Sum of closed convex set and unit ball in normed space

Let $(X, \|.\|)$be a real normed space. Let $A$ be a closed convex suset of $X$ and $\mathbb{B}$ a unit ball in X, i.e.
\mathbb{B}=\{x\in X: \|x\|\leq 1\}.
I would like to ask whether $A+\mathbb{B}$ is still closed.

Solutions Collecting From Web of "Sum of closed convex set and unit ball in normed space"

Edit: Here is a more complete solution.

Proposition. Suppose $X$ is a Banach space. Then $A + \mathbb{B}$ is closed for every closed convex $A$ if and only if $X$ is reflexive.

Proof. One direction I posted previously. If $X$ is reflexive then $\mathbb{B}$ is weakly compact. If $A$ is closed and convex, it is weakly closed. Then by my answer here, applied to the weak topology, $A+\mathbb{B}$ is weakly closed, and in particular norm closed.

For the converse, James’s Theorem asserts that if $X$ is not reflexive, there exists a continuous linear functional $f \in X^*$ which does not attain its norm, i.e. (after rescaling) $\|f\| = 1$ but there is no $x \in \mathbb{B}$ with $f(x) = 1$. Set $A = \{ x : f(x) \ge 1\}$; $A$ is obviously closed and convex. Also, there exist $x_n \in \mathbb{B}$ with $f(x_n) \to 1$; let $y_n = \frac{1}{f(x_n)} x_n$, so that $f(y_n) = 1$ and thus $y_n \in A$. We have $\|y_n + (-x_n)\| = (f(x_n) – \frac{1}{f(x_n)}) \|x_n\| \to 0$, so that $0$ is a limit point of $A + \mathbb{B}$. Now if $0 \in A + \mathbb{B}$, i.e. there are $x \in \mathbb{B}$ and $y \in A$ with $x+y=0$, then $y=-x$ and so $y \in A \cap \mathbb{B}$. Since $y \in A$, we have $f(y) \ge 1$. Since $y \in \mathbb{B}$, $f(y) \le 1$. Hence $f(y)=1$, but by assumption no such $y$ exists.

For a simple example of a functional which does not attain its norm, consider $X = \ell^1$, $f(x) = \sum_n (1 – \frac{1}{n}) f(n)$.

Here is an example of an incomplete normed space where $A + \mathbb{B}$ need not be closed.

Let $X$ be a dense subspace of $C([0,1])$ which does not contain the constants. (For example, take $X$ to be the set of all functions $f(x) = p(x) + e^{x}$, where $p$ is a polynomial with $p(0) = 0$. $X$ is dense because by the Weierstrass theorem we can approximate $e^x-1$ by polynomials vanishing at 0, so the constants are in the closure of $X$, whence by Weierstrass again, the closure of $X$ is dense.) $X$ is still equipped with the uniform norm.

Set $A = \{f \in X : f \ge 1\}$. $A$ is easily seen to be closed and convex in $X$. Since $X$ is dense in $C([0,1])$, for each $n$ we can choose $f_n, g_n \in X$ with $\|f_n – (1 + 1/n)\| < 1/n$, $\|g_n – (-1 + 1/n)\| < 1/n$. Then $f_n \in A$ and $g_n \in \mathbb{B}$. It is easy to see that $f_n + g_n \to 0$ uniformly, so 0 is a limit point of $A+\mathbb{B}$. On the other hand, if $f \in A$ and $g \in \mathbb{B}$, then $f \ge 1$ and $g(x) \ge -1$, so $f+g \equiv 0$ only if $f \equiv 1$ and $g \equiv -1$, which is impossible. Hence $0 \notin A+\mathbb{B}$ and so $A+\mathbb{B}$ is not closed.

One question remains: does this property fail in every incomplete normed space? Another paper of James gives an example of an incomplete normed space where every continuous linear functional attains its norm, so the construction from the non-reflexive case will not carry over.

Let $(e_n)_{n\in\mathbb{N}}$ be an orthonormal basis in a Hilbert space and let $A = \{(1+\frac{1}{n})e_n : n \in \mathbb{N}\}$.

In any topological vector space, the sum of a closed set and a compact set is closed. The only book I have handy is “Topological Spaces” by Claude Berge. In that book, this result is theorem 5 in chapter 9. I assume that other books have the same result.

Note: you don’t need to require that one of the sets be convex, or that one of them is a unit ball, or that the space has a norm.

I think it’s clear that $B$ is closed, because the norm function $X \rightarrow \mathbb{R}$ is continuous, and $B$ is the preimage of the closed unit interval. Also, the addition function $+: X \times X \rightarrow X$ is continuous. For any closed $A$ and $B$, $A + B = +[A \times B]$ is the continuous image of a closed set, and is therefore closed. So the answer is yes.