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# A note on the K-epigraph

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## Abstract

We study the question as to when the closed convex hull of a K-convex map equals its K-epigraph. In particular, we shed light onto the smallest cone K such that a given map has convex and closed K-epigraph, respectively. We apply our findings to several examples in matrix space as well as to convex composite functions.
arXiv:2107.00117v1 [math.OC] 30 Jun 2021
A note on the K-epigraph
Armand Gissler1and Tim Hoheisel2
1Department of Mathematics, ´
Ecole Normale Sup´erieure Paris-Saclay
2Department of Mathematics and Statistics, McGill University, Montr´eal
July 2, 2021
Abstract
We study the question as to when the closed convex hull of a K-convex map equals its K-epigraph.
In particular, we shed light onto the smallest cone Ksuch that a given map has convex and closed
K-epigraph, respectively. We apply our ﬁndings to several examples in matrix space as well as to
convex composite functions.
Keywords: Cone-induced ordering, K-convexity, K-epigraph, convex hull, Fenchel conjugate,
horizon cone, spectral function, convex-composite function, matrix analysis.
MSC2000 Subject Classiﬁcation: 52A41, 65K10, 90C25, 90C46
1 Introduction
Motivation In a recent paper, Burke et al. [8, Corollary 9] show that the closed convex hull of
the set D:= (X, 1
2XXT)|XRn×mis given by
conv D=(X, Y )Rn×m×SnY1
2XXT.
Here, ‘’ is the owner partial ordering [12] on the symmetric matrices Sninduced by the positive
semideﬁnite cone Sn
+via ‘ABif and only if ABSn
+’. At second glance, the set D Rn×m×Sn
is simply the graph of the matrix-valued map F:XRn×m7→ 1
2XXTSn; and conv Din (1)
then appears to be a ‘generalized epigraph’ of Fwhere the partial ordering on the image space
Sn(induced by Sn
+) plays the role of the ordinary ordering of R(induced by R+) for scalar-valued
functions.
More generally, given a map F:E1E2between two (Euclidean) spaces E1and E2and a
cone KE2, we can order E2via ‘yKzif and only if yzK’. In view of the above identity,
the natural question that arises is the following: When is
conv (gph F) = {(x, y)|yKF(x)}(1)
valid? Clearly, this can only hold if the set on the right, which will later be called the K-epigraph of
F, is itself closed and convex, in which case we say that Fis K-closed and K-convex, respectively,
or closed and convex, with respect to (w.r.t.) K, respectively.
Related work The study of K-convexity has a long tradition in convex analysis and is now
part of many textbooks, e.g. [2,19]: Borwein [3] pursued an ambitious program of extending most of
convex analysis to cone convex functions including conjugacy, subdiﬀerential analysis, and duality,
laying out much of the groundwork. Kusraev and Kutateladze [14] take this idea to an even more
general setting by considering convex operators with values in arbitrary ordered vector spaces.
Pennanen [17] develops a deep theory of generalized diﬀerentiation for graph-convex mappings
(these are called convex correspondences in [14]) which contains some results on K-convexity, highly
relevant to our study. One of the most important features of a K-convex map Fis the fact
that the composition gFwith a convex function g, which is increasing with respect to the
ordering induced by K, is convex; a fact that has been well observed and utilized widely in the
literature [4–6, 10, 11, 17].
1
Road map and contributions We start our study in Section 2 with the necessary tools from
convex and variational analysis. In Section 3, we formally introduce and expand on the central
notions of K-convexity and K-closedness. In particular, in Sections 3.1-3.3 we characterize the
functions which are convex w.r.t. a given subspace, half-space and polyhedral cone respectively. In
Section 3.4, we elaborate on Pennanen’s characterization of the dual cone of the smallest closed,
necessarily convex (Proposition 11) cone with respect to which a given Fis convex. We extend
this in Section 3.5 to study the smallest, necessarily closed and convex (Proposition 11) cone with
respect to which Fis convex and closed. Section 4 is fully devoted to the question as to when
(1) holds. Theorem 34 in Section 4.1 provides a characterization for (1), which consitutes one
of the main workhorses for the the rest of Section 4. Section 4.2 is mainly devoted to necessary
conditions for (1). Partly mimicking the scalar case (Theorem 38), in Section 4.2.2 we present
necessary conditions based on aﬃne K-minorization and K-majorization. Section 4.3, in turn,
provides suﬃcient conditions. Section 4.4 presents diﬀerent examples of K-convex maps by which
we illustrate the theory developed in Section 3 and, more importantly, Section 4. In particular, we
apply our ﬁndings to the following maps:
F:XRn×m7→ 1
2XXTSn;
F:XSn
++ X1Sn(inverse matrix);
F:XSn7→ λ(X)1Rn(spectral map);
F:ERmwhere Fiis convex for all i= 1,...,m (component-wise convex).
Most of the criteria worked out in the previous sections for the validity of (1) are brought to bear
directly or indirectly here.
Section 5 taps into the composite framework alluded to above, where, primarily, we study the
following question: given a vector-valued map Fand a (closed, proper) convex function gsuch
that gFis convex, does there exist a cone Ksuch that Fis K-convex and gis increasing in a
K-related ordering?
Notation: In what follows, Edenotes a Euclidean space, i.e. a ﬁnite-dimensional real inner product
space with inner product denoted by ,·i. Given a set SE, we denote its closure, convex hull,
closed convex hull, and convex conical hull by cl S, conv S, conv Sand cone S, respectively. For a
vector uE, we denote its (convex) conical hull by R+u, and R++u={λu |λ > 0}. The indicator
function δS:ER∪ {+∞} of SEis given by δS(x) = 0 if xSand δS(x) = +otherwise.
2 Preliminaries
Throughout we make use of the relative topology for convex sets [18, §6]. The relative interior
ri Cof a convex set CEis its interior in the subspace topology induced by its aﬃne hull
aﬀ C:= {λx + (1 λ)y|λR, x, y C}. For a convex set CEand (any) x0C, the
subspace parallel to Cis par C:= aﬀ Cx0. For convex sets, we have a handy description of the
aﬃne hull.
Lemma 1. Let CEbe a convex. Then aﬀ C={αx βy |α, β >0, α β= 1, x, y C}.
Proof. Set A:= {αx βy |α, β >0, α β= 1, x, y C}. Thus aﬀ CA. Conversely, for
zaﬀ C, there exist λRand x, y Csuch that z=λx + (1 λ)y. If λ(0,1), then zCby
convexity, and hence z= 1 ·z0·zA. If λ > 1, set α:= λ>0, β:= λ1>0, and we get
αβ= 1 and z=αx βy A. Finally, if λ < 0, then 1 λ>0, and thus (β:= λ, α := 1 λ)
zA.
Let f:ER:= R∪ {±∞}. We call fproper if its domain dom f:= {xE|f(x)<+}
is nonempty and fdoesn’t take the value −∞. We say that fis convex if its epigraph epi f:=
{(x, α)E×R|f(x)< α}is convex which coincides with the usual deﬁnition via a secant condi-
tion
f(αx + (1 α)y)αf(x) + (1 α)f(y)x, y dom f, α (0,1),
if fdoes not take the value −∞. Although pointwise convergence is not a suitable for preservation
of many variational properties, see e.g. [19, Chapter 7], it still preserves convexity in the limit.
Lemma 2. Let {fk:ER∪ {+∞}} converge pointwise to f:ER, i.e. fk(x)f(x)for all
xE. If fkis convex for all kE(suﬃciently large), then so is f.
1Here λ(X) = (λ1,...,λn)Tis the vector of eigenvalues of XSnin decreasing order.
2
Proof. For α(0,1) and x, y E, the convexity of fkyields αfk(x) + (1 α)fk(y)>fk(αx + (1
α)y).Passing to the limit k→ ∞ on both sides gives αf + (1 α)f(y)>f(αx + (1 α)y).
We call f:ER {+∞} closed or lower semicontinuous (lsc) if epi fis closed. We set
Γ(E) := {f:ER∪ {+∞} | fproper, convex},
Γ0(E) := {fΓ(E)|fclosed}.
For fΓ(E), its closure cl fΓ0(E) is deﬁned via cl (epi f) = epi (cl f).More generally, given
a convex subset DE, we call f D-closed if epi fis closed in the subspace topology induced by
D×R. We note that D×Ris a metric space, hence closedness is sequential closedness and, in
particular, fis D-closed if and only if
lim inf
k→∞ f(xk)fx)∀{xkD} → ¯xD.
We deﬁne Γ0(D) := {fΓ(E)|f D-closed }.
Lemma 3. Let fΓ(E). Then the following are equivalent:
i) fΓ0(dom f);
ii) f(x) = (cl f)(x)for all xdom f.
Proof. Since fΓ0(E), we have
fΓ0(dom f)epi fis closed in dom f×R
epi f= cl (epi f)dom f×R
epi f= epi (cl f)dom f×R
f(x) = cl f(x)xdom f.
Here the ﬁrst equivalence is simply the deﬁnition of Γ0(dom f). The second is due to the fact that
the closed sets in the dom f×Rsubspace topology are exactly the intersections of closed sets (in
E×R) with dom f×R. The third one is clear as epi (cl f) = cl (epi f), and the fourth one follows
from elementary considerations.
Remark 4. We point out that fΓ0(E)implies that fΓ0(dom f), since the closed set epi f
E×Rintersected with dom f×Ris (trivially) closed in the subspace topology induced by dom f×R.
However, the converse statement is not true. Consider for instance δ(0,1) Γ0((0,1)) \Γ0(R), as
(0,1) ×R+is a closed set in the topology induced by (0,1) ×R, but is not a closed set in R×R.
A nonempty subset KEis called a cone if λx Kfor all λ0 and xK. If the latter only
holds for all λ > 0, we call Kapre-cone. For instance if Kis a cone, then ri Kis a pre-cone,
use e.g. [18, Corollary 6.6.1]. Combining this with the line segment principle [18, Theorem 6.1]
and [18, Theorem 6.3], we ﬁnd the following result.
Lemma 5. Let KEbe a convex cone. Then cl K+ ri Kri K.
The polar cone of a (pre-)cone Kis given by K:= {vE2| ∀uK:hu, vi60},and K
is referred to as the dual cone. Recall that conv K= (K)=: K◦◦ by the bipolar theorem [19,
Corollary 6.21], and that the polarity operation is order reversing. The horizon cone of CEis
given by C:= {uE| ∃{tk} ↓ 0,{xkC}: limk→∞ tkxk=u}.If Cis a nonempty convex set,
then C+C= cl Cand for a cone K, we have K= cl K.
A cone KEinduces and ordering on Evia
yKx:yxKx, y E.
Lemma 6. If Kis a closed and convex cone of E2, then
yKx⇒ hu, xi>hu, yi ∀u∈ −K.
Proof. By the bipolar theorem [19, Corollary 6.21], we have K◦◦ =Kand hence
x>KyxyK=K◦◦ ⇒ hu, x yi>0u∈ −K.
A cone KEis said to be pointed if K(K) = {0}. Such a cone induces a partial ordering.
Lemma 7 (Ordering induced by a pointed cone).Let KEbe pointed. Then
x=yxKyand yKx.
Proof. Straightforward.
3
3K-convexity and K-closedness
We commence this section with the central deﬁnitions of this paper.
Deﬁnition 8 (K-epigraphs, K-convexity and K-closedness).Let KE2be a cone and let F:D
E1E2. Then the K-epigraph of Fis given by
K-epi F={(x, y)D×E2|F(x)Ky} ⊂ E1×E2.(2)
We say that Fis
i) proper if K-epi F6=(i.e. D6=);
ii) K-convex if K-epi Fis convex;
iii) K-closed if K-epi Fis closed.
For DE1convex and KE2a cone, we point out that F:DE2is K-convex if and only if
Kis convex and
αF (x) + (1 α)F(y)>KF(αx + (1 α)y)x, y D, α (0,1).
Moreover, we always have
K-epi F= gph F+{0} × K. (3)
This has, in particular, the following immediate consequence.
Lemma 9. Let F:DE1E2be proper, and K1 K2E2be cones. Then K1-epi F
K2-epi F. In particular, there is at most one cone KE2such that K-epi F=conv (gph F).
In the convex case we can extract the following.
Lemma 10. Let F:DE1E2be proper, and let K1K2E2be convex cones. Then
K2-epi F=K1-epi F+{0} × K2. In particular, if Fis K1-convex, then Fis K2-convex.
Proof. This is due to (3) combined with the fact that K1+K2=K2because K1and K2are convex
cones.
Given a cone KEand its induced ordering, we attach to Ea formal largest element +with
respect to that ordering, and set E:= E∪ {+}. For G:E1E
2its domain is dom G:=
{xE1:G(x)E2}. The graph of Gis given by gph G:= {(x, F (x)) |xdom F}.We adopt
the notions in Deﬁnition 8 via the restriction F:= G|dom G. We record in the next result that
K-closedness and K-convexity requires certain conditions about the underlying cone K.
Proposition 11. Let KE2be a cone, and let F:E1E
2be proper. Then the following hold:
a) If Fis K-closed, then Kis closed.
b) If Fis K-convex, then Kis convex.
Proof. a) Let {ykK} → yand pick xdom F. Then (x, F (x) + yn)K-epi Ffor all kNand
(x, F (x) + yk)(x, F (x) + y)K-epi Fas K-epi Fis closed. Thus, yF(x) + yF(x) = yK.
b) Let y1, y2K, α (0,1). For xdom Fwe hence ﬁnd (x, F (x) + y1)K-epi F, and
(x, F (x) + y2)K-epi F. As K-epi Fis a convex, we have (x, F (x) + αy1+ (1 α)y2)K-epi F,
and consequently αy1+ (1 α)y2K. Thus, Kis convex.
The following proposition shows that a function F:E1E
2is fully determined by its K-epigraph
when Kis a pointed cone.
Proposition 12. Let KE2be a pointed cone, and let F, G :E1E
2be proper. Then
K-epi F=K-epi GF=G.
Proof. Suppose that K-epi F=K-epi G. In particular, for all xdom F, (x, F (x)) K-epi F, so
(x, F (x)) K-epi G, hence xdom Gand F(x)>KG(x). Likewise, for all xdom G, we have
xdom Fand G(x)>KF(x). Thus dom F= dom Gand for any xdom F= dom Gwe have
F(x) = G(x) by Lemma 7.
4
Given FE1:DE2and uE2, we deﬁne the scalarization hu, F i:E1R∪ {+∞} by
hu, F i(x) = hu, F (x)i, x D,
+else. (4)
We adapt this notion for D= do m Fif F:E1E
2where E2is ordered by some cone K. Equipped
with this concept, the following proposition gives a characterization of K-epi F(and gph F) via the
epigraphs (and graphs) of the scalarizations hu, F ifor u-K.
Proposition 13. Let KE2be a closed and convex cone, and let F:E1E
2be proper. Then:
a) K-epi F=Tu-K(id,hu, ·i)1epi hu, F i;
b) If Kis pointed, then gph F=Tu-K(id,hu, ·i)1gph hu, F i.
Proof. We deduce from Lemma 6 that
K-epi F={(x, v)|vKF(x)}
={(x, v)| hu, vi ≥ hu, F (x)i ∀u-K}
=\
u-K
(id,hu, ·i)1epi hu, F i.
Similarly, if Kis pointed, we obtain
gph F={(x, v)|xE1, v =F(x)}
={(x, v)|xE1, v F(x)K(K)}
={(x, v)|xE1, v KF(x)}\{(x, v)|xE1, F (x)Kv}
=\
u-K
(id,hu, ·i)1gph hu, F i.
As an immediate consequence of the latter theorem, one obtains Pennanen’s suﬃcient condition for
K-closedness [17, Lemma 6.2], which unfortunately excludes functions with domains that are not
closed. We therefore provide the following stronger version whose proof is simply a reﬁnement of
Pennanen’s in the next result’s part b). Part a) is a reﬁnement of the scalarization characterization
of K-convexity.
Proposition 14. Let KE2be a closed, convex cone, let F:E1E
2be proper. Then:
a) The following are equivalent:
i) hu, F iis convex for all uri (K);
ii) hu, F iis convex for all u∈ −K;
iii) Fis K-convex.
b) Fis K-closed if hu, F iis lower semicontinuous for all u∈ −K\ {0}and K6=E2.
In particular, if K6=E2and hu, F i ∈ Γ0(E1)for all u∈ −K\ {0}, then Fis K-closed and
K-convex.
Proof. a) ’i)ii)’: Let u∈ −K\ {0}. Then uis a limit {ukri (K)} u, and hence hu, F i
is a pointwise limit of conxex functions huk, F i, hence convex by Lemma 2.
’ii)iii)’: Follows from Proposition 13 a).
’iii)i)’: Follows from Lemma 6.
b) Assume that Fis not K-closed i.e. there exists {(xk, yk)K-epi F} → (x, y )/K-epi F. Then
x /dom For yF(x)/K=K◦◦ . In the latter case there exists u∈ −Ksuch that
hu, yi<hu, F (x)iand hu, yki ≥ hu, F (xk)i ∀kN.(5)
If xdom F, then necessarily u6= 0. On other hand, if x /dom F, since K){0}by
assumption, we can choose u6= 0. All in all, there exists u∈ −K\ {0}such that (5) holds. We
hence obtain
hu, F i(x)>hu, yi= lim inf
k→∞ hu, yki ≥ lim inf
k→∞ hu, F i(xk),
and, consequently, hu, F iis not lsc, which concludes the proof of part b).
5
We close out this preparatory paragraph with the following useful result.
Lemma 15. Let DE1be nonempty, let F:DE2, and let (Ki)iIbe a family of cones of
KiE2. Then \
iI
Ki!-epi F=\
iI
(Ki-epi F).
In particular, if Fis Ki-closed for all iI, then Fis (iIKi)-closed. Moreover, if Fis Ki-convex
for all iIthen Fis (iIKi)-convex. The latter is an equivalence if Kiis convex for all iI.
Proof. For any xdom Fand yE2, we have
(x, y) \
iI
Ki!-epi FyF(x)\
iI
Ki
⇒ ∀iI:yF(x)Ki
⇒ ∀iI: (x, y)Ki-epi F
(x, y)\
iI
Ki-epi F .
The addendum follows from the fact that intersection preserves closedness and convexity, and that
TiIKiKi, so TiIKi-convexity implies Ki-convexity for all iIif these are convex, see
Lemma 10.
3.1 Aﬃne and {0}-convex functions
We extend the notion of aﬃne functions to aﬃne subsets of E, see, e.g., Rockafellar [18, §1] for the
standard case.
Deﬁnition 16. Let AEbe an aﬃne set and let x0A. Then a function F:AE2is said to
be aﬃne if there exists a linear map L: par A7→ E2and a vector bE2such that, for all xA,
we have F(x) = L(xx0) + bfor all xA.
Lemma 17. Let AE1be aﬃne. Then F:AE2is aﬃne if and only if
F(tx + (1 t)y) = tF (x) + (1 t)F(y)x, y A, t (0,1).(6)
Proof. Assume ﬁrst that (6) holds. Discriminating the three cases t[0,1], t > 1 and t < 0, it is
straightforward to show that, in fact, we have
F(tx + (1 t)y) = tF (x) + (1 t)F(y)x, y A, t R.(7)
Now let x0A, i.e. par A=Ax0, and deﬁne L: par AE2by L(x) := F(x+x0)F(x0).
Using (7), we ﬁnd that L(tx + (1 t)y) = tL(x) + (1 t)L(y) for all x, y par Aand tR.
Thus, taking y= 0, as L(0) = 0, gives L(tx) = tL(x) for all xpa r Aand all tR. Hence,
L(x+y) = L(1
2(2x) + 1
2(2y)) = 1
2L(2x) + 1
2L(2y) = L(x) + L(y), for all x, y par A. This implies
that Lis linear. Hence, for all xAand b:= F(x0), we have F(x) = L(xx0) + b. Thus, Fis
aﬃne.
Conversely, if Fis aﬃne, then we can write F=L((·)x0) + bfor some linear map L: par A
E2,x0Aand bE2. Then
F(tx + (1 t)y) = t(L(xx0) + b) + (1 t)(L(yx0) + b) = tF (x) + (1 t)F(y),
for all x, y Aand tR. In particular, this is true for all t(0,1).
Proposition 18. Let DE1be nonempty convex. Then F:DE2is {0}-convex if and only if
there exists an aﬃne function G: aﬀ DE2such that F=G|D.
Proof. First, assume that F:DE2is {0}-convex and zaﬀ D. By Lemma 1, we can write
z=αx βy, for some α, β >0, αβ= 1, and x, y D. Suppose zhas two representations of this
form, i.e. z=αx βy =αxβy, with α, β, α, β >0, αβ=αβ= 1, and x, y, x, yD.
Then ∆ := α+β(= α+β) = 1 + β+β>0. By convexity of D, we ﬁnd that α
x+β
y=
α
x+β
yD. Using the {0}-convexity of F, we have α
F(x) + β
F(y) = α
F(x) + β
F(y).
Multiplying by ∆ and rearranging the above terms, we get
αF (x)βF (y) = αF(x)βF(y).
6
Therefore, we the function G: aﬀ DE2, G(z) = αF (x)βF (y) for zaDgiven by
z=αx βy,α, β >0, αβ= 1, x, y Dis well-deﬁned.
Now, let zand zin Dgiven by z=αxβy and z=αxβy, with α, β , α, β>0, αβ= 1,
αβ= 1, and x, y, x, y D. Let t(0,1) and set p:= (1 t)z+tz, as well as a:= (1 t)α+
and b:= (1 t)β+. Then a, b >0 and ab= 1.
If b= 0, then β=β= 0 and α=α= 1, so z=xDand z=xD, and thus, using the {0}-
convexity of F, we have G(p) = F(p) = F((1t)z+tz) = (1t)F(z)+tF (z) = (1t)G(z)+tG(z).
If b6= 0, then β6= 0 or β6= 0, hence a, b > 0. Then, we have
p=a(1 t)α
ax+
ax
|{z }
D
b(1 t)β
by+
by
|{z }
D
.
Using this, and recalling the fact that a, b >0 and ab= 1, the deﬁnition of Gyields
G(p) = aF (1 t)α
ax+
axbF (1 t)β
by+
by.
As Fis {0}-convex, we thus infer
G(p) = a(1 t)α
aF(x) + a
aF(x)b(1 t)β
bF(y)b
bF(y)
= (1 t) [αF (x)β F (y)]
|{z }
G(z)
+t[αF(x)βF(y)]
|{z }
G(z)
= (1 t)G(z) + tG(z).
All in all, by Lemma 17, Gis aﬃne.
Conversely, if there exists G: aﬀ DE2aﬃne such that F=G|D, then Lemma 17 yields
G(tx + (1 t)y) = tG(x) + (1 t)G(y) for all t(0,1) and x, y D. Hence F(tx + (1 t)y) =
tF (x) + (1 t)F(y) for all x, y D,t(0,1), and as Dis a convex set, Fis {0}-convex.
As a simple corollary we get the following result.
Corollary 19. The {0}-convex functions E1E2are exactly the aﬃne functions.
3.2 Convexity with respect to a (nontrivial) subspace
Using our study on {0}-convexity above, we are now in a position to investigate the functions
which are convex w.r.t. a given (nontrivial) subspace. To this end, observe that for a sub-
space UE, the polar (and the dual) cone of Uequal the orthogonal complement U:=
{uE| hu, yi= 0 yU}.
Lemma 20. Let UE2be a nontrivial subspace and let {e1,...,er}be a basis of U. Then for
F:DE1E2, with DE1convex, the following are equivalent:
i) Fis U-convex;
ii) hu, F iis {0}-convex for all uU;
iii) hei, F iis {0}-convex for all i= 1,...,r.
Proof. ‘i)ii)’: If Fis U-convex and since Uis a subspace, by Proposition 14 we ﬁnd that both
hu, F iand h−u , F iare convex for every uU. Therefore, hu, F iis {0}-convex for all uU.
In turn, if hu, F iis {0}-convex for all uU, then in particular hu, F iis (R+-)convex for all
uU, so we can use Proposition 14 to conclude that Fis U-convex.
‘ii)iii)’: The implication ii)iii) is trivial. In turn, if iii) holds, let uU, i.e. u=Pr
i=1 uiei
for some uiR. By assumption, hei, F i(i= 1,...,r) are {0}-convex. Since hu, F i(x) =
Pr
i=1 uihei, F i(x) for all xdom F, we ﬁnd that hu, F iis {0}-convex.
Proposition 21. Let UE2be a nontrivial subspace and let F:DE1E2with Dconvex.
Then Fis U-convex if and only if there exists an aﬃne map G: aﬀ DE2and a function
H: aﬀ DUsuch that G|D+H|D=F.
7
Proof. Suppose that Fis U-convex and let p:E2Ube the orthogonal pro jection onto U. Deﬁne
H: aﬀ DU, H (x) := (p(F(x)), x D,
0,else.
Set ˆ
G:= FH|D. By Lemma 20 and Proposition 18, hu, F i|Dis the restriction of an aﬃne
function fu: aﬀ DRfor all uU. However, as H(x)U(xE1), we have Du, ˆ
GE=hu, F i
for all uU. Now, let {e1,...,er}be an orthogonal basis of U. Then ˆ
G=Pr
i=1 fei
|Dei. Then
G:= Pr
i=1 feiei: aﬀ DE2is aﬃne and F=G|D+H|D.
Conversely, suppose that there exists a function H: aﬀ DUand an aﬃne map G: aﬀ DE2
such that F=G|D+H|D. Then for uUwe have hu, F i=hu, Gi|D, and thus hu, F iis {0}-
convex, as hu, Giis aﬃne on aﬀ DDand Dis convex. Thus, by Lemma 20, Fis U-convex.
3.3 Convexity with respect to a half-space and polyhedral cones
Every (proper) half-space HEthat is also a cone is of the form H={xE| hw, xi ≥ 0}for
some wE\ {0}. Clearly, His (polyhedral) convex and closed with dual cone H=R+w.
Proposition 22. Let wE2\{0}and let H={xE2| hw, xi>0}be the associated half-space.
Then for F:DE2with DE1(nonempty) convex we have:
a) Fis H-convex if and only if hw, F iis convex.
b) Fis H-closed if hw, F iis lower semicontinuous.
Proof. a) By Proposition 14 a), Fis H-convex if and only if hu, F iis convex for all uri (H) =
R++w. However, htw, F iis convex if and only if hw, F iis convex for all t > 0.
b) This follows with a similar argument as in a) from Proposition 14 b) observing that H6=E2
and H\ {0}=R++w.
Proposition 22 combined with Lemma 15 yields the following result on polyhedral cones.
Corollary 23. Let w1,...,wlE\ {0}and let P=Tl
i=1 Hiwith Hi={x| hwi, xi ≥ 0}. Then
for F:DE2with DE1(nonempty) convex we have:
a) Fis P-convex if and only if hwi, F iis convex for all i= 1,...,l.
b) Fis P-closed if hwi, F iis lower semicontinuous for all i= 1,...,l.
3.4 The smallest closed cone with respect to which Fis convex
The following result ensures the existence of a smallest closed cone KFwith respect to which Fis
convex (hence, KFis also convex by Proposition 11) and characterizes its dual cone.
Proposition 24 (The cone KFand its dual).Let DE1be nonempty and convex and let
F:DE2. Then the following hold:
a) There exists a smallest (nonempty) closed (and convex) cone KFE2with respect to which
Fis convex, i.e if Fis K-convex and Kis closed and convex, then KKF.
b) The dual cone of KFfrom a) is given by K
F={uE2| hu, F iis convex}.
Proof. a) Deﬁne KFas the intersection of all closed and convex cones which respect to which
Fis convex. Then KFis nonempty (as Fis E2-convex and every cone contains 0), closed and
convex, and, by Lemma 15, Fis KF-convex. By construction, there is no smaller cone with these
properties.
b) See [17, Lemma 6.1].
3.5 The smallest (closed) cone with respect to which Fis convex and
closed
We now investigate how the situation changes in comparison to the study in Section 3.4 when we
are looking for the smallest (by Proposition 11 necessarily closed and convex) cone with respect to
which a function Fis convex and closed. Such cone does not need to exist, simply because a given
function F:E1E
2need not be E2-closed. More concretely, for every cone KR, the indicator
F:= δ(0,1] has K-epi F= (0,1] ×K, which is not closed.
8
Proposition 25 (The cone ˆ
KF).Let DE1be nonempty and convex and let F:DE2such
that there exists a (necessarily closed and convex) cone KE2with respect to which Fis closed
and convex. Then there exists a smallest closed and convex cone ˆ
KFE2such that Fis ˆ
KF-closed
and ˆ
KF-convex.
Proof. Follows readily from Lemma 15.
In the spirit of Proposition 24 b), we want to characterize the dual cone of ˆ
KF(if it exists). To
this end, the following lemma is useful.
Lemma 26. Let F:DE1E2with Dconvex (and nonempty). Then the following hold:
a) The set cl {uE2| hu, F i ∈ Γ0(E1)}is either empty or a closed convex cone.
b) The set {uE2| hu, F i ∈ Γ0(D)}is a convex cone (in particular nonempty).
Proof. a) Assume nonemptiness, in which case S:= {uE2| hu, F i Γ0(E1)}is also nonempty.
Then, clearly, Sis convex and, consequently, so is cl S. Moreover, for ucl Sand λ0 there
exist {ukS} → uand {λk>0} → λand with λkukS, hence λu cl S, and thus cl Sis a
closed convex cone.
b) Set K:= {uE2| hu, F i Γ0(D)}. We ﬁrst show that Kis a cone. To this end, note that
h0, F i=δD, whose epigraph D×R+is clearly closed in the topology induced by D×R. Now, for
uKand λ > 0 observe that epi hλu, F i=L1
λ(epi hu, F i) for Lλ: (x, α)7→ (x, α/λ), which is
then closed as the linear preimage of a closed set in the topology induced by D×R. Therefore, Kis a
cone. In order prove that Kis convex, it hence suﬃces to show that K+KK, see [19, Exercise
3.7]. To this end, let u, v Kand take {(xk, αk)epi hu+v, F i} → (x, α)D×R. In
particular, we have (xk, αk− hv, F (xk)i)epi hu, F ifor all kN. Let z:= lim inf khv, F (xk)i.
W.l.o.g. z= limkhv, F (xk)i(otherwise go to subsequence), and by D-closedness of hv , F i, we
ﬁnd (x, z)epi hv , F i, i.e. hv, F (x)i ≤ z. Moreover, by D-closedness of hu, F iwe ﬁnd that
(x, α z)epi hu, F i, hence α≥ hu, F (x)i+z≥ hu, F i(x) + hv , F i(x). Therefore, (x, α)
epi hu+v, F i.
Proposition 27 (The dual cone of ˆ
KF).Let F:DE1E2with Dnonempty and convex. If
ˆ
KF(in the sense of Proposition 25) exists, we have
ˆ
K
F= cl {uE2| hu, F i ∈ Γ0(E1)}= cl {uE2| hu, F i ∈ Γ0(D)}.(8)
Proof. Set ¯
K:= cl ({uE2| hu, F i ∈ Γ0(E1)}).To prove the ﬁrst identity, use [17, Corollary
7.4(ii)], to ﬁnd hu, F i ∈ Γ0(E1) for all all uri (ˆ
K
F). Consequently, also using [18, Theorem
6.2], we have ˆ
K
F¯
K. Now, assume that the converse inclusion were false, and consequently
(by what was already proved) ˆ
K
F(¯
K. Then, by deﬁnition of ¯
K, there must acutally exist
u0∈ {uE2| hu, F i ∈ Γ0(E1)} \ (ˆ
K
F) (since otherwise ˆ
K
F=¯
assumption). In particular, u06= 0 and, by positive homogeneity, we have htu0, F i ∈ Γ0(E1)
for all t > 0, thus hu, F i ∈ Γ0(E1) for all uR+u0\ {0}. Therefore, by Proposition 14, with
L:= (R+u0)( E2, we ﬁnd that Fis L-convex and L-closed. Thus, by Lemma 15, we ﬁnd that
Fis ( ˆ
KFL)-closed and -convex. However, since u0/∈ − ˆ
K
F, we cannot have Lˆ
KF, and hence
ˆ
KFL(ˆ
KF. This contradicts the deﬁnition of ˆ
KFas the smallest closed, convex cone with
respect to which Fis convex and closed, and thus we have proved the ﬁrst identity.
To prove the second identity we ﬁrst note that, from what was proved above and by Remark 4,
we have ˆ
K
F= cl ({uE2| hu, F i ∈ Γ0(E1)})cl {uE2| hu, F i ∈ Γ0(D)}.In order to prove
the converse inclusion, set K:= −{uE2| hu, F i ∈ Γ0(D)}, and observe that by Lemma 26 b)
and [18, Theorem 6.3], respectively, we have
K= cl {uE2| hu, F i ∈ Γ0(D)}and ri (K)⊂ {uE2| hu, F i ∈ Γ0(D)}.(9)
We now show that Fis K-convex: To this end, ﬁrst note that hu, F iis convex for all uri (K),
by (9). Any u∈ −Kis a limit {ukri (K)} → u, and therefore hu, F iis the pointwise
limit of convex functions huk, F i,and hence convex (Lemma 2). Thus, by Proposition 14 a), Fis
K-convex.
We now prove that Fis also K-closed: To this end, let {(xk, yk)K-epi F} → (x, y). In
particular, there exists {vkK}such that F(xk) + vk=ykfor all kN. Moreover, as Kˆ
KF
9
(see above) and ˆ
KF-epi Fis closed (by deﬁnition), we have (x, y)ˆ
KF-epi F, and consequently,
xD. Thus we can use the fact that, by (9), hu , F iis D-closed for all uri (K), and hence
hu, yi= lim
k→∞ hu, yki= lim
k→∞ hu, F (xk)i+hu, vki
|{z }
>0
>lim inf
k→∞ hu, F i(xk)>hu, F i(x)
for all uri (K). Now, every u∈ −Kis a limit {uk∈ −ri (K)} → uwith huk, yi ≥ huk, F (x)i,
and hence hu, yi ≥ hu, F (x)i. As u∈ −Kwas arbitrary, this shows that yKF(x), and thus
(x, y)K-epi F, which shows that K-epi Fis closed and hence Fis K-closed (and K-convex as
proved earlier). Since Kˆ
KFit follows that ˆ
KF=K, which concludes the proof.
The natural question as to when closures in the above result are superﬂuous is addressed after the
following auxiliary result.
Lemma 28. Let {e1,...,en} ⊂ Ebe an orthonormal system and AEan aﬃne set such
that par A= span {e1,...,en}. Then, for all r > 0and all xA, we have Br
n(x)A
conv {x±rei|i= 1,...,n}.
Proof. Let yBr
n(x)A, i.e. y=x+Pn
i=1 yieifor some y1,...,ynR(i= 1,...,n) and
kyxk26r2
n2.As e1,...,enis an orthonormal system, we have
n
X
i=1
y2
i=
n
X
i=1
yiei
2
=kyxk2r2
n2.(10)
On the other hand, we have
y=x+
n
X
i=1
yiei
=x+1
2 1
n
X
i=1
|yi|
r!(re1re1)
| {z }
=0
+
n
X
i=1
|yi|
r(sgn(yi)rei)
=1
2 1
n
X
i=1
|yi|
r!(x+re1) + 1
2 1
n
X
i=1
|yi|
r!(xre1) +
n
X
i=1
|yi|
r(x+ sgn(yi)rei).
In view of (10), |yi|6r
nfor all i= 1,...,n, and thus yconv {x±rei|i= 1,...,n}, which
concludes the proof.
Corollary 29. Let F:DE1E2with Dnonempty and convex and assume that ˆ
KFexists.
Then:
a) ˆ
K
F={uE2| hu, F i ∈ Γ0(D)}if and only if {uE2| hu, F i ∈ Γ0(D)}is closed.
b) If, for every sequence {xkD} → xD(and every xD), there exists vri (ˆ
K
F)for
which {hv, F i(xk)}does not tend to +, then {uE2| hu, F i Γ0(D)}is closed.
c) If, for all xD\ri D, there exists a neighborhood Nxof x, and a continuous ˆ
KF-majorant2
of Fon NxD, then {uE2| hu, F i ∈ Γ0(D)}is closed.
In particular, ˆ
K
F={uE2| hu, F i ∈ Γ0(D)}if Dis relatively open (e.g. aﬃne).
Proof. a) Follows readily from Proposition 27.
b) Denote K={uE2| hu, F i ∈ Γ0(D)}, which is a convex cone by Lemma 26 b). By
Proposition 27, ˆ
K
F= cl K. Consider ucl K. Then hu, F iis convex by Lemma 2 and proper
with dom hu, F i=D.
Now let xDand {xkD} → xD. Then, by assumption, there exists vri (ˆ
K
F) = ri K
such that {hv, F i(xk)}is uniformly bounded away from +, hence w.l.o.g. we can assume that
hv, F i(xk)r < +. As hv, F i ∈ Γ0(D), we have −∞ <hv, F i(x)r < +. Using Lemma
5 (and that ri Kis a pre-cone), we have u+tv ri K, and hence hu+v, F i ∈ Γ0(D) for all t > 0.
Thus
lim inf
k→∞ hu, F i(xk) = lim inf
k→∞ hu+tv, F i(xk)thv, F i(xk)
= lim inf
k→∞ hu+tv, F i(xk)tr
>hu+tv, F i(x)tr.
2We refer the reader to Deﬁnition 39 for a formal introduction.
10
Letting t0 gives lim inf k→∞ hu, F i(xk)>hu, F i(x), hence hu , F i ∈ Γ0(D) as desired.
c) Let {xkD} → xD. We will show that hv, F i(xk) does not tend to +, for any v
ri (ˆ
K
F), which by b) then gives the desired conclusion.
First, suppose that xri Dand set A:= aﬀ D. Let {e1,...,en} ⊂ E1be an orthonormal
system such that A=x+ span {e1,...,en}. Now, xri Dimplies that there exists ε > 0 such
that x±εeiDfor all i= 1,...,n. As Dis convex, we have conv {x±εei|i= 1,...,n} ⊂ D.
As xkx, then xkBε
n(x) for kNlarge enough. As xkD, we have xkAfor all kN.
Hence, , by Lemma 28, we have xkBε
n(x)Aconv {x±εei|i= 1,...,n} ⊂ Dfor kN
large enough. Now let vri (ˆ
K
F) = ri {uE2| hu, F i ∈ Γ0(D)}. Then, hv, F iis convex,
so for klarge enough, we have hv, F i(xk)6max {hv, F i(x±εei)|i= 1,...,n}<+. Hence
hv, F i(xk) does not tend to +.
In turn, for xD\ri Dlet Gxbe the continuous ˆ
KF-majorant of Fon Nxand let vri (ˆ
K
F).
Then, as Gx(y)>KF(y) for all y∈ Nx, hence hv, Gxi(y)>hv, F i(y) for all y∈ Nx. Since
{xkD} → x, we have that xk∈ Nxfor ksuﬃciently large, and thus hv, F i(xk)6hv, Gxi(xk).
However, Gxis continuous, so hv, Gxiis continuous as well, hence hv, Gxi(xk)→ hv, Gxi(x)<
+, thus hv, F i(xk) does not tend to +.
We close out this section by clarifying the question as to when KFand ˆ
KFcoincide.
Proposition 30 (KF=ˆ
KF).Let DE1be nonempty and convex and let F:DE2. Then
KF=ˆ
KFif and only if Fis KF-closed.
Proof. By deﬁnition of the respective cones, we always have ˆ
KFKF. But if Fis KF-closed
then, ˆ
KFKF, by deﬁnition of ˆ
KF, and hence equality holds.
In turn, if Fis not KF-closed, then KF6=ˆ
KF, since Fis ˆ
KF-closed by deﬁnition.
4 When is conv (gph F) = K-epi F?
This section is devoted to our main question as to when the closed convex hull of the graph of a
function equals its K-epigraph.
4.1 A characterization via the horizon cone
We commence this subsection with the central link between the graph and the K-epigraph of a
function. To obtain an elegant proof we brieﬂy tap into Fenchel conjugacy [18]. To this end, realize
that every set SEis uniquely determined through its indicator function δS:ER∪ {+∞}
which is paired in duality with the support function σS:ER∪ {+∞},σS(x) = supyShx, yi
via the conjugacy relations δ
S=σS=σconv S, hence σ
S=δconv S, and thus conv δS=δconv S.
Proposition 31. Let KE2be a closed, convex cone and let F:E1E
2be proper, K-closed
and K-convex. Then
K-epi F= cl conv (gph F) + {0} × K.
Proof. Using (3) and the set-additivity for support functions we have
σK-epi F=σgph F+{0K=σgph F+σ{0K=σconv (gph F)+δE1×K.
Moreover, the assumptions on Fimply that K-epi Fis closed and convex, and hence
δK-epi F=σconv (gph F)+δE1×K= cl δconv (gph F)δ{0K=δcl conv (gph F)+{0K.
Here the second identity uses [18, Theorem 16.4], while the third holds due to the identity δAδB=
δA+Bfor any two sets.
We are now in a position to state our ﬁrst main theorem.
Theorem 32. Let KE2be a closed convex cone and let F:E1E
2be K-convex and K-closed.
Then
K-epi F=conv (gph F)⇒ {0} × K[conv (gph F)].
11
Proof. Suppose that K-epi F= conv (gph F). It follows from Proposition 31 that
conv (gph F) + {0} × Kcl conv (gph F) + {0} × K= conv (gph F).
Taking the horizon cone on both sides and using [19, Exercise 3.12], yields {0K[conv (gph F)].
Now suppose that {0} × K[conv (gph F)]. Then
conv (gph F) + {0} × Kconv (gph F) + [conv (gph F)]= conv (gph F).
where the last identity uses e.g. [19, Theorem 3.6]. Therefore, again using Proposition 31, we obtain
K-epi F= cl (conv (gph F) + {0} × K)conv (gph F)K-epi F .
We will frequently make us of the following trivial observation.
Remark 33. We observe that the closure operation in [conv (gph F)]is superﬂuous, i.e.
[conv (gph F)]= [conv (gph F)].
We immediately obtain the following suﬃcient condition.
Corollary 34. Let KE2be a closed, convex cone and let F:E1E
2be K-convex and K-closed
such that {0} × Kconv (gph F). Then K-epi F= conv (gph F).
Proof. Combine Theorem 34 with [19, Exercise 3.11] and the fact that the horizon operation pre-
serves inclusion.
Combining Theorem 34 with Lemma 9 yields the following result.
Corollary 35. Let Kbe a cone of E2and let F:E1E
2be proper, and deﬁne the closed, convex
cone KF:= {uE2|(0, u)[conv (gph F)]}.Then
K-epi F=conv (gph F)K=KF=ˆ
KF.
Proof. First, let K=KF=ˆ
KF. By deﬁnition of KF=Kwe hence have {0K[conv (gph F)].
From Theorem 34 we thus conclude that K-epi F= conv (gph F).
In turn, assume that K-epi F=conv (gph F). Then by Theorem 34 we have KKF, and
hence conv (gph F) = K-epi FKF-epi F. In addition, {0KF[conv (gph F)], by deﬁnition
of KF. Hence, using (3) and the horizon property of convex sets, we have
KF-epi F= gph F+{0} × KFconv (gph F) + [conv (gph F)]= conv (gph F).
Thus, KF-epi F=conv (gph F) = K-epi Fand hence, by Lemma 9, we already have KF=
K. Moreover, as Fis K-convex and K-closed, we have ˆ
KFK, thus ˆ
KF-epi FK-epi F=
conv (gph F). Using the fact that conv (gph F)ˆ
KF-epi F(as ˆ
KF-epi Fis a closed convex set
containing gph F), we conclude that ˆ
KF-epi F=K-epi F, hence, again by Lemma 9, K=ˆ
KF.
4.2 Necessary conditions
In this subsection we discuss necessary conditions for conv (gph F) = K-epi F.
4.2.1 Necessary conditions on the dual cone
Proposition 36. Let KE2be a cone and F:E1E
2proper such that K-epi F=conv (gph F).
Then K⊂ − ˆ
K
F.
Proof. Let uK\ {0}. By Theorem 34, we have that (0, u)[conv (gph F)]. By Remark 33
there hence exist {(xk, yk)conv (gph F)}and {λk}↓0 such that λk(xk, yk)(0, u). With
κ:= dim E1×E2and Carath´eodory’s theorem [18, Theorem 17.1], for i= 1,...,κ+ 1, we ﬁnd
sequences {xi
kdom F}kand {αi
k}ksuch that Pκ+1
i=1 αi
k= 1 for all kNas well as
xk=
κ+1
X
i=1
αi
kxi
kand yk=
κ+1
X
i=1
αi
k,j F(xi
k,j )kN.
12
Now let t0. Then tk:= tλk
kuk20 and
tkxk0 and tk
κ+1
X
i=1
αi
khu, F i(xi
k) = thλkyk, ui
kuk2t.
Thus, for t0, we have (0, t)[conv (gph hu, F i)]= [conv (gph hu, F i)]. Hence {0} × R+
[conv (gph hu, F i)], so by Theorem 34, epi hu, F i= conv (gph hu, F i). This means that epi hu, F i
is closed and convex so that hu, F i ∈ Γ0(E1). Therefore K\ {0} ∈ {uE2| hu, F i ∈ Γ0(E1)},
hence Kcl ({uE2| hu, F i ∈ Γ0(E1)}) = ˆ
K
F, by Proposition 27.
We readily derive the following necessary condition on the dual cone.
Corollary 37. Under the assumptions of Proposition 36, we have K⊂ −K.
Proof. By Corollary 35, K=ˆ
KF, and by Proposition 36, K⊂ − ˆ
K
F=K.
4.2.2 Aﬃne majorization and minorization
For motivational purposes, we start this subsection with the scalar case (K=R+), where the
question whether the closed convex hull of the graph of a function equals its K-epigraph can
be fully answered via aﬃne majorization. The proof relies, in essence, on a standard separation
argument.
Theorem 38 (The scalar case).Let fΓ0(E). Then epi f=conv (gph f)if and only if fdoes
not have an aﬃne majorant on its domain.
Proof. Suppose that there exists (¯x, ¯
t)epi f\conv (gph f). In particular, ¯xdom f, and by
strong separation [18, Corollary 11.4.2], there exists (s, α)E1×Rsuch that
hs, ¯xi+α¯
t > sup
(x,t)conv (gph f)hs, xi+αt. (11)
Choosing (x, t) := (¯x, f x)), we ﬁnd that α(¯
tfx)) >0,and hence, α > 0. It then follows from
(11) with xdom fand t=f(x) that hs/α, ¯xxi+¯
t > f(x).Thus fis majorized on its domain
by the aﬃne map x7→ −hs/α, xi+hs/α, ¯xi+¯
t, which proves one direction.
To prove the converse implication, suppose now that fhas an aﬃne majorant on its domain,
i.e. there exists (a, β )E×R, such that f(x)≤ ha, xi+β=: g(x) for all xdom f. Now pick
¯xdom f. Then (¯x, g( ¯x)+1) epi f, and it hence suﬃces to show that (¯x, g x)+1) 6∈ conv (gph F).
Assume, by contradiction, that (¯x, g x) + 1) conv (gph F). Then with κ:= dim E×R, by
Carath´eodory’s theorem [18, Theorem 17.1], for i= 1,...,κ+ 1 there exist sequences {ti,k 0}kN
and {xk
idom f}kNsuch that Pκ+1
i=1 ti,k = 1 and Pκ+1
i=1 ti,k(xk
i, f (xk
i)) k→∞ x, g(¯x) + 1).
Consequently
gx) + 1 = lim
k→∞
κ+1
X
i=1
ti,kf(xk
i)lim
k→∞
κ+1
X
i=1
ti,kg(xk
i) = lim
k→∞ *a,
κ+1
X
i=1
ti,kxk
i++β=gx),
which is the desired contradiction and thus concludes the proof.
The questions as to what can be said when fin the above result is only proper and convex, but
not necessarily closed is answered as the opening to Section 4.3.
To start our analysis of the vector-valued case we now formally introduce the notion of K-
minorants and -majorants, respectively.
Deﬁnition 39 (K-minorants/-majorants).Let KE1be a cone, and let F:E1E
2be proper.
A function G:E1E2is said to be:
aK-majorant of Fon Sdom Fif
G(x)F(x)KxS,
aK-minorant of Fon Sdom Fif
F(x)G(x)KxS.
For S= dom F, we say that Gis a K-minorant of F.
13
Naturally, in view of the scalar case from Theorem 38, we are mainly interested in the case where
Gis an aﬃne function.
For a function F:E1E
2, we record that, for a pointed, closed, convex cone Ksuch that
K-epi F=conv (gph F), there cannot exist both an aﬃne K-majorant on the domain and an aﬃne
K-minorant of F.
Proposition 40. Let {0}(KE2be a closed, convex, pointed cone and let F:E1E
2be
proper. If K-epi F=conv (gph F), then Fcannot have both an aﬃne K-majorant on its domain
and an aﬃne K-minorant.
Proof. Assume, by contradiction, that there exist T , L :E1E2be linear and u, w E2be such
that
T(x) + u>KF(x)>KL(x) + w, xdom F.
We thus ﬁnd that gph F⊂ {(x, y)dom F×E2|T(x)y∈ −u+K, y L(x)w+K}:= C.
As Kis closed and convex, so is C, and we hence deduce that conv (gph F)C. Now pick
xdom Fand vK\ {0}. Then (x, F (x) + tv)K-epi Ffor all t > 0. Since K-epi F=
conv (gph F)C, it follows that
T(x)F(x)tv ∈ −u+Kand F(x) + tv L(x)w+Kt > 0.
Dividing by tand letting t+we get vK(K) = {0},which contradicts the choice of v,
and therefore proves the statement.
It is well known that a proper, convex function posseses an aﬃne (R+-)minorant [1, Theorem 9.20].
In the vector-valued setting, we can fall back on this result to get aﬃne K-minorants for proper
K-convex functions when Kis a particular polyhedral cone.
Proposition 41. Let K={xE2| hbi, xi ≥ 0i= 1,...,m}with b1,...,bmlinearly indepen-
dent. If F:E1E
2is K-convex and proper, then Fhas an aﬃne K-minorant.
Proof. It holds that K= cone {b1,...,bm}, see e.g. [19, Lemma 6.45]. This cone is pointed by
linear independence of {b1, . . . , bm}, hence Khas nonempty interior, see [19, Exercise 6.22]. Now,
in view of Proposition 14 a), for all i= 1,...,m, the functions hbi, F i:E1R{+∞} are proper,
convex and hence, see e.g. [1, Theorem 9.20], there exist (ci, δi)E1×R(i= 1,...,m) such that
hbi, F (x)i ≥ hci, xi+δixE1, i = 1,...,m. (12)
Now, let A:E1E2be linear such that A(bi) = cifor all i= 1,...,m and let w∈ −int K. Then
hw, bii<0 for all i= 1, . . . , m, cf. [19, Exercise 6.22]. By positive homogeneity (and since int K
is a pre-cone), there hence exists ¯w∈ −int Kwith h¯w, bii< δifor all i= 1,...,m. Finally, with
¯
L:= Ait hence follows
hbi, F (x)i ≥ hci, xi+δibi,¯
L(x)+h¯w, bii ∀xE1, i = 1,...,m.
Therefore, for all xdom F, we have F(x)(¯
L(x) + ¯w)K, and x7→ ¯
L(x) + ¯wis the desired
aﬃne K-minorant.
4.3 Suﬃcient conditions
In this subsection we are primarily concerned with suﬃcient conditions. We start with some
considerations in the scalar case.
Lemma 42. Let fΓ(E). Then the following hold:
a) conv (gph cl f) = conv (gph f|ri (dom f)).
b) φ:ERis an aﬃne majorant of cl fon dom (cl f)if and only if φis aﬃne majorant of f
on ri (dom f).
c) If fis (dom f)-closed (hence fΓ0(dom f)), then φ:ERis an aﬃne majorant of cl f
on dom (cl f)if and only if φis aﬃne majorant of fon dom f.
Proof. a) As f(x) = cl f(x) for all xri (dom f) ( [18, Theorem 7.4]), we have gph f|ri (dom f)
gph cl f, and hence conv (gph f|ri (dom f))conv (gph f). To prove the converse inclusion let
(x, cl f(x)) gph cl f. Invoking [18, Theorem 7.5] (and [18, Theorem 6.1]), we ﬁnd a sequence
{xkri (dom f)} → xwith f(xk)cl f(x). Therefore, gph cl fcl (gph f|ri (dom f))
14
conv (gph f|ri (dom f)), and hence, the desired inclusion follows by applying the conv -operator on
both sides.
b) If φis an aﬃne majorant of cl fon dom (cl f), then φis an aﬃne majorant of cl fon
ri (dom f)dom (cl f), and hence an aﬃne majorant of fon ri (dom f), since fand cl fcoincide
on ri (dom f). In turn, if φis an aﬃne ma jorant of fon ri (dom f), then for all xdom (cl f),
since ri (dom (cl f)) = ri (dom f) (see [18, Corollary 7.4.1]), by [18, Theorem 7.5] (and [18, Theorem
6.1]), there exists {xkri (dom (cl f))} → xwith limkf(xk) = cl f(xk). However φ(xk)>f(xk)
and φis continuous so φ(x)>cl f(x), thus φis an aﬃne majorant of cl fon dom cl f.
c) By Lemma 3, f(x) = cl f(x) for all xdom f. Therefore, if φ(x)cl f(x) for all x
dom (cl f)dom f, then φ(x)f(x) for all xdom f. In turn, if φis an aﬃne minorant of fon
dom fri (dom f), then b) shows that φis an aﬃne minorant of cl fon dom (cl f).
We record some immediate consequences of the foregoing result.
Corollary 43. Let fΓ(E). Then the following are equivalent:
i) conv (gph f|ri (dom f)) = cl (epi f);
ii) fhas no aﬃne majorant on ri (dom f);
iii) {0} × R+[conv (gph f|ri (dom f))].
Proof. Observe that epi (cl f) = cl (epi f), hence by Lemma 42 a) we have
i)conv (gph cl f) = epi (cl f).(13)
’i)ii)’: By Lemma 42 b), we have that ii) is equivalent to saying that cl fhas no aﬃne minorant
on its domain. Therefore, the desired equivalence follows with (13) from Proposition 38 applied to
cl fΓ0(E).
’i)iii)’: Apply Theorem 34 to cl fand use (13).
Corollary 44. Let fΓ0(dom f). Then the following are equivalent:
i) conv (gph f) = cl (epi f);
ii) fhas no aﬃne majorants on its domain;
iii) {0} × R+[conv (gph f)].
Proof. We observe that fΓ(E) (by deﬁnition of Γ0(dom f)), and that f(x) = cl f(x) for all x
dom f, by Lemma 3. In addition, by Lemma 42, we have conv (gph f|ri (dom f)) = conv (gph cl f).
Thus, we have
conv (gph f)conv (gph cl f) = conv (gph f|ri (dom f))conv (gph f),
and hence conv (gph f) = conv (gph f|ri (dom f)). Consequently
i)conv (gph f|ri (dom f)) = cl (epi f),
and
iii)⇒ {0} × R+[conv (gph f|ri (dom f))].
Moreover, with Lemma 42 we ﬁnd that
ii)fhas no aﬃne ma jorant on ri (dom f).
Therefore, the claimed equivalences follow from Corollary 43.
We now establish suﬃcient conditions in the vector-valued case, building on the results provided
above. We start with the most general result, and then successively tighten the assumptions to
obtain (weaker but) more handy conditions.
Lemma 45. Let KE2be a (nontrivial) closed, convex cone with K⊂ −K, and let F:E1E
2
be proper, K-closed and K-convex. Assume that the following hold:
i) There is a nonempty set LKrge Fsuch that cone L=K;
ii) For every uL, there exists a convex set Cudom Fsuch that F(Cu)R+u;
15
iii) For all uLwe have fu:= hu, F i+δCuΓ0(Cu)and it has no aﬃne majorant on its
domain.
Then K-epi F=conv (gph F). In particular, K=ˆ
KF.
Proof. Let uL\ {0}, and let us prove that (0, u)[conv (gph F)]. As the latter is a cone, we
may assume w.l.o.g that kuk= 1. By iii), fuΓ0(Cu), hence Corollary 44 yields that {0}× R+
[conv (gph fu)]. As conv (gph fu) is convex, we hence have (0, t) + conv (gph fu)conv (gph fu)
for all t0.
Now, let (x, r)conv (gph fu). Hence, with κ:= dim E×R, there there exist convex com-
binations x=Pκ+1
i=1 αixiand r=Pκ+1
u=1 αihu, F i(xi) with xiCu(i= 1,...,κ + 1). By ii),
there exists γ>0 such that Pκ+1
i=1 αiF(xi) = γu, and thus γ=hu, γui=r. Consequently,
(x, ru)conv (gph F).
Moreover, we have (x, r +t)conv (gph fu) for all t0, hence
x= lim
n→∞
κ+1
X
i=1
αi,n
txi,n
tand r+t= lim
n→∞
κ+1
X
i=1
αi,n
thu, F i(xi,n
t)
for certain xi,n
tCuand αi,n
t0 (i= 1,...,κ+ 1) with Pκ+1
i=1 αi,n = 1 for all nN. However,
by ii), there exist γn
t>0 (nN) such that
κ+1
X
i=1
αi,n
tF(xi,n
t) = γn
tunN.
Thus
γn
t=*u,
κ+1
X
i=1
αi,n
tF(xi,n
t)+=
κ+1
X
i=1
αi,n
thu, F i(xi,n
t)r+t.
Thus, (x, ru +tu) = (x, (r+t)u)conv (gph F) for all t>0. Now, for all (¯x, ¯y)conv (gph F),
and for all s>0, we have
x, ¯y+su) = lim
ε0(1 ε)(¯x, ¯y) + ε(x, ru + (s/ε)u).
However, by what was argued above, (x, ru + (s/ε)u)conv (gph F) for all ε > 0, and we picked
x, ¯y)conv (gph F)conv (gph F). By convexity and closedness , it follows that
x, ¯y+su)conv (gph F)x, ¯y)conv (gph F), s 0.
Thus (0, u)[conv (gph F)]for all uL\ {0}as desired. For u= 0 this is trivially true, hence,
altogether we ﬁnd that {0}×L[conv (gph F)]. But as [conv (gph F)]is a closed convex cone,
we have then cone ({0} × L) = {0} × K[conv (gph F)],cf. i). Thus, by Theorem 34, we have
K-epi F= conv (gph F).
We record an immediate consequence.
Proposition 46. Let K6={0}be a closed, convex cone such that K⊂ −Kand such that F:E1
E
2is proper, K-convex and K-closed. Assume that K= cone (b1,...,bN)for b1,...,bnrge F,
and that, for any i= 1, . . . , N , there exists a nonempty convex set CiF1(R+bi)such that,
for all i= 1,...,N, the function fbi:= hbi, F i+δCbiis Cbi-closed and does not have any aﬃne
majorant on Cbi. Then K-epi F=conv (gph F).
Proof. Apply Lemma 45, with L={b1,...,bN}.
To wrap up this section we want to provide a simpliﬁed version of Lemma 45 with more restrictive,
but less arduous assumptions. To this end, we need the following lemma.
Lemma 47. Let KE2be a closed, convex cone with K⊂ −K, and let F:E1E
2be proper,
K-convex and K-closed. Then hu, F i ∈ Γ0(E1)for all uri K.
Proof. We observe from [17, Corollary 7.4(ii)] that hu, F i ∈ Γ0(E1) for all uri (K). Thus if
ri Kri (K) there is nothing to prove.
Hence, we only need to consider the case ri K*ri (K). Since we assume that K⊂ −K, by
the deﬁnition of the relative topology, this can only hold, if aﬀ K aﬀ (K). We note that both
of these sets contain 0, and hence are subspaces of E2. In particular, the orthogonal projection
16
p:E2E2onto aﬀ K(which is ordered by K) is a linear self-adjoint operator. We deﬁne
G:E1(aﬀ K)by
G(x) := (p(F(x)), x dom F,
+,else.
Note that for all α(0,1) and x, y dom F, we have αF (x) + (1 α)F(y)F(αx+ (1 α)y)K.
Hence, as Kaﬀ Kand by linearity of p, for all α(0,1) and x, y dom F= dom G, we have
αF (x) + (1 α)F(y)F(αx + (1 α)y) = αG(x) + (1 α)G(y)G(αx + (1 α)y).
Therefore, Gis K-convex. Moreover, if we denote D:= dom Gand H:= FG:Daﬀ (K),
then
αH(x) + (1 α)H(y)H(αx + (1 α)y) = 0 x, y D, α (0,1).
Hence, His {0}-convex, and by Proposition 18, there exists an aﬃne function ˆ
H:E1aﬀ Ksuch
that ˆ
H|D=H.
Now, let {(xk, zk)K-epi G]} → (x, z)E1×aﬀ K. Then, for all kN,xkdom G, and
there exists vkKsuch that
zk=G(xk) + vk=F(xk)H(xk) + vn=F(xk)ˆ
H(xk) + vk.
As ˆ
H:E1aﬀ Kis aﬃne, it is continuous, so ˆ
H(xk)ˆ
H(x), and thus F(xk) + vkz+ˆ
H(x).
Therefore {(xk, F (xk) + vk)K-epi F} → (x, z +ˆ
H(x)). As Fis K-closed, we have (x, z +ˆ
H(x))
K-epi F, thus xdom F= dom G,ˆ
H(x) = H(x), and z+H(x)F(x)K, so zG(x)Kand
(x, z)K-epi G. This proves that Gis K-closed.
Let Kbe the dual cone of Kin aﬀ K. As K⊂ −Kby assumption, we consequently obtain
K⊂ −Kaﬀ K=Kaﬀ K. Hence, ri Kri K. Moreover, as G:E1(aﬀ K)is proper,
K-closed and K-convex, by [17, Corollary 7.4(ii)], we have hu, Gi ∈ Γ0(E1) for all uri K. But
for any uri Kaﬀ K, as pis self-adjoint, we have hu, Gi=hu, p(F)i=hu, F i. Thus, for any
uri Kwe have hu, F i ∈ Γ0(E1).
Proposition 48. Let KE2be a proper, closed, convex cone such that K⊂ −Kand let
F:E1E
2be proper, K-convex and K-closed with ri Krge F. Moreover, assume that, for any
uri K, there exists a nonempty convex set CuF1(R+u)such that fu:= hu, F i+δCudoes
not have any aﬃne majorant on its domain. Then, K-epi F=conv (gph F).
Proof. By Lemma 47, for all uri K, we have hu, F i ∈ Γ0(E1), hence fuΓ0(Cu). Applying
Lemma 45 with L= ri Kyields the desired result.
4.4 Examples
In this section we put our ﬁndings from the previous sections to the test on various examples of K-
convex functions. Throughout, we equip the matrix space Rn×mwith the Frobenius inner product
,·i :Rn×m×Rn×mR,hX, Y i= tr (XTY). In particular, on the space of symmetric matrices
Sn, the transposition is superﬂuous.
4.4.1 F:X7→ 1
2XXT
We consider the function
F:Rn×mSn, F (X) = 1
2XXT.(14)
It plays a central role in study of the matrix-fractional [7–9] and variational Gram functions [10,13].
Proposition 49. Let Fbe given by (14). Then the following hold:
a) ˆ
KF=KF=Sn