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Analysis & Optimization
Problem Set 3
June 6
1. (a) The answer is 2. We can approach this problem from any of multiple perspectives. First, recall
the inequality12
a2
b2
ab ≤
+
∀a, b ∈ R.
2
2
If x > 0, then by taking a = x1/2 and b = x−1/2 , we see that
x
1
+
,
2 2x
1≤
i.e.:
2≤x+
1
x
∀x > 0.
Since x + x1 = 2 if x = 1, this inequality will not hold for 2 replaced by any larger number. So the
answer is 2.
We can also take the usual perspective when faced with minimizing a function, which is to use
calculus: define f : (0, ∞) → (0, ∞) by x 7→ x + x1 . Then f 0 (x) = 1 − x−2 , so f 0 (x) = 0 if and
only if x = 1. Since f 00 (x) = 2x−3 > 0, f is convex and so 1 is in fact the global minimum. So the
answer is f (1) = 2.
Last (and probably also least), we can note as before that f 0 (x) = 0 if and only if x = 1.
Furthermore, if x < 12 , then
1
1
f (x) = x + > > 2,
x
x
and if x > 2 then
1
f (x) = x + > x > 2.
x
Since f (1) = 2, we can disregard the set (0, 12 ) ∪ (2, ∞) in our search for a global minimum of
f. Since [ 12 , 2] is compact and f is continuous on this set, f has a global minimum on it. Since
f ( 21 ) = 52 = f (2) and 1 is the only interior local minimum (and f (1) = 2 < 25 ), we can conclude
that the global minimum of f is at 1 and the answer is f (1) = 2.
(b) Let x, y, x0 , y 0 > 0 with xy, x0 y 0 ≥ 1. Then x ≥
xy 0 + x0 y ≥
1
y
and x0 ≥
1
y0 ,
and so
y0
y
1
y0
+ 0 =
+ 0 ≥ 2,
y
y
y
y
y
by part (a).
1 Sometimes
2 If
this is apparently referred to as Cauchy’s inequality, though in reality it doesn’t really seem deserving of a name.
you are not familiar with this identity, it can be obtained by noting that 0 ≤ (a − b)2 = a2 + b2 − 2ab ∀a, b ∈ R.
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(c) For convenience of notation define the set Ω := {(x, y) ∈ R2 : x > 0, xy ≥ 1}. Let (x, y), (x0 , y 0 ) ∈ Ω
and λ ∈ [0, 1], so that
λx + (1 − λ)x0 λy + (1 − λ)y 0 = λ2 xy + (1 − λ)2 x0 y 0 + λ(1 − λ) xy 0 + x0 y
≥ λ2 + (1 − λ)2 + 2λ(1 − λ)
= 1,
where in the inequality we used that xy, x0 y 0 ≥ 1 and part (b). Clearly we also have
λx + (1 − λ)x0 > 0,
and so λ(x, y) + (1 − λ)(x0 , y 0 ) ∈ Ω. So Ω is convex.
(d) The set is not convex; for example, we have (1, 1) and (−1, −1) in the set, but (0, 0), a point on
the line segment connecting (1, 1) and (−1, −1), is not in the set.
(e) The set is not convex; for example, we have (1, 0) and (−1, 0) in the set, but (0, 0), a point on the
line segment connecting (1, 1) and (−1, −1), is not in the set.
2. (a) Note that x > 1 and y > 2 imply xy ≥ 1; consequently this set is actually
{(x, y) : x > 1, y > 2} = {(x, y) : x > 1} ∩ {(x, y) : y > 2}.
Since this is the intersection of open sets, it is open (and
√ not closed). It is not bounded, since for
any n > 2, (n, n) is an element of the set with norm n 2, which is arbitrarily large.
(b) This set is
{(x, y) : x ≥ 1} ∩ {(x, y) : y ≥ 1} ∩ {(x, y) : x + y ≤ 3},
an intersection of closed sets that is consequently closed (and not open) itself. It is bounded, since
(x, y) in the set implies x2 + y 2 ≤ x2 + y 2 + 2xy = (x + y)2 ≤ 9, where we have used that x and
y are both positive.
(c) As before this set is easily expressed as the intersection of closed sets, and so itp
is itself closed (and
not open), and is also bounded since it is given that (x, y, z) in the set implies x2 + y 2 + z 2 ≤ 2.
3. Note first that f is the sum of convex functions, and is consequently convex itself.3 Consequently, any
local minimum of f (if one exists) is necessarily also a global minimum. We compute
∇f (x, y) = (ex+y + α, ex+y − 2e−2y + β).
If f has a local minimum at (x, y), then ∇f (x, y) = 0, so that α = − exp(x + y) < 0; furthermore
β = 2e−2y − ex+y > −ex+y = α.
Thus if f has a local minimum, we must have α < 0 and β > α. We might then conjecture that these
conditions are in fact sufficient for f to have a local minimum. Suppose then that α < 0 and β > α.
A quick calculation shows that
β−α
1
∇f ln |α| + 21 ln β−α
= 0.
2 , − 2 ln 2
Since f is convex, having ∇f vanish at a point is enough to guarantee a local minimum.
In summary, f has a local minimum if and only if α < 0 and β > α, and it is in fact a global minimum.
3 If
we hadn’t been savvy enough to note that a sum of convex functions is convex, we could have computed that
„ x+y
«
e
ex+y
Hess f (x, y) =
x+y
x+y
−2y
e
e
+ 4e
is positive definite for any choice of (x, y), since ex+y > 0 and det Hess f (x, y) = 4ex−y > 0. (Consequently f is convex.)
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4. Let A be a symmetric positive semidefinite matrix. By the spectral theorem, there is an orthogonal
matrix P and a diagonal matrix D such that A = P DP −1 = P DP T . Since the diagonal elements of
D are the eigenvalues of A,4 the diagonal elements of D are nonnegative.5 We can then define (letting
d1 , . . . , dn denote the diagonal entries of D) the matrix

√
d1 · · ·
0
.
..  ,
..
e := 
D

 ..
.
√.
dn
0
···
e 2 = D, as can be seen from a direct computation. We then see that a
and we furthermore note that D
e −1 , since
square root of A is given by P DP
e −1 P DP
e −1 = P D
e DP
e −1 = P DP −1 = A.
P DP
4 If you are unfamiliar with this fact, note that Av = λv if and only if P DP −1 v = λv, if and only if DP −1 v = λP −1 v, so
that v is an eigenvector of A with eigenvalue λ if and only if P −1 v is an eigenvector of D with eigenvalue λ. This shows that
the eigenvalues of A are the same as the eigenvalues of D; the conclusion follows from noting that the eigenvectors of a diagonal
matrix are simply its diagonal elements.
5 Recall that a matrix is positive semidefinite if and only if all of its eigenvalues are nonnegative.
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