2
$\begingroup$

I have a general question about how can we characterize the weak and weak$^\ast$ topologies and continuous maps between then in normed spaces:

Let's start with a normed space $X$. The weak topology on $X$ is the smallest topology such that every $x^\ast\in X^\ast$ is a continuous linear functional. Here $X^\ast$ is the topological dual of $X$, therefore previously equipped with the operator norm topology. Similarly, the weak$^\ast$ topology is the weakest topology on $X^\ast$ such that every linear functional $J_X(x):X^\ast\rightarrow\mathbb{C}$ is continuous, where $J_X:X\rightarrow X^{\ast\ast}$ is the canonical embedding. Both these definitions are a bit unpractical when dealing with claims about continuity, so can we characterize continuity by sequential continuity, i.e., a map $T:X\rightarrow Y$ is weakly continuous (respectively, weak$^\ast$ continuous) iff $Tx_j\xrightarrow{w} Tx$ (respectively, $Tx_j\xrightarrow{w^\ast}Tx$) for every weakly convergent (respectively, weak$^\ast$ convergent) sequence $x_j\xrightarrow{w}x$? Of course this would be true for nets, but usually sequences are easier to work with.

I know generally sequences are not enough to characterize a topology, however I've seen this notion of continuity being used around and it got me wondering if that's the notion of weak and weak$^\ast$ topology that is being referred to in this context.

$\endgroup$
4
  • 1
    $\begingroup$ I'm not entirely clear on your goal in asking this question, but my general reaction is to not think in terms of sequential continuity, unless there's a special reason. The "inverse-image-of-open-set-is-open" is a more robust notion of continuity for most purposes, I think. The "weak" and/or "weak-dual" topologies are eminently describable in terms of local basis at $0$ of open sets... $\endgroup$ Aug 7 at 1:19
  • $\begingroup$ ... and, maybe, the sequential-continuity and the specifics of the weak and weak-dual topologies are inadvertently obscuring the simple general ideas of topologies on vector spaces. ? $\endgroup$ Aug 7 at 1:44
  • $\begingroup$ Let me give you an example where I'm not sure if I am using the right criteria: when showing that the adjoint map $T^\ast:Y^\ast\rightarrow X^\ast$ of a linear map $T:X\rightarrow Y$ between normed spaces $X, Y$, where $T^\ast x^\ast=x^\ast\circ T$, is weak$^\ast$-continuous the argument would be to show that $T^\ast x^\ast_j\xrightarrow{w^\ast} T^\ast x^\ast$ for every net that satisfies $x^\ast_j\xrightarrow{w^\ast} x^\ast$ in $Y^\ast$. Would it be possible to use only sequences here? Why? $\endgroup$ Aug 7 at 3:07
  • $\begingroup$ I'm aware of the description of both topologies using local basis, and also that there is a seminorm criteria for continuity that is reminiscent of the norm criteria when dealing more generally with LCS, but it would be nice to have a sequential characterization to justify this kind of argument. $\endgroup$ Aug 7 at 3:07
0
$\begingroup$

Sequences are not enough to characterize the weak topology or weak continuity.

Indeed, recall that the normed space $\ell^1$ has this peculiar property that weak convergence of sequences is equivalent to strong convergence of sequences. However, weak topology is not equal to the topology of the norm (in infinite dimension it never is since the open unit ball is not weakly open).

Similarly, the function $\|\cdot\|_1 : \ell^1 \to \Bbb{R}$ is strongly continuous, weakly sequentially continuous but not weakly continuous. Indeed, otherwise for any weakly convergent net $x_j \xrightarrow{w} x$ we would have $$\|x_j - x\| \to 0 \implies x_j \xrightarrow{s} x$$ which would imply that the topologies are equal, but they aren't.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.