Convex analysis and optimization: Solutions

191 Pages·2005·0.903 MB·English

by  Bertsekas D.P., Nedic A., Ozdaglar A.E.

Corrections for the book CONVEX ANALYSIS AND OPTI-
MIZATION, Athena Scientific, 2003, by Dimitri P. Bertsekas
Last Changed: 5/3/04
p. 3 (+22) Change “as the union of the closures of all line segments” to
“as the closure of the union of all line segments”
p. 37 (-2) Change “Every x” to “Every x(cid:1)=0”
p. 38 (+1) Change “Every x in” to “Every x∈/ X that belongs to”
p. 38 (+19) Change “i.e.,” to “with x ,...,x ∈(cid:3)n and m≥2, i.e.,”
1 m
p. 63 (+4, +6, +7, +19) Change four times “c(cid:1)y” to “a(cid:1)y”
p. 67 (+3) Change “y ∈AC” to “y ∈AC”
p. 70 (+9) Change “[BeN02]” to “[NeB02]”
p. 110 (+3 after the figure caption) Change “... does not belong to
the interior of C” to “... does not belong to the interior of C and hence
does not belong to the interior of cl(C) [cf. Prop. 1.4.3(b)]”
(cid:1) (cid:2) (cid:1) (cid:2)
p. 148 (-8) Change “ x|r(x)≤γ ” to “ z |r(z)≤γ ”
p. 213 (-6) Change “remaining vectors v , j (cid:1)= i.” to “vectors v with
j j
v (cid:1)=v .”
j i
p. 219 (+3) Change “f :C (cid:6)→(cid:3)” to “f :(cid:3)n (cid:6)→(cid:3)”
i i
p. 265 (+10) Change “d/(cid:8)d(cid:8)” to “−d/(cid:8)d(cid:8)”
p. 268 (-3) Change “j ∈A(x∗)” to “j ∈/ A(x∗)”
p. 338 (+17) Change “Section 5.2” to “Section 5.3”
p. 384 (+6) Change “convex, possibly nonsmooth functions” to “smooth
functions, and convex (possibly nonsmooth) functions”
p. 446 (+6 and +8) Interchange “... constrained problem (7.16)” and
“... penalized problem (7.19)”
p. 458 (+13) Change “... as well real-valued” to “... as well as real-
valued”
p. 458 (-10) Change “We will focus on this ... dual functions.” to “In
thiscase,thedualproblemcanbesolvedusinggradient-likealgorithmsfor
differentiable optimization (see e.g., Bertsekas [Ber99a]).”
1
Convex Analysis and
Optimization
Chapter 1 Solutions
Dimitri P. Bertsekas
with
Angelia Nedi´c and Asuman E. Ozdaglar
Massachusetts Institute of Technology
Athena Scientific, Belmont, Massachusetts
http://www.athenasc.com
LAST UPDATE March 24, 2004
CHAPTER 1: SOLUTION MANUAL
1.1
Assume that C is convex. Then, clearly (λ +λ )C ⊂ λ C+λ C; this is true
1 2 1 2
even if C is not convex. To show the reverse inclusion, note that a vector x in
λ C+λ C is of the form x = λ x +λ x , where x ,x ∈ C. By convexity of
1 2 1 1 2 2 1 2
C, we have
λ λ
1 x + 2 x ∈C,
λ +λ 1 λ +λ 2
1 2 1 2
and it follows that
x=λ x +λ x ∈(λ +λ )C.
1 1 2 2 1 2
Hence λ C+λ C ⊂(λ +λ )C.
1 2 1 2
ForacounterexamplewhenC isnotconvex,letC beasetin(cid:3)n consisting
oftwovectors,0andx(cid:4)=0,andletλ =λ =1. ThenevidentlyC isnotconvex,
1 2
and (λ +λ )C =2C ={0,2x} while λ C+λ C =C+C ={0,x,2x}, showing
1 2 1 2
that (λ +λ )C (cid:4)=λ C+λ C.
1 2 1 2
1.2 (Properties of Cones)
(a) Let x∈∩i∈ICi and let α be a positive scalar. Since x∈Ci for all i∈I and
eachCi isacone,thevectorαxbelongstoCi foralli∈I. Hence,αx∈∩i∈ICi,
showing that ∩i∈ICi is a cone.
(b) Let x∈C ×C and let α be a positive scalar. Then x=(x ,x ) for some
1 2 1 2
x ∈ C and x ∈ C , and since C and C are cones, it follows that αx ∈ C
1 1 2 2 1 2 1 1
and αx ∈ C . Hence, αx = (αx ,αx ) ∈ C ×C , showing that C ×C is a
2 2 1 2 1 2 1 2
cone.
(c) Let x∈C +C and let α be a positive scalar. Then, x=x +x for some
1 2 1 2
x ∈C and x ∈C , and since C and C are cones, αx ∈C and αx ∈C .
1 1 2 2 1 2 1 1 2 2
Hence, αx=αx +αx ∈C +C , showing that C +C is a cone.
1 2 1 2 1 2
(d) Let x ∈ cl(C) and let α be a positive scalar. Then, there exists a sequence
{x }⊂C such that x →x, and since C is a cone, αx ∈C for all k. Further-
k k k
more, αx →αx, implying that αx∈cl(C). Hence, cl(C) is a cone.
k
(e)FirstweprovethatA·Cisacone,whereAisalineartransformationandA·C
is the image of C under A. Let z ∈A·C and let α be a positive scalar. Then,
Ax = z for some x ∈ C, and since C is a cone, αx ∈ C. Because A(αx) = αz,
the vector αz is in A·C, showing that A·C is a cone.
Next we prove that the inverse image A−1·C of C under A is a cone. Let
x∈A−1·C and let α be a positive scalar. Then Ax∈C, and since C is a cone,
αAx ∈ C. Thus, the vector A(αx) is in C, implying that αx ∈ A−1 ·C, and
showing that A−1·C is a cone.
2
1.3 (Lower Semicontinuity under Composition)
(a) Let {x } ⊂ (cid:3)n be a seque(cid:1)nce of (cid:2)vectors converging to some x ∈ (cid:3)n. By
k
continuity of f, it follows that f(x ) ⊂ (cid:3)m converges to f(x) ∈ (cid:3)m, so that
k
by lower semicontinuity of g, we have
(cid:3) (cid:4) (cid:3) (cid:4)
liminfg f(x ) ≥g f(x) .
k
k→∞
Hence, h is lower semicontinuous.
(b) Assume, to arrive at a contradiction, that h is not lower semicontinuous at
some x ∈ (cid:3)n. Then, there exists a sequence {x } ⊂ (cid:3)n converging to x such
k
that (cid:3) (cid:4) (cid:3) (cid:4)
liminfg f(x ) <g f(x) .
k
k→∞
Let {xk}K be a subsequence attaining the above limit inferior, i.e.,
(cid:3) (cid:4) (cid:3) (cid:4) (cid:3) (cid:4)
lim g f(x ) =liminfg f(x ) <g f(x) . (1.1)
k k
k→∞, k∈K k→∞
Without loss of generality, we may assume that
(cid:3) (cid:4) (cid:3) (cid:4)
g f(x ) <g f(x) , ∀ k∈K.
k
Since g is monotonically nondecreasing, it follows that
f(x )<f(x), ∀ k∈K,
k
whichtogetherwiththefact{xk}K →xandthelowersemicontinuityoff yields
f(x)≤ liminf f(x )≤ limsup f(x )≤f(x),
k k
k→∞, k∈K k→∞, k∈K
(cid:1) (cid:2)
showing that f(xk) K → f(x). By our choice of the sequence {xk}K and by
lower semicontinuity of g, it follows that
(cid:3) (cid:4) (cid:3) (cid:4) (cid:3) (cid:4)
lim g f(x ) = liminf g f(x ) ≥g f(x) ,
k k
k→∞, k∈K k→∞, k∈K
contradicting Eq. (1.1). Hence, h is lower semicontinuous.
As an example showing that the assumption that g is monotonically non-
decreasing is essential, consider the functions
(cid:5)
0 if x≤0,
f(x)=
1 if x>0,
and g(x)=−x. Then (cid:3) (cid:4) (cid:5)
0 if x≤0,
g f(x) =
−1 if x>0,
which is not lower semicontinuous at 0.
3
1.4 (Convexity under Composition)
(a) Let x,y∈C and let α∈[0,1]. Then we have
(cid:3) (cid:4) (cid:6) (cid:3) (cid:4)(cid:7)
h αx+(1−α)y =g f αx+(1−α)y
(cid:3) (cid:4)
≤g αf(x)+(1−α)f(y)
(cid:3) (cid:4) (cid:3) (cid:4)
≤αg f(x) +(1−α)g f(y)
=αh(x)+(1−α)h(y),
where the first inequality above follows from the convexity of f and the mono-
tonicity of g, while the second inequality follows from the convexity of g. If g
is monotonically increasing and f is strictly convex, then the first inequality in
the preceding relation is strict whenever x(cid:4)=y and α∈(0,1), showing that h is
strictly convex.
(b)Letx,y∈(cid:3)n andletα∈[0,1]. Then,bythedefinitionsofhandf,wehave
(cid:3) (cid:4) (cid:6) (cid:3) (cid:4)(cid:7)
h αx+(1−α)y =g f αx+(1−α)y
(cid:6) (cid:3) (cid:4) (cid:3) (cid:4)(cid:7)
=g f αx+(1−α)y ,...,f αx+(1−α)y
1 m
(cid:3) (cid:4)
≤g αf (x)+(1−α)f (y),...,αf (x)+(1−α)f (y)
(cid:6) (cid:3)1 1(cid:4) m(cid:3) m(cid:4)(cid:7)
=g α f (x),...,f (x) +(1−α) f (y),...,f (y)
1 m 1 m
(cid:3) (cid:4) (cid:3) (cid:4)
≤αg f (x),...,f (x) +(1−α)g f (y),...,f (y)
(cid:3) 1 (cid:4) m (cid:3) (cid:4) 1 m
=αg f(x) +(1−α)g f(y)
=αh(x)+(1−α)h(y),
where the first inequality follows by convexity of each f and monotonicity of g,
i
while the second inequality follows by convexity of g.
1.5 (Examples of Convex Functions)
(a) It can be seen that f is twice continuously differentiable over X and its
1
Hessian matrix is given by
⎡ ⎤
1−n 1 ··· 1
⎢ x21 x1x2 x1xn ⎥
⎢ 1 1−n ··· 1 ⎥
∇2f1(x)= f1n(2x)⎢⎢⎣ x2x1 x22 .. x2xn ⎥⎥⎦
.
1 1 ··· 1−n
xnx1 x1x2 x2n
for all x = (x ,...,x ) ∈ X. From this, direct computation shows that for all
1 n
z=(z ,...,z )∈(cid:3)n and x=(x ,...,x )∈X, we have
1 n 1 n
(cid:14)(cid:14) (cid:16) (cid:16)
(cid:15)n 2 (cid:15)n (cid:6) (cid:7)
z(cid:4)∇2f (x)z= f1(x) zi −n zi 2 .
1 n2 x x
i i
i=1 i=1
4
Note that this quadratic form is nonnegative for all z ∈ (cid:3)n and x ∈ X, since
f (x)<0, and for any real numbers α ,...,α , we have
1 1 n
(α +···+α )2 ≤n(α2+···+α2),
1 n 1 n
inviewofthefactthat2α α ≤α2+α2. Hence,∇2f (x)ispositivesemidefinite
j k j k 1
for all x∈X, and it follows from Prop. 1.2.6(a) that f is convex.
1
(b) We show that the Hessian of f is positive semidefinite at all x ∈ (cid:3)n. Let
2
β(x)=ex1 +···+exn. Then a straightforward calculation yields
(cid:15)n (cid:15)n
z(cid:4)∇2f (x)z= 1 e(xi+xj)(z −z )2 ≥0, ∀ z∈(cid:3)n.
2 β(x)2 i j
i=1 j=1
Hence by Prop. 1.2.6, f is convex.
2 (cid:3) (cid:4)
(b) The function f (x) = (cid:11)x(cid:11)p can be viewed as a composition g f(x) of the
3
scalarfunctiong(t)=tpwithp≥1andthefunctionf(x)=(cid:11)x(cid:11). Inthiscase,gis
convexandmonotonicallyincreasingoverthenonnegativeaxis,thesetofvalues
that f can take, while f is convex over (cid:3)n (since any vector norm is convex,
see the discussion preceding Prop. 1.2.4). Using Exercise 1.4, it follows that the
function f (x)=(cid:11)x(cid:11)p is convex over (cid:3)n.
3 (cid:3) (cid:4)
(c) The function f (x) = 1 can be viewed as a composition g h(x) of the
4 f(x)
function g(t)=−1 for t<0 and the function h(x)=−f(x) for x∈(cid:3)n. In this
t
case, the g is convex and monotonically increasing in the set {t|t<0}, while h
is convex over (cid:3)n. Using Exercise 1.4, it follows that the function f (x)= 1
4 f(x)
is convex over (cid:3)n.
(cid:3) (cid:4)
(d) The function f (x)=αf(x)+β can be viewed as a composition g f(x) of
5
thefunctiong(t)=αt+β,wheret∈(cid:3),andthefunctionf(x)forx∈(cid:3)n. Inthis
case, g is convex and monotonically increasing over (cid:3) (since α ≥ 0), while f is
convexover(cid:3)n. UsingExercise1.4,itfollowsthatthefunctionf (x)=αf(x)+β
5
is convex over (cid:3)n.
(cid:3) (cid:4)
(cid:4)
(e) The function f (x)=eβxAx can be viewed as a composition g f(x) of the
6
function g(t) = eβt for t ∈ (cid:3) and the function f(x) = x(cid:4)Ax for x ∈ (cid:3)n. In this
case,g isconvexandmonotonicallyincreasingover(cid:3),whilef isconvexover(cid:3)n
(sinceAispositivesemidefinite). UsingExercise1.4,itfollowsthatthefunction
(cid:4)
f (x)=eβxAx is convex over (cid:3)n.
6
(f) This part is straightforward using the definition of a convex function.
1.6 (Ascent/Descent Behavior of a Convex Function)
(a)Letx ,x ,x bethreescalarssuchthatx <x <x . Thenwecanwritex
1 2 3 1 2 3 2
as a convex combination of x and x as follows
1 3
x −x x −x
x = 3 2x + 2 1x ,
2 x −x 1 x −x 3
3 1 3 1
5
so that by convexity of f, we obtain
x −x x −x
f(x )≤ 3 2f(x )+ 2 1f(x ).
2 x −x 1 x −x 3
3 1 3 1
This relation and the fact
x −x x −x
f(x )= 3 2f(x )+ 2 1f(x ),
2 x −x 2 x −x 2
3 1 3 1
imply that
(cid:3) (cid:4) (cid:3) (cid:4)
x −x x −x
3 2 f(x )−f(x ) ≤ 2 1 f(x )−f(x ) .
x −x 2 1 x −x 3 2
3 1 3 1
Bymultiplyingtheprecedingrelationwithx −x andbydividingitwith(x −
3 1 3
x )(x −x ), we obtain
2 2 1
f(x )−f(x ) f(x )−f(x )
2 1 ≤ 3 2 .
x −x x −x
2 1 3 2
(b) Let {x } be an increasing scalar sequence, i.e., x < x < x < ···. Then
k 1 2 3
according to part (a), we have for all k
f(x )−f(x ) f(x )−f(x ) f(x )−f(x )
2 1 ≤ 3 2 ≤···≤ k+1 k . (1.2)
x −x x −x x −x
2 1 3 2 k+1 k
(cid:3) (cid:4)
Since f(xk)−f(xk−1) /(xk−xk−1) is monotonically nondecreasing, we have
f(xk)−f(xk−1) →γ, (1.3)
xk−xk−1
where γ is either a real number or ∞. Furthermore,
f(x )−f(x )
k+1 k ≤γ, ∀ k. (1.4)
x −x
k+1 k
We now show that γ is independent of the sequence {x }. Let {y } be
k j
any increasing scalar sequence. For each j, choose x such that y < x and
kj j kj
x <x <···<x ,sothatwehavey <y <x <x . Bypart(a),
k1 k2 kj j j+1 kj+1 kj+2
it follows that
f(yj+1)−f(yj) ≤ f(xkj+2)−f(xkj+1),
y −y x −x
j+1 j kj+2 kj+1
and letting j →∞ yields
f(y )−f(y )
lim j+1 j ≤γ.
j→∞ yj+1−yj
Similarly, by exchanging the roles of {x } and {y }, we can show that
k j
f(y )−f(y )
lim j+1 j ≥γ.
j→∞ yj+1−yj
6
Thus the limit in Eq. (1.3) is independent of the choice for {x }, and Eqs. (1.2)
k
and (1.4) hold for any increasing scalar sequence {x }.
k
We consider separately each of the three possibilities γ < 0,γ = 0, and
γ > 0. First, suppose that γ < 0, and let {x } be any increasing sequence. By
k
using Eq. (1.4), we obtain
(cid:15)k−1 f(x )−f(x )
f(x )= j+1 j (x −x )+f(x )
k x −x j+1 j 1
j+1 j
j=1
(cid:15)k−1
≤ γ(x −x )+f(x )
j+1 j 1
j=1
=γ(x −x )+f(x ),
k 1 1
and since γ < 0 and x → ∞, it follows that f(x ) → −∞. To show that f
k k
decreasesmonotonically,pickanyxandy withx<y,andconsiderthesequence
x =x, x =y, and x =y+k for all k≥3. By using Eq. (1.4) with k=1, we
1 2 k
have
f(y)−f(x)
≤γ <0,
y−x
sothatf(y)−f(x)<0. Hencef decreasesmonotonicallyto−∞,corresponding
to case (1).
Suppose now that γ =0, and let {x } be any increasing sequence. Then,
k
byEq.(1.4),wehavef(x )−f(x )≤0forallk. Iff(x )−f(x )<0forall
k+1 k k+1 k
k, then f decreases monotonically. To show this, pick any x and y with x < y,
and consider a new sequence given by y1 = x, y2 = y, and yk = xK+k−3 for all
k ≥ 3, where K is large enough so that y < x . By using Eqs. (1.2) and (1.4)
K
with {y }, we have
k
f(y)−f(x) f(x )−f(x )
≤ K+1 K <0,
y−x x −x
K+1 K
implying that f(y)−f(x) < 0. Hence f decreases monotonically, and it may
decreaseto−∞ortoafinitevalue,correspondingtocases(1)or(2),respectively.
If for some K we have f(x )−f(x )=0, then by Eqs. (1.2) and (1.4)
K+1 K
where γ = 0, we obtain f(x ) = f(x ) for all k ≥ K. To show that f stays at
k K
thevaluef(x )forallx≥x , chooseanyxsuchthatx>x , anddefine{y }
K K K k
as y1 =xK, y2 =x, and yk =xN+k−3 for all k≥3, where N is large enough so
that x<x . By using Eqs. (1.2) and (1.4) with {y }, we have
N k
f(x)−f(x ) f(x )−f(x)
K ≤ N ≤0,
x−x x −x
K N
so that f(x) ≤ f(x ) and f(x ) ≤ f(x). Since f(x ) = f(x ), we have
K N K N
f(x)=f(x ). Hence f(x)=f(x ) for all x≥x , corresponding to case (3).
K K K
(cid:3) Finally,supp(cid:4)osethatγ >0,andlet{x }beanyincreasingsequence. Since
k
f(xk)−f(xk−1) /(xk −xk−1) is nondecreasing and tends to γ [cf. Eqs. (1.3)
and (1.4)], there is a positive integer K and a positive scalar (cid:5) with (cid:5) < γ such
that
(cid:5)≤ f(xk)−f(xk−1), ∀ k≥K. (1.5)
xk−xk−1
7
Therefore, for all k>K
(cid:15)k−1 f(x )−f(x )
f(x )= j+1 j (x −x )+f(x )≥(cid:5)(x −x )+f(x ),
k x −x j+1 j K k K K
j+1 j
j=K
implying that f(x )→∞. To show that f(x) increases monotonically to ∞ for
k
allx≥x ,pickanyx<y satisfyingx <x<y,andconsiderasequencegiven
K K
byy1 =xK,y2 =x,y3 =y,andyk =xN+k−4 fork≥4,whereN islargeenough
so that y<x . By using Eq. (1.5) with {y }, we have
N k
f(y)−f(x)
(cid:5)≤ .
y−x
Thus f(x) increases monotonically to ∞ for all x ≥ x , corresponding to case
K
(4) with x=x .
K
1.7 (Characterization of Differentiable Convex Functions)
If f is convex, then by Prop. 1.2.5(a), we have
f(y)≥f(x)+∇f(x)(cid:4)(y−x), ∀ x,y∈C.
By exchanging the roles of x and y in this relation, we obtain
f(x)≥f(y)+∇f(y)(cid:4)(x−y), ∀ x,y∈C,
and by adding the preceding two inequalities, it follows that
(cid:3) (cid:4)
(cid:4)
∇f(y)−∇f(x) (x−y)≥0. (1.6)
Conversely, let Eq. (1.6) hold, and let x and y be two points in C. Define
the function h:(cid:3)(cid:13)→(cid:3) by
(cid:3) (cid:4)
h(t)=f x+t(y−x) .
Consider some t,t(cid:4) ∈ [0,1] such that t < t(cid:4). By convexity of C, we have that
x+t(y−x) and x+t(cid:4)(y−x) belong to C. Using the chain rule and Eq. (1.6),
we have
(cid:6) (cid:7)
dh(t(cid:4)) − dh(t) (t(cid:4)−t)
dt dt
(cid:6) (cid:3) (cid:4) (cid:3) (cid:4)(cid:7)(cid:4)
= ∇f x+t(cid:4)(y−x) −∇f x+t(y−x) (y−x)(t(cid:4)−t)
≥0.
Thus, dh/dt is nondecreasing on [0,1] and for any t∈(0,1), we have
(cid:17) (cid:17)
h(t)−h(0) 1 t dh(τ) 1 1 dh(τ) h(1)−h(t)
= dτ ≤h(t)≤ dτ = .
t t dτ 1−t dτ 1−t
0 t
Equivalently,
th(1)+(1−t)h(0)≥h(t),
and from the definition of h, we obtain
(cid:3) (cid:4)
tf(y)+(1−t)f(x)≥f ty+(1−t)x .
Since this inequality has been proved for arbitrary t ∈ [0,1] and x,y ∈ C, we
conclude that f is convex.
8
1.8 (Characterization of Twice Continuously Differentiable
Convex Functions)
Suppose that f :(cid:3)n (cid:13)→(cid:3) is convex over C. We first show that for all x∈ri(C)
and y ∈ S, we have y(cid:4)∇2f(x)y ≥ 0. Assume to arrive at a contradiction, that
there exists some x∈ri(C) such that for some y∈S, we have
y(cid:4)∇2f(x)y<0.
Withoutlossofgenerality,wemayassumethat(cid:11)y(cid:11)=1. Usingthecontinuityof
∇2f, we see that there is an open ball B(x,(cid:5)) centered at x¯ with radius (cid:5) such
that B(x,(cid:5))∩aff(C)⊂C [since x∈ri(C)], and
y(cid:4)∇2f(x)y<0, ∀ x∈B(x,(cid:5)). (1.7)
By Prop. 1.1.13(a), for all positive scalars α with α<(cid:5), we have
f(x¯+αy)=f(x¯)+α∇f(x¯)(cid:4)y+ 1y(cid:4)∇2f(x¯+α¯y)y,
2
for some α¯ ∈[0,α]. Furthermore, (cid:11)(x+αy)−x(cid:11)≤(cid:5) [since (cid:11)y(cid:11)=1 and α¯ <(cid:5)].
Hence, from Eq. (1.7), it follows that
f(x¯+αy)<f(x¯)+α∇f(x¯)(cid:4)y, ∀ α∈[0,(cid:5)).
Ontheotherhand,bythechoiceof(cid:5)andtheassumptionthaty∈S,thevectors
x¯+αy are in C for all α with α ∈ [0,(cid:5)), which is a contradiction in view of
the convexity of f over C. Hence, we have y(cid:4)∇2f(x)y ≥ 0 for all y ∈ S and all
x∈ri(C).
Next,letxbeapointinC thatisnotintherelativeinteriorofC. Then,by
the Line Segment Principle, there is a sequence {x }⊂ri(C) such that x →x.
k k
As seen above, y(cid:4)∇2f(x )y ≥0 for all y ∈S and all k, which together with the
k
continuity of ∇2f implies that
y(cid:4)∇2f(x)y= lim y(cid:4)∇2f(x )y≥0, ∀ y∈S.
k
k→∞
It follows that y(cid:4)∇2f(x)y≥0 for all x∈C and y∈S.
Conversely, assume that y(cid:4)∇2f(x)y≥0 for all x∈C and y∈S. By Prop.
1.1.13(a), for all x,z∈C we have
(cid:3) (cid:4)
f(z)=f(x)+(z−x)(cid:4)∇f(x)+ 1(z−x)(cid:4)∇2f x+α(z−x) (z−x)
2
for some α ∈ [0,1]. Since x,z ∈ C, we have that (z−x) ∈ S, and using the
convexity of C and our assumption, it follows that
f(z)≥f(x)+(z−x)(cid:4)∇f(x), ∀ x,z∈C.
From Prop. 1.2.5(a), we conclude that f is convex over C.
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