J. Aust. Math. Soc. 81 (2006), 49–61
STRICT CONVEXITY OF SETS IN ANALYTIC TERMS
LEONI DALLA and TELEMACHOS HATZIAFRATIS
(Received 11 July 2004; revised 21 March 2005)
Communicated by A. Rubinov
Abstract
We compare the geometric concept of strict convexity of open subsets of Rn with the analytic concept of
2-strict convexity, which is based on the defining functions of the set, and we do this by introducing the
class of 2N -strictly convex sets. We also describe an exhaustion process of convex sets by a sequence of
2-strictly convex sets.
2000 Mathematics subject classification: primary 52A20.
Keywords and phrases: strictly convex set, 2N -strictly convex set, defining function, exhaustion of a
convex set.
1. Introduction
K
K
K
Let
.Rn / be the set of convex compact subsets of Rn with non-empty interior.
Then .Rn /, equipped with the Hausdorff metric, is a complete metric space. A set
in .Rn / is called strictly convex if its boundary does not contain any line segments.
Klee [1] proved that the subset of
.Rn /, which contains the sets which are not
strictly convex, is a set of first category in .Rn /. Zamfirescu [4, 3] improved this
result by proving that the above set is ¦ -porous. In view of these results we can say
that the set of strictly convex sets is not only dense in .Rn /, but is a large set. On the
other hand, there is another concept of strict convexity—this is what we call 2-strict
convexity—defined in terms of a defining function of D. More precisely, if D ⊂ Rn
is a bounded open set with C 2 boundary and ² : Rn → R is a C 2 function so that
K
K
K
D = {x ∈ Rn : ².x/ < 0};
@ D = {x ∈ Rn : ².x/ = 0};
c 2006 Australian Mathematical Society 1446-8107/06 $A2:00 + 0:00
49
50
Leoni Dalla and Telemachos Hatziafratis
[2]
and d².x/ 6 = 0 for each x ∈ @ D; then D is said to be 2-strictly convex if for every
x ∈ @ D and every y = .y1 ; : : : ; yn / ∈ Rn \ {0},
n
X
@².x/
j=1
@xj
yj = 0
implies that
X @ 2 ².x/
y j yk > 0:
@ x j @ xk
1≤ j;k≤n
This concept of 2-strict convexity plays an important role in complex analysis. For
example, 2-strict convexity affects the solvability of the inhomogeneous CauchyRiemann equations in the domain, with Lipschitz estimates (see Range [2]). Thus, in
a sense, it is natural to ask what is the relation between the geometric and analytic
strict convexity. It is easy to show that if a set is 2-strictly convex, then it is also strictly
convex in the geometric sense. The converse does not hold however, and in order to
deal with this question we generalize 2-strict convexity and we obtain the concept of
2N -strictly convexity. Then, using this, we state and prove Theorem 2.4. We will also
describe a process of exhausting an arbitrary convex set by 2-strictly convex sets with
smooth boundary. The main results are Theorems 2.4 and 3.5 and their corollaries,
and the various lemmas that we present, are needed in their proof. Although some of
the ideas involved in these lemmas are essentially known, we include them here for
completeness.
2. Strictly convex sets
Let us first recall the definition of defining functions. Let D ⊂ Rn be a bounded
open set with C 1 boundary. A C 1 function ² : Rn → R is said to be a defining function
of D if D = {x ∈ Rn : ².x/ < 0}, @ D = {x ∈ Rn : ².x/ = 0}, and d².x/ 6 = 0
for each x ∈ @ D. If the boundary of D is assumed to be C ∞ , then we can choose
a defining function ² : Rn → R to be C ∞ (see [2]). Also if the boundary of D is
assumed to be real analytic, then we can choose a C ∞ defining function ² : Rn → R
which is moreover real analytic in a neighborhood of @ D.
We will also use the following notation about higher order differentials. For m ∈ N,
x ∈ Rn , and y = .y1 ; : : : ; yn / ∈ Rn
dm ².x/.y/ =
X
1≤ j1 ;:::; jm ≤n
@ m ².x/
y j · · · y jm ;
@ x j1 · · · @ x jm 1
defined in the case ² is of class C m . In particular, d1 ².x/ = d².x/.
LEMMA 2.1. Let D ⊂ Rn be a bounded open set with C 1 boundary and ² : Rn → R
a C 1 defining function of it. Then D is convex if and only if d1 ².x/.x − y/ > 0 for
every x ∈ @ D and y ∈ D.
[3]
Strict convexity of sets in analytic terms
51
PROOF. Assume that D is convex. Suppose that for some x ∈ @ D and y ∈ D,
d1 ².x/.x − y/ < 0:
By the convexity of D, r .t/ = t x + .1 − t/y ∈ D, for t ∈ [0; 1/. Define
h.t/ = ².r .t//;
for t ∈ [0; 1]:
Then h is a C 1 function, h.1/ = ².x/ = 0 and h.t/ < 0 for t ∈ [0; 1/. Also
h 0 .1/ = d1 ².x/.x − y/ < 0:
Therefore there is an " with 0 < " < 1 such that h 0 .t/ < 0 for t ∈ [1 − "; 1]. It
follows that h is strictly decreasing in the interval [1 − "; 1]. So h.1 − "/ > h.1/ = 0,
which contradicts the fact that h.t/ < 0 for t ∈ [0; 1/.
This proves that
(1)
d1 ².x/.x − y/ ≥ 0
for every x ∈ @ D and y ∈ D:
Now suppose that for some x ∈ @ D and y ∈ D, d1 ².x/.x − y/ = 0. Since D is
assumed open, there is a Ž > 0 so that the ball
B.y; Ž/ = {y + su : s ∈ .−Ž; Ž/ and u ∈ Rn with |u| = 1} ⊂ D:
Let u ∈ Rn with |u| = 1. Then y + su ∈ D, for s ∈ .−Ž; Ž/. By (1),
d1 ².x/ x − .y + su/ ≥ 0:
Since d1 ².x/.x − y/ = 0, it follows that sd1 ².x/.u/ ≤ 0, for s ∈ .−Ž; Ž/. Hence
d1 ².x/.u/ = 0, for every u ∈ Rn with |u| = 1. This contradicts the assumption that
d².x/ 6 = 0 and completes the proof that d1 ².x/.x − y/ > 0 for every x ∈ @ D and
y ∈ D.
Conversely, assume that d1 ².x/.x − y/ > 0 for every x ∈ @ D and y ∈ D. Let
y; z ∈ D, y 6 = z, and suppose that for some x ∈ [y; z], x ∈ @ D. Let us write
x = y + t .z − y/ = z + s.y − z/;
for some t; s ∈ .0; 1/:
Then
d1 ².x/.x − y/ > 0
and
d1 ².x/.x − z/ > 0;
d1 ².x/.z − y/ > 0
and
d1 ².x/.y − z/ > 0;
which implies that
which is impossible.
S If x 6 ∈ D
S then ².x/ > 0. Since
Thus for x ∈ [y; z], either x ∈ D or x 6 ∈ D.
².y/ < 0, there exists a − ∈ .0; 1/ so that ² y + − .x − y/ = 0. Then y + − .x − y/ ∈
@ D ∩ [y; z], which has already been excluded. This proves that D is convex.
52
Leoni Dalla and Telemachos Hatziafratis
[4]
LEMMA 2.2. Let D ⊂ Rn be a bounded open and convex set with C ∞ boundary,
² : Rn → R a C ∞ defining function of D, x ∈ @ D, and y ∈ Rn . If dm ².x/.y/ = 0 for
m = 1; 2; : : : ; 2k, then d2k+1 ².x/.y/ = 0.
PROOF. Since dm ².x/.sy/ = s m dm ².x/.y/, for every m, we may assume that
|y| = 1. If d2k+1 ².x/.y/ > 0, then there is an " > 0 such that d2k+1 ².z/.y/ > 0 for
z ∈ B.x; "/. By Taylor’s theorem, for 0 < |t| < ",
².x + t y/ = ².x/ +
=
2k
X
1
1
dm ².x/.t y/ +
d2k+1 ².zt /.t y/
m!
.2k + 1/!
m=1
1
t 2k+1 d2k+1 ².zt /.y/;
.2k + 1/!
for some zt ∈ [x; x + t y] ⊂ B.x; "/. It follows that for t ∈ .−"; 0/, ².x + t y/ < 0,
and therefore x + t y ∈ D. Then, by Lemma 2.1,
d1 ².x/ x − .x + t y/ = d1 ².x/.−t y/ > 0:
This is a contradiction since, by assumption, d1 ².x/.y/ = 0. This shows that
d2k+1 ².x/.y/ > 0 cannot hold.
Similarly we prove that the inequality d2k+1 ².x/.y/ < 0 cannot hold either, and
the equation d2k+1 ².x/.y/ = 0 follows.
LEMMA 2.3. Let D ⊂ Rn be a bounded open and convex set with C ∞ boundary,
² : Rn → R a C ∞ defining function of D, x ∈ @ D and y ∈ Rn . If dm ².x/.y/ = 0 for
m = 1; 2; : : : ; 2k − 1, and d2k ².x/.y/ 6 = 0, then d2k ².x/.y/ > 0.
PROOF. If ` is the straight line ` = {r .t/ = x + t y : t ∈ R}, then the intersection
S is a line segment, that is, ` ∩ D
S = [a; b] for some a; b ∈ Rn , and x ∈ [a; b].
`∩ D
We consider the following cases.
(i) a = b. Then the function h.t/ = ².r .t// for t ∈ R has the property h.t/ > 0
for t 6 = 0 and h.0/ = 0. Thus h has a minimum at t = 0. Since
(
dm ².x/.y/ = 0; for m = 1; 2; : : : ; 2k − 1;
.m/
h .0/ =
d2k ².x/.y/ 6 = 0; for m = 2k;
it follows that h .2k/ .0/ = d2k ².x/.y/ > 0.
(ii) a 6 = b and [a; b] ⊂ @ D. In this case ².x + t y/ = 0 for every t ∈ [−; ¦ ], where
−; ¦ ∈ R with − < ¦ and 0 ∈ [−; ¦ ]. Therefore,
d 2k d2k ².x/.y/ = 2k ².x + t y/ = 0:
dt
t=0
Thus this case cannot occur.
[5]
Strict convexity of sets in analytic terms
53
S = D ∪ @ D, there is a t0 6 = 0
(iii) a 6 = b and [a; b] 6 ⊂ @ D. Since [a; b] ⊂ D
such that x + t0 y ∈ D. By Lemma 2.1, d1 ².x/ x − .x + t0 y/ = d1 ².x/.−t0 y/ > 0.
However, this implies that d1 ².t0 y/ 6 = 0 and therefore this case cannot occur either.
This completes the proof of the lemma.
DEFINITION. Let D ⊂ Rn be a bounded open set with C ∞ boundary and ² : Rn →
R a C ∞ defining function of it. For x ∈ @ D, let T .x/ = {y ∈ Rn : d1 ².x/.y/ = 0}
and
d j ².x/.y/ = 0; for j = 1; 2; : : : ; 2m − 1;
n
S2m .x/ = y ∈ R \ {0} :
d j ².x/.y/ > 0; for j = 2m:
The set D is called 2k-strictly convex at the point x if for every y ∈ T .x/ \ {0}, there
exists a positive integer m.y/ ≤ k such that y ∈ S2m.y/ .x/, and k = max{m.y/ :
y ∈ T .x/}. The set D is called 2N -strictly convex if for each x ∈ @ D, there is
k.x/ ≤ N such that D is 2k.x/-strictly convex at x, and N = max{k.x/ : x ∈ @ D}.
If D is 2-strictly convex, then N = 1, and therefore k.x/ = 1 for every x ∈ @ D.
Hence for x ∈ @ D, m.y/ = 1 for every y ∈ T .x/ − {0}, that is,
d2 ².x/.y/ > 0
when
d1 ².x/.y/ = 0 and y 6 = 0:
In some sense, the 2-strictly convex sets are the most strictly convex sets.
REMARK 1. For each fixed x ∈ @ D, the sets T .x/ and S2m .x/ do not depend on the
defining function ², that is, they depend only on the set D. This can be justified as
follows. Let ½ : Rn → R be another C ∞ defining function of D. Then there is a C ∞
function h, defined in a neighborhood W of @ D, so that h > 0 and ½ = h² in W (see
[2, page 51]). Thus d1 ½.x/.y/ = h.x/d1 ².x/.y/ if ².x/ = 0, and the independence
of T .x/ of the defining function, follows.
To show that the sets S2m .x/ are also independent of the defining function, it suffices
to prove that
(2)
dk ½.x/.y/ = h.x/dk ².x/.y/;
if x ∈ @ D
d j ².x/.y/ = 0
for j = 1; 2; : : : ; k − 1:
and
Let us prove this for k = 2. A straightforward computation shows that
X
X
@ 2½
@ 2²
.x/yi1 yi2 = h.x/
.x/yi1 yi2
@ xi1 @ xi2
@ xi1 @ xi2
1≤i 1 ;i2 ≤n
1≤i 1 ;i2 ≤n
! n
!
n
X
X @²
@h
+2
.x/yi1
.x/yi2
@ xi1
@ xi2
i =1
i =1
1
+ ².x/
X
1≤i 1 ;i2 ≤n
2
@ h
.x/yi1 yi2 ;
@ xi1 @ xi2
2
54
Leoni Dalla and Telemachos Hatziafratis
[6]
and therefore d2 ½.x/.y/ = h.x/d2 ².x/.y/, if ².x/ = 0 and d1 ².x/.y/ = 0.
The proof of (2) in the general case is a computation, which is simplified if we
consider the functions
l.t/ = ½.x + t y/
and observe that
and r .t/ = ².x + t y/;
d j l d j ½.x/.y/ = j dt t=0
and
t ∈ R;
d j r d j ².x/.y/ = j :
dt t=0
Thus (2) follows from the formula
k d k .r / X k d j d k− j r
=
;
j dt j dt k− j
dt k
j=0
where .t/ = h.x + t y/.
Notice also that (2) implies that for each k,
{x ∈ @ D : d j ².x/.y/ = 0 for j = 1; : : : ; k − 1}
= {x ∈ @ D : d j ½.x/.y/ = 0 for j = 1; : : : ; k − 1}:
THEOREM 2.4. Let D ⊂ Rn be a bounded open set with real analytic boundary.
The following are equivalent:
(i) D is connected and 2N -strictly convex, for some N ∈ N.
(ii) D is convex and its boundary does not contain line segments.
PROOF. (i) implies (ii). Let  = {.x; y/ ∈ D × D : .1 − t/x + t y ∈ D for every
t ∈ [0; 1]}. Then  is an open subset of D×D. We claim that  is also closed in D×D,
for otherwise there would exist a sequence .x ¹ ; y¹ / ∈ , .x¹ ; y¹ / → .x; y/ ∈ D × D
and .x; y/ 6 ∈ . Then for some − ∈ .0; 1/, z = .1 − − /x + − y 6 ∈ D. Since x ¹ → x,
S for t ∈ [0; 1].
y¹ → y, and .1 − t/x¹ + t y¹ ∈ D, it follows that .1 − t/x + t y ∈ D
Hence h.t/ = ² .1 − t/x + t y ≤ 0 for t ∈ [0; 1], and h.− / = ².z/ = 0, since
z ∈ @ D. Throughout this proof, ² : Rn → R is a C ∞ defining function of D which
is real analytic in a neighborhood of @ D. Thus h.t/ takes its maximum at t = − , and
therefore h 0 .− / = d1 ².z/.y − x/ = 0. By assumption, there is an m ≤ N such that
d j ².z/.y − x/ = 0;
j = 1; 2; : : : ; 2m − 1;
and
d2m ².z/.y − x/ > 0:
In other words,
h . j/ .− / = 0 for j = 1; 2; : : : ; 2m − 1;
and
h .2m/ .− / > 0;
and this is a contradiction since h.t/ takes its maximum at t = − . This proves that 
is closed in D × D.
[7]
Strict convexity of sets in analytic terms
55
Since D is assumed to be connected, it follows that  = D × D, that is, D is
convex.
Now we will show that @ D does not contain line segments. Assume, to reach a
contradiction, that [x; y] ⊂ @ D and x 6 = y. Then
h.t/ = ² .1 − t/x + t y ; for t ∈ [0; 1];
and therefore h .m/ .0/ = dm ².x/.y − x/ = 0 for every m ∈ N. However, this
contradicts the assumption that D is 2N -strictly convex.
(ii) implies (i). Let x ∈ @ D and y ∈ T .x/ \ {0}. If d2m ².x/.y/ = 0 for every
m ∈ N, then by Lemma 2.2, d2m+1 ².x/.y/ = 0 for every m ∈ N. Since ² is assumed
to be real analytic in a neighborhood of @ D,
².x + t y/ = ².x/ +
∞
X
dm ².x/.t y/ = 0; for t ∈ .−"; "/;
m=1
where " > 0. However, this implies that x + t y ∈ @ D for t ∈ .−"; "/, that is, @ D
contains line segments, and this is a contradiction.
It follows that there exists l.y/ so that d2l.y/ ².x/.y/ 6 = 0. Let m.y/ be the smallest
integer with this property, that is, d2m.y/ ².x/.y/ 6 = 0. If m.y/ = 1 then, by Lemma 2.3,
d2 ².x/.y/ > 0. If m.y/ ≥ 2, then d2 j ².x/.y/ = 0 for j = 1; : : : ; m.y/ − 1, and
by Lemma 2.2, d2 j+1 ².x/.y/ = 0 for j = 1; : : : ; m.y/ − 1. Hence by Lemma 2.3,
d2m.y/ ².x/.y/ > 0. This shows that for every y ∈ T .x/ \ {0} there is m.y/ such that
y ∈ S2m.y/ .x/.
Let k.x/ = sup{m.y/ : y ∈ T .x/ \ {0}}. We will prove that k.x/ < ∞. To reach
a contradiction assume that k.x/ = ∞. Since m.y=|y|/ = m.y/ for y ∈ T .x/ \ {0},
there exists a sequence y¹ ∈ T .x/ \ {0} with |y¹ | = 1 and m.y¹ / → ∞. By the
compactness of the set {y : y ∈ T .x/ \ {0} and |y| = 1}, we may assume that
y¹ → y ∈ T .x/ \ {0} with |y| = 1. Let m.y/ be such that y ∈ S2m.y/ .x/. Then
d2m.y/ ².x/.y/ > 0; and therefore d2m.y/ ².x/.y¹ / > 0 for ¹ ≥ ¹0 , where ¹0 ∈ N. By
the choice of m.y¹ /, m.y¹ / ≤ m.y/ for ¹ ≥ ¹0 , which contradicts m.y¹ / → ∞.
So far we proved that for each x ∈ @ D there is k.x/ so that D is 2k.x/-strictly
convex at x. Let N = sup{k.x/ : x ∈ @ D}. It remains to show that N < ∞.
To reach a contradiction, assume that N = ∞. By the compactness of @ D, there
is a sequence x¹ ∈ @ D such that x¹ → x ∈ @ D and k.x¹ / → ∞. Let y¹ ∈ T .x ¹ /
with |y¹ | = 1 and k.x ¹ / = m x¹ .y¹ /, where m x¹ .y¹ / is the m.y¹ /-integer associated to
y¹ ∈ T .x ¹ /. Passing to a subsequence, we may assume that y¹ → y with |y| = 1.
Since y¹ ∈ T .x ¹ /, d1 ².x ¹ /.y¹ / = 0, and therefore, by continuity, d1 ².x/.y/ = 0, that
is, y ∈ T .x/ \ {0}. Then d2m x .y/ ².x/.y/ > 0, and therefore d2m x .y/ ².x ¹ /.y¹ / > 0, for
¹ ≥ ¹0 , where ¹0 ∈ N. By the choice of m x¹ .y¹ /, m x¹ .y¹ / ≤ m x .y/ for ¹ ≥ ¹0 , and
consequently k.x ¹ / = m x¹ .y¹ / ≤ m x .y/, which contradicts k.x ¹ / → ∞.
This proves that N ∈ N and that D is 2N -strictly convex.
56
Leoni Dalla and Telemachos Hatziafratis
[8]
COROLLARY 2.5. Let D ⊂ Rn be a bounded open set with real analytic boundary.
Then the following are equivalent:
(i) D is connected and 2N -strictly convex for some N ∈ N.
S \ {x}, d1 ².x/.x − y/ > 0.
(ii) For every x ∈ @ D and y ∈ D
PROOF. (i) implies (ii). By Theorem 2.4, D is convex and its boundary does not
contain line segments. If y ∈ D, d1 ².x/.x − y/ > 0 by Lemma 2.1. If y ∈ @ D, then
x + t .y − x/ ∈ D for every t ∈ .0; 1/. Then, by Lemma 2.1,
d1 ².x/ x − t .y − x/ = td1 ².x/.x − y/ > 0;
and (ii) follows.
(ii) implies (i). By Lemma 2.1, D is convex. We claim that @ D does not contain line
segments. If [x; y] ⊂ @ D with x 6 = y, then ² .1 − t/x + t y/ = 0 for t ∈ [0; 1], and
this would imply that d1 .x/.x − y/ = 0. Hence @ D does not contain line segments,
and (i) follows from Theorem 2.4.
REMARK 2. By examining the proof of Theorem 2.4 we see that if @ D is of class
C 2N and D is 2N -strictly convex then D is convex and its boundary does not contain
line segments, that is, one direction of the theorem does not require real analyticity
of the boundary (C 2N is enough). However, for the other direction real analyticity is
necessary as seen by the example of the set {.x; y/ ∈ R2 : y > .x/}, where
(
2
e−1=x if x 6 = 0;
.x/ =
0
if x = 0;
examining its boundary locally at the point .0; 0/.
3. Exhaustion of convex sets by 2-strictly convex sets
The notation A ⊂⊂ B that we will be using, is defined as follows: A ⊂⊂ B if and
only if there is a compact set E so that A ⊂ E ⊂ B.
LEMMA 3.1. Let D ⊂ Rn be an open bounded and convex set. Then there is a
continuous convex function ½ : D → R such that for every a ∈ Rn ,
{x ∈ D : ½.x/ < a} ⊂⊂ D:
PROOF. We may assume without loss of generality that 0 ∈ D. Let us consider
Minkowski’s functional
1
½ D .x/ = inf t > 0 : x ∈ D ;
t
[9]
Strict convexity of sets in analytic terms
57
defined for x ∈ Rn . Then ½ D : Rn → R is a continuous convex function with
the properties D = {x ∈ Rn : ½ D .x/ < 1}, @ D = {x ∈ Rn : ½ D .x/ = 1}, and
S = {x ∈ Rn : ½ D .x/ > 1}. Define
Rn \ D
½.x/ =
1
;
1 − ½ D .x/
for x ∈ D:
It is straightforward to check that the function ½ : D → R has the required properties.
LEMMA 3.2. Let f : K × G → R be a continuous function, where G ⊂ Rn is
a closed rectangle and K is a compact set. Then for every " > 0 there exist points
¾ j ∈ G and positive numbers c j , j = 1; 2; : : : ; M, such that
M
Z
X
c j f .x; ¾ j / −
f .x; y /dy < "
G
for x ∈ K :
j=1
More generally this holds in the case where G ⊂ Rn is a compact set whose boundary
has measure zero.
PROOF. This follows from the uniform continuity of f and the definition of the
integral.
LEMMA 3.3. Let D ⊂ Rn be an open set and f : D → R a continuous
function.
R
∞
n
Let us also consider ∈ C .R /, ≥ 0, .x/ = 0 for |x| ≥ 1, and .x/ d x = 1.
For sufficiently small " > 0, set D" = {x ∈ D : dist.x; @ D/ > "} and define the
function
Z
1
x−y
f " .x/ =
f .y/ n dy; x ∈ D" :
"
"
y∈Rn
Then f " is well-defined and C ∞ in D" , and f " converges to f uniformly on compact
subsets of D as " → 0+ .
Moreover, if the function f is convex in D, then f " is convex in D" too.
PROOF. Firstly the existence of functions like is quite standard, but we can also
give a specific example as follows: For t ∈ R, set
(
.t/ =
and define e
.x/ =
exp −1=.t + 1/2 exp −1=.t − 1/2
if |t| < 1;
0
if |t| ≥ 1;
R
e x.
e d
.|x|2 / and = =
58
Leoni Dalla and Telemachos Hatziafratis
Now, since supp./ ⊂ B.0; 1/,
Z
1
x−y
f " .x/ =
f .y/ n dy;
"
"
y∈B.x;"/
[10]
x ∈ D" :
and therefore f " is well-defined for x ∈ D" .
Furthermore, if a ∈ D" , there is a sufficiently small r > 0, such that B.a; r / ⊂ D
and for x close to a,
Z
1
x−y
f " .x/ =
f .y/ n dy:
"
"
y∈B.a;r/
It follows that f " is C ∞ in D" .
Next, changing the variable in the integral which defines f " , we obtain
Z
f " .x/ =
f .x − "y/.y/dy; for x ∈ D" :
y∈B.0;1/
Also if K is a compact subset of D, then the set K " = {x ∈ Rn : dist.x; K / ≤ "} is a
compact subset of D, for " > 0 and sufficiently small.
By observing that the function f .u/.y/ for .u; y/ ∈ K " × B.0; 1/ is uniformly
continuous, and writing
Z
f " .x/ − f .x/ =
f .x − "y/ − f .x/ .y/ dy;
y∈B.0;1/
R
which follows from the assumption d x = 1, we obtain that f " converges to f ,
uniformly on K , as " → 0+ .
Finally, it follows from Lemma 3.2 and the fact that the uniform limit of convex
functions is also convex, that each f " is convex, if f is a convex function. Indeed, f "
is a − uniform on compact subsets of D" − limit of functions of the form
x −→
M
X
f .x − "y j /c j ;
with c j > 0;
j=1
which are convex functions.
A C 2 -function ¼ : W → R, defined in an open set W ⊂ Rn , is 2-strictly convex if
for every x ∈ W ,
d2 ¼.x/.y/ =
X @ 2 ¼.x/
y j yk > 0
@ x j @ xk
1≤ j;k≤n
for y ∈ Rn \ {0}:
In the following we will use the term strictly convex function to mean 2-strictly convex
function.
[11]
Strict convexity of sets in analytic terms
59
LEMMA 3.4. Let ¼ be a strictly convex C 2 -function in a neighborhood of a compact
set K ⊂ Rn . Then there exists an " > 0 with the following property: If g is a
C 2 -function in a neighborhood of K and |@ 2 g=@ x j @ xk | < " on K for every j , k, then
the function ¼ + g is strictly convex in a neighborhood of K .
PROOF. Define
1
" = 2 min
n
(
)
X @ 2 ¼.x/
y j yk : x ∈ K and y ∈ Rn with |y| = 1 :
@
x
@
x
j
k
1≤ j;k≤n
Then " > 0 and it is straightforward to check that it has the required property.
THEOREM 3.5. Let D ⊂ Rn be a bounded open and convex set, K ⊂ D a compact
convex subset, and U ⊂ D an open neighborhood of K . Then there exists a C ∞ strictly
convex function ¼ : D → R such that for every a ∈ R, {x ∈ D : ¼.x/ < a} ⊂⊂ D,
and moreover ¼ < 0 on K and ¼ > 0 on D \ U .
PROOF. By Lemma 3.1, there is a continuous convex function ½ : D → R such
that for every a ∈ R, {x ∈ D : ½.x/ < a} ⊂⊂ D. Adding a negative constant to the
function ½, if necessary, we may assume that ½ < 0 on K .
Now consider the set K 0 = {x ∈ D : ½.x/ ≤ 0}. Since K is assumed to be
convex, for every x ∈ K 0 ∩ .D \ U /, there is an affine function u x (that is, a function
of the form u x .t/ = c0 + c1 t1 + · · · + cn tn , with ci constants) so that u x .x/ > 0 while
u x < 0 on K . By the compactness of the set K 0 ∩ .D \ U /, we may choose affine
functions u 1 ; : : : ; u M in such a way that max.u 1 ; : : : ; u M / > 0 on K 0 ∩ .D \ U / and
max.u 1 ; : : : ; u M / < 0 on K .
Setting u = max.½; u 1 ; : : : ; u M /, we have u < 0 on K , u > 0 on D \ U , and of
course u is continuous and convex on D. Also, {x ∈ D : u.x/ < a} ⊂⊂ D for every
a ∈ R.
In order to construct a C ∞ and strictly convex function with these properties, we
consider the sets D j = {x ∈ D : u.x/ < j } for j = 0; 1; 2; : : : , and we construct a
sequence of functions ¼ j , j = 1; 2; : : : , such that
Sj ,
(1) ¼ j is C ∞ and strictly convex in a neighborhood of D
(2) ¼ j < 0 on K ,
S j ,and
(3) ¼ j ≥ u on D
(4) ¼ j = ¼ j−1 on D j−2 , for j ≥ 2.
Then, setting ¼ = ¼ j on D j−2 , we obtain the desired function.
First we construct ¼1 . Let us choose " > 0 so that u + " < 0 on K . By Lemma 3.3,
S1 , so that |e
there is a C ∞ and convex function e
¼1 in a neighborhood of D
¼1 − u| < "=2
2
S1 . Then, for Ž > 0, the function e
on D
¼1 .x/+Ž|x| is strictly convex in a neighborhood
60
Leoni Dalla and Telemachos Hatziafratis
[12]
S1 , and if Ž is sufficiently small, the function ¼1 .x/ = e
of D
¼1 .x/ + Ž|x|2 + "=2 satisfies
the required conditions.
Next let us assume that the functions ¼1 ; : : : ; ¼ j−1 have been constructed for j ≥ 2.
S j−1 , there is a C ∞ function b
Since ¼ j−1 is C ∞ in a neighborhood of D
¼ j−1 : Rn → R
S j−1 . By Lemma 3.3, there is a C ∞ and
such that b
¼ j−1 = ¼ j−1 in a neighborhood of D
S
S j . Let Ž > 0
convex function e
¼ j in a neighborhood of D j , so that |e
¼ j − u| < 1=2 on D
∞
be sufficiently small so that the strictly convex C
b
¼ j .x/ = e
¼ j .x/ + Ž|x|2 ;
S j , satisfies |b
S j . Then (and in
defined in a neighborhood of D
¼ j − u| < 1=2 on D
combination with the definition of the sets D j )
S j−2
b
¼ j < j − 3=2 on D
S j \ D j−1 :
and b
¼ j > j − 3=2 on D
Now let us take a C ∞ and convex function : R → R with .t/ = 0 for t ≤ j − 3=2
and 0 > 0 for t > j − 3=2. For example, we may choose
Z t

exp − 1=.s − j + 3=2/2 ds if t > j − 3=2;
.t/ =
 s=0
0
if t ≤ j − 3=2:
S j−2 and ◦ b
S j \ D j−1 . Moreover ◦ b
Then ◦ b
¼ j = 0 on D
¼ j > 0 on D
¼ j is convex
S
S
in a neighborhood of D j and strictly convex in a neighborhood of D j \ D j−1 . Here
Sj , b
we used the facts that b
¼ j is strictly convex in a neighborhood of D
¼ j > j − 3=2
0
S
on D j \ D j−1 and that .t/ > 0 for t > j − 3=2, to conclude that ◦ b
¼ j is strictly
S j \ D j−1 .
convex in a neighborhood of D
Now we claim that for a sufficiently large C > 0, the function ¼ j = b
¼ j−1 +C ◦b
¼j
satisfies the conditions (1)–(4). First let us observe that (4) and (2) are satisfied for
S j−2 , we obtain (4), and
every C > 0. Indeed, since ◦ b
¼ j = 0 and b
¼ j−1 = ¼ j−1 on D
(2) follows, from K ⊂ D0 ⊂ D j−2 , because ¼ j−1 satisfies (2). That (3) is satisfied
S j−1 ,
for C > 0, sufficiently large, follows from the facts that b
¼ j−1 .= ¼ j−1 / ≥ u on D
S j , and ◦ b
S j \ D j−1 .
◦b
¼ j ≥ 0 on D
¼ j > 0 on D
It remains to show that (1) also holds, if C > 0 and sufficiently large. Since the
function b
¼ j−1 .= ¼ j−1 / is strictly convex and ◦ b
¼ j is convex in a neighborhood
S j−1 , it follows that ¼ j is strictly convex in a neighborhood of D
S j−1 for C > 0.
of D
S
Moreover, since ◦ b
¼ j is strictly convex in a neighborhood of D j \ D j−1 , it follows
from Lemma 3.4, that
1
¼j = C ◦ b
¼j + b
¼ j−1
C
S j \ D j−1 , if C > 0 is sufficiently large, so
is also strictly convex in a neighborhood of D
that 1=C is small enough. Here we also used the fact that the second order derivatives
of the function b
¼ j−1 are all bounded on a compact set.
[13]
Strict convexity of sets in analytic terms
61
This completes the proof of the theorem.
COROLLARY 3.6. Let D ⊂ Rn be an open convex set. Then D can be exhausted by a
sequence of 2-strictly convex sets with C ∞ boundary, that is, there exists a sequence D j
of 2-strictly convex sets with C ∞ boundary such that D j ⊂⊂ D j+1 ⊂⊂ D and
S
D j = D.
PROOF. Assume that D is bounded. According to Theorem 3.5, there exists a C ∞
and strictly convex function ¼ : D → R such that for a ∈ R,
{x ∈ D : ¼.x/ < a} ⊂⊂ D:
However, the function ¼, being strictly convex, has at most one critical point in D.
Therefore, the sets D j = {x ∈ D : ¼.x/ < j } have C ∞ boundary, for j ∈ N and
sufficiently large, and, since ¼ is a strictly convex function, they are 2-strictly convex.
Now it is clear that these sets have the required properties.
S
Finally if D is unbounded we may write D = ∞
j=1 G j , where G j ⊂⊂ G j+1 ⊂⊂ D
are open convex sets, and choose the 2-strictly convex sets D j with C ∞ boundary such
that G j ⊂⊂ D j ⊂⊂ G j+1 .
References
[1] V. Klee, ‘Some new results on smoothness and rotundity in normed linear spaces’, Math. Ann. 139
(1959), 51–63.
[2] R. M. Range, Holomorphic functions and integral representations in several complex variables,
Graduate Texts in Mathematics 108 (Springer, New York, 1986).
[3] T. Zamfirescu, ‘Nearly all convex bodies are smooth and strictly convex’, Monatsh. Math. 103
(1987), 57–62.
, ‘Baire categories in convexity’, Atti Sem. Mat. Fis. Univ. Modena 39 (1991), 139–164.
[4]
The University of Athens
Department of Mathematics
Panepistemiopolis 15784
Athens
Greece
e-mail: [email protected], [email protected]