I'm working on the following question:
Show that the closed unit ball $B(0,1)$ in a normed space is also weakly closed.
I think I want to show that $$\lim_{n\rightarrow \infty}f(x_n) \in f(B(0,1)) \subset \mathbb{R}$$ for all $f \in X^*$
This is strange to me though because for each $f$, I will get a different image in $\mathbb{R}$. This correct though right? I'm checking that the limit lives in the image, so long as the image is created by an element of $X^*$.
EDIT: This question is not a duplicate of the linked question. In fact, I have an answer already. I wanted to get input on this idea that "I'm checking if the limit lives in ALL images formed from elements of $X^*$" is a way to think about this problem because I've never heard it posed that way before.
First, you cannot show weak closedness by weakly converging sequences.
Second, (sequential) closedness usually means that limits of converging sequences of elements from the closed set are in the closed set as well. For weak convergence, this has nothing to do with any functionals. So you idea does not work.