sets are not compact. In particular closed balls are not compact. So if $(x_{n})$ is a bounded sequence in such a space then we cannot say that $\exists$ a convergent subsequence, thus can we use some different notion of convergence to get some limit that we can find useful?

Direct Method of Calculus of Variations

We are interested in minimizing some functional $E : A \to [0, \infty)$ .
We want to take $(x_{n}) \subset A$ such that $E (x_{n}) \to inf_{A} E$ . Our aim is to ensure $| | x_{n} | |$ bounded.
We hope that $(x_{n})$ has a subsequence that converges in some way for some $x^{*}$ and show that $x^{*} \in A$ and $E (x^{*}) = \underset{x \in A}{Min} E (x)$ .

Hahn-Banach Theorem

Let $(X, | | \cdot | |_{X})$ be a Banach space, we denote all linear continuous functions on $X$ by $(X^{*}, | | \cdot | |_{X^{*}})$ is large. We look to a subspace $U \subset X$ and add a continuous linear functional only defined on $U$ , i.e. , $u^{*} : U \to K$ . Then $\exists$ of an extension of $u^{*}$ such that:

\begin{aligned} x^{*} : & X \to K \\ {x^{*} |}_{U} & = u^{*} \\ | | x^{*} | |_{X^{*}} & = | | u^{*} | |_{U^{*}} \end{aligned}

Example

Let $(X, | | \cdot | |_{X})$ be a normed space.
a) $\forall x \in X, x \neq 0 \exists x^{*} \in X^{*}$ such that $| | x^{*} | |_{X^{*}} = 1$ and $x^{*} (x) = | | x | |_{X}$
b) $X^{*}$ separates points of $X$ i.e. for $x_{1}, x_{2} \in X$ $x_{1} \neq x_{2} \exists x^{*} \in X^{*}$ such that $x^{*} (x_{1}) \neq x^{*} (x_{2})$

Proof

b) Use (a)
Let $x = x_{1} - x_{2}$ then $\exists x^{*}$ such that $x^{*} = | | x | |_{X} \neq 0$ then as $x^{*}$ is linear we get $x^{*} (x) = x^{*} (x_{1}) - x^{*} (x_{2}) \neq 0 ⟹ x^{*} (x_{1}) \neq x^{*} (x_{2})$ .

Corollary

$\forall x, y \in X$ with $x \neq y ⟹ \exists f \in X^{*}$ such that $f (x) \neq f (y)$ .

Definition

Let $X$ be a normed space, $(x_{n})$ a sequence in $X$ . We say $(x_{n})$ converges weakly to $x \in X$ if $\forall f \in X^{*}$ we have $f (x_{n}) \to f (x)$ .
We denote weak convergence either with $x_{n} \overset{w}{\to} x$ or $x_{n} ⇀ x$ .

Example

$l^{2} (C)$ with $e^{(n)}$ standard basis.

e_{k}^{(n)} = δ_{k}^{n}

We claim that $e^{n} ⇀ 0$ . Its clear that $e^{n} ↛ 0$

Proof

$\forall f \in (l^{2})^{*}$ , by Riesz representation theorem:

f (x_{n}) = \sum x_{k} y_{k}

so $f (e^{(n)}) = y_{n} \to 0$ as $\sum_{k = 1}^{\infty} (y_{k})^{2} < \infty$ .

What could be wrong with $L^{2} (R)$ ?

Example

Let $g_{n} = χ_{[0, \frac{1}{n}]} \sqrt{n}$ . Notice that $| | g_{n} | |_{L^{2}} = \int_{0}^{\frac{1}{n}} {\sqrt{n}}^{2} d x = 1 ↛ 0$ .
We claim $g_{n} ⇀ 0$ .

Proof

Let $T \in (L^{2})^{*}$ so $\exists f \in L^{2}$ such that $T (g) = \int f g$ . We have:

\begin{aligned} | T (g_{n}) | & = \int_{0}^{\frac{1}{n}} g_{n} \cdot f \\ \overset{Cauchy Schwarz}{\leq} (\int_{0}^{\frac{1}{n}} g_{n}^{2})^{\frac{1}{2}} (\int_{0}^{\frac{1}{n}} f^{2})^{\frac{1}{2}} \\ \leq {(\int_{R} f^{2} χ_{[0, \frac{1}{n}]})}^{\frac{1}{2}} \\ = 0 (By dominated convergence theorem) \end{aligned}