Weak convergence came up in my PDE class and I'm trying to understand it even though I lack background in topology and functional analysis. Please check if my understanding is correct.
Definition: Let $X$ be a real Banach space with norm $\left\lVert . \right\lVert_{X}.$ A sequence $(x_n)_{n \in \mathbb{N}}$ converges weakly to $x \in X,$ if for all continuous linear functionals $f: X \rightarrow \mathbb{R}, f(x_n) \rightarrow f(x).$
Question: Why is this equivalent to convergence in the weak topology?
$\bullet$ Weak topology: The coarsest topology so that all continuous linear functionals on $X$ are still continuous. Note that we started with some topology on $X$ with respect to which we identify the linear continuous functionals. In a Banach space, we start with the norm-induced topology (Please confirm if my understanding is correct, this is not explicitly stated in my course but I guess it is obvious).
$\bullet$ Convergence wrt a topology $\mathcal T$ on $X$: $(x_n)_{n \in \mathbb{N}} \subset X$ converges to $x \in X,$ if for every open set $\mathcal O \in \mathcal T,$ such that $x \in \mathcal O,$ there is $N \in \mathbb{N},$ so that $x_n \in \mathcal O$ $ \forall n \geq N.$
Is my intuition below correct?
The weak topology consists of sets $(\phi_{1})^{-1}(A_1), …, (\phi_{n})^{-1}(A_n),$ where $\phi_{k}$ is a linear continuous functional and $A_k$ an open set in $\mathbb{R}.$ (see link at the end of my post) If $x_n$ converges to $x$ wrt to the weak topology, then for each set of the above form, if $x \in (\phi_{1})^{-1}(A_1), …, (\phi_{n})^{-1}(A_n),$ $x_n \in (\phi_{1})^{-1}(A_1), …, (\phi_{n})^{-1}(A_n) \forall n \geq N$ for some $N \in \mathbb{N}.$ Using the definition of preimage, it means that $ \phi_{j}(x) \in A_j $ for some $j \in [n]$ implies that $\phi_{j}(x_n) \in A_j \forall n$ large enough. And we would have to do this for every single continuous linear functional. That's why the convergence wrt to the weak topology amounts to the definition I started with.
Is this intuition correct?
Some intuition about weak topology – Was useful for understanding the weak topology.
Best Answer
I rewrote your question, keeping and adjusting the correct statements, and adding a paragraph and Proposition.
Definition: Let $X$ be a real Banach space with norm $\left\lVert . \right\lVert_{X}.$ A sequence $(x_n)_{n \in \mathbb{N}}$ converges weakly to $x \in X,$ if for each continuous linear functional $f: X \rightarrow \mathbb{R}$, the sequence $f(x_n)$ converges to $f(x).$
$\bullet$ Weak topology: The coarsest topology so that all continuous linear functionals on $X$ are still continuous. Note that we started with some topology on $X$ with respect to which we identify the linear continuous functionals. In a Banach space, we start with the norm-induced topology.
It is easy to check that the weak topology has a subbase consisting of sets $f^{-1}(A)$, where $f:X\to \mathbb R$ is a continuous linear functional and $A$ is an open subset of $\mathbb{R}$.
$\bullet$ Convergence wrt a topology $\mathcal T$ on $X$: a sequence $(x_n)_{n \in \mathbb{N}}$ of points of $X$ converges to $x \in X,$ if for every open set $\mathcal O \in \mathcal T,$ such that $x \in \mathcal O,$ there is $N \in \mathbb{N},$ so that $x_n \in \mathcal O$ $ \forall n \geq N.$
Proposition. Let $X$ be a Banach space, $x\in X$ be any point, and $(x_n)_{n\in\mathbb N}$ be any sequence of points of $X$. Then $(x_n)_{n\in\mathbb N}$ converges to $x$ in the weak topology $\mathcal T$ on $X$ iff $(x_n)_{n\in\mathbb N}$ weakly converges to $x$.
Proof.
$(\Rightarrow)$ Suppose that the sequence $(x_n)_{n\in\mathbb N}$ converges to the point $x$ in the weak topology on $X$. Let $f:X\rightarrow \mathbb{R}$ be any continuous linear functional. Let $A$ be any neighborhood of the point $f(x)$ in $\mathbb R$. Then $f^{-1}(A)\in\mathcal T$, so there exists $N \in \mathbb{N}$ such that $x_n \in f^{-1}(A)$ for each natural $n\ge N$. Therefore $f(x_n)\in A$ for each $n\ge N$. It follows that the sequence $(f(x_n))_{n\in\mathbb N}$ converges to the point $f(x)$.
$(\Leftarrow)$ Suppose that the sequence $(x_n)_{n\in\mathbb N}$ weakly converges to the point $x$. Let $B\in\mathcal T$ be any neighborhood of the point $x$. There exist a natural number $K$ and for each $k\in\{1,\dots,K\}$ open set $A_k\in\mathbb R$ and a continuous linear functional $f_k:X\to\mathbb R$ such that $x\in\bigcap_{k=1}^K f^{-1}_k(A_k)\subset B$. Let $k\in\{1,\dots,K\}$ be any number. Since the sequence $(f_k(x_n))_{n\in\mathbb N}$ converges to $f_k(x)$, there exists $N_k \in \mathbb{N}$ such that $f_k(x_n)\in A_k$ for each natural $n\ge N_k$. Put $N=\max\{N_k:k\in\{1,\dots,K\}\}$. Let $n\ge N$ be any natural number and $k\in\{1,\dots,K\}$ be any number. Then $f_k(x_n)\in A_k$. It follows that $x_n\in \bigcap_{k=1}^K f^{-1}_k(A_k)\subset B$. Therefore the sequence $(x_n)_{n\in\mathbb N}$ converges to the point $x$ in the weak topology of the space $X$. $\square$