Let A be a closed, bounded, and convex subset of a Banach space X.
Suppose V is a convex and open subset of X containing A, (A⊂V). Is there an open neighborhood of zero, say U in X such that A+U⊆V?
"Consider case where bdV (the boundary of V) is weakly closed"
I know that When A is weakly compact, the answer is yes (which is the case in all reflexive spaces), but the general case is unclear to me!
I am personally not positive about the answer "yes" in general cases. But I am guessing this property holds true in any Asplund spaces.
What do you guys think?
I don't see what is your argument in reflexive case. Consider as V the open unit ball of l2 and let A be the set of those x = (x1,x2,...) which satisfy the condition $\sum_{n=1}^\infty \frac{n+1}{n} |x_n|^2 \le 1$. This A is a closed ellipsoid contained in V, with zero distance to the unit sphere (of course, the distance is not attained). So, there is no such a neighborhood of zero U in X for which A+U⊆V.
I don't see what is your argument in reflexive case. Consider as V the open unit ball of l2 and let A be the set of those x = (x1,x2,...) which satisfy the condition $\sum_{n=1}^\infty \frac{n+1}{n} |x_n|^2 \le 1$. This A is a closed ellipsoid contained in V, with zero distance to the unit sphere (of course, the distance is not attained). So, there is no such a neighborhood of zero U in X for which A+U⊆V.
Thanks Prof Vladimir, for clarification and your smart example. About reflexive argument, in my original question the set V had a special structure (interior of a polyhedral cone) so I proved this property for this special case which was easy, but in that proof I only applied the property of being weakly closed of the bdV (boundary of V). And I wrongly thought this is the case for boundary of all convex sets, so I thought we have this property for all open, convex V in reflexive spaces ! (which is incorrect)
So still I have no idea about this question, where V has a weakly closed boundary.
Or in more specific case when V is a interior of a polyhedral cone.
Thanks again!
In non-reflexive space, weak closedness of the boundary of V does not help. Let us use the James theorem (in every non-reflexive space X there is a linear continuous functional f of norm 1, that does not attain its norm). Consider A being the closed unit ball of X, and take as V the set of those x, where f(x) < 1. In this example A is a subset of V, but the distance of A to the (weakly closed) boundary {x: f(x) = 1} of V is zero.
Thanks! therefore this property characterizes reflexive spaces. (considering V has weakly closed boundary)
one more thing,
Can we say that a Banach space is reflexive iff every two disjoint convex, bounded, closed subsets have positive distance ?
In another word, in your construction can we find a big enough closed ball centered zero, say rB, in X such that {x: f(x)=1} \cap rB still has zero distance to B ? (Although seems trivial !)
Thanks again, and sorry for my frequent questions.
It may be a little bit off line, but the title "How do convex, closed, bounded sets behave in Banach spaces?" reminded me of a theorem of A. Grothendieck which says that in a Banach space a set is compact if and only if it is a subset of a closed convex hull of a sequence convergent to zero.
I do not have references to this theorem at hand, but am positive that I would dig it out.
Dear Ashkan,
In your previous message, if you take arbitrary r > 1 then {x: f(x)=1} \cap rB still has zero distance to B.
Concerning the post by Roman Sznajder: this can be found in Lindenstrauss, Tzafriri, Volume 1, Proposition 1.e.2.
Best, Vladimir
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