Definition 2.1. Let f : IR → IR be a real valued function.
(a) f is said to be SOC-monotone of order n if the corresponding vector-valued function fsoc satisfies the following:
x Kn y =⇒ fsoc(x) Kn fsoc(y). (2.1) We say f is SOC-monotone if f is SOC-monotone of all order n.
(b) f is said to be SOC-convex of order n if the corresponding vector-valued function fsoc satisfies the following:
fsoc (1 − λ)x + λy Kn (1 − λ)fsoc(x) + λfsoc(y), (2.2) for all x, y ∈ IRn and 0 ≤ λ ≤ 1. We say f is SOC-convex if f is SOC-convex of all order n.
Remark 2.1. The SOC-concavity is defined in a similar way. We elaborate more about the concepts of SOC-convexity and SOC-monotonicity in this remark.
39
1. A function f is SOC-convex of order 1 is the same as f being a convex function. If a function f is SOC-convex of order n, then f is SOC-convex of order m for any m ≤ n, see Figure 2.1(a).
2. A function f is SOC-monotone of order 1 is the same as f being an increasing function. If a function f is SOC-monotone of order n, then f is SOC-monotone of order m for any m ≤ n, see Figure 2.1(b).
3. If f is continuous, then the condition (2.2) can be replaced by the more special condition:
fsoc x + y 2
Kn 1
2 fsoc(x) + fsoc(y) . (2.3) 4. It is clear that the set of SOC-monotone functions and the set of SOC-convex functions are both closed under positive linear combinations and under pointwise limits.
SOC-convex SOC-convex of order n+1
SOC-convex of order n SOC-convex of order 3 SOC-convex of order 2
......
(a) SOC-convex functions
SOC-monotone SOC-monotone of order n+1
SOC-monotone of order n SOC-monotone of order 3 SOC-monotone of order 2
......
(b) SOC-monotone functions
Figure 2.1: The concepts of SOC-convex and SOC-monotone functions
Proposition 2.1. Let f : IR → IR be f (t) = α + βt. Then, (a) f is SOC-monotone on IR for every α ∈ IR and β ≥ 0;
(b) f is SOC-convex on IR for all α, β ∈ IR.
Proof. The proof is straightforward by checking that Definition 2.1 is satisfied. Proposition 2.2. (a) Let f : IR → IR be f (t) = t2, then f is SOC-convex on IR.
(b) Hence, the function g(t) = α + βt + γt2 is SOC-convex on IR for all α, β ∈ IR and γ ≥ 0.
Proof. (a) For any x, y ∈ IRn, we have 1
2 fsoc(x) + fsoc(y) − fsoc x + y 2
= x2+ y2
2 − x + y 2
2
= 1
4(x − y)2 Kn 0, which says (2.3) is satisfied. Since f is continuous, it implies that f is SOC-convex.
(b) It is an immediate consequence of part(a).
Example 2.1. The function f (t) = t2 is not SOC-monotone on IR.
Solution. Taking x = (1, 0), y = (−2, 0), then x − y = (3, 0) Kn 0. But, x2− y2 = (1, 0) − (4, 0) = (−3, 0) 6Kn 0,
which violates (2.1).
As mentioned in Section 1.2, if f is defined on a subset J ⊆ IR, fsoc is defined on its corresponding set given as in (1.11), i.e.,
S = {x ∈ IRn| λi(x) ∈ J, i = 1, 2.} ⊆ IRn.
In addition, from Proposition 2.2(a), it indicates that f (t) = t2 is also SOC-convex on the smaller interval [0, ∞). These observations raise a natural question. Is f (t) = t2 SOC-monotone on the interval [0, ∞) although it is not SOC-SOC-monotone on IR? The answer is no! Indeed, it is true only for n = 2, but, false for n ≥ 3. We illustrate this in the next example.
Example 2.2. (a) The function f (t) = t2 is SOC-monotone of order 2 on [0, ∞).
(b) However, f (t) = t2 is not SOC-monotone of order n ≥ 3 on [0, ∞).
Solution. (a) Suppose that x = (x1, x2) K2 y = (y1, y2) K2 0. Then, we have the following inequalities:
|x2| ≤ x1, |y2| ≤ y1, |x2 − y2| ≤ x1− y1, which implies
x1− x2 ≥ y1 − y2 ≥ 0,
x1+ x2 ≥ y1+ y2 ≥ 0. (2.4)
The goal is to show that fsoc(x) − fsoc(y) = (x21+ x22− y12− y22, 2x1x2− 2y1y2)
K2 0, which suffices to verify that x21+ x22− y12− y22 ≥ |2x1x2− 2y1y2|. This can be seen by
x21+ x22− y21− y22 −
2x1x2− 2y1y2
= x21+ x22− y12− y22− (2x1x2− 2y1y2), if x1x2− y1y2 ≥ 0 x21+ x22− y12− y22− (2y1y2− 2x1x2), if x1x2− y1y2 ≤ 0
= (x1− x2)2− (y1− y2)2, if x1x2− y1y2 ≥ 0 (x1+ x2)2− (y1+ y2)2, if x1x2− y1y2 ≤ 0
≥ 0 ,
where the inequalities are true due to the inequalities (2.4).
(b) From Remark 2.1, we only need to provide a counterexample for case of n = 3 to show that f (t) = t2 is not SOC-monotone on the interval [0, ∞). Take x = (3, 1, −2) ∈ K3 and y = (1, 1, 0) ∈ K3. It is clear that x − y = (2, 0, −2)
K3 0. But, x2 − y2 = (14, 6, −12) − (2, 2, 0) = (12, 4, −12) 6K3 0.
Now we look at the function f (t) = t3. As expected, f (t) = t3 is not SOC-convex.
However, it is true that f (t) = t3 is SOC-convex on [0, ∞) for n = 2, whereas false for n ≥ 3. Besides, we will see f (t) = t3 is neither SOC-monotone on IR nor SOC-monotone on the interval [0, ∞). Nonetheless, it is true that it is SOC-monotone on the interval [0, ∞) for n = 2. The following two examples demonstrate what we have just said.
Example 2.3. (a) The function f (t) = t3 is not SOC-convex on IR.
(b) However, f (t) = t3 is SOC-convex of order 2 on [0, ∞).
(c) Moreover, f (t) = t3 is not SOC-convex of order n ≥ 3 on [0, ∞).
Solution. (a) Taking x = (0, −2), y = (1, 0) gives 1
2 fsoc(x) + fsoc(y) − fsoc x + y 2
=
−9 8, −9
4
6K2 0,
which says f (t) = t3 is not SOC-convex on IR.
(b) It suffices to show that fsoc x+y2
K2
1
2 fsoc(x)+fsoc(y) for any x, y
K2 0. Suppose that x = (x1, x2)
K2 0 and y = (y1, y2)
K2 0, then we have
x3 = x31+ 3x1x22, 3x21x2+ x32, y3 = y13+ 3y1y22, 3y21y2+ y23, which yields
fsoc(x+y2 ) = 18 (x1+ y1)3+ 3(x1 + y1)(x2 + y2)2, 3(x1+ y1)2(x2+ y2) + (x2+ y2)3,
1
2 fsoc(x) + fsoc(y) = 12 x31+ y13+ 3x1x22+ 3y1y22, x32+ y32 + 3x21x2+ 3y12y2.
After simplifications, we denote 12 fsoc(x) + fsoc(y) − fsoc(x+y2 ) := 18(Ξ1, Ξ2), where
Ξ1 = 4x31+ 4y13+ 12x1x22+ 12y1y23− (x1+ y1)3− 3(x1+ y1)(x2+ y2)2, Ξ2 = 4x32+ 4y23+ 12x21x2+ 12y21y2− (x2+ y2)3− 3(x1+ y1)2(x2+ y2).
We want to show that Ξ1 ≥ |Ξ2|, for which we discuss two cases. First, if Ξ2 ≥ 0, then Ξ1− |Ξ2|
= (4x31+ 12x1x22− 12x21x2− 4x32) + (4y31+ 12y1y22− 12y12y2− 4y23)
− (x1 + y1)3+ 3(x1+ y1)(x2+ y2)2− 3(x1+ y1)2(x2+ y2) − (x2+ y2)3
= 4(x1− x2)3+ 4(y1− y2)3− (x1+ y1) − (x2+ y2)3
= 4(x1− x2)3+ 4(y1− y2)3− (x1− x2) + (y1− y2)3
= 3(x1− x2)3+ 3(y1− y2)3− 3(x1− x2)2(y1− y2) − 3(x1− x2)(y1− y2)2
= 3 (x1− x2) + (y1− y2)
(x1− x2)2− (x1− x2)(y1− y2) + (y1− y2)2
−3(x1− x2)(y1− y2) (x1− x2) + (y1− y2)
= 3 (x1− x2) + (y1− y2)
(x1− x2) − (y1− y2)2
≥ 0,
where the inequality is true since x, y ∈ K2. Similarly, if Ξ2 ≤ 0, we also have Ξ1 − |Ξ2|
= (4x31+ 12x1x22+ 12x21x2+ 4x32) + (4y13+ 12y1y22+ 12y12y2+ 4y23)
− (x1+ y1)3+ 3(x1+ y1)(x2+ y2)2 + 3(x1+ y1)2(x2+ y2) + (x2+ y2)3
= 4(x1+ x2)3+ 4(y1+ y2)3− (x1+ y1) + (x2+ y2)3
= 4(x1+ x2)3+ 4(y1+ y2)3− (x1+ x2) + (y1+ y2)3
= 3(x1+ x2)3+ 3(y1+ y2)3− 3(x1+ x2)2(y1+ y2) − 3(x1+ x2)(y1+ y2)2
= 3 (x1+ x2) + (y1+ y2)
(x1+ x2)2− (x1+ x2)(y1+ y2) + (y1+ y2)2
−3(x1+ x2)(y1+ y2) (x1+ x2) + (y1+ y2)
= 3 (x1+ x2) + (y1+ y2)
(x1+ x2) − (y1+ y2)2
≥ 0,
where the inequality is true since x, y ∈ K2. Thus, we have verified that f (t) = t3 is SOC-convex on [0, ∞) for n = 2.
(c) Again, by Remark 2.1, we only need to provide a counterexample for case of n = 3.
To see this, we take x = (2, 1, −1), y = (1, 1, 0) K3 0. Then, we have 1
2 fsoc(x) + fsoc(y) − fsoc x + y 2
= (3, 1, −3) 6
K3 0, which implies f (t) = t3 is not even SOC-convex on the interval [0, ∞). Example 2.4. (a) The function f (t) = t3 is not SOC-monotone on IR.
(b) However, f (t) = t3 is SOC-monotone of order 2 on [0, ∞).
(c) Moreover, f (t) = t3 is not SOC-monotone of order n ≥ 3 on [0, ∞).
Solution. To see (a) and (c), let x = (2, 1, −1) K3 0 and y = (1, 1, 0)
K3 0. It is clear that x
K3 y. But, we have fsoc(x) = x3 = (20, 14, −14) and fsoc(y) = y3 = (4, 4, 0), which gives fsoc(x) − fsoc(y) = (16, 10, −14) 6K3 0. Thus, we show that f (t) = t3 is not even SOC-monotone on the interval [0, ∞).
To see (b), let x = (x1, x2) K2 y = (y1, y2) K2 0, which means
|x2| ≤ x1, |y2| ≤ y1, |x2 − y2| ≤ x1− y1.
Then, it leads to the inequalities (2.4) again. On the other hand, we know fsoc(x) = x3 = x31 + 3x1x22, 3x21x2+ x32,
fsoc(y) = y3 = y13+ 3y1y22, 3y21y2+ y23.
For convenience, we denote fsoc(x) − fsoc(y) := (Ξ1, Ξ2), where
Ξ1 = x31− y13+ 3x1x22− 3y1y22, Ξ2 = x32− y23+ 3x21x2− 3y12y2.
We wish to prove that fsoc(x) − fsoc(y) = x3− y3 K2 0, which suffices to show Ξ1 ≥ |Ξ2|.
This is true because
x31− y13+ 3x1x22− 3y1y22−
x32− y32+ 3x21x2− 3y12y2
= x31− y13+ 3x1x22− 3y1y22− (x32− y23+ 3x21x2− 3y12y2), if Ξ2 ≥ 0 x31− y13+ 3x1x22− 3y1y22+ (x32− y23+ 3x21x2− 3y21y2), if Ξ2 ≤ 0
= (x1− x2)3− (y1− y2)3, if Ξ2 ≥ 0 (x1+ x2)3− (y1+ y2)3, if Ξ2 ≤ 0
≥ 0,
where the inequalities are due to the inequalities (2.4).
Hence, we complete the verification.
Now, we move to another simple function f (t) = 1/t. We will prove that −1t is SOC-monotone on the interval (0, ∞) and 1t is SOC-convex on the interval (0, ∞) as well. For the proof, we need the following technical lemmas.
Lemma 2.1. Suppose that a, b, c, d ∈ IR. For any a ≥ b > 0 and c ≥ d > 0, there holds
a b
c d
≥ a + c b + d
Proof. The proof follows from ac(b + d) − bd(a + c) = ab(c − d) + cd(a − b) ≥ 0. Lemma 2.2. For any x = (x1, x2) ∈ Kn and y = (y1, y2) ∈ Kn, we have
(a) (x1+ y1)2− ky2k2 ≥ 4x1py12− ky2k2. (b) x1+ y1− ky2k2
≥ 4x1 y1− ky2k.
(c) x1+ y1+ ky2k2
≥ 4x1 y1+ ky2k.
(d) x1y1− hx2, y2i ≥px21− kx2k2py12− ky2k2.
(e) (x1 + y1)2− kx2+ y2k2 ≥ 4px21− kx2k2py12− ky2k2. Proof. (a) The proof follows from
(x1+ y1)2− ky2k2 = x21+ (y12− ky2k2) + 2x1y1
≥ 2x1
q
y12− ky2k2+ 2x1y1
≥ 2x1
q
y12− ky2k2+ 2x1
q
y12− ky2k2
= 4x1
q
y12− ky2k2,
where the first inequality is true due to the fact that a + b ≥ 2√
ab for any positive numbers a and b.
(b) The proof follows from
(x1+ y1 − ky2k)2− 4x1(y1− ky2k)
= x21+ y21+ ky2k2− 2x1y1− 2y1ky2k + 2x1ky2k
= (x1− y1+ ky2k)2 ≥ 0.
(c) Similarly, the proof follows from
(x1+ y1 + ky2k)2− 4x1(y1+ ky2k)
= x21+ y21+ ky2k2− 2x1y1+ 2y1ky2k − 2x1ky2k
= (x1− y1− ky2k)2 ≥ 0.
(d) From (1.8), we know that x1y1− hx2, y2i ≥ x1y1− kx2k ky2k ≥ 0, and (x1y1− kx2k ky2k)2− x21 − kx2k2
y21− ky2k2
= x21ky2k2+ y12kx2k2− 2x1y1kx2k ky2k
= (x1ky2k − y1kx2k)2 ≥ 0.
Hence, we obtain x1y1− hx2, y2i ≥ x1y1− kx2k ky2k ≥px21− kx2k2py12− ky2k2,
(e) The proof follows from
(x1+ y1)2− kx2+ y2k2
= x21− kx2k2 + y21− ky2k2 + 2 (x1y1− hx2, y2i)
≥ 2 q
(x21− kx2k2)(y12− ky2k2) + 2 (x1y1 − hx2, y2i)
≥ 2 q
(x21− kx2k2)(y12− ky2k2) + 2 q
(x21− kx2k2)(y12− ky2k2)
= 4 q
(x21− kx2k2)(y12− ky2k2), where the first inequality is true since a + b ≥ 2√
ab for all positive a, b and the second inequality is from part(d).
The inequalities in Lemma 2.2(d)-(e) can be achieved by applying Proposition 1.8(b).
Next proposition is an important feature of the SOC-function corresponding to f (t) = 1t which is very useful in the subsequent analysis and also similar to the operator setting.
Proposition 2.3. Let f : (0, ∞) → (0, ∞) be f (t) = 1t. Then, (a) −f is SOC-monotone on (0, ∞);
(b) f is SOC-convex on (0, ∞).
Proof. (a) It suffices to show that x Kn y Kn 0 implies fsoc(x) = x−1 Kn y−1 = fsoc(y). For any x = (x1, x2) ∈ Kn and y = (y1, y2) ∈ Kn, we know that y−1 =
1
det(y)(y1, −y2) and x−1 = det(x)1 (x1, −x2), which imply fsoc(y) − fsoc(x) = y−1− x−1
=
y1
det(y) − x1
det(x), x2
det(x) − y2 det(y)
= 1
det(x) det(y) det(x)y1− det(y)x1, det(y)x2− det(x)y2.
To complete the proof, we need to verify two things.
(1) First, we have to show that det(x)y1− det(y)x1 ≥ 0. Applying Lemma 2.1 yields det(x)
det(y) = x21− kx2k2
y21 − ky2k2 = x1+ kx2k y1+ ky2k
x1− kx2k y1− ky2k
≥ 2x1 2y1 = x1
y1.
Then, cross multiplying gives det(x)y1 ≥ det(y)x1, which says det(x)y1− det(y)x1 ≥ 0.
(2) Secondly, we need to argue that k det(y)x2 − det(x)y2k ≤ det(x)y1− det(y)x1. This
is true by
(det(x)y1− det(y)x1)2− k det(y)x2− det(x)y2k2
= (det(x))2y21− 2 det(x) det(y)x1y1+ (det(y))2x21
− (det(y))2kx2k2− 2 det(x) det(y)hx2, y2i + (det(x))2ky2k2
= (det(x))2(y12− ky2k2) + (det(y))2(x21− kx2k2)
−2 det(x) det(y)(x1y1− hx2, y2i)
= (det(x))2det(y) + (det(y))2det(x) − 2 det(x) det(y)(x1y1− hx2, y2i)
= det(x) det(y) det(x) + det(y) − 2x1y1+ 2hx2, y2i
= det(x) det(y) (x21 − kx2k2) + (y12− ky2k2) − 2x1y1+ 2hx2, y2i
= det(x) det(y) (x1 − y1)2− (kx2k2+ ky2k2− 2hx2, y2i)
= det(x) det(y) (x1 − y1)2− (kx2− y2k2)
≥ 0,
where the last step holds by the inequality (1.8).
Thus, from all the above, we prove y−1− x−1 ∈ Kn, that is, y−1 Kn x−1. (b) For any x Kn 0 and y Kn 0, using (1.8) again, there hold
x1− kx2k > 0, y1− ky2k > 0, |hx2, y2i| ≤ kx2k · ky2k ≤ x1y1. From x−1 = det(x)1 (x1, −x2) and y−1 = det(y)1 (y1, −y2), we also have
1
2 fsoc(x) + fsoc(y) = 1 2
x1
det(x) + y1
det(y), − x2
det(x) − y2
det(y)
, and
fsoc x + y 2
= x + y 2
−1
= 2
det(x + y) x1+ y1, −(x2+ y2).
For convenience, we denote 12 fsoc(x) + fsoc(y) − fsoc x+y2 := 12(Ξ1, Ξ2), where Ξ1 ∈ IR and Ξ2 ∈ IRn−1 are given by
Ξ1 =
x1
det(x) + y1 det(y)
− 4(x1+ y1) det(x + y), Ξ2 = 4(x2+ y2)
det(x + y)−
x2
det(x) + y2 det(y)
.
Again, in order to prove f is SOC-convex, it suffices to verify two things: Ξ1 ≥ 0 and kΞ2k ≤ Ξ1.
(1) First, we verify that Ξ1 ≥ 0. In fact, if we define the function g(x) := x1
x21 − kx2k2 = x1 det(x),
then we observe that
g x + y 2
≤ 1 2
g(x) + g(y)
⇐⇒ Ξ1 ≥ 0.
Hence, to prove Ξ1 ≥ 0, it is equivalent to verifying g is convex on int(Kn). Since int(Kn) is a convex set, it is sufficient to argue that ∇2g(x) is a positive semidefinite matrix.
From direct computations, we have
∇2g(x) = 1 (x21− kx2k2)3
2x31+ 6x1kx2k2 −(6x21+ 2kx2k2)xT2
−(6x21+ 2kx2k2)x2 2x1 (x21− kx2k2)I + 4x2xT2
.
Let ∇2g(x) be viewed as the matrix A B BT C
given as in Lemma 1.1 (here A is a scalar). Then, we have
AC − BTB
= 2x1 2x31+ 6x1kx2k2
(x21− kx2k2)I + 4x2xT2 − 6x21+ 2kx2k22
x2xT2
= 4x41+ 12x21kx2k2
x21− kx2k2 I − 20x41− 24x21kx2k2+ 4kx2k4 x2xT2
= 4x41+ 12x21kx2k2
x21− kx2k2 I − 4 5x21− kx2k2
x21− kx2k2 x2xT2
= x21− kx2k2
4x41+ 12x21kx2k2 I − 4 5x21− kx2k2 x2xT2
= x21− kx2k2 M,
where we denote the whole matrix in the big parenthesis of the last second equality by M . It can be verified that x2xT2 is positive semidefinite with only one nonzero eigenvalue kx2k2. Hence, all the eigenvalues of the matrix M are (4x41 + 12x21kx2k2 − 20x21kx2k2+ 4kx2k4) and 4x41 + 12x21kx2k2 with multiplicity of n − 2, which are all positive since
4x41+ 12x21kx2k2− 20x21kx2k2+ 4kx2k4
= 4x41− 8x21kx2k2+ 4kx2k4
= 4 x21− kx2k2
> 0.
Thus, by Lemma 1.1, we see that ∇2g(x) is positive definite and hence is positive semidef-inite. This means g is convex on int(Kn), which says Ξ1 ≥ 0.
(2) It remains to show that Ξ21 − kΞ2k2 ≥ 0 : Ξ21− kΞ2k2
=
x21
det(x)2 + 2x1y1
det(x) det(y) + y12 det(y)2
− 8(x1 + y1) det(x + y)
x1
det(x)+ y1 det(y)
+ 16
det(x + y)2
x21+ 2x1y1+ y12
−
4(x2+ y2) det(x + y)−
x2
det(x) + y2 det(y)
2
=
x21
det(x)2 + 2x1y1
det(x) det(y) + y12 det(y)2
− 8(x1 + y1) det(x + y)
x1
det(x)+ y1
det(y)
+ 16
det(x + y)2
x21+ 2x1y1+ y12
−
16
det(x + y)2
kx2k2+ 2hx2, y2i + ky2k2
−8
x2+ y2
det(x + y) , x2
det(x) + y2 det(y)
+
kx2k2
det(x)2 + 2hx2, y2i
det(x) det(y) + ky2k2 det(y)2
= x21− kx2k2
det(x)2 +2(x1y1− hx2, y2i)
det(x) det(y) +y21− ky2k2 det(y)2
+ 16
det(x + y)2
x21− kx2k2 + 2 x1y1− hx2, y2i + y12− ky2k2
−8
x21− kx2k2
det(x + y) det(x)+ x1y1− hx2, y2i
det(x + y) det(x)+ x1y1− hx2, y2i
det(x + y) det(y) + y21− ky2k2 det(x + y) det(y)
= x21− kx2k2
1
det(x)2 + 16
det(x + y)2 − 8
det(x + y) det(x)
+ y12− ky2k2
1
det(y)2 + 16
det(x + y)2 − 8
det(x + y) det(y)
+2 x1y1− hx2, y2i
1
det(x) det(y) + 16
det(x + y)2 − 4
det(x + y)det(x)− 4
det(x + y) det(y)
= x21− kx2k2 det(x + y) − 4 det(x) det(x) det(x + y)
2
+ y12− ky2k2 det(x + y) − 4 det(y) det(y) det(x + y)
2
+2 x1y1− hx2, y2i (det(x + y) − 4 det(x))(det(x + y) − 4 det(y)) det(x) det(y) det(x + y)2
.
Now applying the facts that det(x) = x21 − kx2k2, det(y) = y21 − ky2k2, and det(x + y) − det(x) − det(y) = 2(x1y1− hx2, y2i), we can simplify the last equality (after a lot of algebra simplifications) and obtain
Ξ21− kΞ2k2 = det(x + y) − 2 det(x) − 2 det(y)2 det(x) det(y) det(x + y) ≥ 0.
Hence, we prove that fsoc x+y2 Kn 12 fsoc(x) + fsoc(y), which says the function f (t) = 1
t is SOC-convex on the interval (0, ∞).
Proposition 2.4. (a) The function f (t) = 1+tt is SOC-monotone on (0, ∞).
(b) For any λ > 0, the function f (t) = λ+tt is SOC-monotone on (0, ∞).
Proof. (a) Let g(t) = −1t and h(t) = 1 + t. Then, we see that g is SOC-monotone on (0, ∞) by Proposition 2.3, while h is SOC-monotone on IR by Proposition 2.2. Since f (t) = 1 −1+t1 = h(g(1 + t)), the result follows from the fact that the composition of two SOC-monotone functions is also SOC-monotone, see Proposition 2.9.
(b) Similarly, let g(t) = 1+tt and h(t) = λt, then both functions are SOC-monotone by part(a). Since f (t) = g(h(t)), the result is true by the same reason as in part(a).
Proposition 2.5. Let Lx be defined as in (1.21). For any x Kn 0 and y Kn 0, we have Lx Ly ⇐⇒ L−1y L−1x ⇐⇒ Ly−1 Lx−1.
Proof. By the property of Lx that x Kn y ⇐⇒ Lx Ly, and Proposition 2.3(a), then proof follows.
Next, we examine another simple function f (t) = √
t. We will see that it is SOC-monotone on the interval [0, ∞), and −√
t is SOC-convex on [0, ∞).
Proposition 2.6. Let f : [0, ∞) → [0, ∞) be f (t) =√
t. Then, (a) f is SOC-monotone on [0, ∞);
(b) −f is SOC-convex on [0, ∞).
Proof. (a) This is a consequence of Property 1.3(b).
(b) To show −f is SOC-convex, it is enough to prove that fsoc x+y2 Kn fsoc(x)+f2 soc(y), which is equivalent to verifying that x+y2 1/2
Kn
√x+√ y
2 , for all x, y ∈ Kn. Since x + y Kn 0, by Property 1.3(e), it is sufficient to show that x+y2 Kn √x+√y
2
2
. This can be seen by x+y2 −√x+√y
2
2
= (
√x−√ y)2
4 Kn 0. Thus, we complete the proof.
Proposition 2.7. Let f : [0, ∞) → [0, ∞) be f (t) = tr where 0 ≤ r ≤ 1. Then, (a) f is SOC-monotone on [0, ∞);
(b) −f is SOC-convex on [0, ∞).
Proof. (a) Let r be a dyadic rational, i.e., a number of the form r = 2mn, where n is any positive integer and 1 ≤ m ≤ 2n. It is enough to prove the assertion is true for such r since the dyadic rational numbers are dense in [0, 1]. We will claim this by induction on n. Let x, y ∈ Kn with x Kn y, then by Property 1.3(b) we have x1/2 Kn y1/2. Therefore, part(a) is true when n = 1. Suppose it is also true for all dyadic rational m
2j, in which 1 ≤ j ≤ n − 1. Now let r = 2mn with m ≤ 2n. By induction hypothesis, we know x2n−1m Kn y2n−1m . Then, by applying Property 1.3(b), we obtain
x2n−1m
1/2
Kn
y2n−1m
1/2
, which says x2nm Kn y2nm. Thus, we have shown that x Kn y Kn 0 implies xr Kn yr, for all dyadic rational r in [0, 1]. Then, the desired result follows.
(b) The proof is similar to the above arguments. First, we observe that
x + y 2
−
√ x +√
y 2
2
=
√ x −√
y 2
2
Kn 0, which implies x+y2 1/2
Kn 12 √ x +√
y
by Property 1.3(b). Hence, we show that the assertion is true when n = 1. By induction hypothesis, suppose x+y2 m
2n−1 Kn
x
m 2n−1+y
m 2n−1
2
. Then, we have
x + y 2
m
2n−1
− x2nm + y2nm 2
2
Kn x2n−1m + y2n−1m 2
!
− x2nm + y2nm 2
2
= x2nm − y2nm 2
2
Kn 0,
which implies x+y2 2nm
Kn
x2nm+y2nm 2
by Property 1.3(b). Following the same argu-ments about dyadic rationals in part(a) yields the desired result.
From all the above examples, we observe that f being monotone does not imply f is SOC-monotone. Likewise, f being convex does not guarantee that f is SOC-convex.
Now, we move onto some famous functions which are used very often for NCP (nonlinear complementarity problem), SDCP, and SOCCP. It would be interesting to know about the SOC-convexity and SOC-monotonicity of these functions. First, we will look at the Fischer-Burmeister function, φFB : IRn× IRn → IRn, given by
φFB(x, y) = (x2+ y2)1/2− (x + y), (2.5) which is a well-known merit function for complementarity problem, see [87, 139]. Here, (·)2 and (·)1/2 are defined through Jordan product introduced as in (1.5) in Chapter 1.
For SOCCP, it has been shown that squared norm of φFB, i.e.,
ψFB(x, y) = kφFB(x, y)k2, (2.6)
is continuously differentiable (see [48]) whereas ψFB is only shown differentiable for SDCP (see [145]). In addition, φFB is proved to have semismoothness and Lipschitz continuity in the paper [142] for both cases of SOCCP and SDCP. For more details regarding further properties of these functions associated with SOC and the roles they play in the solutions methods, please refer to [47, 119–121]. In NCP setting, φFB is a convex function, so we may wish to have an analogy for SOCCP. Unfortunately, as shown below, it does not behave like an SOC-convex function.
Example 2.5. Let φFB be defined as in (2.5) and ψFB defined as in (2.6).
(a) The function ρ(x, y) = (x2+ y2)1/2 does not satisfy (2.2).
(b) The Fischer-Burmeister function φFB does not satisfy (2.2).
(c) The function ψFB : IRn× IRn→ IR is not convex.
Solution. (a) A counterexample occurs when taking x = (7, 2, 6), y = (4, 2, 5) and u = (1, 3, 5), v = (1, 7, 6).
(b) Suppose that it satisfies (2.2). Then, we will have ρ satisfies (2.2) by ρ(x, y) = φFB(x, y) + (x + y), which is a contradiction to part(a). Thus, φFB does not satisfy (2.2).
(c) Let x = (1, −2), y = (1, −1) and u = (0, −1), v = (1, −1). Then, we have
φFB(x, y) = −3 +√ 13
2 , 7 −√ 13 2
!
=⇒ ψFB(x, y) = kφFB(x, y)k2 = 21 − 5√ 13.
φFB(u, v) = −1 +√ 5
2 , 5 −√ 5 2
!
=⇒ ψFB(u, v) = kφFB(u, v)k2 = 9 − 3√ 5.
Thus, 12 ψFB(x, y) + ψFB(u, v) = 12(30 − 5√
13 − 3√
5) ≈ 2.632.
On the other hand, let (ˆx, ˆy) := 12(x, y) + 12(u, v), that is, ˆx = (12, −32) and ˆy = (1, −1).
Indeed, we have ˆx2+ ˆy2 = 92, −72 and hence (ˆx2+ ˆy2)1/2 = 1+2
√ 2 2 ,1−2
√ 2
2 , which implies ψFB(ˆx, ˆy) = kφFB(ˆx, ˆy)k2 = 14 − 8√
2 ≈ 2.686. Therefore, we obtain ψFB 1
2(x, y) + 1 2(u, v)
> 1
2ψFB(x, y) + 1
2ψFB(u, v), which shows ψFB is not convex.
Another function based on the Fischer-Burmeister function is ψ1 : IRn× IRn → IR, given by
ψ1(x, y) := k[φFB(x, y)]+k2, (2.7) where φFB is the Fischer-Burmeister function given as in (2.5). In the NCP case, it is known that ψ1 is convex. It has been an open question whether this is still true for
SDCP and SOCCP (see Question 3 on page 182 of [145]). In fact, Qi and Chen [128]
gave the negative answer for the SDCP case. Here we provide an answer to the question for SOCCP: ψ1 is not convex in the SOCCP case.
Example 2.6. Let φFB be defined as in (2.5) and ψ1 defined as in (2.7).
(a) The function [φFB(x, y)]+ = [(x2+ y2)1/2− (x + y)]+ does not satisfy (2.2).
(b) The function ψ1 is not convex.
Solution. (a) Let x = (2, 1, −1), y = (1, 1, 0) and u = (1, −2, 5), v = (−1, 5, 0). For simplicity, we denote φ1(x, y) := [φFB(x, y)]+. Then, by direct computations, we obtain
1
2φ1(x, y) + 1
2φ1(u, v) − φ1 1
2(x, y) + 1 2(u, v)
= 1.0794, 0.4071, −1.0563 6K3 0,
which says φ1 does not satisfy (2.2).
(b) Let x = (17, 5, 16), y = (20, −3, 15) and u = (2, 3, 3), v = (9, −7, 2). Then, it can be easily verified that 12ψ1(x, y) + 12ψ1(u, v) − ψ1 12(x, y) + 12(u, v) < 0, which implies ψ1 is not convex.
Example 2.7. (a) The function f (t) = |t| is not SOC-monotone on IR.
(b) The function f (t) = |t| is not SOC-convex on IR.
(c) The function f (t) = [t]+ is not SOC-monotone on IR.
(d) The function f (t) = [t]+ is not SOC-convex on IR.
Solution. To see (a), let x = (1, 0), y = (−2, 0). It is clear that x
K2 y. Besides, we have x2 = (1, 0), y2 = (4, 0) which yields |x| = (1, 0) and |y| = (2, 0). But, |x| − |y| = (−1, 0) 6
K2 0.
To see (b), let x = (1, 1, 1), y = (−1, 1, 0). In fact, we have |x| = √
2,√1
2,√1
2
, |y| = (1, −1, 0), and |x + y| = (√
5, 0, 0). Therefore,
|x| + |y| − |x + y| =
√
2 + 1 −√
5, −1 + 1
√2, 1
√2
6K3 0,
which says fsoc x+y2 6
K3
1
2 fsoc(x) + fsoc(y) . Thus, f (t) = |t| is not SOC-convex on IR.
To see (c) and (d), just follows (a) and (b) and the facts that [t]+ = 12(t + |t|) where t ∈ IR, and Property 1.2(f): [x]+ = 12(x + |x|) where x ∈ IRn.
To close this section, we check with one popular smoothing function, f (t) = 1
2
√
t2+ 4 + t ,
which was proposed by Chen and Harker [38], Kanzow [84], and Smale [138]; and is called the CHKS function. Its corresponding SOC-function is defined by
fsoc(x) = 1 2
(x2+ 4e)12 + x ,
where e = (1, 0, · · · , 0). The function f (t) is convex and monotone, so we may also wish to know whether it is SOC-convex or SOC-monotone or not. Unfortunately, it is neither SOC-convex nor SOC-monotone for n ≥ 3, though it is both SOC-convex and SOC-monotone for n = 2. The following example demonstrates what we have just said.
Example 2.8. Let f : IR → IR be f (t) =
√t2+ 4 + t
2 . Then, (a) f is not SOC-monotone of order n ≥ 3 on IR;
(b) however, f is SOC-monotone of order 2 on IR;
(c) f is not SOC-convex of order n ≥ 3 on IR;
(d) however, f is SOC-convex of order 2 on IR.
Solution. Again, by Remark 2.1, taking x = (2, 1, −1) and y = (1, 1, 0) gives a coun-terexample for both (a) and (c).
To see (b) and (d), it follows by direct verifications as what we have done before.