For one reason or the other I work in a reflexive infinite dimensional Banach space $X$ and I am interested in curves $$ \gamma:[0,1]\to X $$ that are only weakly continuous.
I have no idea how to visualise their behaviour.
My best understanding of how bad things can go is that we have sequences on the unit sphere that converge to any point inside the unit ball. From this observation it is clear that I can always build this behaviour on one point, but how much can I push it? Even a countable set seems doable, but I would still get a locally strongly continuous curve.
In general that means that I could expect curves to wildly jump inside and outside of some fixed ball. I would be interested in one such example if there is one. In such constructions one things tends to happen, and it seems that the derivative ought to explode.
We know that the weak topology in a reflexive Banach space $X$ can be metrized locally when the space $X$ is separable, and thus we obtain a notion of Lipschitz continuous curve and absolutely continuous curve when considered with respect this metric.
So my question are:
- Can you show an example of a weakly continuous curve that is not strongly continuous in at least a dense set if not more, and/or
- Are there examples of weakly continuous curves in separable reflexive Banach spaces that are not strongly continuous but are absolutely continuous or Lipschitz with respect to the metrization of some ball?
A very simple way to get an example of a weakly continuous curve that is not strongly continuous is to just take a weakly convergent sequence that is not strongly convergent and interpolate it to an entire curve. Specifically, suppose $(x_n)$ is a sequence in $X$ that converges to a point $x$ weakly but not strongly. Define $\gamma:[0,1]\to X$ by $\gamma(1/n)=x_n$, $\gamma(0)=x$, and $\gamma(t)$ interpolates linearly between $x_n$ and $x_{n+1}$ as $t$ goes from $1/n$ to $1/(n+1)$. This is clearly not strongly continuous since $x_n$ does not converge to $x$ strongly. It is clearly weakly continuous everywhere except possibly at $0$. To see it is weakly continuous at $0$, let $U$ be a weak neighborhood of $x$, which we may assume is convex. Since $x_n\to x$ weakly, for some $N$ we have $x_n\in U$ for all $n\geq N$. Convexity of $U$ then implies that $\gamma(t)\in U$ for all $t\leq 1/N$.
Less explicitly, the Hahn-Mazurkiewicz theorem says that any connected, locally connected compact metrizable space is a continuous image of $[0,1]$. In particular, this includes a closed ball with the weak topology in any separable reflexive Banach space $X$. So, there is a weakly continuous curve $\gamma:[0,1]\to X$ whose image is the entire closed unit ball. When $X$ is infinite-dimensional such a curve will have to be rather wildly strongly discontinuous, since the unit ball is so far from being strongly compact.