Intereting Posts

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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.

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**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.

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