$ z= 1
z= 0
C ˆ
rec(C)
Fig. 5.5.Homogenization of a convex set.
Proof. Denote the set on the right-hand side byD. Letd∈rec(C). Ifx∈C, then xn :=x+nd∈C, (xn,1)/n ∈ K(C), and (xn,1)/n →(d,0) ∈ K(C), provingD⊆K(C).
To prove the reverse inclusionK(C)⊆D, let (d, t)∈K(C) be such that tn(xn,1) →(d, t). Ift >0, then letting d=tx, we have tn(xn,1) →t(x,1).
Thus, xn → x ∈ C, since C is closed, and consequently (d, t) = t(x,1) ∈ K(C)⊆D.
If, however,t = 0, thentnxn →dandtn→0; we claim thatd∈rec(C).
Let x ∈ C and s > 0 be arbitrary. We have dn := tn(xn −x) → d, xn = x+dn/tn ∈C, and for large enoughnsuch that 0≤stn ≤1,
x+sdn = (1−stn)x+stn
x+dn
tn
= (1−stn)x+stnxn ∈C.
This givesx+sdn→x+sd∈C, and proves the claim. ut
5.7 Continuity of Convex Functions
There are no continuity assumptions made in the definition of a convex func- tion. However, it turns out that a convex function is continuous in the relative interior of its domain, even Lipschitz continuous under a mild boundedness assumption. In finite dimensions, this boundedness assumption is not needed, because it is automatically satisfied. These constitute some of the most basic results on convex functions.
Letf :E→R∪{∞}be a convex function on a normed linear spaceE, and define L= aff(dom(f)). Since f is always +∞off L, in questions regarding the continuity off, it makes sense to considerf only onL, since otherwisef
can never be continuous at any point whenLis a proper subset of E. Thus, in this section, we consider convex functions with full-dimensional domains.
Lemma 5.42.Letf :E→R∪ {∞}be a convex function in a normed linear spaceE such that dom(f)has a nonempty interior.
0
thenf is bounded from above in a neighborhood of every point of int(dom(f)) (with the bound depending on the point).
Proof. LetBr0(x0)∈int(dom(f)) be such thatf(x)≤a <∞forx∈Br0(x0), and letx∈int(dom(f)) be an arbitrary point. There exists γ >1 such that w:=x0+γ(x−x0)∈dom(f).
We claim that Br(x) ∈ dom(f), where r = (γ−1)r0/γ. Geometrically, this follows from the fact that the convex hull of w and Br0(x0) forms a truncated conic region that encloses a ball atxwhose radius can be computed using similarity. Analytically, ifkv−xk ≤r, consider the pointudefined by the equation w =:u+γ(v−u). Then 0 = (1−γ)(u−x0) +γ(v−x) and ku−x0k ≤rγ/(γ−1) =r0, so thatu∈dom(f). This proves the claim.
Since v = w/γ+ ((γ−1)/γ)u ∈ dom(f), using the convexity of f, we obtain
f(v) =f 1
γw+γ−1 γ u
≤ 1
γf(w) +γ−1
γ f(u)≤ 1
γf(w) +γ−1 γ a=:b.
Thusf is bounded from above by the constantbonBr(x). ut Theorem 5.43.Letf :E→R∪{∞}be a convex function on a normed linear spaceE. Thenf is continuous inri(dom(f))if and only if there exists a point x0∈ri(dom(f))such thatf is bounded from above in a relative neighborhood ofx0 (that is, a neighborhood of x0 in the relative topology of aff(dom(f))).
Proof. DefineC = dom(f) and assume that int(dom(f))6=∅, by restricting f to aff(dom(f)) if necessary. If f is continuous at x0 ∈ int(C), then there exists a neighborhoodx0∈N⊆C such that|f(x)−f(x0)| ≤1 onN, andf is bounded from above onN by the constantf(x0) + 1.
Conversely, assume that f(x)≤a <∞ forxin a neighborhoodN ofx0. We may assume, if necessary by considering the functionx7→f(x+x0)−f(x0) on the setC−x0, thatx0= 0∈int(C) andf(0) = 0. Consider the symmetric neighborhoodS =N∩(−N) of 0, and pick any 0< ε <1. If z ∈εS, then
±z/ε∈S, and we have
−εa≤f(z)≤(1−ε)f(0) +εf(z/ε)≤εa,
where the second inequality follows from the convexity of f and the first inequality from
0 =f(0) =f 1
1 +εz+ ε
1 +ε(−z/ε)
≤ 1
1 +εf(z) + ε
1 +εf(−z/ε)≤ 1
1 +εf(z) + ε 1 +εa.
∈int(dom(f)), Iffis bounded from above in a neighborhood of some pointx
5.7 Continuity of Convex Functions 137 This proves the continuity off at x0= 0. By virtue of Lemma 5.42, f is bounded from above on a neighborhood of every point ofx∈int(C), so that
f is continuous on int(C). ut
It is possible to strengthen the continuity result above.
Definition 5.44.A function f :X →R on a metric spaceX is called Lip- schitz continuous if there exists a constantK >0such that
|f(x)−f(y)| ≤Kd(x, y) for all x, y∈X.
The function f is called locally Lipschitz continuous if each point x∈X has a neighborhood on whichf is Lipschitz continuous.
Theorem 5.45.Let f : E → R∪ {∞} be a convex function on a normed linear spaceE. If there exists a pointx0∈ri(dom(f))such thatf is bounded from above in a relative neighborhood of x0, then the function f is locally Lipschitz continuous on ri(dom(f)).
Proof. We may again assume that int(dom(f)) 6= ∅. By virtue of Theo- rem 5.43,f is continuous in a neighborhoodBr0(x0). Suppose that
m≤f(x)≤M on Br0(x0),
and pick 0< r < r0. We claim thatf is Lipschitz continuous inBr(x0). Let v1, v2 ∈Br(x0), and assume for now that kv2−v1k ≤r0−r. The function g(w) := f(v1+w)−f(v1) is convex and finite-valued, g(0) = 0, and g is bounded from above in the ball N := Br0−r(0) by the constant M −m.
It follows from the proof of Theorem 5.43 that |g(w)| ≤ ε(M −m) on εN. Consequently, sincev2−v1∈(kv2−v1k/(r0−r))N, we have
|f(v2)−f(v1)|=|g(v2−v1)| ≤ M−m
r0−r kv2−v1k. (5.6) Now ifv1, v2∈Brare arbitrary, we can partition the interval [v1, v2] intoN points{uk}N1 in such a way thatu1=v1,uN =v2, andkuk−uk−1k ≤r0−r.
Applying (5.6) to each pair (uk−1, uk),k = 2, . . . , N, and adding the results gives
|f(v2)−f(v1)| ≤
N
X
k=2
|f(uk)−f(uk−1)| ≤
N
X
k=2
M −m
r0−r kuk−uk−1k
=M −m
r0−r kv2−v1k,
where the equality follows since each vectoruk−uk−1 has the same direction v2−v1. This proves the claim and the theorem. ut
In infinite-dimensional vector spaces, the boundedness assumption on the functionf is really necessary, where even the continuity of linear functionals is tied up with boundedness. However, the situation is different in the finite- dimensional case.
Lemma 5.46.Let f : E → R∪ {∞} be a convex function on a finite- dimensional normed linear space E. If x ∈ ri(dom(f)), then f is bounded from above in a relative neighborhood ofx.
Proof. Letn= dim(E). As in the proof of Theorem 5.43, we may assume that int(dom(f))6=∅. If x∈int(dom(f)), then
x∈N = int(∆) = n+1
X
i=1
λiui:λi>0,
n+1
X
i=1
λi= 1
,
where {ui}n+11 are affinely independent vectors in dom(f), so that ∆ = co({ui}n+11 ) is an n-simplex contained in dom(f). If y = Pn+1
i=1 λiui ∈ ∆, then
f(y) =fn+1X
i=1
λiui
≤
n+1
X
1=1
λif(ui)≤max{f(u1), . . . , f(un+1)},
proving our claim. ut
Corollary 5.47.Let f : E → R∪ {∞} be a convex function on a finite- dimensional normed linear space E. If we consider f as a function f : aff(C)→R∪ {∞}, thenf is locally Lipschitz continuous on ri(C).