Re: continuos linear image of open sets
From: Dan Luecking (Look-In-Sig_at_uark.edu)
Date: 03/28/05
- Next message: Geometry and Topology Journal: "Paper published by Algebraic and Geometric Topology"
- Previous message: Dan Luecking: "Re: Weak Topology"
- In reply to: David C. Ullrich: "Re: continuos linear image of open sets"
- Messages sorted by: [ date ] [ thread ]
Date: Mon, 28 Mar 2005 20:30:07 +0000 (UTC)
On Fri, 25 Mar 2005 14:30:04 +0000 (UTC), "David C. Ullrich"
<ullrich@math.okstate.edu> wrote:
>On Thu, 24 Mar 2005 16:00:11 +0000 (UTC), "G. A. Edgar"
><edgar@math.ohio-state.edu.invalid> wrote:
>
>>In article <d1t8nc$e80$1@news.ks.uiuc.edu>, Jose Gascon
>><jogascon@una.edu.ve> wrote:
>>
>>> It is well known that if T:X->X, X Banach space, is a continuos linear
>>> operator and surjective T maps open sets to open sets. What about if
>>> you drop the hypothesis that T is surjective, is T(O), O open in X,
>>> Borel?
>>
>>No.
>>
>>I will provide an example of a bounded linear operator T:X -> Y
>>from one Banach space to another such that the image of
>>the closed unit ball is not a Borel set. It then follows that the
>>image of the open ball is not Borel, since any closed
>>ball is a countable intersection of open balls. Taking
>>a direct sum X+Y and mapping (x,y) -> (0,Tx) we get an
>>example mapping a Banach space to itself.
>>
>>Let Y be l^2, say, nicely rotund. Let S be the unit sphere in Y,
>>S = { y : ||y|| = 1 }, a closed set. Let Q be a nasty subset
>>of S, non-Borel. Let X = l^1(Q), the space of functions
>>f : Q -> R such that ||f|| = sum_{q in Q} |f(q)| converges.
>>Define T : X -> Y by T(f) = sum_{q in Q} f(q) q, the sum converges
>>absolutely. T is linear and ||T|| = 1. Now I claim the closed ball
>>B = { f in X : ||f|| <= 1 } maps onto a non-Borel set in Y.
>>Indeed, the intersection of T(B) with the closed set S is exactly
>>Q,
>
>I'm not sure whether I see why this is so. It's not true,
>quite, that the only way to get f(x) in S (with ||x||=1)
>is to take x such that x(q_0) = 1 for some q_0 in Q and
>f(q) = 0 for all other q in Q; for example it could
>happen that q_0 and -q_0 are both in Q, in which case
>f(q_0) = 1/2 and f(-q_0) = -1/2 does it.
>
>(Of course that's not a counterexample to what you
>claim, just a counterexample to what I thought at
>first was the reason it was true. And in any case
>it's easy to ensure that Q intersect -Q is empty,
>which eliminates the problem.)
Well T(B) = T(-B), so surely (T(B) intersect S) contains (Q union -Q).
In the complex case, it would also contain e^{i\theta}Q (but that's as
far as it goes).
It is quite possible (in the complex case) for Q to be non-Borel and
contained in a one-dimensional subspace, so the range of T is a
hyperplane and T(B) is the unit ball of that hyperplane.
So you have to start with a Q that is nonBorel and closed under
q |-> e^{i\theta}q for all \theta.
Dan
-- Dan Luecking Department of Mathematical Sciences University of Arkansas Fayetteville, Arkansas 72701 To reply by email, change Look-In-Sig to luecking
- Next message: Geometry and Topology Journal: "Paper published by Algebraic and Geometric Topology"
- Previous message: Dan Luecking: "Re: Weak Topology"
- In reply to: David C. Ullrich: "Re: continuos linear image of open sets"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
|
