Show that any convex subset of a Banach space X is closed with respect to the norm if and only if it is closed in the weak topology

Question

How can it be proved? In a way which could be understood by a undergraduate math student. Specially if and "only if".

Answer 1

In the sequel, $E$ is our Banach space and $E'$ its dual. I consider $E$ a real vector space. The "strong" topology is the one induced by some fixed norm $\Vert\cdot\Vert$.

The "if" part:

Let's denote $B_{w}:=B(a,u_{1},\dots,u_{n},R):=\left\{x\in E\,\vert\,|u_{i}(x-a)|<R,\,i=1,\dots,n\right\}$ where $R>0$, $u_{i}\in E'$ and $a\in E$ (for some $n\in\mathbb{N}\setminus\{0,1\}$). The weak topology on $E$ is defined by

$$\mathcal{T}_{w}(E):=\left\{A\,\vert\,A=\cup B_{w}\right\}$$

(in other words, $A$ is open if $A$ is a union of some $B_{w}$). One can prove that it is indeed a topology.

Let $x$ be in $B(a,u_{1},\dots,u_{n},R)$ and $y\in E$ such that $\Vert x-y\Vert<\epsilon$. We want to prove that there exists such an epsilon $\epsilon$ such that $y\in B(a,u_{1},\dots,u_{n},R)$. In other words, we want to prove that any (strongly) open ball is contained in some $B_{w}$. By triangular inequality and by linearity of $u_{i}$ for all $i=1,\dots,n$, we have $$|u_{i}(y-a)|\le \underbrace{|u_{i}(x-a)|}_{<R-\delta}+|u_{i}(y-x)|$$ for some $\delta>0$. We have

$$|u_{i}(y-x)|\le\Vert u_{i}\Vert_{\text{op}}\cdot\Vert y-x\Vert<\Vert u_{i}\Vert\epsilon<\delta$$ for some suitable $\epsilon$, e.g. $\epsilon=\frac{\delta}{2u}$ where $u:=\max\left\{\Vert u_{i}\Vert_{\text{op}}\,\vert\,i=1,\dots,n\right\}$. Then, we have

$$|u_{i}(y-a)|<R-\delta+\delta=R$$

for any $i=1,\dots,n$. The "strong" topology is finer than the weak one, so that any closed set w.r.t. the weak topology is a closed set w.r.t. the "strong" topology.

The "only if" part:

Let $C$ be a "strongly" closed and convex subset. Define $U=E\setminus C$ and $z\in U$. The Hahn-Banach theorem* provides the existence of some $u\in E'$ and $\alpha\in\mathbb{R}$ such that

$$u(z)<\alpha<u(y)$$

for all $y\in C$. Then, $\left\{x\in E\,\vert\,u(x)<\alpha\right\}$ is a "weakly" open set included in $U$ and containing $z$. It is true for any $z\in U$, so that $U$ is "weakly" open and $C$ is "weakly" closed.

*One version of the Hahn-Banach theorem can be expressed as follows: Let $V$ be a real vector space, and $A,B\subset V$ two non-empty convex sets. If $A\cap B=\emptyset$ and $\text{int}(A)\neq\emptyset$, then there exists an hyperplane separating $A$ and $B$, i.e. $f\in V'$ and $\alpha\in\mathbb{R}$ such that

$$f(a)<\alpha<f(b)$$

for all $a\in A$ and for all $b\in B$.