Skip to main content

Section 5.6 Weak Topology

In a Banach space, in particular in spaces of infinite dimension, we find a large variety of notions of convergence. The default notion of convergence would be using the norm in a Banach space, i.e. a sequence \(\{x_n\}\subset X\) converges strongly to \(x\in X\text{,}\) if and only if
\begin{equation*} \|x_n- x\|\to 0, \qquad \mbox{as} \quad n\to \infty. \end{equation*}
A weaker form of convergence is the following definition, which uses the dual space as the judge of convergence.

Definition 5.6.1.
Let \(X\) be a Banach space. A sequence \(\{x_n\}\subset X\)converges weakly to \(x\text{,}\) which we denote as
\begin{equation*} x_n \rightharpoonup x, \end{equation*}
if \(f(x_n)\to f(x)\) for all \(f\in X^*\) as \(n\to \infty.\)
We see immediately that weak convergence is weaker than strong convergence in the next Lemma:
Each \(f\in X^*\) is continuous by definition, hence as \(x_n\to x\) we automatically have \(f(x_n) \to f(x)\text{,}\) which implies weak convergence.
On the other hand, a weak convergent sequence does not necessarily have to be strongly convergent as the following example shows.
[weak does not imply strong] Let \(H\) be a separable Hilbert space with countable orthonormal basis \(\{e_i\}\text{.}\) We will show that \(e_i \rightharpoonup 0\) as \(i\to\infty\text{.}\) We test with elements from \(H^*\text{.}\) By the Riesz Representation Theorem 5.5.1 each \(f\in H^*\) has a unique representation \(x_f\in H\) such that \(f(x) = (x_f, x)\) for all \(x\in H\) and \(\|f\|=\|x_f\|\text{.}\) In particular \(f(e_i) = (x_f, e_i)\text{.}\) The norm of \(x_f\) can be expressed as
\begin{equation*} \|f\|^2 = \|x_f\|^2 = \sum_{i=1}^\infty |(x_f, e_i) |^2, \end{equation*}
which is a convergent series. Hence \(|(x_f, e_i)|\to 0\) for \(i\to \infty\) and we have weak convergence to \(0\text{.}\)

On the other hand \(\|e_i - 0 \|=\|e_i\| = 1 \) for all \(i\text{,}\) hence \(e_i\) does not converge strongly to \(0\text{.}\)
Assume \(x_n\rightharpoonup x\) and \(x_n \rightharpoonup y\text{.}\) Then \(f(x) = f(y)\) for all \(f\in H^*\) and by Corollary 5.2.11 we find \(x=y\text{,}\) i.e. uniqueness.

To show boundedness we consider a test function \(f\in X^*\text{.}\) Then \(\{f(x_n)\}\) is a convergent sequence in \(\RR\text{,}\) hence it is bounded \(|f(x_n)|\leq C_f\) for all \(n\text{.}\) Now we associate \(x_n\) with an element \(G_n\in X^{**}\) as \(G_n(f) = f(x_n)\) for all \(f\in X^*\text{.}\) Then \(|G_n(f)|\leq C_f\) for all \(n\text{.}\) By the Uniform Boundedness Principle Theorem 4.4.4 it follows that \(\|G_n\|_{**}\) is bounded. But \(\|G_n\|_{**} = \|x_n\|\) by definition, hence \(\{x_n\}\) is bounded.
For each of \(f\in X^*\) we have \(f(x_n)\to f(x)\text{.}\) Now consider \(g\in Y^*\text{.}\) Then \(gA\in X^*\) and
\begin{equation*} g(Ax_n) = (gA)(x_n) \to (gA)(x) = g(Ax), \quad \mbox{for} \quad n\to \infty, \end{equation*}
hence we have already weak convergence \(Ax_n\rightharpoonup Ax\text{.}\) Now \(\{x_n\}\) is bounded and \(A\) is compact. Hence \(\{A x_n\}\) has a convergent subsequence which converges strongly in \(Y\text{.}\) But such a strong convergent sequence is also weakly convergent, and weak limits are unique. Hence we must have that \(Ax_n\to Ax\) in \(Y\text{.}\)