# Is the sum of two closed convex subsets of a real normed space closed?

Let $K_1,K_2$ closed convex sets of a real normed space $V$, both containing 0. Is $K_1+K_2$ closed? Can you give-me a counterexample?

I aim to prove the following result:

Let $K_1,...,K_n$ closed convex sets of a normed space $V$, and let $c_1,...c_n$ positive real numbers. Prove that, if $x\in V$ can not be of the form $x=c_1x_1+\cdots c_nx_n$ for $x_i\in K_i$, $i=1,...,n$, then there exist $f\in V^*$ (the dual of $V$) such that $f (x)>1$ and $f (y)\leq \frac{1}{c_i}$ for all $y\in K$ and $i=1,...,n$.

I can prove it if $c_1K_1+\cdots+c_nK_n$ is closed, but couldn't prove if this is true.

This is not true even for closed subspaces of a Hilbert space. Consider the space of square-summable sequences $\ell^2$ and Let $$K_1 = \{x\in \ell^2 : x_{2k} = 0 \ \forall k\}, \quad K_2 = \{x\in \ell^2 : x_{2k-1} = k\,x_{2k} \ \forall k\}$$ It is easy to see that every eventually-zero sequence $y$ can be written as the sum of an element of $K_1$ and an element of $K_2$; just solve the systems $$(t, 0) + (ks, s) = (y_{2k-1}, y_{2k}) \tag{1}$$ for each $k$, which yields $$s = y_{2k},\quad t = y_{2k-1}- ky_{2k} \tag{2}$$ Therefore, $K_1+K_2$ is dense in $\ell^2$.

But $K_1+K_2$ does not contain the element $y = (1/k : k\in\mathbb{N})$, because formulas $(2)$ result in sequences that do not even tend to $0$, let alone being square-summable.

Another way to see this is that every element of $K_2$ satisfies $\lim_{k\to\infty} k x_{2k} = 0$, and so does (trivially) every element of $K_1$, so the property passes to the sum $K_1+K_2$.

HINT:

An example of two convex closed subsets of $\mathbb{R}^2$ whose sum is not closed.

Let $$K_1 =\{(x,y) \ | \ x,y > -1 \textrm{ and } (x+1)(y+1)\ge 1 \}$$ $$K_2 =\{(x,0) \ | \ x \le 0 \}$$

One checks that $K_1$, $K_2$ are closed and convex, and their sum is $$K_1 + K_2 = \{(x,y)\ | \ y>-1\}$$.

• But in this case, $0\notin K_1$ Commented Sep 20, 2017 at 13:03
• @Filburt: oh, didn't notice that. Basically, they should not be disjoint. Commented Sep 20, 2017 at 13:21
• @Filburt: you can check what I added Commented Sep 20, 2017 at 13:43