The subspace generated by a convex and closed subset of a Banach space Let $X$ be a Banach space and $C$ a norm-closed convex subset of $X$. Is the linear subspace generated by $C$ 
 norm-closed in $X$?
 A: No. Let $X=L^2(0,1)$. Define $C=\{u\in X: \ |u(x)|\le 1\}$. This set is closed and convex. Its convex hull contains only bounded functions. The closure of the convex hull is equal to $X$.
To see this, let $u\in L^2(0,1)\setminus L^\infty(0,1)$, hence $u$ is not in the linear hull of $C$. Define $u_n = \max(-n,\min (u(x),n))$. Then $n^{-1} u_n\in C$ and $u_n\to u$ by the dominated convergence theorem.
A: Another example would be the Hilbert cube $C$, that is the set of all $x= (x_n) \in l^2_{\{1,2,\ldots\}}$ so that $|x_n|\le \frac{1}{n}$ for all $n\ge 1$. $C$ is convex and compact, but the span of $C$ consists of sequences $x_n$ such that 
$(n x_n)$ is bounded, so it's not all of the space. For instance $(\frac{1}{\sqrt{n} \log (n+1)})$ is not in the span. However, the span is dense, because ot contains all basis elements $e_n$.
Note that if $C$ is convex and symmetric then the span of $C$ equals $\cup_{n\ge 0} nC$. If that is space is closed then $C$ must contain a neighborhood of the origin in it. That would not happen in general. 
$C$ convex and symmetric can be described by by a family of continuous functionals $(\phi_i)_{i\in I}$:
$$C=\{x \in X\ | \ |\phi_i(x)|\le 1\  \forall i\in I\}$$
Then the span of $C$ is 
$$\{ x \in X \ | \ \exists M\  \forall i\in I \ |\phi_i(x)|\le M \ \}$$
If $(\psi_j)_{j\in J}$ are all the continuous functionals such that 
$\psi_j$ are $0$ on $C$ then the closure of the span of $C$ is 
$$\{x\in X\ | \ \forall j \in J\  \psi_j(x)=0\}$$
So in general the span is not closed. 
