A NEW PROOF OF A LEMMA BY PHELPS
ANTONIA E. CARDWELL
Received 7 December 2005; Revised 3 May 2006; Accepted 7 May 2006
We give a different proof of a lemma by Phelps (1960) which asserts, roughly speaking, that if two norm-one functionals f andghave their hyperplanes f−1(0) andg−1(0) suf- ficiently close together, then eitherf−gorf+gmust be small. We also extend this result to a complex Banach space.
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In 1960 in [2], Phelps proved the following lemma.
Lemma 1. Suppose that Eis a real normed linear space and that>0. If f,g∈S∗ are such that f−1(0)∩ᐁ⊂g−1[−/2,/2], then eitherf−g ≤orf +g ≤. (Here,ᐁ represents the unit ball ofEandS∗is the unit sphere ofE∗.)
This lemma was then used the following year as a crucial step in the proof of the well-known Bishop-Phelps theorem [1] that every Banach space is subreflexive; in other words, every functional on a Banach spaceEcan be approximated by a norm-attaining functional on the same space. The original proof of this lemma uses the Hahn-Banach theorem and is therefore fairly abstract.
In this note, we present an alternate proof forLemma 1stated above. This proof gives a geometric argument while extending the lemma to a complex Banach space.Lemma 1is shown to be a special case when the bound ofonf ±gis replaced by 5. This replacement does not affect the fundamental conclusion ofLemma 1.
We now state the extended lemma.
Lemma 2. LetXbe a complex Banach space and letbe such that 0<<1/2. Letϕ,ψ∈ X∗,ϕ = ψ =1. Suppose that for allx∈X withx ≤1 andϕ(x)=0, it holds that
|ψ(x)| ≤. Then there is some complex numberαsuch that|α| =1 andϕ−αψ ≤5. It will be shown that ifϕandψare real-valued functionals on a real Banach spaceX, thenαwill in fact be either 1 or−1, thus proving the amended original result.
We now proveLemma 2.
Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 28063, Pages1–3
DOI10.1155/IJMMS/2006/28063
2 A new proof of a lemma by Phelps
Proof. Lete∈Xbe such thate =1 and|ϕ(e)| ≥1−/4. We will first show that|ψ(e)|
≥1−(5/2). To see this, let f ∈X such thatf =1 and |ψ(f)| ≥1−/4. Let k= 1−/4 and let t=ϕ(f)/ϕ(e). Then 0≤ |t| ≤1/(1−/4)=1/k≤8/7 and if we take w=(k/(k+ 1))(f−te), thenw ≤(k/(k+ 1))(f+|t|e)≤(k/(k+ 1))(1 + 1/k)=1.
Moreover,
ϕ(w)= k k+ 1
ϕ(f)−ϕ(f)
ϕ(e)ϕ(e)=0 (1)
so we have
≥ψ(w)= k
k+ 1ψ(f)−tψ(e)
≥ k
k+ 1ψ(f)− |t|ψ(e)
≥ k
k+ 1ψ(f)− |t|ψ(e).
(2)
Thus 1
kψ(e)≥ |t|ψ(e)≥ψ(f)−k+ 1 k ≥
1−
4
−k+ 1
k =k−k+ 1
k . (3) This gives
ψ(e)≥k2−(k+ 1)=
1− 4
2
−
2− 4
=1−
2+
2
16−2+
2
4 ≥1−5 2
(4) as required. Notice that, ifϕandψare real valued, the above still holds.
Now, there exist β,γ∈C such that |β| = |γ| =1, βϕ(e)∈[1−/4, 1]⊂R, and γψ(e)∈[1−5/2, 1]⊂R; and so|βϕ(e)−γψ(e)| ≤5/2.
Letx∈X be such thatx ≤1 and writex=λe+y, whereλ=ϕ(x)/ϕ(e) and y= x−λe. Then |λ| ≤ |ϕ(x)|/|ϕ(e)| ≤1/(1−/4)≤8/7, y ≤ x+|λ|e ≤15/7, and ϕ(y)=ϕ(x)−(ϕ(x)/ϕ(e))ϕ(e)=0. So, by hypothesis,|ψ((7/15)y)| ≤, that is,|ψ(y)| ≤ (15/7). Then, if we takeα=γ/β, we have|α| =1 and
ϕ(x)−αψ(x)= 1
|β|βϕ(x)−γψ(x)=βλϕ(e) +βϕ(y)−γλψ(e)−γψ(y)
≤ |λ|βϕ(e)−γψ(e)+|γ|ψ(y)≤8 7·
5
2+ 1·15 7 =5.
(5)
Butxwas an arbitrary element of the unit ball ofX, so we haveϕ−αψ ≤5.
Notice that ifXis a real Banach space, the above argument still holds. Also, ifϕand ψare real valued, we can choosee∈X such thatϕ(e)≥1−/4, givingβ=1, and then from the claim, eitherγ=1 orγ= −1. So eitherα=1 orα= −1, yielding Phelps’ result,
up to a constant.
Antonia E. Cardwell 3 References
[1] E. Bishop and R. R. Phelps, A proof that every Banach space is subreflexive, Bulletin of the Amer- ican Mathematical Society 67 (1961), 97–98.
[2] R. R. Phelps, A representation theorem for bounded convex sets, Proceedings of the American Mathematical Society 11 (1960), no. 6, 976–983.
Antonia E. Cardwell: Department of Mathematics, Millersville University of Pennsylvania, Millersville, PA 17551-0302, USA
E-mail address:[email protected]
