EE5138R: Solutions to Problem Set 3
Assigned: 30/01/15
Due: 06/02/15
1. BV Problem 3.1
Solution:
(a) This is just the definition of convexity with λ = (b − x)/(b − a).
(b) We obtain the first inequality by subtracting f (a) from both sides of the inequality in part (a).
The second inequality follows from subtracting f (b). Geometrically, the inequalities mean that
the slope of the line segment between (a, f (a)) and (b, f (b)) is larger than the slope of the segment
between (a, f (a)) and (x, f (x)), and smaller than the slope of the segment between (x, f (x)) and
(b, f (b)).
(c) This follows from part (b) by taking the limit for x → a on both sides of the first inequality, and
by taking the limit for x → b on both sides of the second inequality.
(d) From part (c),
f 0 (b) − f 0 (a)
≥0
b−a
and taking the limit as b → a shows that f 00 (a) ≥ 0.
2. BV Problem 3.5
Hint: Use the differentiability of f and the first-order condition for convexity.
Solution: The function F is differentiable with
Z x
1
f (x)
F 0 (x) = − 2
f (t) dt +
x 0
x
Z x
2
2f (x) f 0 (x)
F 00 (x) = 3
f (t) dt −
+
x 0
x2
x
Z x
2
(f (t) − f (x) − f 0 (x)(t − x)) dt
= 3
x 0
Convexity now follows from the fact that
f (t) ≥ f (x) + f 0 (x)(t − x)
for all x, t ∈ dom f , which implies that F 00 (x) ≥ 0.
3. (Optional) BV Problem 3.13
Solution: The negative entropy is strictly convex and differentiable on Rn++ and so
f (u) > f (v) + ∇f (v)T (u − v)
1
for all u, v ∈ Rn++ with u 6= v. Evaluating both sides of the inequality, we obtain
X
X
X
ui log ui >
vi log vi +
(log vi + 1)(ui − vi )
i
i
=
X
i
ui log vi + 1T (u − v)
i
Re-arranging this inequality gives the desired result.
4. BV Problem 3.16
Solution:
(a) f (x) = ex − 1 with dom f = R. This function is strictly convex, and therefore quasiconvex. Also
quasiconcave but not concave.
(b) f (x1 , x2 ) = x1 x2 with dom f = R2++ . The Hessian of f is
0 1
2
∇ f (x) =
,
1 0
which is neither positive semidefinite nor negative semidefinite. Therefore, f is neither convex
nor concave. It is quasiconcave, since its superlevel sets
{(x1 , x2 ) ∈ R2++ |x1 x2 ≥ 0}
are convex. It is not quasiconvex.
(c) f (x1 , x2 ) = 1/(x1 x2 ) with dom f = R2++ . The Hessian of f is
1
2/x21
1/(x1 x2 )
2
0
∇ f (x) =
2/x22
x1 x2 1/(x1 x2 )
Therefore, f is convex and quasiconvex. It is not quasiconcave or concave.
(d) f (x1 , x2 ) = x1 /x2 with dom f = R2++ . The Hessian of f is
0
−1/x22
2
∇ f (x) =
−1/x22 2x1 /x32
which is not positive or negative semidefinite. Therefore, f is not convex or concave. It is quasiconvex and quasiconcave (i.e., quasilinear), since the sublevel and super- level sets are halfspaces.
(e) f (x1 , x2 ) = x21 /x2 with dom f = R2++ . f is convex, as mentioned on page 72 and worked out in
class. Therefore, f is convex and quasiconvex. It is not concave or quasiconcave (see the figure in
the book).
1−α
(f) f (x1 , x2 ) = xα
where 0 ≤ α ≤ 1 with dom f = R2++ . The Hessian of f is
1 x2
α(α − 1)x1α−2 x1−α
α(α − 1)xα−1
x−α
2
2
1
2
∇ f (x) =
−α−1
α(α − 1)xα−1
x−α
(1 − α)(−α)xα
1 x2
1
2
−1/x21
1/(x1 x2 )
1−α
= α(1 − α)xα
1 x2
1/(x1 x2 )
−1/x22
T
1/x1
1/x1
1−α
= −α(1 − α)xα
x
1 2
−1/x2 −1/x2
Hence,
−∇2 f (x) 0
We conclude that f is concave and quasiconcave. It is not convex or quasiconvex.
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5. BV Problem 3.17
Solution: The first derivatives of f are given by
!(1−p)/p
1−p
X p
∂f
f (x)
p−1
=
xi
xi =
∂xi
xi
i
The second derivatives are
1−p
∂2f
=
∂xi ∂xj
xi
f (x)
xi
−p f (x)
xj
1−p
=
1−p
f (x)
f (x)2
xi xj
1−p
for i 6= j and
∂2f
1−p
=
2
∂xi
f (x)
f (x)2
x2i
1−p
1−p
−
xi
f (x)
xi
1−p
.
We need to show that

!2 
X yi f (x)1−p
X y 2 f (x)2−p
1
−
p
i

−
≤0
y T ∇2 f (x)y =
1−p
f (x)
x
x2−p
i
i
i
i
This follows by applying the Cauchy-Schwarz inequality aT b ≤ kakkbk with
−p/2
1−p/2
f (x)
f (x)
ai =
,
bi = yi
xi
xi
P 2
and noting that i ai = 1.
6. (Optional) BV Problem 3.18
Solution:
(a) Define g(t) = f (Z + tV ), where Z 0 and V ∈ Sn . We have
g(t) = tr((Z + tV )−1 )
= tr(Z −1 (I + tZ −1/2 V Z −1/2 )−1 )
= tr(Z −1 Q(I + tΛ)−1 QT )
= tr(QT Z −1 Q(I + tΛ)−1 )
X
=
(QT ZQ)ii (1 + tλi )−1
i
where we used the eigenvalue decomposition Z −1/2 V Z −1/2 = QΛQT . In the last equality we
express g as a positive weighted sum of convex functions 1/(1 + tλi ), hence it is convex.
(b) Define g(t) = f (Z + tV ), where Z 0 and V ∈ Sn . We have
g(t) = (det(Z + tV ))1/n
= (detZ 1/2 det(I + tZ −1/2 V Z −1/2 )detZ 1/2 )1/n
!1/n
Y
1/n
= (detZ)
(1 + tλi )
i
where λi , i = 1, . . . , n are the eigenvalues of Z −1/2 V Z −1/2 . From the last equality, we see
Q that g
is a concave function of t on {t|Z + tV 0}, since detZ > 0 and the geometric mean ( i xi )1/n
is concave on Rn++ .
3
7. (Reverse Jensen’s inequality) Suppose f is convex, λ1 > 0, λi ≤ 0, i = 2, . . . , n, and
let x1 , . . . , xn ∈ dom f . Show that the inequality
!
n
n
X
X
f
λ i xi ≥
λi f (xi )
i=1
Pn
i=1
λi = 1 and
i=1
always holds.
Hint: Draw a picture for the n = 2 case first. For the general case, express x1 as a convex combination
of λ1 x1 + . . . + λn xn and x2 , . . . , xn , and use Jensen’s inequality.
Solution: Let
x1 = µ1
n
X
!
λ i xi
+ µ2 x2 + . . . + µn xn
i=1
for some µi ≥ 0 and
f
µ1
n
X
Pn
µi = 1. Then applying Jensen’s inequality, we have
!
!
n
X
λi xi + µ2 f (x2 ) + . . . + µn f (xn )
+ µ2 x2 + . . . + µn xn ≤ µ1 f
i=1
!
λi xi
i=1
i=1
Now set
µ1 =
1
,
λ1
µi = −
λi
, ∀ i = 2, . . . , n
λ1
so µi ≥ 0 for each i = 1, . . . , n and
n
X
µi =
i=1
1
λ2
λn
1
−
− ... −
=
(1 − λ2 − . . . − λn ) = 1
λ1
λ1
λ1
λ1
Pn
λi = 1. Plugging these choices of {µi }ni=1 into Jensen’s inequality, we
!
n
X
1
λn
λ2
f (x1 ) ≤
f
f (xn )
λi xi − f (x2 ) − . . . −
λ1
λ
λ1
1
i=1
because we are given that
obtain
i=1
which upon rearrangements, yields the desired reverse Jensen’s inequality.
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