This inequality can be proved as follows: For eachy, y′ ∈F0,
f0(y) +f0(y′) =f0(y+y′)p(y+y′)p(y−x1) +p(y′+x1), hence f0(y)−p(y −x1) p(y′ +x1)−f0(y′), which implies the desired inequality. ⊓⊔
LetF be a flat in a linear spaceE andA⊂F. The following set is called thecoreofAin F:
coreFA=
x∈A∀y∈F, ∃δ >0 such that
|t|δ⇒(1−t)x+ty∈A ,
where |t|δ can be replaced by −δt 0 (or 0 tδ). Each point of coreFAis called acore pointofAin F. In the case thatAis convex,
x∈coreFA⇔ ∀y∈F, ∃δ >0 such that (1 +δ)x−δy∈A
⇔ ∀y∈F, ∃δ >0 such that (1−δ)x+δy∈A.
When F = E, we can omit the phrase “in E” and simply write coreA by removing the subscript E. By definition, A ⊂ B ⊂ F implies coreFA ⊂ coreFB. We also have the following fact:
Fact For eachA⊂F,coreFA=∅if and only if flA=F.
Indeed, the “if” part is trivial. To show the “only if” part, letx∈coreFA.
For eachy∈F, we haveδ >0 such thatz= (1 +δ)x−δy∈A. Then,y= δ−1(1 +δ)x−δ−1z∈flA. Note that flA⊂F becauseA⊂F. Consequently, flA=F.
2.3.2 Proposition For every convex setA ⊂E, coreflAA = rintA, which is also convex. Hence, coreA =∅ implies coreA = rintA and core coreA = coreA.
Proof. Because coreflAA ⊂ rintA by definition, it suffices to show that rintA ⊂coreflAA. For each x ∈ rintA and y ∈ flA, we need to find some s >0 such that (1 +s)x−sy∈A. This can be done using the same proof of the inclusionC∩flCx⊂Cx in Proposition 2.2.5(7). ⊓⊔
Remark 5.WhenAis a finite-dimensional convex set, coreFA=∅if and only ifF = flAaccording to Propositions 2.3.2 and 2.2.6. However, this does not hold for an infinite-dimensional convex set. For example, consider the convex set INf in RN. Then,RNf = flINf and coreRNfINf = rintINf =∅.
With regard to convex sets defined by a non-negative sublinear functional, we have the following proposition:
2.3 The Hahn–Banach Extension Theorem 89 2.3.3 Proposition Letp:E→Rbe a non-negative sublinear functional of a linear subspace E. Then,
p−1([0,1)) = corep−1([0,1)) = corep−1(I).
Proof. The inclusion corep−1([0,1))⊂corep−1(I) is obvious.
Letx∈p−1([0,1)). For eachy∈E, we can chooseδ >0 so thatδp(x−y)<
1−p(x). Then,
0p((1 +δ)x−δy) =p(x+δ(x−y))p(x) +δp(x−y)<1, i.e.,x∈corep−1([0,1)). Hence,p−1([0,1))⊂corep−1([0,1)).
Ifp(x)1, thenx∈corep−1(I) because
p((1 +t)x−t0) = (1 +t)p(x)>1 for anyt >0.
This means that corep−1(I)⊂p−1([0,1)). ⊓⊔
For eachA⊂E with0∈coreA, the Minkowski functionalpA:E → R+ can be defined as follows:
pA(x) = inf
s >0x∈sA
= inf
s >0s−1x∈A . Then, for eachx∈Eand t >0,
pA(tx) = inf
s >0s−1tx∈A
= inf
ts >0(ts)−1tx∈A
=tinf
s >0s−1x∈A
=tpA(x),
i.e., pA satisfies (SL2). In the above, pA(tx) = pt−1A(x). Then, it follows thatpt−1A=tpA for eacht >0. Replacingt byt−1, we have
ptA=t−1pA for eacht >0.
IFA⊂E is convex, the Minkowski functionalpA has the following desir-able properties:
2.3.4 Proposition LetA ⊂E be a convex set with 0∈coreA. Then, the Minkowski functionalpA is sublinear and
rintA= coreA=p−A1([0,1))⊂A⊂p−A1(I) = rclA, so∂A=p−A1(1). Moreover,
pA(x) = 0 ⇔ R+x⊂A.
In order thatpA is a norm onE, it is necessary and sufficient thatR+x⊂A ifx=0andtA⊂A if |t|<1.
Proof. First, we prove thatpA is sublinear. As already observed, pA satisfies (SL2). To show that pA satisfies (SL1), let x, y ∈ E. Since A is convex, we have
s−1x, t−1y∈A ⇒ (s+t)−1(x+y) = s
s+ts−1x+ t
s+tt−1y∈A, which implies thatpA(x+y)pA(x) +pA(y).
The first equality rintA= coreAhas been stated in Proposition 2.3.2. It easily follows from the definitions that coreA⊂p−A1([0,1))⊂A⊂p−A1(I) and p−A1(1)⊂rclA, so p−A1(I)⊂rclA. By Propositions 2.3.2 and 2.3.3, we have
coreA= core coreA⊂corep−1A ([0,1)) =p−1A ([0,1))⊂coreA,
which means the second equality coreA = p−1A ([0,1)). To obtain the third equalityp−1A (I) = rclA, it remains to show that rclA⊂p−1A (I). Letx∈rclA.
Since0∈rintA, it follows from Proposition 2.2.3 thats−1x∈rintC⊂Cfor eachs >1, which implies thatpA(x)1, i.e.,x∈p−A1(I).
By definition,pA(x) = 0 if and only iftx∈Afor an arbitrarily larget >0, which means thatR+x⊂Abecause Ais convex.
BecausepAis sublinear,pAis a norm if and only ifpA(x)= 0 andpA(x) = pA(−x) for every x ∈E\ {0}. Because pA(x)= 0 if and only if R+x ⊂A, it remains to show that pA(x) =pA(−x) for everyx∈E\ {0}if and only if tA⊂Awhenever|t|<1.
Assume that pA(x) = pA(−x) for eachx∈E. Ifx∈A and |t|<1 then pA(tx) =pA(|t|x) =|t|pA(x)<1, which implies thattx∈A. Hence, tA⊂A whenever|t|<1.
Conversely, assume thattA ⊂ A whenever |t| < 1. For each s > pA(x), r−1x ∈ A for some 0 < r < s, and we have s−1(−x) = (−s−1r)r−1x ∈ A, hence pA(−x) pA(x). Replacing x with −x, we have pA(x) pA(−x).
Therefore,pA(x) =pA(−x). ⊓⊔
When the Minkowski functional pA is a norm on E, we call it the Minkowski norm. In this case, rclA, rintA, and ∂A are the unit closed ball, the unit open ball, and the unit sphere, respectively, of the normed lin-ear space E = (E, pA). Then, rclA and rintA are symmetric about 0, i.e., rclA=−rclAand rintA=−rintA. We should note that a convex setA⊂E is symmetric about 0 if and only if tA⊂ A whenever |t| 1, (in the next section,Ais said to be circled).
A subsetW ⊂E is called a wedgeifx+y ∈W for eachx, y ∈W and tx∈W for eachx∈W,t0, or equivalently,W is convex andtW⊂W for everyt0. Note that ifA⊂E is convex thenR+Ais a wedge. For a wedge W ⊂E, the following statements are true:
(1) 0∈coreW ⇔W =E;
(2) W =E, x∈coreW ⇒ −x∈W.
2.3 The Hahn–Banach Extension Theorem 91 A coneC⊂E is a wedge withC∩(−C) ={0}. Each translation of a cone is also called acone.
Using the Hahn–Banach Extension Theorem, we can prove the following separation theorem:
2.3.5 Theorem (Separation Theorem) Let A and B be convex sets in E such that coreA = ∅ and (coreA)∩B = ∅. Then, there exists a linear functional f :E→Rsuch that f(x)< f(y)for everyx∈coreAandy ∈B, and supf(A)inff(B).
Proof. Recall that coreA= rintA(Proposition 2.3.2). For a linear functional f :E →R, iff(x)< f(y) for every x∈coreA andy ∈B, then supf(A) inff(B). Indeed, letx∈A,y∈B,v∈coreA, and 0t <1. Since (1−t)v+ tx∈coreAby Proposition 2.2.3, we have
(1−t)f(v) +tf(x) =f((1−t)v+tx)< f(y), where the left side tends tof(x) ast→1, and hencef(x)f(y).
Note thatW =R+(A−B) is a wedge. Moreover, (coreA)−B⊂coreW. Indeed, let x ∈ coreA and y ∈ B. For each z ∈ E, choose δ > 0 so that (1 +δ)x−δ(y+z)∈A. Then,
(1 +δ)(x−y)−δz = (1 +δ)x−δ(y+z)−y∈A−B⊂W.
Therefore, it suffices to construct a linear functional f : E → R such that f(coreW)⊂(−∞,0).
Now, we shall show thatW ∩(B−coreA) =∅. Assume that there exist x0∈A,x1∈coreA, y0, y1∈B, andt00 such thatt0(x0−y0) =y1−x1. Note that rintx0, x1 ⊂rintA= coreA by Proposition 2.2.3. Hence,
t0
t0+ 1x0+ 1
t0+ 1x1= t0
t0+ 1y0+ 1
t0+ 1y1∈(coreA)∩B, which contradicts the fact that (coreA)∩B=∅.
Take v0 ∈ (coreA)−B ⊂ coreW. Then, note that −v0 ∈ W. For each x ∈ E, we have δ > 0 such that (1 +δ)v0−δ(−x) ∈ W, which implies x+δ−1(1 +δ)v0∈W. Then, we can definep:E→Rby
p(x) = inf
t0x+tv0∈W .
Because W is a wedge, we see that pis sublinear. Since−v0∈W, it follows thatp(s(−v0)) =sandp(sv0) = 0 for everys0. Applying the Hahn–Banach Extension Theorem 2.3.1, we can obtain a linear functionalf :E →R such thatf(s(−v0)) =sfor eachs∈Randf(x)p(x) for everyx∈E(see Figure 2.3). For eachz∈coreW, we haveδ >0 such that (1 +δ)z−δ(z+v0)∈W, i.e., z−δv0 ∈ W. Accordingly, (z−δv0) +tv0 ∈W for every t 0, which meansp(z−δv0) = 0. Thus, we have
f(z)< f(z) +δ=f(z−δv0)p(z−δv0) = 0. ⊓⊔
x
x+p(x)v0
v0 W p(x)
f(x)
−v0
f−1(0)
E E×Ê the graph ofp
the graph off Fig. 2.3. The graphs ofpandf
Remark 6.Using the Hahn–Banach Extension Theorem, we have proved the Separation Theorem. Conversely, the Hahn–Banach Extension Theorem can be derived from the Separation Theorem. Indeed, under the assumption of the Hahn–Banach Extension Theorem 2.3.1, we define
A=
(x, t)∈E×Rt > p(x)
and B=
(x, f(x))∈E×Rx∈F , where B = Gr(f) is the graph of f. Then, A and B are disjoint convex sets in E×R. It is straightforward to show that coreA = A = ∅. By the Separation Theorem 2.3.5, we have a linear functionalϕ:E×R→Rsuch that A⊂ϕ−1((−∞, r]) andB⊂ϕ−1([r,∞)) for somer∈R. Then,r0 because 0 =ϕ(0,0)∈ϕ(B). If ϕ(z)<0 for some z∈B, then ϕ(tz) = tϕ(z)< r for sufficiently larget >0. This is a contradiction becausetz∈B. Ifϕ(z)>0 for some z∈B, then−z∈B andϕ(−z) =−ϕ(z)<0, which is a contradiction.
Therefore, B ⊂ ϕ−1(0). Note that ϕ(0,1) < 0 because (0,1) ∈ A. Since ϕ(x, t) =ϕ(x,0) +tϕ(0,1) for eachx∈E, we haveϕ({x} ×R) =R. Observe that ({x}×R)∩ϕ−1(0) is a singleton. Then,f extends to the linear functional f˜: E →R whose graph is ϕ−1(0), i.e., (x,f˜(x)) ∈ ϕ−1(0) for each x∈ E.
Sinceϕ−1(0)⊂(E×R)\A, it follows that ˜f(x)p(x) for everyx∈E.
The Separation Theorem 2.3.5 can also be obtained as a corollary of the following two theorems, where we do not use the Hahn–Banach Extension Theorem 2.3.1.
2.3.6 Theorem For each pair of disjoint non-empty convex sets A, B ⊂E, there exists a pair of disjoint convex setsA, B⊂E such thatA⊂A, B⊂B, andA∪B=E.
Proof. LetP be the collection of pairs (C, D) of disjoint convex sets such that A ⊂C andB ⊂D. For (C, D),(C′, D′)∈ P, we define (C, D)(C′, D′) if C ⊂C′ andD ⊂D′. Then, it is easy to see that P = (P,) is an inductive ordered set. Due to Zorn’s Lemma,P has a maximal element (A, B).
2.3 The Hahn–Banach Extension Theorem 93 To show thatA∪B =E, assume the contrary, i.e., there exists a point v0∈E\(A∪B). By the maximality of ( A, B), we can obtain two points
x∈A∩ B∪ {v0} and y∈B∩ A∪ {v0}.
Then, x∈ v0, y1 for somey1 ∈B and y ∈ v0, x1 for some x1 ∈A. Note that x∈rintv0, y1andy∈rintv0, x1. Consider the triangle v0, x1, y1. It is easy to see that x1, x and y1, y meet at a point v1. Since x1, x ⊂ A andy1, y ⊂B, it follows that v1∈A∩B, which is a contradiction. ⊓⊔ 2.3.7 Theorem For each pair of disjoint non-empty convex setsC, D ⊂E withC∪D=E,rclC∩rclD is a hyperplane if rclC∩rclD=E.
Proof. First, we show that rclC∩rclD=∂C=∂D. To prove that∂C⊂∂D, letx∈∂C. It suffices to findy∈C such that
(1−t)x+ty∈C for 0< t1 and (1 +t)x−ty∈E\C=D fort >0.
To this end, take y′, y′′ ∈C such that (1−t)x+ty′ ∈C for 0< t1 and (1 +t)x−ty′′ ∈C fort >0. Then, y= 12y′+12y′′∈C is the desired point.
Indeed, for each 0< t1,
(1−t)x+ty= (1−t)x+12ty′+12ty′′
= (1−12t) 1−t
1−12tx+
1 2t 1−12ty′
+12ty′′∈C.
Moreover, note that
(1−s)((1 +t)x−ty) +sy′
= (1−s)(1 +t)x−12(1−s)ty′−12(1−s)ty′′+sy′.
For eacht >0, lets=t/(2 +t)∈(0,1). Then, (1−s)t = 2s. Therefore, we have
(1−s)((1 +t)x−ty) +sy′= (1 +s)x−sy′′∈C,
which means that (1 +t)x−ty ∈ C. Similarly, we have ∂D ⊂ ∂C. Hence,
∂C=∂D. Since rintC∩rintD=∅, it follows that rclC∩rclD=∂C=∂D.
Next, we show that∂Cis a flat. It suffices to show that ifx, y ∈∂C and t >0, thenx′= (1 +t)x−ty∈∂C. Ifx′∈∂C, thenx′∈rintCorx′∈rintD.
In this case,x∈rintx′, y ⊂rintCorx∈rintx′, y ⊂rintDby Proposition 2.2.3. This is a contradiction. Therefore,x′∈∂C.
It remains to show that if ∂C = E then ∂C is a hyperplane. We have v ∈ E\ ∂C. It suffices to prove that E = fl(∂C ∪ {v}). Without loss of generality, we may assume that v ∈ rintC. On the other hand, ∂C = ∅ because C = E. Let z ∈ ∂C. Then, w = z−(v −z) = 2z−v ∈ rintD.
y′′ x∈∂C y′
y
(1 +t)x−ty∈D (1−s)((1 +t)x−ty) +sy′
(1−t)x+ty∈C D C
Fig. 2.4. ∂C⊂∂D
Otherwise, w ∈ rclC, from which, using Proposition 2.2.3, it would follow that z= 12v+12w∈rintv, w ⊂rintC, which is a contradiction.
For eachx∈E\∂C,x∈rintC or x∈rintD. Whenx∈rintC, let s= sup
t∈I(1−t)x+tw∈C . x∈rintC v
w=z−(v−z) ∂C
z y= (1−s)x+sw C
D
Fig. 2.5. The casex∈rintC
Then,y= (1−s)x+sw∈∂C, which implies that x= 1
1−sy− 2s
1−sz+ s
1−sv∈fl(∂C∪ {v}).
In the case thatx∈rintD, let s= sup
t∈I(1−t)x+tv∈D .
x∈rintD v
∂C z y= (1−s)x+sv C
D
Fig. 2.6.The casex∈rintD
Then,y= (1−s)x+sv∈∂D=∂C, which implies that