Thedimensionof a convex setC⊂Eis defined by the dimension of the flat hull flC, i.e., dimC = dim flC. Concerning the flat hull of a convex set, we have the following proposition:
2.2.1 Proposition For each convex setC⊂E, flC=
(1−t)x+tyx, y∈C, t∈R . Proof. Each z ∈ flC can be written z = n
i=1tixi, where xi ∈ C and n
i=1ti = 1. We may assume that t1 · · · tn ∈ R\ {0}. If t1 0 then z ∈C. Otherwise,tk <0 and tk+1 >0 for some k= 1, . . . , n−1. Then, we havet=n−k
i=1 tk+i >0, where 1−t=k
i=1ti<0. Let x=
k
i=1
(1−t)−1tixi, y=
n−k
i=1
t−1tk+ixk+i∈C.
Then,z= (1−t)x+ty. Accordingly, we have flC⊂
(1−t)x+tyx, y∈C, t∈R . The converse inclusion is obvious. ⊓⊔
The smallest convex set containingA⊂E is called theconvex hullofA and is denoted by A. We simply writev1, . . . , vn={v1, . . . , vn}. Then,
∆n−1=e1, . . . ,en. Observe that
v1, . . . , vn= ni=1z(i)viz∈∆n−1 and A= x1, . . . , xnn∈N, x1, . . . , xn∈A
. For each two non-empty subsetsA, B⊂E,
A∪B=
(1−t)x+tyx∈ A, y∈ B, t∈I and aA+bB=aA+bB fora, b∈R.
The second equality can be proved as follows: BecauseaA+bB is convex andaA+bB⊂aA+bB, we haveaA+bB ⊂aA+bB. To show thataA+bB ⊂ aA+bB, let x∈ Aand y∈ B. Then, x=
n
i=1tixiandy=mj=1sjyjfor somexi∈A,yj∈B, andti, sj>0 with
n i=1ti=
m
j=1sj= 1. Sinceaxi+byj∈aA+bBand
n i=1
m
j=1tisj= 1, it follows that
ax+by=
n
i=1
ti(axi+by) =
n
i=1
ti
m
j=1
sj(axi+byj)
=
n
i=1 m
j=1
tisj(axi+byj)∈ aA+bB.
2.2 Convex Sets 79 Let C and C′ be non-empty convex sets in the linear spaces E and E′, respectively. A function f : C → C′ is said to beaffine (or linear in the affine sense) if
f((1−t)x+ty) = (1−t)f(x) +tf(y) for eachx, y∈C andt∈I.
As in the definition of a flat,Ican be replaced byR, i.e., x, y∈C, t∈R, (1−t)x+ty∈C
⇒f((1−t)x+ty) = (1−t)f(x) +tf(y).
Indeed, let z = (1−t)x+ty ∈ C in the above expression. When t <0, consider
x= 1
1−tz+ −t 1−ty, 1
1−t∈I, −t
1−t = 1− 1 1−t. Whent >1, consider
y= 1
tz+t−1 t x, 1
t ∈I, t−1
t = 1−1 t. As is easily seen,f :C→C′ is affine if and only if
f ni=1z(i)vi
=
n
i=1
z(i)f(vi)
for eachn∈N,vi∈C andz∈∆n−1, which is equivalent to the following:
vi ∈C, ti∈R,
n
i=1
tivi∈C,
n
i=1
ti= 1⇒f ni=1tivi
=
n
i=1
tif(vi).
For every affine functionf :C→E′ of a convex setC⊂E into another linear spaceE′, the imagef(C) is also convex.
2.2.2 Proposition Let C and D be non-empty convex sets in the linear spaces E and E′, respectively. Every affine function f : C → D uniquely extends to an affine function f˜: flC → flD. Moreover, iff is injective (or surjective) then so isf˜.
Proof. Let C0 be a maximal affinely independent subset of C. Then, flC = flC0. Due to Proposition 2.1.5, f|C0 uniquely extends to an affine function f˜: flC→flD. From the above remark, we can see that ˜f|C=f.
Iff is injective, we show that ˜f is also injective. By the definition of ˜f in the proof of Proposition 2.1.5, it suffices to show thatf(C0) is affinely inde-pendent. Assume thatf(C0) is not affinely independent, i.e., there are distinct pointsv1, . . . , vn∈C0andt1, . . . , tn ∈R\{0}such thatn
i=1tif(vi) =0and
n
i=1ti= 0. Without loss of generality, it can be assumed thatt1, . . . , tk>0 and tk+1, . . . , tn <0. Note that 1 < k < nand k
i=1ti =−n
j=k+1tj >0.
Let
x=
k
i=1
ti
svi and y=
n
j=k+1
−tj
svj, wheres=
k
i=1
ti>0.
Then,x, y ∈Candf(x) =f(y) because f(x)−f(y) =1
s
n
i=1
tif(vi) =0.
Since f is injective, we have x = y. Hence, it follows that k
i=1tivi =
−n
j=k+1tjvj, i.e., n
i=1tivi = 0. Because C0 is affinely independent, t1=· · ·=tn= 0, which is a contradiction.
Finally, we show that iff is surjective then so is ˜f. By Proposition 2.2.1, eachz∈flDcan be written as follows:
z= (1−t)y+ty′, y, y′∈D, t∈R.
Since f is surjective, we have x, x′ ∈C such that f(x) = y and f(x′) =y′. Then, (1−t)x+tx′∈flC and
f˜((1−t)x+tx′) = (1−t)y+ty′=z.
Therefore, ˜f is also surjective. ⊓⊔
LetC be a convex set in a linear spaceE. The following set is called the radial interior ofC:
rintC=
x∈C∀y∈C, ∃δ >0 such that (1 +δ)x−δy ∈C .3 In the caseC=v1, . . . , vn, observe that
rintv1, . . . , vn= ni=1z(i)vi
z∈∆n−1∩(0,∞)n . Indeed, letx0 =
n
i=1n−1vi ∈ v1, . . . , vn. For eachx∈rintv1, . . . , vn, we have y ∈ v1, . . . , vn such that x ∈ x0, y, i.e.,x = (1−t)x0 +ty for somet ∈(0,1). Then,y =
n
i=1z(i)vi for some z ∈ ∆n−1. It follows thatx=ni=1((1−t)n−1+tz(i))vi, whereni=1((1−t)n−1+tz(i)) = 1 and (1−t)n−1+tz(i) >0 for alli = 1, . . . , n. Thus, xis a point of the rightside set. Conversely, it is straightforward to prove that each point of the rightside set belongs tov1, . . . , vn.
In particular, rintv1, v2 = {(1 −t)v1 +tv2 | 0 < t < 1}, and hence rintv1, v2=v1, v2 \ {v1, v2}ifv1=v2. The radial interior ofCcan also be defined as
rintC=
x∈C∀y∈C, ∃z∈C such thatx∈rinty, z . For eachx∈C, the following subsetCx⊂C is called thefaceofC atx:4
3In K¨othe’s book, rintCis denoted byCi and called thealgebraic kernelofC.
4The faceCxis a little differently defined than thesupporting facetofCthrough xin K¨othe’s book.
2.2 Convex Sets 81 Cx=
y∈C∃δ >0 such that (1 +δ)x−δy∈C
=
y∈C∃z∈C such thatx∈rinty, z . By an easy observation, we have
rintC={x∈C|Cx=C}, i.e., x∈rintC⇔Cx=C.
WhenCx={x}, we callxanextreme point ofC. It is said thatx∈E is linearly accessiblefromC if there is some y∈C such that
rintx, y ⊂C (i.e.,x, y \ {x} ⊂C).
Theradial closurerclC ofCis the set of all linearly accessible points from C.5It should be noted that rclC⊂flC by Proposition 2.2.1, hence fl rclC= flC. Consequently, we have the following inclusions:
rintC⊂C⊂rclC⊂flC.
The set∂C= rclC\rintCis called the radial boundaryofC.
Remark 1.Note that A ⊂ B implies rclA ⊂ rclB, but it does not imply rintA ⊂ rintB. For example, consider A = In × {0} ⊂ B = In+1. Then, A∩rintB=∅.
For the Hilbert cubeQ= [−1,1]N, we have rintQ=
x∈Qsupi∈N|x(i)|<1
(−1,1)N. Observe that rint[−1,1]Nf = (−1,1)Nf but rintINf =∅, where
[−1,1]Nf =RNf∩[−1,1]N, (−1,1)Nf =RNf ∩(−1,1)N, and INf =RNf∩IN.6 As is easily observed,INf = rcl(INf \ {0}). It will be shown in Remark 3 that INf\ {0}= rclCfor some convex setC⊂RNf.
Remark 2.The unit closed ball Bc0 of the Banach spacec0 has no extreme points. In fact, everyx∈Bc0is the midpoint of two distinct pointsy, z∈Bc0, i.e., x = 12y+ 12z. For example, choose n ∈ N so that |x(n)| < 12 and let y, z ∈ Bc0 such that y(i) = z(i) = x(i) for i = n, y(n) = x(n) + 12, and z(n) =x(n)−12.
2.2.3 Proposition LetC⊂E be a convex set. Ifx∈rintC,y ∈rclC, and 0t <1, then(1−t)x+ty∈rintC, i.e., x, y \ {y} ⊂rintC.
Proof. For eachz∈C, we have to findv∈Cand 0< s <1 such that (1−t)x+ty= (1−s)z+sv∈rintz, v.
Takew∈Cso that rintw, y ⊂C, and choose 0< r <1 so that
y
(1−t)x+ty
z
w x
z′
w′
v=t1y+t2w+t3w′+t4z′
u= t1
t1+t2
y+ t2
t1+t2
w u′= t3
t3+t4
w′+ t4
t3+t4
z′
Fig. 2.1.(1−t)x+ty∈rintC
z′= (1 +r)x−rz, w′= (1 +r)x−rw∈C.
The desiredvis to be written as
v=t1y+t2w+t3w′+t4z′= (t1+t2)u+ (t3+t4)u′∈C, wheret1+t2+t3+t4= 1,t1, t2, t3, t4>0,
u= t1
t1+t2
y+ t1
t1+t2
w, u′= t3
t3+t4
w′+ t4
t3+t4
z′ ∈C.
Then, we have
(1−s)z+sv= (1−s)z+s(t1y+t2w+t3w′+t4z′)
=st1y+s(t2−t3r)w+s(t3+t4)(1 +r)x+ (1−s−st4r)z.
To obtain (1−s)z+sv= (1−t)x+ty, it is enough to findt1, t2, t3, t4 >0 and 0 < s < 1 satisfying the simultaneous equations: st1 = t, t2 = t3r, s(t3+t4)(1 +r) = 1−t, and 1−s=st4r, i.e.,
(∗) t1= t
s, t4= 1−s
rs , t3=1
r − 1 +rt
(1 +r)rs, t2= 1− 1 +rt (1 +r)s. Sincet1, t4<1 and 0< t2 (< t3), it is necessary to satisfy
max t, 1
1 +r, 1 +rt 1 +r
!
< s <1.
We can take such ansbecause the left side of the above inequality is less than 1. Then, we can definet1, t2, t3, t4>0 as in (∗), which satisfiest1+t2+t3+t4= 1. Thus, we have the desiredv=t1y+t2w+t3w′+t4z′∈C. ⊓⊔
5In K¨othe’s book, rclCis denoted byCaand called thealgebraic hullofC.
6It is known that [−1,1]f ≈If.
2.2 Convex Sets 83 Although we verified in Remark 1 that A ⊂ B does not imply rintA ⊂ rintB in general, we do have the following corollary:
2.2.4 Corollary LetAandB be non-empty convex sets inE. IfA⊂B and A∩rintB=∅, thenrintA⊂rintB.
Proof. Let x ∈ A∩rintB. For each y ∈ rintA, we have z ∈ A such that y ∈ rintx, z. Since rintx, z ⊂rintB by Proposition 2.2.3, it follows that y∈rintB. ⊓⊔
2.2.5 Proposition For each convex set C ⊂ E, the following statements hold:
(1) BothrintC andrclC are convex;
(2) rint rintC= rintC⊂rint rclC;
(3) rintC=∅ ⇒rint rclC= rintC, rcl rintC= rcl rclC= rclC, in which case∂rintC=∂rclC=∂C;
(4) rintC=∅ ⇒flC= fl rintC;
(5) rintC=∅, rclC= flC⇒rintC=C= flC;
(6) ∂C=∅ ⇔ ∅ =CflC;
(7) Cx is convex andCx=C∩flCxforx∈C;
(8) x∈rintCx forx∈C, hence(Cx)x=Cx; (9) (Cx)y =Cy forx∈C andy∈Cx; (10) Cx=Cy forx∈C andy∈rintCx.
Proof. (1): To prove the convexity of rintC, we can apply Proposition 2.2.3.
It is now quite straightforward to show the convexity of rclC.
(2): To show rintC⊂rint rintC, we can apply Proposition 2.2.3. Because rint(rintC)⊂rintC by Corollary 2.2.4, we have rint rintC= rintC.
For eachx∈rintC andy∈rclC, 12x+12y∈rintC by Proposition 2.2.3.
Then, we haveδ >0 such that (1+δ)x−δ(12x+12y)∈C, i.e., (1+12δ)x−12δy∈ C. Hence,x∈rint rclC.
(3): Let x0 ∈rintC. For eachx∈rint rclC, we havey ∈rclC such that x∈rintx0, y, which implies thatx∈rintCby Proposition 2.2.3. Combining this with (2) yields rint rclC= rintC.
We now havex0 ∈rintC = rint rclC. Ifx∈ rcl rclC, then rintx0, x ⊂ rint rclC= rintCby Proposition 2.2.3, which means thatx∈rcl rintC. Since rcl rintC⊂rclC⊂rcl rclC, we have rcl rintC= rclC= rcl rclC.
(4): Letx0∈rintC. For each x∈C, 12x+12x0 ∈fl rintC by Proposition 2.2.3. Then, it follows from Proposition 2.2.1 that x= 2(12x+12x0)−x0 ∈ fl rintC. Accordingly, we have C ⊂ fl rintC, which implies flC ⊂ fl rintC.
Since fl rintC⊂flC, we have flC= fl rintC.
(5): Let x0 ∈ rintC. For each x ∈ flC, 2x−x0 ∈ flC = rclC. Then, x=12x0+12(2x−x0)∈rintC⊂C by Proposition 2.2.3.
(6): Assume∅ =CflC. Then, we havex∈flC\C, which can be written asx= (1 +t)y−tz for somey=z∈C andt >0 by Proposition 2.2.1. Let
s= inf
t >0(1 +t)y−tz∈C 0.
Then, (1 +s)y−sz∈rclC\rintC=∂C.
WhenC= flC, i.e.,C is a flat, we have rclC= rintC=C by definition, which means∂C=∅. Therefore, ∂C=∅implies∅ =CflC.
(7): First, we show thatCx is convex. For each y, z∈Cx, we can choose δ >0 so that (1 +δ)x−δy∈C and (1 +δ)x−δz∈C. Then, for eacht∈I,
(1 +δ)x−δ
(1−t)y+tz
= (1−t)
(1 +δ)x−δy +t
(1 +δ)x−δz
∈C, which means (1−t)y+tz ∈Cx.
BecauseCx⊂C∩flCx, it remains to showC∩flCx⊂Cx. By Proposition 2.2.1, each y ∈ C ∩flCx can be written as y = (1 −t)y′ +ty′′ for some y′, y′′ ∈ Cx and t ∈ R. Because of the convexity of Cx, we have y ∈ Cx if t ∈I. Then, we may assume thatt < 0 (ift > 1, exchange y′ withy′′). We haveδ >0 such thatz′= (1 +δ)x−δy′ ∈C. Observe that
(1 +s)x−sy= (1 +s) δ
1 +δy′+ 1 1 +δz′
−s
(1−t)y′+ty′′
=
(1 +s)δ
1 +δ −s(1−t)
y′+1 +s
1 +δz′−sty′′.
Lets=δ/(1−t−tδ)>0. Then, since 1 +s= (1−t)(1 +δ)/(1−t−tδ), it follows that
(1 +s)x−sy= 1−t
1−t−tδz′+ −tδ
1−t−tδy′′∈C, which implies thaty∈Cx.
Cx∋y′′ x
y= (1−t)y′+ty′′
Cx∋y′
(1 +s)x−sy
z′= (1 +δ)x−δy′∈C Fig. 2.2.C∩flCx⊂Cx
(8): From the definition of rintCx, it easily follows thatx∈rintCx.
2.2 Convex Sets 85 (9): BecauseCx ⊂ C, we have (Cx)y ⊂ Cy by definition. We will show that Cy ⊂Cx, which impliesCy = (Cy)y ⊂(Cx)y by (8) and the definition.
For eachz∈Cy, chooseδ1>0 so thatu= (1 +δ1)y−δ1z∈C. On the other hand, sincey∈Cx, we haveδ2>0 such thatv= (1 +δ2)x−δ2y∈C. Then,
(1 +δ1)(1 +δ2) 1 +δ1+δ2
x− δ1δ2
1 +δ1+δ2
z= 1 +δ1
1 +δ1+δ2
v+ δ2
1 +δ1+δ2
u∈C, which means thatz∈Cx.
(10): Sincey∈rintCx, we have (Cx)y =Cx. On the other hand, (Cx)y = Cy by (9). ⊓⊔
Remark 3.It should be noted that, in general, rcl rclC= rclC. For example, letCbe the convex set inRNf defined as follows:
C=
x∈INf ∃k∈N such that
i∈Nx(i)k−1, x(i)= 0 at leastk manyi∈N
.
It is easy to see that 0∈rclC, i.e., rclC ⊂INf \ {0}. For eachx∈INf \ {0}, choose k ∈Nso that k−1
i∈Nx(i), and let y ∈C such thaty(i) =k−2 forikandy(i) = 0 fori > k. If 0< t1, then (1−t)x+ty∈C because (1−t)x(i) +ty(i)= 0 for at leastkmanyi∈Nand
i∈N
(1−t)x(i) +ty(i)
= (1−t)
i∈N
x(i) +t
i∈N
y(i)k−1. Therefore, rclC = INf \ {0}. As observed in Remark 1, rcl
INf \ {0}
= INf. Hence, we have rcl rclC= rclC. It should also be noted that rintC=∅.
In the finite-dimensional case, we have the following proposition:
2.2.6 Proposition Every non-empty finite-dimensional convex set C has a non-empty radial interior, i.e.,rintC=∅, and therefore
rcl rintC= rcl rclC= rclC and ∂rintC=∂rclC=∂C.
Proof. We have a maximal affinely independent finite subset{v1, . . . , vn} ⊂C.
Then,v0=n
i=1n−1vi∈rintC. Indeed, sinceC⊂fl{v1, . . . , vn}, eachx∈C can be written as x=n
i=1tivi, wheren
i=1ti= 1. Observe that (1 +δ)v0−δx= (1 +δ)
n
i=1
n−1vi−δ
n
i=1
tivi
=
n
i=1
(n−1+δ(n−1−ti))vi.
When v0 = x, we have s = min{n−1−ti | i = 1, . . . , n} < 0. Let δ = 1/(−sn) > 0. Then, n−1+δ(n−1−ti) 0 for every i = 1, . . . , n, which implies that (1 +δ)v0−δx∈C. ⊓⊔
2.2.7 Additional Results for Convex Sets (1) For every two convex setsC andD,
(C∩D)x=Cx∩Dx for eachx∈C∩D.
(2) For every two convex setsC andD with rintC∩rintD=∅, rint(C∩D) = rintC∩rintD.
In general, rintC∩rintD⊂rint(C∩D).
Sketch of Proof.To show that rint(C∩D)⊂rintC∩rintD, letx0 ∈ rintC∩rintD. For each x ∈ rint(C∩D), take y ∈ C∩D so that x∈rintx0, y. Since rintx0, y ⊂rintCby Proposition 2.2.3, it follows thatx∈rintC. Hence, rint(C∩D)⊂rintC. Similarly, we have rint(C∩ D)⊂rintD.
(3) Let C and D be convex sets in the linear spacesE andE′, respectively.
Then,C×Dis also convex,
rint(C×D) = rintC×rintD and rcl(C×D) = rclC×rclD.
Moreover, (C×D)(x,y)=Cx×Dy for each (x, y)∈C×D.
(4) Let f :C →E′ be an affine function of a convex set C in a linear space Einto another linear spaceE′, andD be a convex set inE′. Then,f(C) andf−1(D) are convex and
f−1(D)x=Cx∩f−1(Df(x)) for eachx∈f−1(D) (⊂C).
In particular,Cx⊂f−1(f(C)f(x)) (i.e.,f(Cx)⊂f(C)f(x)) for eachx∈C.
Whenf is injective,f(Cx) =f(C)f(x)for eachx∈C.
Sketch of Proof. It is easy to see that f(f−1(D)x) ⊂ Df(x), hence f−1(D)x⊂f−1(Df(x)). Also,f−1(D)x⊂Cxbecausef−1(D)⊂C. Ac-cordingly,f−1(D)x⊂Cx∩f−1(Df(x)). To prove the converse inclusion, for eachy∈f−1(Df(x))∩Cx, chooseδ >0 so that (1+δ)f(x)−δf(y)∈ Dand (1 +δ)x−δy∈C. Then, (1 +δ)x−δy∈f−1(D).
(5) For every (bounded) subsetAof a normed linear spaceE= (E, · ), the following hold:
(i) x−ysupz∈Ax−z for eachx∈E andy∈ A; (ii) diamA= diamA.
Sketch of Proof.(i): Writey=ni=1z(i)xifor somex1, . . . , xn∈Aand z∈∆n−1.
(ii): For eachx, y∈ A, x−y sup
z∈A
x−z sup
z∈A
sup
z′∈A
z−z′= diamA.
Remark 4.In (2) above, rint(C∩D)= rintC∩rintDin general. Consider the case thatC∩D=∅ but rintC∩rintD=∅.
In (4) above,f(Cx)=f(C)f(x) in general. For instance, letC={(s, t)∈ R2| |s| ≤t≤1} ⊂R2. Then, pr1(C) = [−1,1], pr1(C0) ={0}, and pr1(C)0= pr1(C).