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Section 5.7 Weak\(^*\) Topology

Weak\(^*\) convergence is often perceived of something very strange. I hope to convince you that this is not true. It is a close cousin to weak convergence with the only difference that test functions are not chosen from the dual but from the pre-dual. Let us look at this in detail.

Definition 5.7.1.
Let \(X\) be a Banach space. A sequence \(\{f_n\}\subset X^*\) is weak\(^*\) convergent, denoted by
\begin{equation*} f_n \stackrel{*}{\rightharpoonup} f, \end{equation*}
if \(f_n(x) \to f(x) \) for all \(x\in X\text{.}\)

Compared to weak convergence in Definition 5.6.1, it seems we simply switched the roles of \(x\) and \(f\text{.}\) That's a good way to see it. In applications, you need to understand if you are currently working in the function space \(X\) or in its dual space \(X^*\text{.}\) In \(X\text{,}\) you would use weak convergence and in \(X^*\) you would by default, use weak\(^*\) convergence.

However, as \(X^*\) is again a Banach space, we can use test functions in \(X^{**}\) and still talk about weak convergence in \(X^*\text{.}\) If the Banach space is reflexive \(X^{**} = X\text{,}\) then chosing test functions in \(X\) or in \(X^{**}\) is the same, hence then weak convergence = weak\(^*\) convergence.

Weak\(^*\) convergence has one special gem, which does not hold in general for weak convergence, and that is a compactness result.
The proof uses a very classical diagonal sequence argument. Let \(\{ x_k\}\subset X\) denote a countable dense subset of \(X\text{.}\) For the first element \(x_1\) we consider the sequence \(\{ f_n (x_1)\}\text{.}\)
  • \(\{f_n(x_1)\}\) is a bounded sequence in \(\RR\) and it has a convergent subsequence \(\{f_{n_{1j}}(x_1)\}_j\text{.}\)
We take this subsequence and evaluate it at \(x_2\) and so forth. \\
  • \(\{f_{n_{1j}}(x_2)\}\) is a bounded sequence in \(\RR\) and it has a convergent subsequence \(\{f_{n_{2j}}(x_2)\}_j\text{.}\)
  • \(\displaystyle \vdots\)
  • \(\{f_{n_{kj}}(x_{k+1})\}\) is a bounded sequence in \(\RR\) and it has a convergent subsequence \(\{f_{n_{(k+1)j}}(x_{k+1})\}_j\text{.}\)
\noindent We obtain nested sequences
\begin{equation} \{f_n\} \supseteq \underbrace{ \underbrace{\underbrace{\{f_{n_{1j}}\}}_{\mbox{\footnotesize convergent at } x_1} \supseteq\{f_{n_{2j}}\}}_{\mbox{\footnotesize convergent at } x_1, x_2} \supseteq \cdots \supseteq \{f_{n_{kj}}\}}_{\mbox{\footnotesize convergent at } x_1, \dots, x_k} \supseteq \cdots \tag{5.7.1} \end{equation}

Hence we define a diagonal sequence \(g_j := f_{n_{jj}}\) such that \(\{g_j(x_k)\} \) converges for each \(x_k\text{.}\) In \(\ep-\delta\) notation this means that for each \(\ep>0\) we can find an index \(J>0\) such that
\begin{equation*} |g_j(x_n) - g_l(x_n) |<\ep, \quad \mbox{for all } j,l > J. \end{equation*}
Now for \(x\in X\) we can always find an \(x_k\in X\) with \(\|x-x_k\|<\ep\text{.}\) By assumption all \(g_j\) are bounded and continuous, hence
\begin{equation*} |g_j(x_k)-g_j(x) |\leq \|g_j\|_{op} \|x_k - x\|\leq M\ep, \end{equation*}
where \(M\) denotes the bound on the \(f_n\text{.}\) Taking all these estimates together we find
\begin{align*} |g_j(x) - g_l(x) |\amp\leq\amp |g_j(x) - g_j(x_k)| + |g_j(x_k) - g_l(x_k)| + |g_l(x_k) - g_l(x) |\\ \amp\leq \amp M\ep + \ep + M\ep = (2M+1) \ep. \end{align*}
Then \(\{g_j(x)\}_j\) is a Cauchy sequence in \(\RR\) for each \(x\in X\text{.}\) Hence, by definition, \(\{g_j\}\) is weak\(^*\) convergent \(g_j \stackrel{*}{\rightharpoonup} g\text{.}\) It is now easy to show that the limit \(g\) is linear and bounded, hence \(g\in X^*\text{.}\)

A direct consequence is the reflexive-weak compactness result.
We know that \((X^*)^*=X\text{,}\) hence by Alaoglu a bounded sequence in \(X\) is a bounded sequence in the dual of the dual \((X^*)^*\text{,}\) and has a weak\(^*\) convergent subsequence with test functions in \(X^*\text{.}\) But this is weak convergence in \(X\text{.}\)
Let \(\{f_k\}\subset L^p(\Omega)\) be bounded for a smooth bounded domain \(\Omega\) and \(1<p<\infty\text{.}\) Then
\begin{equation*} (L^p(\Omega))^* = L^q(\Omega), \quad \frac{1}{p}+\frac{1}{q} =1, \quad (L^p(\Omega))^{**} = L^p(\Omega). \end{equation*}
Hence \(\{f_k\}\) has a weak convergent subsequence \(\{f_{n_k}\}\text{.}\) This means there exists an \(f\in L^p(\Omega)\) such that
\begin{equation*} \int_\Omega g f_{n_k} dx \to \int_\Omega g f dx \quad \mbox{ as } k\to \infty, \quad \mbox{for all } g\in L^q(\Omega). \end{equation*}

The spaces \(L^1\) and \(L^\infty\) are not reflexive, hence there is a real difference between weak and weak\(^*\) convergence. If we consider a bounded sequence \(\{f_n\}\subset L^\infty(\Omega)\text{,}\) then we understand \(L^\infty(\Omega)= (L^1(\Omega))^*\) as dual space of \(L^1\text{.}\) Hence \(\{f_n\}\) is a bounded sequence in \((L^1(\Omega))^*\) and from Alaoglu's Theorem 5.7.3 we get a weak\(^*\) convergent subsequence \(\{f_{n_k}\}\) such that
\begin{equation*} \int_\Omega f_{n_k} g dx \to \int_\Omega f g dx \quad \mbox{for all } g\in L^1(\Omega). \end{equation*}