Finite sum of convex closed sets.

Question

In a normed space, is the finite sum of closed convex subsets which containing zero closed convex?

Answer 1

I assume you are talking about sum ($A+B = \{a+b \; : a \in A, b \in B\}$) and not union.

No, in infinite dimensions the sum of two closed convex subsets need not be closed. For example, in a Banach space the sum of two closed subspaces need not be closed. See e.g. my answer here for a counterexample in a Hilbert space.

EDIT: This is also true for $\mathbb R^2$. Consider $A = \{(x,y): y \ge e^x-1 \}$ and $B$ the nonnegative $x$ axis $\{(x,0): x \ge 0\}$. For any $y > -1$, $(0,y) \in A + B$ because we can write it as $(-x, y) + (x,y)$ for some $x > 0$ such that $e^{-x} < 1 + y$, but $(0,-1) \notin A + B$.