What does it mean for a sequence $\{f_n\}_{n=1}^\infty\subseteq H$ to converge weakly? I know it means that it converges in the weak topology and I've read a few definitions of weak topology which all seemed quite confusing and it seems like I'm missing something important here.

  • 3
    $\begingroup$ Why don't you state what you know about weak convergence first? $\endgroup$ – Batman Mar 1 '14 at 5:22
  • $\begingroup$ Normally it means that you've given up on vector convergence. :) $\endgroup$ – DisintegratingByParts Mar 1 '14 at 15:06

In Banach space $X$ a sequence $\{f_n\}$ converges weakly to $f$ if $$ \varphi(f_n)\to\varphi(f), $$ for all $\varphi\in X^*$, where $X^*$ is the dual of $X$.

In the case of Hilbert space $H$, every element of the dual space is realized by an element of $H$ (Riesz Representation Theorem). Thus $f_n\to f$ weakly if and only if $$ \langle f_n,\varphi\rangle\to\langle f,\varphi\rangle, $$ for all $\varphi\in H$.


Let $\{f_n\}_{n=1}^\infty$ be a sequence in $H$, and let $f\in H$. Then:

$f_n\to f$ in the weak topology if and only if for all $\phi\in H^*$ one has the convergence $\phi(f_n)\to \phi(f)$ as $n\to \infty$.

Why is this? Let's first look at the direction ($\Rightarrow$). So we assume $f_n\to f$ in the weak topology, and want to prove $\phi(f_n)\to \phi(f)$. To this end, let $U\subseteq \mathbb{R}$ (or $\mathbb{C}$, if you are working over the complex numbers) be a neighborhood of $\phi(f)$. By the definition of the weak topology, the linear map $\phi\colon H\to \mathbb{R}$ when $H$ is equipped with the weak topology; thus, in particular, $\phi^{-1}(U)$ is a weak open neighborhood of $f$ in $H$. Since $f_n\to f$ in the weak topology, it follows that $f_n\in \phi^{-1}(U)$ for all large enough $n$. Therefore $\phi(f_n)\in U$ for all large enough $n$. This proves that $\phi(f_n)\to \phi(f)$.

Now let's look at the direction ($\Leftarrow$). So we assume that $\phi(f_n)\to \phi(f)$ for all $\phi\in H^*$, and want to prove that $f_n\to f$ in the weak topology. To this end, let $V$ be a weak open neighborhood of $f$. Since the weak topology on $H$ has for a subbasis all sets of the form $\phi^{-1}(U)$ where $U\subseteq \mathbb{R}$ is open and $\phi\in H^*$, it follows that we can find a weak neighborhood $V'\subseteq V$ of $f$ of the form $$V' = \phi_1^{-1}(U_1)\cap \cdots\cap \phi_m^{-1}(U_m)$$ where $U_i\subseteq \mathbb{R}$ are open and $\phi_i\in H^*$. By our assumption, we know that $\phi_i(f_n)\to \phi_i(f)\in U_i$ for each $i$, and therefore $\phi_i(f_n)\in U_i$ for large enough $n$ for each $i = 1,\ldots, m$. That is to say, for large $n$, we have $f_n\in V'$. This proves that $f_n\to f$ in the weak topology.

I hope this answer was helpful, but since you haven't said what you know or exactly where you're confused, it might just be completely useless. In any case, if you have specific questions about any of the details here, please feel free to ask.

Edit: Just an extra remark. If $f_n\to f$ in the strong topology, then certainly $\phi(f_n)\to \phi(f)$ for every $\phi\in H^*$, since such $\phi$ are continuous, and thus $f_n\to f$ in the weak topology as well. The converse does not hold. For instance, if $\{e_n\}_{n=1}^\infty$ is any orthonormal basis in $H$, then the sequence $e_n$ converges weakly to $0$ (why?), but does not converge to $0$.


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