Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer
Thank you for your interesting arguments. I am waiting the counterexample of you. – blindman Oct 16 '12 at 6:08
@blindman: Ok, I found one. – Nate Eldredge Oct 16 '12 at 6:23
I'm interested to know the answer to the Banach space version of this question. Maybe it would be worth posting another question? – Nick Alger Oct 16 '12 at 7:06
Nate Eldredge: Dear Sir. I have the same opinion with Nick Alger. I am waiting the Banach version of your counterexample. – blindman Oct 16 '12 at 8:44
@NickAlger: I've addressed it. – Nate Eldredge Oct 16 '12 at 19:38

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

share|cite|improve this answer
Dear Sir. Thank you for your anwser. Here $A$ is not convex. – blindman Oct 16 '12 at 3:15

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.

share|cite|improve this answer
This is true, but doesn't answer the question, as $A$, $\mathbb{B}$ could both be non-compact. – Nate Eldredge Oct 16 '12 at 4:33
@bubba: Be careful. $\mathbb{B}$ is not compact in infinite dimensional space $X$. – blindman Oct 16 '12 at 6:11
@blindman. Oops. It's been about 30 years since I looked at a topology book, and I see that my memory isn't as good as I thought. Apologies. – bubba Oct 16 '12 at 9:34

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.

share|cite|improve this answer
The continuous image of a closed set is not in general closed. Consider $f :\mathbb{R} \to \mathbb{R}$ with $f(x) = \tan^{-1} x$; $\mathbb{R}$ is closed but $f(\mathbb{R}) = (-\pi/2, \pi/2)$ is not. – Nate Eldredge Oct 16 '12 at 3:57
I just deleted a completely wrong answer... sorry. – coffeemath Oct 16 '12 at 4:00
Very good, thank you @Nate. I wonder if we can save that argument somehow, though. If we assume that $A$ is compact, we can say that the image of a compact set is compact, right? And in a metric space, compact sets are closed. – Hew Wolff Oct 16 '12 at 4:04
@HewWolff: Yes, if $A$ is compact then this is true. See my answer here, which as I mention works for any topological group. – Nate Eldredge Oct 16 '12 at 4:07
B is compact, so I don't think we need to require that A is also compact. – bubba Oct 16 '12 at 4:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.