Take the 2-minute tour ×
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.

Let $(X,\|\cdot\|)$ be a Banach space and $A,B,C\subset X$ closed bounded non-empty convex subsets. Let $+$ denote the Minkowski symbol for addition.

Does the $+$ satisfy: $$A+C\subset B+C\implies A\subset B$$

share|improve this question
add comment

1 Answer 1

It does. Suppose $a\in A$ with $a\notin B$. Since $B$ is closed and not empty, there is $b\in B$ with minimal distance from $A$. Project everything onto $a-b$, and denote the projected objects by primes. Since the sets are closed and convex, their projections are closed intervals. $B'$ is all on one side of $a'$, since the line segment from $b$ to a point whose projection is on the other side of $a'$ would contain points closer to $a$ than $b$. Thus $a'+C'\not\subset B'+C'$, hence $a+C\not\subset B+C$ and thus $A+C\not\subset B+C$.

[Edit:]

As t.b. pointed out, this doesn't work in general, but as D. Thomine pointed out, it can be fixed using the Hahn–Banach theorem. Again, suppose $a\in A$ with $a\notin B$. Since $\{a\}$ is convex and compact and $B$ is convex and closed, there is a continuous linear map $\lambda:X\to\mathbb R$ strictly separating the two, so there is $s\in\mathbb R$ such that $\lambda(a)\lt s\lt\lambda(b)$ for all $b\in B$. Since $C$ is bounded and $\lambda$ is continuous, $\lambda(C)$ is bounded. Thus $\lambda(a)+\lambda(C)\not\subset\lambda(B)+\lambda(C)$, hence $a+C\not\subset B+C$ and thus $A+C\not\subset B+C$.

Note that only the boundedness of $C$, not that of $A$ or $B$ has been used.

share|improve this answer
    
How do you get that $b \in B$ with minimal distance from $A$ and how exactly do you project? I think this argument works fine if you assume the space to be uniformly convex (hence +1) but I don't see how to do this in full generality. Maybe I'm missing something simple here. –  t.b. Jul 25 '12 at 13:46
2  
@t.b. : in this case I think we can work with Hanh-Banach. Find an hyperplane which strictly separates $a$ and $B$; its direction will play the role of $a-b$ in Joriki's argument. –  D. Thomine Jul 25 '12 at 13:57
    
@D.Thomine Yes, that works. Thanks! –  t.b. Jul 25 '12 at 14:06
    
There is now a follow-up question here where it is claimed that the present question isn't properly answered. Maybe you want to add a line about the fix suggested by D. Thomine? –  t.b. Aug 14 '12 at 20:24
    
@t.b.: Thanks for notifying me -- sorry I couldn't respond immediately; I travelled to Palestine the day you wrote and it took me a while to get net access. I've completed the proof now. I had also failed to point out where I used the boundedness of $C$ in the first version; perhaps you can check that I got the details right this time? –  joriki Aug 18 '12 at 4:48
show 1 more comment

Your Answer

 
discard

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.