For convex set，weak closed if and only if strong closed I know the proof needs to use Hahn-Banach theorem，so the property "Convex" can be used. But I wonder how can I understand this theorem more vividly. Could  somebody give me your understanding about the convex property here.  Thanks！
 A: When looking at this theorem, one way is obvious: since the weak topology is weaker than the norm topology, any weakly-closed set must be norm-closed.  When we look at the other direction we choose a point $x \in X\setminus C$ where $X$ is our Banach space and $C$ is our norm-closed convex set.
The key idea here is that if you have two convex sets that do not intersect, then they "bend away" from each other (this is the geometric idea that $C$ contains the segment between any two of its points completely is making exact) and that allows us to fit a separating hyperplane between them -- that's where the Hahn-Banach theorem comes in.  The convex sets need to be closed, since otherwise their closures might "close up" the space between them and squeeze the hyperplane out.  In this case, $C$ is a closed convex set by hypothesis, and $\{x\}$ is a closed, convex (compact!) set in the norm-topology so we can apply the Hahn-Banach theorem to get a hyperplane and some $\alpha$ so that $x^*(x) < \alpha < x^*(c) \quad \forall c\in C$ and any functional $x^* \in X^*$.  Then $(x^*)^{-1}((\alpha, \infty))$ is a weakly open set that contains $x$ and doesn't intersect $C$ completing the proof.
It's worth noting that although the norm- and weak-topologies normally only coincide for finite-dimensional spaces this theorem gives us an idea that the topologies might coincide on specific sets in infinite dimensional spaces (which is true), which can be useful at times.
A: Apart from the issue about convexity, I think one of the crucial things about the proof is the following:
$(X,\|\cdot\|)^{\ast}=(X,w)^{\ast}$, where the former is the set of all continuous linear functionals with respect to the normed topology, where the later is the set of all continuous linear functionals with respect to the weak topology.
So a normed-continuous linear functional is also weakly-continuous and vice versa.
When one uses the Hahn-Banach Theorem, one picks for continuous linear functional to separate the points, because of the fact that $(X,\|\cdot\|)^{\ast}=(X,w)^{\ast}$, picking the normed or the weak topololgy one does not matter, so that is the reason why Hahn-Banach Theorem does the same effect to the norm-closure and weak-closure, so to speak.
Convexity is just for the condition of Hahn-Banach Theorem being utilized, personally I think the fact that $(X,\|\cdot\|)^{\ast}=(X,w)^{\ast}$ is much more crucial.
