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Let $A$ be a closed, convex, set in a Banach space $X$, and let $B$ be a closed, bounded, convex set in $X$. Assume that $A \cap B = \emptyset$. Set $C = A- B$. Prove that $C$ is closed, and convex.

So proving $C$ is convex is not too hard, however I am having issues proving it is closed. According to existing convex geometry literature, this is a well-known result, that requires the assumption of $B$ being bounded. However, I am not sure how to use it. I assume the proof would begin by picking $c_n \in C$ such that $c_n \to c$, $c \in X$, and showing that in fact that $c \in C$. We would write: $$ c_n = a_n - b_n $$ and somehow show that $c = a -b$ for some $a,b \in C$, presumably $a = \lim_{n \to \infty} a_n$, $b = \lim_{n \to \infty} b_n$. But the existence of the limit for $c$ says nothing about the limits of $a_n,b_n$ (it can be deduced that $a_n$ is norm bounded using the norm bound on $b_n$), but I am truly at a loss on how to proceed. A hint would be appreciated.

rubikscube09
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  • Are you sure $X$ is a Banach space? Off the top of my head, I can come up with a proof for reflexive spaces, using the fact that the unit ball is weakly compact. I can also come up with a counterexample if $B$ is not bounded. However, for arbitrary Banach spaces I do not yet see the answer. – SmileyCraft Jan 19 '19 at 18:52
  • So the book I am using (Brezis'-Functional Analysis) actually does this in the context of Hilbert spaces! I realize now that he perhaps would intend on me decomposing $c_n = P_A(c_n) - P_B(c_n)$ and then passing to the limit (not sure if this works either because it does not use boundedness). But if you have a more general proof (in reflexive spaces) I would by all means be happy to see it. – rubikscube09 Jan 19 '19 at 18:55
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    Brezis Exercise 5.9. (For future readers coming from this reference.) – Tera Dec 10 '20 at 15:59

1 Answers1

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If $X$ is reflexive, then the theorem would hold. We find that $b_n$ must have a weak point of accumulation $b$ by the Banach-Alaoglu theorem and the Eberlein-Šmulian theorem. Then $b\in B$ and $c+b\in A$ by A convex subset of a Banach space is closed if and only if it is weakly closed and $c=c+b-b$.

That $B$ is bounded is really necessary for this theorem. Consider the following example.

Let $X=\mathbb{R}^2$ and $A=\{(x,y):x\geq0,xy\geq1\}$ and $B=\{(x,y):x\geq0,xy\leq-1\}$. Then $A$ and $B$ are disjoint, closed, convex and bounded. However $A-B=\{(x,y):y>0\}$ is not closed.

That $X$ is reflexive is also necessary for this theorem. Consider the following example.

Let $X=\ell^1$ and $A=\{x\in X:x_n\geq0,\sum x_n=1\}$. Define $T\in B(X)$ by $(Tx)_n=(1+\frac1n)x_n$ and let $B=T(A)$. Then $A$ and $B$ are disjoint, closed, convex and bounded. So $0\not\in A-B$. However $e_n-T(e_n)\to0$, so $A-B$ is not closed.

That $A$ and $B$ are convex is also necessary. Consider the following example.

Let $X=\ell^2$ and let $A$ be the closed unit ball and let $B=\{(1+\frac1n)e_n\}$. Then $A$ and $B$ are disjoint, closed and bounded, and $A$ is convex. So $0\not\in A-B$. However $e_n-(1+\frac1n)e_n\to0$, so $A-B$ is not closed.

That $A$ and $B$ are convex is, however, not necessary in the finite dimensional case. We then find that $b_n$ must have a point of accumulation $b$ by the Bolzano-Weierstrass theorem. Then $b\in B$ and $c+b\in A$ and $c=c+b-b$.

SmileyCraft
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  • Nice counterexample! – gerw Jan 19 '19 at 19:44
  • Thank you, I understand the use of Alaoglu's theorem to extract a weak limit $b$, however would this have to phrased in the language of nets because compact $\not \implies$ sequentially compact in general? – rubikscube09 Jan 19 '19 at 19:48
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    @rubikscube09 Good point. I looked it up, and it follows from the Eberlein Smulian theorem. – SmileyCraft Jan 19 '19 at 19:51
  • Right, I had seen that result before but technically we have not proved it in class yet. I will attempt to write the result in terms of nets rather than sequences. – rubikscube09 Jan 19 '19 at 19:53
  • Also a point of clarification: if weak closedness is equivalent to strong closedness for convex sets, then is it necessarily true that the strong and weak closures of a convex set coincide? I feel this should be true but it is weird to think about. – rubikscube09 Jan 19 '19 at 20:04
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    You have a bit of a typo, but that is a brilliant question worthy of its own post :) – SmileyCraft Jan 19 '19 at 20:05
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    Nevermind, it's rather silly. Weakly closed implies strongly closed, so $\mathrm{cl}(K) \subset \mathrm{cl-wk}(K)$ as $\mathrm{cl-wk}(K)$ is a closed set containing $K$ . Meanwhile, strongly closed + convex implies weakly closed, so $\mathrm{cl-wk}(K) \subset \mathrm{cl}(K)$ because $\mathrm{cl}(K)$ is a weakly closed set containing $K$. – rubikscube09 Jan 19 '19 at 20:20