I nternat. J. Math. & Math. Sci.
(1978) 459-477
Vol
459
TANGENT CONES, STARSHAPE AND CONVEXITY
J. M. BORWEIN
Department of Mathematics
Dalhousie University
Halifax, Nova Scotia
Canada, B3H 4H8
(Received March 28, 1978)
ABSTRACT.
In the last few years various infinite dimensional extensions to
Krasnoselski’s Theorem on starshaped sets [14] have been made.
with a paper by Edelstein and Keener
These began
[8] and have culmdnated in the papers
by Borwein, Edelstein and O’Brien [3] [4] by Edelstein, Keener and O’Brien
[9] and finally by O’Brien [16].
Unrelatedly, Borwein and O’Brien [5] posed a question which arises in
optimization
[2] [II] of when a closed
set is pseudoconvex at all its mere-
bers.
In this paper we show that these two questions can be handled simultaneously through a slight refinement of the powerful central result in
[16] with attendant strengthening of the results in [5] [16].
This in turn
leads to some interesting characterizations of convexity, starshape and of
various functional conditions.
460
J. M. BORWEIN
KEY WORDS AND PHRASES.
fun n
Tange cones, pseudo tangent cones, starshape, convety,
conditions.
AMS (MOS) SUBJECT CLASSIFICATION (1970) CODES.
i.
52A30, 52A35, 90C30.
INTRODUCTION
We will suppose throughout that X is a (real Hausdorff) locally convex
(topological vector) space with continuous linear dual X’.
If A
will require in fact that X be a Banach space.
A
0
A ri
A
and co A
c
Most of our results
X is an arbitrary set,
denote respectively the closure, interior, interior
relative to the closed affine span of A
and the convex hull of A.
Aff(A)
and span(A) will denote the affine and linear spans of A respectively.
[7] and [12].
other usage is in accord with
In accord with previous work [7] [12] we will say x
line segment
[ x,a] (= co{x,a})
of cardinality a
If
A.
is contained in
A.
a
If
A
has visibility
of cardinality
card(A)
a
a in A if the
sees
We say
A
(a-visibility) if every subset of
simultaneously seen by some point
starshape.
Any
we say
A has k-visibility for every natural number
k
a
is
A has
we say
A has
We define
finite visibility.
{xeA: [x,a]cA}
A(a)
star
(i.I)
and
star A
Then
A has starshape if and only if
the requirement that
{starA(a)
families of cardinality
point
[9]
a.
aeA}
n(starA(a)
aeA}
star A
x’
#
and a-visibility is
have nonempty intersection for sub-
We also recall that
if there is a nonzero
(1.2)
X’
a e A
such that
is said to be a cone
x’ (a)
sup{x’ (x): x
starA(a) }.
Krasnoselski
n+l
[14] showed that
if
A
n
is a compact subset of R
cone points can be seen by some point in
A
then
A
and every
is starshaped.
This
TANGENT CONES, STARSHAPE AND CONVEXITY
was generalized in [4] to show that if
A
461
is a closed relatively weakly compact
subset of a Frchet space with finite visibility it has starshape.
The proof
method here is entirely different from Krasnoselski’s but recently O’Brien [16]
has shown that if
A
is a closed relatively weakly compact subset of a Banach
space and every finite set of cone points is visible in
A
A
that
is star-
shaped.
Another object of our investigations will be the tangent cone to
A
at
[19] defined by
a
{heX: h
T(A,a)
lira t
n
(an-A)
a
n
e A
Here convergence is in the initial topology on
n
R
or a smooth manifold in
T(A,a)
T(A,a)
co
in general
if and only if
[2].
Also if
a
a
0
A
If
(1.3)
is a convex set
Following Guignard
In general,
[II]
A
to
we call
at
a
(1.4)
a e
n
In R
is nonempty.
is an isolated point of
A
> 0}
n
T(A,a)
It is always a closed convex cone and if
{0}
X.
t
the pseudo-tangent cone
P(A,a)
P(A,a)
a
n
is the standard tangent cone.
it is a closed cone but need not be convex.
the closed convex hull of
a
A.
This is not true
P(A,a)= X
These cones have proven useful in optimization for the formulation of
necessary and sufficient conditions for constrained optima to exist ([1], [2]
[ii],[19])
convex at
When looking at sufficiency one says that a set
a
A
is
pseudo-
if
A- a
Borwein and O’Brien
5] called
pseudoconvex at all its members.
P(A,a)
(1.5)
a closed set pseudoconvex if it is
They showed among other things that a closed
462
J.M. BORWEIN
bounded subset of a super reflexive Banach space [7] is pseudoconvex if and
only if it is convex.
P(A,a) + a and
PA(a)
If we let
n{PA(a): aEA}
P(A)
(1.6)
A c P(A)
pseudoconvexity is the requirement that
Finally we will examlne
one other entity
F{P(A,a)
R(A)
aEA}
(1.7)
which again is clearly a nonempty convex cone.
CONVEXITY AND STARSHAPE
2.
We begln with some elementary relationships that will be essential to our
development.
PROPOSITION I.
In any locally convex space
starA(a
(i)
PA(a)
star(A)
(ii)
and
P(A).
c
Since (ii) follows from (i) and the relevant definitions it suffices to show (i).
Let
x
lira
n(an
starA(a)_
-a)
1
a =--x
n
n
Then
x-a
E
T(A,a)
In particular, if
a
E
+ n-I
n
Thus x
star A
A
a
A and since an
/
a
PA(a)
is pseudoconvex at
implication fails trivally as is shown by
A
{I:
n
a
(The reverse
nEN} u {0} and a
It follows that a closed convex set is pseudoconvex since
star A
0).
A in this
case.
Let
(a)
X
span A
and
x/I Ixl 12
l(x)
(b)
(Recall that
We will say a norm closed set
A
is nice if
is a closed subspace of a weakly compactly generated Banach space
I(A-x)
X
is relatively
weakly compact for all x
is weakly compactly generated if
X
span K
not in
A
for some weakly
TANGENT CONES, STARSHAPE AND CONVEXITY
K [7]).
compact set
463
The nice sets turn out to be exactly the boundedly rel-
atively weakly compact sets as the next Lemma shows.
LEMMA I.
A
is boundedly relatively weakly compact exactly when
is relatively weakly compact for each
PROOF.
A
A
Since
is, we may assume
Then x
n
a
n
/llanl 12.
A.
x
is boundedly relatively weakly compact exactly when
x
x
{l(an)}
{Xn
Let
0
Now either the sequence
sequence and equivalently
{x
be a sequence in I(A)
{a
has an unbounded sub-
n
has a subsequence convergent to zero, or
n
{a
being bounded has a weakly convergent subsequence (call it
limit
(because
{llanll 2}
Since we may assume
x
A)
0
is similar.
l(A-x)
we see that
{x }
n
again) with
n
converges to some positive
converges weakly to
n
{a
Since, by the Eberlein-Smulian Theroem [7]
a
-I x.
The converse
weak compactness and
weak sequential compactness agree, we are done.
Moreover, any boundedly relatively weakly compact set
compactly generated.
the unit ball in
X.
B
Let
and set
B
K
z n
n
n
U{K
K
Each
B
n
=A n nB
To see this let
n
n
n
for each
A has span A weakly
n
N
where
K
is also.
Thus the weak closure of
a weakly compactly generated space and since
desired property.
is
N}.
is relatively weakly compact and another application of the
Smulian Theorem shows that
B
A
It follows that any closed set
span K
A
span A
K
Eberlein-
generates
has the
in a reflexive Banach
space is nice as is any weakly compact set in a normed space (we may embed it
in its completion without changing its closure) and any boundedly relatively
weakly compact set in a Banach space.
is provided by
A
{kb: beB, k
>
0}
A non-trivial example of the later class
where
B
is any closed relatively weakly
J. M. BORWEIN
464
An examination of the proof of the
compact set not containing the origin.
central result in [16] shows that it holds in the following form.
[16].
PROPOSITION 2.
X
[x,a]
a e A
and, for some
Suppose
A
a e U
interior, a smooth point
U
is a smooth point of
u
THEOREM I.
f(u)
with
u
at
A
If
a
A
such that
U
>
U}
u
x’(x)
if there is a unique support functional
is a nice closed subset of a Banach space
n{P(A,a) + a: aeA}
(2.1)
P(A)
By Proposition 1 (ii) it suffices to show that given any
PROOF.
star A
with
i
star A
x
X’
x’
U
{a}
x’ (a) --sup{x’(u):
Recall that
for
Then there exists a convex set
A n U
(ii)
closed set in a Banach space
is a nice
and a functional
(i)
f
A
PA(a)
x
for some
[x,a]
such that
A.
Let
a
h e T(A,a)
Suppose
Proposition 2.
A
a
star A
x
Since
we can find
be the smooth point guaranteed by
h
Then if
is nonzero we can find a
sequence
a
/
n
a
a
n
e
A
a
#
n
a
A(h)
It suffices to show that
t
n
> 0
{+
with
h
n
%h: % > 0}
tn(an-a)
and
U
0
are disjoint as then
they can be separated by a linear functional which supports
a
is a smooth point this functional can be taken to be
that
x’(a + h)
As
PA(a)
a
+
co
T(A,a)
=>
x’(a)
>
x’(x)
it is clear that
h e T(A,x)
x k
PA(a).
h
/
x’
U
at
a
Since
whence it follows
TANGENT CONVES, STARSHAPE AND CONVEXITY
0
A(h) o U
It remains to show that
and some % > 0
+
n
n
This contradicts
A n U
0
__l)u
t
t
n
a
t >
and
n
n
(1
n
1
%
Then assuming, as we may, that
n
%h
If not
465
U
0
1
for
n > nA
we have
0
U
n
|
since
So
U
A n U
is convex and
0 <
< 1
t
{a}
n
and one is done.
0
Before proceeding to derive consequences of this result let us introduce
We will say that
the notion of a proper point.
P(A,a) # X
An application of the Hahn-Banach theorem shows that if
starA(a
proper it is a cone point (but not conversely) since
A
note that by Theorem i each nice closed subset
A
is a proper point of
a
X
a
if
is
We
P(A,a)
has a proper point.
In
fact, a simpleadapttion of the proof of Proposition 2 in [16] will show that in
this case the proper points are dense in the boundary of
THEOREM 2.
Suppose
a weakly compact subset of
(i)
W
X
Then
PROOF.
A
such that either
F(m)
{PA(ai):
is convex and has dimension
members intersects in
QA(a)
I
=<
i _< m]
intersects in
and every finite family with
n
star A n W
PA(a)
n W
#
Then in (i)
{QA(a):
compact with the finite intersection property and hence
#
W, or
W
is starshaped and
We let
W is
is a nice closed set in a Banach space and
every finite family
(ii)
n+l
A
A
n{QA(a):
a
A}
P(A)
W
star A
W
a
A}
are weakly
466
J. M. BORWEIN
Case (ii) follows similarly from Helly’s Theorem [18]
on applying Theorem i
since the
QA(a)
are now convex subsets of
In particular if
A
R
n.
So if every finite family of proper points can be seen in
Similar remarks apply to (ii).
A
co A
W
is relatively weakly compact we may let
(i) holds.
Thus as special cases, Theorem 2 includes
Krasnoselski’s theorem and the result in [!6] with cone points replaced by the
smaller set of proper points.
In this framework the naturalness of using cone
(proper) points is obvious since the improper points play no role in defining
P(A)
Finally, we observe that if
or in any intersection property.
X
is
separable a closed set is starshaped if every countable collection of
{PA(a):
a e A}
has intersection.
This of course is guaranteed by countable
visibility.
At least in incomplete normed space or Frchet space, Theorem 1 can fail
since it is then possible for there to be no proper points in which case
It is easy to show [5] that if
only if
a
A
is a support point for
is pseudoconvex at
co A
a
a
P(A)
X
is proper if and
Since in the above mentioned cases
closed, bounded convex sets with no support points exist [5], [13], the Theorem
fails for these sets.
We turn now to psuedoconvexity of
and since
P(A)
COROLLARY I.
star A
A
Since this is equivalent to
is guaranteed by Theorem
we have
A
star
A
A
P(A)
and so:
A nice closed set in a Banach space is convex if and only
if it is pseudoconvex.
Thus to try and find a pseudoconvex non-convex subset in Banach space one
must look for badly non weakly compact subsets of non reflexive spaces.
In a
normed space however, any non-convex set with no proper points is pseudoconvex.
If we take the union of two separated copies of a convex set with no proper
TANGENT CONES
STARSHAPE AND CONVEXITY
467
S is disconnected and pseudoconvex and both Corollary 1
points the ensuing set
and Theorem 2 (i) fail for
S
The only other condition which one seems to be able to impose on a pseudoconvex set to make it convex is the following:
A.
A pseudoconvex set
PROPOSITION 3.
a
ta
t
+ (l-(-t)a 2
I
=I[a
a
4
X
e S
a
4
u
t
+
a3]:
is not convex and that
A
for some
U
in
u e
U}.
relatively open set
S
A
Suppose
S
+ U
3
A
A with support functional
exist and
A
x’ (a t
<
and
a
a
e
3
A
A
lie in
2
ri
and
and choose a
Since
A- a
x’
Thus
tx’(a I) + (l-t)x’(a 2)
However, the line segment through
follows that
and (ii)
Consider the set of segments
A
is a proper point in the boundary of
is a support of
#
is a relatively open set and thus some point
x’(a t
3.
a
I
Let
[0 I]
t e
A with
a
ri
A
its proper points are dense in the relative boundary of
PROOF.
A
(i)
is convex if
x’ (a 4)
a
t
and
a
4
=>
p(A,a4)
4
#
x’ (a 4)
extends into
ri
A
This contradiction means that
a
whence it
t
fails to
is convex.
UNBOUNDEDNESS OF STAR A
We now turn to conditions which further specify the form that
R(A)
In any locally convex space
R(eA + x)
P(A) + R(A)
PROOF.
If
x
A
takes.
R(A)
We first examine the set
PROPOSITION 4.
star
R(A)
x
P(A,a)
X
x e X
(3.1)
> 0
(3.2)
P(A).
P(A, om)
P(A+x,a+x)
Thus
J. M. BORWEIN
468
x e
R(A) + P(A)
eA+x}
Thus
So
is a convex cone.
+
p
e
+ %h
C
cone of
C
If
P(A,a)
h
is said to recede in direction
The set of all directions
If
h
if
is called the recession
is a nice closed starshaped set in a Banach space then
A
is the recession cone of a star
PROOF.
since
[17].
rec C
THEOREM 3.
R(A)
-> O
for
P(A,a) +
and (3.2) holds.
P(A)
C
p
then
P(A,a) + a
c
P(A)
R(A)
0
Since
R(A)
r
P(A,a) + P(A,a) + a
Recall that a convex set
c
p e P(A)
A
a
r
+ r
p
and (3.1) holds.
R(A+x)
Conversely if
P(A,a)
r
a
a’
n{P(A+x,a’):
star A
s
s
+
and
th e star A
A
R(A)
h
t > 0
we have for any
P(A) + R(A)
+ R(A)
star A
P(A)
where the containments follow from Proposition I, (3.2) and (2.1) in that order.
Thus
star A recedes in direction
recession direction,
star A
s
s
+
nh
h
R(A)
and
and
rec C
h
Conversely, if
is a
n e N
P(A,a) + a
Va A
and
h
P(A,a)
because
lim--- (s-a)
n
is a closed cone.
P(A,a)
Thus
a
A
h e R(A)
It is well known that a finite dimensional closed convex set is bounded if
and only if it has no recession directions.
Theorem 3 says that in finite
dimensions a closed starshaped set has an unbounded star exactly when
R(A)
non trivial.
This can be guaranteed by requiring that for every finite set
{al
A
I
0 < t <
,a
m
one can find a point
This "finite
ray"
h
# 0
with
th
+ a.I
A
for
<
condition imposes the finite intersection
is
i < m,
TANGENT CONES, STARSHAPE AND CONVEXITY
property on
{P(A,a):
a e A} n S
where
lxll
{x:
$
R(A) # 0
dimensions this last set is compact,
P(A) #
Implicit in Theorem 3 is the proof that, if
P(A)
cone for
If
star A
it is possible that
#
469
i}.
,
Since in finite
R(A)
is the recession
R(A) is nonzero as is
shown by
{(x,y): x
A
R(A)
which has
R(A)
consider
(x,y): x
R, y
=>
0
y
o
i} u {(x,y): x
or
0}
=>
0
o
=< y =<
I}
It seems reasonable therefore to
as a general cone of recession directions for any closed set.
This
is at least in part born out by the next proposition.
We recall that a functional
sup{x’ (x):
x’(c)
C}
c
x’
strongly exposes a convex set
x’(c
and if whenever
x’ (c)
n
and
c
n
C
at c
C
c
if
n
We recall that a Banach space has the Radon-Nikodym property [7] if and only if
every closed bounded convex set is the convex hull of its strongly exposed points.
These spaces include dual spaces which are subspaces of weakly compactly generated
spaces [7].
PROPOSITION 5.
R(A)
0
(i)
A
is closed relatively weakly compact in a Banach space.
(ii)
A
is closed and bounded in a Radon-Nikodym space.
(iii)
A
is closed bounded and convex in a Banach space.
(iv)
A
is compact in a locally convex space.
PROOF.
C
The following all imply
Let
a
A
and let
is weakly compact and thus
functional
space [6]).
x’
Let
x
C
co(A-a)
and suppose
can be separated from
In case(i)
C
x
C by a strongly exposing
(the strongly exposing functionals are a dense subset in the dual
x’
expose
c
strongly.
Then
c
A- a
as
A
is norm
470
J. M. BORWEIN
closed
c
+
a
A
a
x’(x)
Thus
x’(x)
P(A,)
x
is a cone
> 0
while
and
>
x’(c) _> x’(a-a)
R(A)
x’(a-a-c)
x’(aR(A)
In particular
Yae A
C
a e A
< o
and because
C
It follows that
R(A)
is a bounded set and
Case (ii) follows similarily since now the strongly
0
exposing functionals of any closed bounded convex set are dense in the dual [7].
Case (iii) follows from the Bishop-Phelps Theorem since now
density of the support functionals suffices.
C
A- a
In case (iv) any separating
functional is a support functional and supports
C
at an extreme point of
by the Krein-Milman Theorem [12].
Milman’s converse [12, page 132]
that the extreme point belongs to
A
a
and
C
implies
and one proceeds as above.
It is not clear that parts of (i) and (ii) can be deduced from Proposition 2
and it remains unanswered as to whether a closed bounded set in a Banach space
can have
R(A) # 0.
We next collect a number of characterizations of
PROPOSITION 6.
0 e star A
and
A
(ii)
0 e star A
and
star A
PROOF
span A
R(A). <--
<=>
R(A)
A
<=>
R(A)
A
X
is a convex cone.
<=> star
A
is a convex cone.
is a subspace.
By Theorem 3 and the hypotheses,
(i) =>
A
A
R(A)
under one roof:
is a nice closed set in a Banach space
(i)
(iii)
Thus
A
If
star A
R(A)
star A
Since
A
+ R(A)
star AC A
is a convex cone
0 e star A
A.
Also
A
R(A)
TANGENT CONES, STARSHAPE AND CONVEXITY
since
star
(iii)
R(A)
A + A
=>
(ii) is analogous.
X
span A
X
We may assume
P(A)
X
star A
star A
A
X
R(A)
aff A
star A
0
is nonempty but
translation as can the case that
imply
and then, by Theorem i, since
which implies star A
The case in which
471
In general
can be handled by
A
R(A)
does not
is a cone as is shown by
IYl
{(x’,y):
A
< x
x
=>
i}
in
R
2
4. SPECIAL SETS
There are cases in which the individual structure of the pseudotangent cones
can be specified more closely.
is somehow connected to a
spaces
X
Specifically one has some extra information if
Frchet
g
between Banach
Y
and
PROPOSITION 6.
(+/-)
(ii)
g’(x)P(A,x)
B
If
continuously differentiable at
g’(x)-ip(B,g(x))
PROOF
differentiable function
A
(i)
g(X))o
set, A
is a closed convex
x
with
P(A,x)
and
This is proven in
(ii)
P(g(A)
g (x)
g
-i
(B)
and
g
is
surjective, then
P(g(A),g(x))
P(B,g(x))
[Ii].
Under the resularity conditions imposed on
g
it is essentially
proven in [I0] that
g(x + (y))
has a solution
to
y
as
> 0
(y)
g(x)
g’(x)y
for all sufficiently small
Suppose that
goes to zero.
that there is a sequence
h
m
h
with
g’(x)h
y
g’(x)h
m
%
(ey) converges
and that
m
P(B,g(x)).
(bm-g(x))
One can show
b
m
e B
%
m
> 0.
472
J. M. BORWEIN
It suffices to show that
7k +
g’(x)
For
I
g(x)
(x + qb(
x)
P(A,x)
e
m
k
m
Then
m
n-__In g(x)
e B
k
h
and so
h /%
g(x+())
exists and
k e P(A,x)
k
Let
(g,(x)k + g(x)) +
(nk-)
sufficiently large
n
n
h
B
Since
P(A x)
The conclusions of
I(B),x)
is a subspace whenever
(ii) now follows from (i).
B
It follows that if
x
differentiable at
star A
x e g
If every
is a regular point.
R
is a smooth surface in
(B)
A
(whenever
is an affine subspace
n
-i
(g
is regular
is continuously
is surjective) it follows by using (2.1) that
g’(x)
and
x
P(g
is a subspace,
In particular, if
is nice)
it has such a star.
g
-i
(0)
Our next proposition substantiates
the intuition that a nontrivial smooth surface cannot simultaneously be bounded
Indeed much more is true.
and starshaped.
PROPOSITION 7.
A
space or
is closed and bounded in a Radon-Nikodym
star A
is relatively weakly compact in a Banach space and
Either there is some point
nonempty.
A
A
Suppose either
a e A
P(A,a)
with
is
not a subspace, or
is singleton.
As in the proof of Proposition 5 (i), (ii)
PROOF.
exposed point
X
a
A
0 e
star
x’ P(A,a 0)
if,
find
a
[0, =)
0e
star A
a
# a0
2
with
>__
0
x’(al-a0)
A- a
> 0
a2
a0
and, since
P(A,a 0)
P(A,a 0)
0
e
h e P(A,a
for all
which implies
then
such that
Vae A, a#a
0
x’(h)
A we see that
x’
and a strongly exposing functional
(a 0) < x (a)
It follows that
there is a strongly
p(A,a0)
0)
al
If
a
# a0,
a
On the other hand
is not a subspace.
and again unless
Again since
P(A,a 0)
lies in
0
{a
x’ (a 2
a e
0}
A we can
a
0)
> 0
TANGENT CONVES, STARSHAPE AND CONVEXITY
P(A, a0)
If
is not a subspace.
is unbounded the proposition is certainly not true as is shown by
A
(x,y)I
A
473
0
x
or
0
y
star A
cones are subspaces and
R
in
2
In any case, if all the pseudotangent
is bounded and non empty it is a singleton.
and Proposition 6 (ii), we have in
As immediate corollaries to Theorem
reflexive spaces
COROLLARY 2.
(i)
If in addition
A
regular on
0,)
Since
g(x)
and
and
g
is always regular
(4.3) when A
5.
(-,-I]
we get
X
B
g(A)
g
and
is
is closed
x2-1
(4.3)
A}
Ixl
{x:
A
Indeed, when
=>
I}
[i,=)
g(X)
From (4.3) we may deduce that if
g(X)
and
g
is
on the right hand
we get
x
and when
is closed
since this is the right hand side of
Y
Y
FUNCTIONAL CHARACTERIZATIONS
We first recall certain classes of functions.
f(tx+(l-t)y)
convex if
Similarily,
x,y
(B)
(4.2)
we know that the entities
star A
side of (4.2) fail to intersect.
x =-I
-i
A}
x
g(A) {g(x) + g’(x)P(A,x): x
star
b
g’(x)-ip(B,g(x)):
{x +
star A
For example, if
g
A
regular at every point of
(ii)
A
is a closed convex set
B
If
in
X
f: X
and
Y
=< tf(x)
S
if
is affine
0 < t <
is said to be S-quasiconvex at
(x-x)
+ (l-t)f(y)
We will also say
x
x,y
f(tx + (l-t)y)
in
X
tf(x)
if
f(x)
f: X + R
f(x)e- S
is star-like at
0
and
/
Y,
implies that
x
is
=< t <
(l-t)f(y)
is a convex cone in a space
S
If
for
R u{}
A function f" X
f" X
for
/
Y
f’(x).
if, for every x
X,
J.M. BORWEIN
474
(x-x,f(x)-f(x))
Here
all
f
we say
x
>= f (x)}
{(x,y): y
Epif
THEOREM 4.
P(Epif,(x,f(x))
is the epigraph of
If (5.1) holds for
f
is star-like.
X
If
is continuous and starlike, it
f
is reflexive and
is convex.
PROOF.
Note first that (5.1) is a condition close
Once we show that in fact
Epif.
Theorem 1 that
end we observe that since
is on the form
is pseudoconvex it will follow from
Epif
is convex and hence
Epif
Thus
is a convex function.
To this
Epif
is trivially pseudoconvex at any point
Eplf
P(Eplf, (x,f(x))
(0,s)
s > 0
We note now that if
not of this form.
f
is continuous any point on the boundary of
f
(x,f(x))
o pseudoconvexity of
since
(o s)
li. (n (x-x)
n (f (x)
+ n1
f())
s
x
T(Epif
f(x)
and so
(x-x, f(x) + s
P(Epif, (x,f(x))
because (5.1) holds and
dition says
x e X
Epif
is pseudoconvex at
Epif
x
X
is a convex cone.
(x,f(x))
s _> 0
This last con-
and since this holds for each
is convex.
In the next corollary
COROLLARY 3.
then
limsup
f (x+th)-f (x)
is the upper one
t->0+
Suppose
f(x+ h)
x e X
d+f(x,h)
f
sided derivative of
for each
P(Epif, (x,f(x)))
fCx))
f(x)
f
f
is continuous and satisfies
=> d+f(x,h)
is convex.
Yhe X
(5.2)
TANGENT CONES, STARSHAPE AND CONVEXITY
PROOF.
s
/
n
0
We show (5.2) implies (5.1).
Let
h e X
4 75
and let
s
n
> 0
be chosen so that
f
d+f(x,h)
Then, setting xn
+
x
s h
n
lira
n-=
t
(X+Snh)- f (x)
s
1
s
n
n
we have
n
(tn (Xn-X) tn(f (Xn)-f (x))
and, since
f(Xn.)
(h,d+f(x,h))
f(x)
(0,f(x+h)-f(x)-d+f(x,h))
(h,f(x+h)-f(x))
/
As before,
P(Epif, (x,f(x))
P(Epif, (x,f(x))
P(Epif, (x,f(x))
d+f (x,h))
(h,
and we deduce that
which since
h
The inequality (5.2), of course, easily implies convexity if
If
d+f(x,h)
h
exists and is convex in
COROLLARY 4.
(5.1).
is arbitrary is
f: X + Y
has a closed pseudoconvex graph,
f
is
affine.
PROOF.
By Theorem I, the graph of
immediately that
f
f
is a convex set.
is affine.
sets"
We finish by giving a condition for the "level
pseudoconvex at a point.
conditions
and
Y
and
{x: f(x)-f(x0)
PROOF.
of a function to be
This is again relevant for sufficiency of optimality
[2], [ii]
THEOREM 5.
X
This implies
Suppose
f
f
is a continuous mapping between two Banach spaces
x
is S-quasiconvex at
S}
is pseudoconvex at
We must show that
A
f
-I
If
0
x
x
0
is a regular point,
0
(f(x0)-S)
is pseudoconvex at
x
0
J.M. BORWEIN
476
The S-quasiconvexity says that
f
Since
x
-i
x
0
+ f’(x0) -i (-S)
is regular, Proposition 6 (ii) implies that
0
A
In case
(f(x0)-S)
y
R
S
and
x
0
+ P(f -i (S-f(x0)),x 0)
[0, =)
for regularity continuous differentiability
f’ (x0) # 0
is unnecessary and it suffices that
reduces to the usual notion of quasiconvexity at
X
Since S-quasiconvexity
x
[15] we
0
derive (assuming
is reflexive)
COROLLARY 5.
nonzero when
PROOF.
of
PA(X0)
{x: f(x)
Suppose
f(x0)
f(x)
Since
f: X
f
<, f(Xo)}
/
Then
R
is quasiconvex at
{x: f(x) <_ f(x
0)}
{x" f (x)
f (x
is continuous,
x
and
f’ (x)
is
is a convex set.
0)
contains the boundary
By hypothesis, therefore, this later set is pseudo-
convex and in turn, convex.
In particular, a quasiconvex function with a nonzero derivative globally
has all its level sets convex.
One can derive this last result directly with-
out asking for a non vanishing derivative.
ACKNOWLEDGEMENT.
Partially supported on NRC A-4493.
TANGENT CONES, STARSHAPE AND CONVEXITY
477
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