We may verify that the domain S of fsoc is open in H if and only if J is open in IR. Also, S is always convex since, for any x = xe+ x0e, y = ye+ y0e ∈ S and β ∈ [0, 1],
λ1[βx + (1 − β)y] = βx0+ (1 − β)y0 − kβxe+ (1 − β)yek ≥ min{λ1(x), λ1(y)}, λ2[βx + (1 − β)y] = βx0+ (1 − β)y0 + kβxe+ (1 − β)yek ≤ max{λ2(x), λ2(y)}, which implies that βx + (1 − β)y ∈ S. Thus, fsoc(βx + (1 − β)y) is well defined.
Throughout this section, all differentiability means Fr´echet differentiability. If F : H → H is (twice) differentiable at x ∈ H, we denote by F0(x) (F00(x)) the first-order F-derivative (the second-order F-derivative) of F at x. In addition, we use Cn(J ) and C∞(J ) to denote the set of n times and infinite times continuously differentiable real functions on J , respectively. When f ∈ C1(J ), we denote by f[1] the function on J × J defined by
f[1](λ, µ) :=
( f (λ)−f (µ)
λ−µ if λ 6= µ,
f0(λ) if λ = µ, (2.45)
and when f ∈ C2(J ), denote by f[2] the function on J × J × J defined by f[2](τ1, τ2, τ3) := f[1](τ1, τ2) − f[1](τ1, τ3)
τ2− τ3 (2.46)
if τ1, τ2, τ3 are distinct, and for other values of τ1, τ2, τ3, f[2] is defined by continuity; e.g., f[2](τ1, τ1, τ3) = f (τ3) − f (τ1) − f0(τ1)(τ3− τ1)
(τ3− τ1)2 , f[2](τ1, τ1, τ1) = 1
2f00(τ1).
For a linear operator L from H into H, we write L ≥ 0 (respectively, L > 0) to mean that L is positive semidefinite (respectively, positive definite), i.e., hh, Lhi ≥ 0 for any h ∈ H (respectively, hh, Lhi > 0 for any 0 6= h ∈ H).
Lemma 2.5. Let B := {z ∈ hei⊥ | kzk ≤ 1}. Then, for any given u ∈ hei⊥ with kuk = 1 and θ, λ ∈ R, the following results hold.
(a) θ + λhu, zi ≥ 0 for any z ∈ B if and only if θ ≥ |λ|.
(b) θ − kλzk2 ≥ (θ − λ2)hu, zi2 for any z ∈ B if and only if θ − λ2 ≥ 0.
Proof. (a) Suppose that θ + λhu, zi ≥ 0 for any z ∈ B. If λ = 0, then θ ≥ |λ| clearly holds. If λ 6= 0, take z = −sign(λ)u. Since kuk = 1, we have z ∈ B, and consequently, θ + λhu, zi ≥ 0 reduces to θ − |λ| ≥ 0. Conversely, if θ ≥ |λ|, then using the Cauchy-Schwartz Inequality yields θ + λhu, zi ≥ 0 for any z ∈ B.
(b) Suppose that θ−kλzk2 ≥ (θ−λ2)hu, zi2for any z ∈ B. Then, we must have θ−λ2 ≥ 0.
If not, for those z ∈ B with kzk = 1 but hu, zi 6= kukkzk, it holds that (θ − λ2)hu, zi2 > (θ − λ2)kuk2kzk2 = θ − kλzk2,
which contradicts the given assumption. Conversely, if θ − λ2 ≥ 0, the Cauchy-Schwartz inequality implies that (θ − λ2)hu, zi2 ≤ θ − kλzk2 for any z ∈ B.
Lemma 2.6. For any given a, b, c ∈ R and x = xe+ x0e with xe 6= 0, the inequality akhek2− hhe, xei2 + bh0+ hxe, hei2
+ ch0− hxe, hei2
≥ 0 (2.47)
holds for all h = he+ h0e ∈ H if and only if a ≥ 0, b ≥ 0 and c ≥ 0.
Proof. Suppose that (2.47) holds for all h = he+ h0e ∈ H. By letting he = xe, h0 = 1 and he = −xe, h0 = 1, respectively, we get b ≥ q0 and c ≥ 0 from (2.47). If a ≥ 0 does not hold, then by taking he = q
b+c+1
|a|
ze
kzek with hze, xei = 0 and h0 = 1, (2.47) gives a contradiction −1 ≥ 0. Conversely, if a ≥ 0, b ≥ 0 and c ≥ 0, then (2.47) clearly holds for all h ∈ H.
Lemma 2.7. Let f ∈ C2(J ) and ue ∈ hei⊥ with kuek = 1. For any h = he+ h0e ∈ H, define
µ1(h) := h0− hue, hei
√2 , µ2(h) := h0+ hue, hei
√2 , µ(h) :=pkhek2− hue, hei2. Then, for any given a, d ∈ IR and λ1, λ2 ∈ J, the following inequality
4f00(λ1)f00(λ2)µ1(h)2µ2(h)2+ 2(a − d)f00(λ2)µ2(h)2µ(h)2 +2 (a + d) f00(λ1)µ1(h)2µ(h)2+ a2− d2 µ(h)4
−2 [(a − d) µ1(h) + (a + d) µ2(h)]2µ(h)2 ≥ 0 (2.48) holds for all h = he+ h0e ∈ H if and only if
a2− d2 ≥ 0, f00(λ2)(a − d) ≥ (a + d)2 and f00(λ1)(a + d) ≥ (a − d)2. (2.49) Proof. Suppose that (2.48) holds for all h = he+ h0e ∈ H. Taking h0 = 0 and he 6= 0 with hhe, uei = 0, we have µ1(h) = 0, µ2(h) = 0 and µ(h) = khek > 0, and then (2.48) gives a2− d2 ≥ 0. Taking he 6= 0 such that |hue, hei| < khek and h0 = hue, hei 6= 0, we have µ1(h) = 0, µ2(h) = √
2h0 and µ(h) > 0, and then (2.48) reduces to the following inequality
4(a − d)f00(λ2) − (a + d)2 h20+ (a2− d2)(khek2− h20) ≥ 0.
This implies that (a − d)f00(λ2) − (a + d)2 ≥ 0. If not, by letting h0 be sufficiently close to khek, the last inequality yields a contradiction. Similarly, taking h with he6= 0 satisfying
|hue, hei| < khek and h0 = −hue, hei, we get f00(λ1)(a + d) ≥ (a − d)2 from (2.48).
Next, suppose that (2.49) holds. Then, the inequalities f00(λ2)(a − d) ≥ (a + d)2 and f00(λ1)(a + d) ≥ (a − d)2 imply that the left-hand side of (2.48) is greater than
4f00(λ1)f00(λ2)µ1(h)2µ2(h)2− 4(a2− d2)µ1(h)µ2(h)µ(h)2+ a2− d2 µ(h)4,
which is obviously nonnegative if µ1(h)µ2(h) ≤ 0. Now assume that µ1(h)µ2(h) > 0. If a2− d2 = 0, then the last expression is clearly nonnegative, and if a2− d2 > 0, then the last two inequalities in (2.49) imply that f00(λ1)f00(λ2) ≥ (a2− d2) > 0, and therefore,
4f00(λ1)f00(λ2)µ1(h)2µ2(h)2 − 4(a2− d2)µ1(h)µ2(h)µ(h)2+ a2− d2 µ(h)4
≥ 4(a2− d2)µ1(h)2µ2(h)2− 4(a2− d2)µ1(h)µ2(h)µ(h)2+ a2− d2 µ(h)4
= (a2− d2)2µ1(h)µ2(h) − µ(h)22
≥ 0.
Thus, we prove that inequality (2.48) holds. The proof is complete.
To proceed, we introduce the regularization of a locally integrable real function. Let ϕ be a real function of class C∞ with the following properties: ϕ ≥ 0, ϕ is even, the support supp ϕ = [−1, 1], and R
IRϕ(t)dt = 1. For each ε > 0, let ϕε(t) = 1εϕ(tε). Then, supp ϕε = [−ε, ε] and ϕε has all the properties of ϕ listed above. If f is a locally integrable real function, we define its regularization of order ε as the function
fε(s) :=
Z
f (s − t)ϕε(t)dt = Z
f (s − εt)ϕ(t)dt. (2.50) Note that fε is a C∞ function for each ε > 0, and limε→0fε(x) = f (x) if f is continuous.
Lemma 2.8. For any given f : J → IR with J open, let fsoc : S → H be defined by (1.9).
(a) fsoc is continuous on S if and only if f is continuous on J .
(b) fsoc is (continuously) differentiable on S if and only if f is (continuously) differen-tiable on J . Also, when f is differendifferen-tiable on J , for any x = xe + x0e ∈ S and v = ve+ v0e ∈ H,
(fsoc)0(x)v =
f0(x0)v if xe = 0;
(b1(x) − a0(x))hxe, veixe+ c1(x)v0xe
+a0(x)ve+ b1(x)v0e + c1(x)hxe, veie if xe 6= 0,
(2.51)
where
a0(x) = f (λ2(x)) − f (λ1(x)) λ2(x) − λ1(x) , b1(x) = f0(λ2(x)) + f0(λ1(x))
2 ,
c1(x) = f0(λ2(x)) − f0(λ1(x))
2 .
(c) If f is differentiable on J , then for any given x ∈ S and all v ∈ H, (fsoc)0(x)e = (f0)soc(x) and he, (fsoc)0(x)vi = hv, (f0)soc(x)i .
(d) If f0 is nonnegative (respectively, positive) on J , then for each x ∈ S, (fsoc)0(x) ≥ 0 (respectively, (fsoc)0(x) > 0).
Proof. (a) Suppose that fsoc is continuous. Let Ω be the set composed of those x = te with t ∈ J . Clearly, Ω ⊆ S, and fsoc is continuous on Ω. Noting that fsoc(x) = f (t)e for any x ∈ Ω, it follows that f is continuous on J . Conversely, if f is continuous on J , then fsoc is continuous at any x = xe+ x0e ∈ S with xe 6= 0 since λi(x) and ui(x) for i = 1, 2 are continuous at such points. Next, let x = xe + x0e be an arbitrary element from S with xe = 0, and we prove that fsoc is continuous at x. Indeed, for any z = ze+ z0e ∈ S sufficiently close to x, it is not hard to verify that
kfsoc(z) − fsoc(x)k ≤ |f (λ2(z)) − f (x0)|
2 + |f (λ1(z)) − f (x0)|
2 +|f (λ2(z)) − f (λ1(z))|
2 .
Since f is continuous on J , and λ1(z), λ2(z) → x0 as z → x, it follows that f (λ1(z)) → f (x0) and f (λ2(z)) → f (x0) as z → x.
The last two equations imply that fsoc is continuous at x.
(b) When fsoc is (continuously) differentiable, using the similar arguments as in part(a) can show that f is (continuously) differentiable. Next, assume that f is differentiable.
Fix any x = xe + x0e ∈ S. We first consider the case where xe 6= 0. Since λi(x) for i = 1, 2 and kxxe
ek are continuously differentiable at such x, it follows that f (λi(x)) and ui(x) are differentiable and continuously differentiable, respectively, at x. Then, fsoc is differentiable at such x by the definition of fsoc. Also, an elementary computation shows that
[λi(x)]0v = hv, ei + (−1)ihxe, v − hv, eiei
kxek = v0+ (−1)ihxe, vei
kxek , (2.52)
xe kxek
0
v = v − hv, eie
kxek − hxe, v − hv, eieixe
kxek3 = ve
kxek − hxe, veixe
kxek3 (2.53) for any v = ve+ v0e ∈ H, and consequently,
[f (λi(x))]0v = f0(λi(x))
v0+ (−1)ihxe, vei kxek
, [ui(x)]0v = 1
2(−1)i
ve
kxek − hxe, veixe
kxek3
.
Together with the definition of fsoc, we calculate that (fsoc)0(x)v is equal to f0(λ1(x))
2
v0−hxe, vei kxek
e − xe kxek
−f (λ1(x)) 2
ve
kxek − hxe, veixe kxek3
+f0(λ2(x)) 2
v0+ hxe, vei kxek
e + xe kxek
+f (λ2(x)) 2
ve
kxek − hxe, veixe kxek3
= b1(x)v0e + c1(x) hxe, vei e + c1(x)v0xe+ b1(x)hxe, veixe
+a0(x)ve− a0(x)hxe, veixe,
where λ2(x) − λ1(x) = 2kxek is used for the last equality. Thus, we obtain (2.51) for xe 6= 0. We next consider the case where xe= 0. Under this case, for any v = ve+v0e ∈ H,
fsoc(x + v) − fsoc(x)
= f (x0+ v0− kvek)
2 (e − ve) + f (x0+ v0+ kvek)
2 (e + ve) − f (x0)e
= f0(x0)(v0− kvek)
2 e +f0(x0)(v0+ kvek)
2 e
+f0(x0)(v0+ kvek)
2 ve− f0(x0)(v0 − kvek)
2 ve+ o(kvk)
= f0(x0)(v0e + kvekve) + o(kvk), where ve= kvve
ek if ve6= 0, and otherwise ve is an arbitrary unit vector from hei⊥. Hence, fsoc(x + v) − fsoc(x) − f0(x0)v
= o(kvk).
This shows that fsoc is differentiable at such x with (fsoc)0(x)v = f0(x0)v.
Assume that f is continuously differentiable. From (2.51), it is easy to see that (fsoc)0(x) is continuous at every x with xe6= 0. We next argue that (fsoc)0(x) is continuous at every x with xe= 0. Fix any x = x0e with x0 ∈ J. For any z = ze+ z0e with ze6= 0, we have
k(fsoc)0(z)v − (fsoc)0(x)vk
≤ |b1(z) − a0(z)|kvek + |b1(z) − f0(x0)||v0| (2.54) +|a0(z) − f0(x0)|kvek + |c1(z)|(|v0| + kvek).
Since f is continuously differentiable on J and λ2(z) → x0, λ1(z) → x0 as z → x, we have
a0(z) → f0(x0), b1(z) → f0(x0) and c1(z) → 0.
Together with equation (2.54), we obtain that (fsoc)0(z) → (fsoc)0(x) as z → x.
(c) The result is direct by the definition of fsoc and a simple computation from (2.51).
(d) Suppose that f0(t) ≥ 0 for all t ∈ J . Fix any x = xe + x0e ∈ S. If xe = 0, the result is direct. It remains to consider the case xe 6= 0. Since f0(t) ≥ 0 for all t ∈ J , we have b1(x) ≥ 0, b1(x) − c1(x) = f0(λ1(x)) ≥ 0, b1(x) + c1(x) = f0(λ2(x)) ≥ 0 and a0(x) ≥ 0. From part(b) and the definitions of b1(x) and c1(x), it follows that for any h = he+ h0e ∈ H,
hh, (fsoc)0(x)hi = (b1(x) − a0(x))hxe, hei2+ 2c1(x)h0hxe, hei + b1(x)h20+ a0(x)khek2
= a0(x)khek2− hxe, hei2 + 1
2(b1(x) − c1(x)) [h0 − hxe, hei]2 +1
2(b1(x) + c1(x)) [h0+ hxe, hei]2 ≥ 0.
This implies that the operator (fsoc)0(x) is positive semidefinite. Particularly, if f0(t) > 0 for all t ∈ J , we have that hh, (fsoc)0(x)hi > 0 for all h 6= 0. The proof is complete. Lemma 2.8(d) shows that the differential operator (fsoc)0(x) corresponding to a dif-ferentiable nondecreasing f is positive semidefinite. Therefore, the differential operator (fsoc)0(x) associated with a differentiable SOC-monotone function is also positive semidef-inite.
Proposition 2.19. Assume that f ∈ C1(J ) with J being an open interval in IR. Then, f is SOC-monotone if and only if (fsoc)0(x)h ∈ K for any x ∈ S and h ∈ K.
Proof. If f is SOC-monotone, then for any x ∈ S, h ∈ K and t > 0, we have fsoc(x + th) − fsoc(x) Kn 0,
which, by the continuous differentiability of fsoc and the closedness of K, implies that (fsoc)0(x)h Kn 0.
Conversely, for any x, y ∈ S with x Kn y, from the given assumption we have that fsoc(x) − fsoc(y) =
Z 1 0
(fsoc)0(x + t(x − y))(x − y)dt ∈ K.
This shows that fsoc(x) Kn fsoc(y), i.e., f is SOC-monotone. The proof is complete.
Proposition 2.19 shows that the differential operator (fsoc)0(x) associated with a dif-ferentiable SOC-monotone function f leaves K invariant. If, in addition, the linear operator (fsoc)0(x) is bijective, then (fsoc)0(x) belongs to the automorphism group of K.
Such linear operators are important to study the structure of the cone K (see [61]).
Proposition 2.20. Assume that f ∈ C1(J ) with J being an open interval in IR. If f is SOC-monotone, then
(a) fsoc(x) ∈ K for any x ∈ S;
(b) fsoc is a monotone function, that is, hfsoc(x) − fsoc(y), x − yi ≥ 0 for any x, y ∈ S.
Proof. Part(a) is direct by using Proposition 2.19 with h = e and Lemma 2.8(c). By part(a), f0(τ ) ≥ 0 for all τ ∈ J . Together with Lemma 2.8(d), (fsoc)0(x) ≥ 0 for any x ∈ S. Applying the integral mean-value theorem, it then follows that
hfsoc(x) − fsoc(y), x − yi = Z 1
0
hx − y, (fsoc)0(y + t(x − y))(x − y)idt ≥ 0.
This proves the desired result of part (b). The proof is complete.
Note that the converse of Proposition 2.20(a) is not correct. For example, for the function f (t) = −t−2 (t > 0), it is clear that fsoc(x) ∈ K for any x ∈ intK, but it is not SOC-monotone by Example 2.13(b). The following proposition provides another sufficient and necessary characterization for differentiable SOC-monotone functions.
Proposition 2.21. Let f ∈ C1(J ) with J being an open interval in IR. Then, f is SOC-monotone if and only if
f[1](τ1, τ1) f[1](τ1, τ2) f[1](τ2, τ1) f[1](τ2, τ2)
=
"
f0(τ1) f (ττ2)−f (τ1)
2−τ1
f (τ1)−f (τ2)
τ1−τ2 f0(τ2)
#
O, ∀τ1, τ2 ∈ J. (2.55)
Proof. The equality is direct by the definition of f[1] given as in (2.45). It remains to prove that f is SOC-monotone if and only if the inequality in (2.55) holds for any τ1, τ2 ∈ J. Assume that f is SOC-monotone. By Proposition 2.19, (fsoc)0(x)h ∈ K for any x ∈ S and h ∈ K. Fix any x = xe+ x0e ∈ S. It suffices to consider the case where xe 6= 0. Since (fsoc)0(x)h ∈ K for any h ∈ K, we particularly have (fsoc)0(x)(z + e) ∈ K for any z ∈ B, where B is the set defined in Lemma 2.5. From Lemma 2.8(b), it follows that
(fsoc)0(x)(z + e) = [(b1(x) − a0(x)) hxe, zi + c1(x)] xe+ a0(x)z + [b1(x) + c1(x)hxe, zi] e.
This means that (fsoc)0(x)(z + e) ∈ K for any z ∈ B if and only if
b1(x) + c1(x)hxe, zi ≥ 0, (2.56) [b1(x) + c1(x)hxe, zi]2 ≥
(b1(x) − a0(x)) hxe, zi + c1(x)xe+ a0(x)z
2. (2.57) By Lemma 2.5(a), we know that (2.56) holds for any z ∈ B if and only if b1(x) ≥ |c1(x)|.
Since by a simple computation the inequality in (2.57) can be simplified as b1(x)2− c1(x)2− a0(x)2kzk2 ≥b1(x)2− c1(x)2− a0(x)2 hz, xei2, applying Lemma 2.5(b) yields that (2.57) holds for any z ∈ B if and only if
b1(x)2− c1(x)2 − a0(x)2 ≥ 0.
This shows that (fsoc)0(x)(z + e) ∈ K for any z ∈ B if and only if
b1(x) ≥ |c1(x)| and b1(x)2− c1(x)2− a0(x)2 ≥ 0. (2.58) The first condition in (2.58) is equivalent to b1(x) ≥ 0, b1(x) − c1(x) ≥ 0 and b1(x) + c1(x) ≥ 0, which, by the expressions of b1(x) and c1(x) and the arbitrariness of x, is equivalent to f0(τ ) ≥ 0 for all τ ∈ J ; whereas the second condition in (2.58) is equivalent to
f0(τ1)f0(τ2) − f (τ2) − f (τ1) τ2− τ1
2
≥ 0, ∀τ1, τ2 ∈ J.
The two sides show that the inequality in (2.55) holds for all τ1, τ2 ∈ J.
Conversely, if the inequality in (2.55) holds for all τ1, τ2 ∈ J, then from the arguments above we have (fsoc)0(x)(z + e) ∈ K for any x = xe+ x0e ∈ S and z ∈ B. This implies that (fsoc)0(x)h ∈ K for any x ∈ S and h ∈ K. By Proposition 2.19, f is SOC-monotone.
Propositions 2.19 and 2.21 provide the characterizations for continuously differentiable SOC-monotone functions. When f does not belong to C1(J ), one may check the SOC-monotonicity of f by combining the following proposition with Propositions 2.19 and 2.21.
Proposition 2.22. Let f : J → IR be a continuous function on the open interval J , and fε be its regularization defined by (2.50). Then, f is SOC-monotone if and only if fε is SOC-monotone on Jε for every sufficiently small ε > 0, where Jε := (a + ε, b − ε) for J = (a, b).
Proof. Throughout the proof, for every sufficiently small ε > 0, we let Sεbe the set of all x ∈ H whose spectral values λ1(x), λ2(x) belong to Jε. Assume that fεis SOC-monotone on Jεfor every sufficiently small ε > 0. Let x, y be arbitrary vectors from S with x Kn y.
Then, for any sufficiently small ε > 0, we have x + εe, y + εe ∈ Sε and x + εe Kn y + εe.
Using the SOC-monotonicity of fε on Jε yields that fεsoc(x + εe) Kn fεsoc(y + εe).
Taking the limit ε → 0 and using the convergence of fεsoc(x) → fsoc(x) and the continuity of fsoc on S implied by Lemma 2.8(a), we readily obtain that fsoc(x) Kn fsoc(y). This shows that f is SOC-monotone.
Now assume that f is SOC-monotone. Let ε > 0 be an arbitrary sufficiently small real number. Fix any x, y ∈ Sε with x Kn y. Then, for all t ∈ [−1, 1], we have x − tεe, y − tεe ∈ S and x − tεe Kn y − tεe. Therefore, fsoc(x − tεe) Kn fsoc(y − tεe), which is equivalent to
f (λ1− tε) + f (λ2− tε)
2 −f (µ1− tε) + f (µ2− tε) 2
≥
f (λ1− tε) − f (λ2− tε)
2 xe−f (µ1− tε) − f (µ2− tε)
2 ye
.
Together with the definition of fε, it then follows that fε(λ1) + fε(λ2)
2 − fε(µ1) + fε(µ2) 2
=
Z f (λ1 − tε) + f (λ2− tε)
2 − f (µ1− tε) + f (µ2− tε) 2
ϕ(t)dt
≥ Z
f (λ1− ε) − f (λ2− ε)
2 xe− f (µ1− ε) − f (µ2− ε)
2 ye
ϕ(t)dt
≥
Z f (λ1 − ε) − f (λ2− ε)
2 xe− f (µ1− ε) − f (µ2− ε)
2 ye
ϕ(t)dt
=
fε(λ1) − fε(λ2)
2 xe− fε(µ1) − fε(µ2)
2 ye
.
By the definition of fεsoc, this shows that fεsoc(x) Kn fεsoc(y), i.e., fε is SOC-monotone.
From Proposition 2.21 and [21, Theorem V. 3.4], f ∈ C1(J ) is SOC-monotone if and only if it is matrix monotone of order 2. When the continuous f is not in the class C1(J ), the result also holds due to Proposition 2.22 and the fact that f is matrix monotone of order n if and only if fε is matrix monotone of order n. Thus, we have the following main result.
Proposition 2.23. The set of continuous SOC-monotone functions on the open interval J coincides with that of continuous matrix monotone functions of order 2 on J .
Remark 2.2. Combining Proposition 2.23 with L¨owner’s Theorem [103] shows that if f : J → IR is a continuous SOC-monotone function on the open interval J , then f ∈ C1(J ).
We now move to the characterizations of SOC-convex functions, and shows that the continuous f is SOC-convex if and only if it is matrix convex of order 2. First, for the first-order differentiable SOC-convex functions, we have the following characterizations.
Proposition 2.24. Assume that f ∈ C1(J ) with J being an open interval in IR. Then, the following hold.
(a) f is SOC-convex if and only if for any x, y ∈ S,
fsoc(y) − fsoc(x) − (fsoc)0(x)(y − x) Kn 0.
(b) If f is SOC-convex, then (f0)soc is a monotone function on S.
Proof. (a) By following the arguments as in [20, Proposition B.3(a)], the proof can be done easily. We omit the details.
(b) From part(a), it follows that for any x, y ∈ S,
fsoc(x) − fsoc(y) − (fsoc)0(y)(x − y) Kn 0, fsoc(y) − fsoc(x) − (fsoc)0(x)(y − x) Kn 0.
Adding the last two inequalities, we immediately obtain that
(fsoc)0(y) − (fsoc)0(x) (y − x) Kn 0.
Using the self-duality of K and Lemma 2.8(c) then yields
0 ≤e, (fsoc)0(y) − (fsoc)0(x) (y − x) = hy − x, (f0)soc(y) − (f0)soc(x)i . This shows that (f0)soc is monotone. The proof is complete.
To provide sufficient and necessary characterizations for twice differentiable SOC-convex functions, we need the following lemma that offers the second-order differential of fsoc.
Lemma 2.9. For any given f : J → IR with J open, let fsoc : S → H be defined by (1.9).
(a) fsoc is twice (continuously) differentiable on S if and only if f is twice (continuously) differentiable on J . Furthermore, when f is twice differentiable on J , for any given x = xe+ x0e ∈ S and u = ue+ u0e, v = ve+ v0e ∈ H, we have that
(fsoc)00(x)(u, v) = f00(x0)u0v0e + f00(x0)(u0ve+ v0ue) + f00(x0)hue, veie if xe = 0; and otherwise
(fsoc)00(x)(u, v) = (b2(x) − a1(x))u0hxe, veixe+ (c2(x) − 3d(x))hxe, ueihxe, veixe
+d(x)hue, veixe+ hxe, veiue+ hxe, ueive + c2(x)u0v0xe + b2(x) − a1(x)hxe, ueiv0xe+ a1(x) v0ue+ u0ve +b2(x)u0v0e + c2(x)v0hxe, uei + u0hxe, veie
+a1(x)hue, veie + (b2(x) − a1(x))hxe, ueihxe, veie, (2.59) where
c2(x) = f00(λ2(x)) − f00(λ1(x))
2 , b2(x) = f00(λ2(x)) + f00(λ1(x))
2 ,
a1(x) = f0(λ2(x)) − f0(λ1(x))
λ2(x) − λ1(x) , d(x) = b1(x) − a0(x) kxek .
(b) If f is twice differentiable on J , then for any given x ∈ S and u, v ∈ H, (fsoc)00(x)(u, v) = (fsoc)00(x)(v, u),
hu, (fsoc)00(x)(u, v)i = hv, (fsoc)00(x)(u, u)i.
Proof. (a) The first part is direct by the given conditions and Lemma 2.8(b), and we only need to derive the differential formula. Fix any u = ue+ u0e, v = ve+ v0e ∈ H. We first consider the case where xe = 0. Without loss of generality, assume that ue6= 0. For any sufficiently small t > 0, using Lemma 2.8(b) and x + tu = (x0+ tu0) + tue, we have that
(fsoc)0(x + tu)v = [b1(x + tu) − a0(x + tu)] hue, veiue+ c1(x + tu)v0ue +a0(x + tu)ve+ b1(x + tu)v0e + c1(x + tu)hue, veie.
In addition, from Lemma 2.8(b), we also have that (fsoc)0(x)v = f0(x0)v0e + f0(x0)ve. Using the definition of b1(x) and a0(x), and the differentiability of f0 on J , it follows that
limt→0
b1(x + tu)v0e − f0(x0)v0e
t = f00(x0)u0v0e, limt→0
a0(x + tu)ve− f0(x0)ve
t = f00(x0)u0ve, limt→0
b1(x + tu) − a0(x + tu)
t = 0,
limt→0
c1(x + tu)
t = f00(x0)kuek.
Using the above four limits, it is not hard to obtain that
(fsoc)00(x)(u, v) = lim
t→0
(fsoc)0(x + tu)v − (fsoc)0(x)v t
= f00(x0)u0v0e + f00(x0)(u0ve+ v0ue) + f00(x0)hue, veie.
We next consider the case where xe 6= 0. From Lemma 2.8(b), it follows that (fsoc)0(x)v = (b1(x) − a0(x)) hxe, vei xe+ c1(x)v0xe
+a0(x)ve+ b1(x)v0e + c1(x) hxe, vei e, which in turn implies that
(fsoc)00(x)(u, v) = [(b1(x) − a0(x)) hxe, vei xe]0u + [c1(x)v0xe]0u
+ [a0(x)ve+ b1(x)v0e]0u + [c1(x) hxe, vei e]0u. (2.60)
By the expressions of a0(x), b1(x) and c1(x) and equations (2.52)-(2.53), we calculate that (b1(x))0u = f00(λ2(x)) [u0+ hxe, uei]
2 +f00(λ1(x)) [u0− hxe, uei]
2
= b2(x)u0+ c2(x)hxe, uei, (c1(x))0u = c2(x)u0+ b2(x)hxe, uei, (a0(x))0u = f0(λ2(x)) − f0(λ1(x))
λ2(x) − λ1(x) u0+ b1(x) − a0(x)
kxek hxe, uei
= a1(x)u0+ d(x)hxe, uei, (hxe, vei)0u =
1
kxekue− hxe, uei kxek xe, ve
.
Using these equalities and noting that a1(x) = c1(x)/kxek, we obtain that h
b1(x) − a0(x)hxe, veixe
i0
u = h
b2(x) − a1(x)u0+ (c2(x) − d(x))hxe, ueii
hxe, veixe
+(b1(x) − a0(x))
1
kxekue− hxe, uei kxek xe, ve
xe + (b1(x) − a0(x)) hxe, vei
1
kxekue− hxe, uei kxek xe
=h
(b2(x) − a1(x))u0+ (c2(x) − d(x))hxe, ueii
hxe, veixe +d(x)hue, veixe− 2d(x)hxe, veihxe, ueixe+ d(x)hxe, veiue; h
a0(x)ve+ b1(x)v0ei0
u =h
a1(x)u0+ d(x)hxe, ueii ve+h
b2(x)u0 + c2(x)hxe, ueii v0e;
h
c1(x)v0xei0
u =h
c2(x)u0 + b2(x)hxe, ueii
v0xe+ c1(x)v0ue− hxe, ueixe kxek
= h
c2(x)u0+ b2(x)hxe, ueii
v0xe+ a1(x)v0
h
ue− hxe, ueixe
i
; and
h
c1(x)hxe, veiei0
u =h
c2(x)u0 + b2(x)hxe, ueii
hxe, veie + c1(x) ue− hxe, ueixe kxek , ve
e
= c2(x)u0hxe, veie + b2(x) − a1(x)hxe, uei hxe, vei e + a1(x)hue, veie.
Adding the equalities above and using equation (2.60) yields the formula in (2.59).
(b) By the formula in part (a), a simple computation yields the desired result. Proposition 2.25. Assume that f ∈ C2(J ) with J being an open interval in IR. Then, the following hold.
(a) f is SOC-convex if and only if for any x ∈ S and h ∈ H, (fsoc)00(x)(h, h) ∈ K.
(b) f is SOC-convex if and only if f is convex and for any τ1, τ2 ∈ J,
f00(τ2) 2
f (τ2) − f (τ1) − f0(τ1)(τ2− τ1) (τ2− τ1)2
≥ f (τ1) − f (τ2) − f0(τ2)(τ1− τ2) (τ2− τ1)2
2
. (2.61)
(c) f is SOC-convex if and only if f is convex and for any τ1, τ2 ∈ J, 1
4f00(τ1)f00(τ2) (2.62)
≥ f (τ2) − f (τ1) − f0(τ1)(τ2− τ1) (τ2− τ1)2
f (τ1) − f (τ2) − f0(τ2)(τ1− τ2) (τ2− τ1)2
.
(d) f is SOC-convex if and only if for any τ1, τ2 ∈ J and s = τ1, τ2,
f[2](τ2, s, τ2) f[2](τ2, s, τ1) f[2](τ1, s, τ2) f[2](τ1, s, τ1)
O.
Proof. (a) Suppose that f is SOC-convex. Since fsoc is twice continuously differentiable by Lemma 2.9(a), we have for any given x ∈ S, h ∈ H and sufficiently small t > 0,
fsoc(x + th) = fsoc(x) + t(fsoc)0(x)h + 1
2t2(fsoc)00(x)(h, h) + o(t2).
Applying Proposition 2.24(a) yields that 12(fsoc)00(x)(h, h) + o(t2)/t2 Kn 0. Taking the limit t ↓ 0, we obtain (fsoc)00(x)(h, h) ∈ K. Conversely, fix any z ∈ K and x, y ∈ S.
Applying the mean-value theorem for the twice continuously differentiable hfsoc(·), zi at x, we have
hfsoc(y), zi = hfsoc(x), zi + h(fsoc)0(x)(y − x), zi +1
2h(fsoc)00(x + t1(y − x))(y − x, y − x), zi
for some t1 ∈ (0, 1). Since x + t1(y − x) ∈ S, the given assumption implies that hfsoc(y) − fsoc(x) − (fsoc)0(x)(y − x), zi ≥ 0.
This, by the arbitrariness of z in K, implies that fsoc(y)−fsoc(x)−(fsoc)0(x)(y −x) Kn 0.
From Proposition 2.24(a), it then follows that f is SOC-convex.
(b) By part (a), it suffices to prove that (fsoc)00(x)(h, h) ∈ K for any x ∈ S and h ∈ H if and only if f is convex and (2.61) holds. Fix any x = xe+ x0e ∈ S. By the continuity of
(fsoc)00(x), we may assume that xe 6= 0. From Lemma 2.9(a), for any h = he+ h0e ∈ H, (fsoc)00(x)(h, h) =h
c2(x) − 3d(x)hxe, hei2+ 2 b2(x) − a1(x)h0hxe, heii xe +h
c2(x)h20+ d(x)khek2i xe+h
2a1(x)h0+ 2d(x)hxe, heii he +
h
2c2(x)h0hxe, hei + b2(x)h20 + a1(x)khek2i e +(b2(x) − a1(x))hxe, hei2e.
Therefore, (fsoc)00(x)(h, h) ∈ K if and only if the following two inequalities hold:
b2(x) h20+ hxe, hei2 + 2c2(x)h0hxe, hei + a1(x) khek2− hxe, hei2 ≥ 0 (2.63) and
h
b2(x) h20+ hxe, hei2 + 2c2(x)h0hxe, hei + a1(x) khek2− hxe, hei2i2
≥
c2(x)h20 + d(x)khek2 xe+ 2 (b2(x) − a1(x)) h0hxe, heixe + (c2(x) − 3d(x)) hxe, hei2xe+ 2 (a1(x)h0+ d(x)hxe, hei) he
2. (2.64) Observe that the left-hand side of (2.63) can be rewritten as
f00(λ2(x))(h0+ hxe, hei)2
2 +f00(λ1(x))(h0− hxe, hei)2
2 + a1(x)(khek2− hxe, hei2).
From Lemma 2.6, it then follows that (2.63) holds for all h = he+ h0e ∈ H if and only if f00(λ1(x)) ≥ 0, f00(λ2(x)) ≥ 0 and a1(x) ≥ 0. (2.65) In addition, by the definition of b2(x), c2(x) and a1(x), the left-hand side of (2.64) equals
h
f00(λ2(x))µ2(h)2+ f00(λ1(x))µ1(h)2+ a1(x)µ(h)2i2
, (2.66)
where µ1(h), µ2(h) and µ(h) are defined as in Lemma 2.7 with ue replaced by xe. In the following, we use µ1, µ2 and µ to represent µ1(h), µ2(h) and µ(h) respectively. Note that the sum of the first three terms in k · k2 on the right-hand side of (2.64) equals
1
2(c2(x) + b2(x) − a1(x)) (h0+ hxe, hei)2xe
+1
2(c2(x) − b2(x) + a1(x)) (h0− hxe, hei)2xe
+d(x) khek2 − hxe, hei2 xe− 2d(x)hxe, hei2xe
= f00(λ2(x))µ22xe− f00(λ1(x))µ21xe− a1(x) + d(x)µ22xe + a1(x) − d(x)µ21xe+ 2d(x)µ2µ1xe+ d(x)µ2xe
=: E(x, h)xe,
where (µ2− µ1)2 = 2hxe, hei2 is used for the equality, while the last term is (a1(x) − d(x)) (h0− hxe, hei) he+ (a1(x) + d(x)) (h0+ hxe, hei) he
= √
2 (a1(x) − d(x)) µ1he+√
2 (a1(x) + d(x)) µ2he. Thus, we calculate that the right-hand side of (2.64) equals
E(x, h)2+ 2h
a1(x) − d(x)µ1 + a1(x) + d(x)µ2i2
khek2 +2√
2E(x, h)a1(x) − d(x)µ1hxe, hei + 2√
2E(x, h)a1(x) + d(x)µ2hxe, hei
= E(x, h)2+ 2h
a1(x) − d(x)µ1 + a1(x) + d(x)µ2
i2
µ2+ (µ2− µ1)2 2
+2E(x, h)(µ2− µ1)h
(a1(x) − d(x))µ1+ (a1(x) + d(x))µ2i
= h
E(x, h) + (µ2− µ1) [(a1(x) − d(x))µ1+ (a1(x) + d(x))µ2]i2
+2h
(a1(x) − d(x)) µ1+ (a1(x) + d(x)) µ2i2
µ2, (2.67)
where the expressions of µ1, µ2 and µ are used for the first equality. Now substituting the expression of E(x, h) into (2.67) yields that the right-hand side of (2.67) equals h
f00(λ2(x))µ22− f00(λ1(x))µ21+ d(x)µ2i2
+ 2h
(a1(x) − d(x)) µ1+ (a1(x) + d(x)) µ2i2
µ2. Together with equation (2.66), it follows that (2.64) is equivalent to
4f00(λ1(x))f00(λ2(x))µ21µ22+ 2 (a1(x) − d(x)) f00(λ2(x))µ22µ2 +2 (a1(x) + d(x)) f00(λ1(x))µ21µ2+ a1(x)2− d(x)2 µ4
−2 [(a1(x) − d(x)) µ1+ (a1(x) + d(x)) µ2]2µ2 ≥ 0.
By Lemma 2.7, this inequality holds for any h = he+ h0e ∈ H if and only if a1(x)2− d(x)2 ≥ 0, f00(λ2(x)) a1(x) − d(x) ≥ a1(x) + d(x)2
, f00(λ1(x)) a1(x) + d(x) ≥ a1(x) − d(x)2
, which, by the expression of a1(x) and d(x), are respectively equivalent to
f (λ2) − f (λ1) − f0(λ1)(λ2− λ1)
(λ2− λ1)2 · f (λ1) − f (λ2) − f0(λ2)(λ1− λ2) (λ2 − λ1)2 ≥ 0, f00(λ2)
2
f (λ2) − f (λ1) − f0(λ1)(λ2− λ1)
(λ2− λ1)2 ≥ f (λ1) − f (λ2) − f0(λ2)(λ1− λ2) (λ2− λ1)2
2
, (2.68) f00(λ1)
2
f (λ1) − f (λ2) − f0(λ2)(λ1− λ2)
(λ2− λ1)2 ≥ f (λ2) − f (λ1) − f0(λ1)(λ2− λ1) (λ2− λ1)2
2
,
where λ1 = λ1(x) and λ2 = λ2(x). Summing up the discussions above, f is SOC-convex if and only if (2.65) and (2.68) hold. In view of the arbitrariness of x, we have that f is SOC-convex if and only if f is convex and (2.61) holds.
(c) It suffices to prove that (2.61) is equivalent to (2.62). Clearly, (2.61) implies (2.62).
We next prove that (2.62) implies (2.61). Fixing any τ2 ∈ J, we consider g(t) : J → IR defined by
g(t) = f00(τ2)
2 [f (τ2) − f (t) − f0(t)(τ2− t)] − [f (t) − f (τ2) − f0(τ2)(t − τ2)]2 (t − τ2)2
if t 6= τ2, and otherwise g(τ2) = 0. From the proof of [68, Theorem 2.3], we know that (2.61) implies that g(t) attains its global minimum at t = τ2. Consequently, (2.61) follows.
(d) The result is immediate by part(b) and the definition of f[2] given as in (2.46). Propositions 2.24 and 2.25 provide the characterizations for continuously differentiable SOC-convex functions, which extend the corresponding results of [44, Section 4]. When f is not continuously differentiable, the following proposition shows that one may check the SOC-convexity of f by checking that of its regularization fε. Since the proof can be done easily by following that of Proposition 2.22, we omit the details.
Proposition 2.26. Let f : J → IR be a continuous function on the open interval J , and fε be its regularization defined by (2.50). Then, f is SOC-convex if and only if fε is SOC-convex on Jε for every sufficiently small ε > 0, where Jε := (a + ε, b − ε) for J = (a, b).
By [68, Theorem 2.3] and Proposition 2.26, we can obtain the below consequence immediately.
Proposition 2.27. The set of continuous SOC-convex functions on the open interval J coincides with that of continuous matrix convex functions of order 2 on J .
Remark 2.3. Combining Proposition 2.27 with Kraus’ theorem [92] shows that if f : J → IR is a continuous SOC-convex function, then f ∈ C2(J ).
We establish another sufficient and necessary characterization for twice continuously differentiable SOC-convex functions f by the differential operator (fsoc)0.
Proposition 2.28. Let f ∈ C2(J ) with J being an open interval in IR. Then, f is SOC-convex if and only if
x Kn y =⇒ (fsoc)0(x) − (fsoc)0(y) ≥ 0, ∀x, y ∈ S. (2.69)
Proof. Suppose that f is SOC-convex. Fix any x, y ∈ S with x Kn y, and h ∈ H.
Since fsoc is twice continuously differentiable by Lemma 2.9(a), applying the mean-value theorem for the twice continuously differentiable hh, (fsoc)0(·)hi at y, we have
h, (fsoc)0(x) − (fsoc)0(y) h = h, (fsoc)00(y + t1(x − y))(x − y, h)
= x − y, (fsoc)00(y + t1(x − y))(h, h)
(2.70) for some t1 ∈ (0, 1), where Lemma 2.9(b) is used for the second equality. Noting that y + t1(x − y) ∈ S and f is SOC-convex, from Proposition 2.25(a) we have
(fsoc)00(y + t1(x − y))(h, h) ∈ K.
This, together with x−y ∈ K, yields thatx − y, (fsoc)00(x + t1(x − y))(h, h) ≥ 0. Then, from (2.70) and the arbitrariness of h, we have (fsoc)0(x) − (fsoc)0(y) ≥ 0.
Conversely, assume that the implication in (2.69) holds for any x, y ∈ S. For any fixed u ∈ K, clearly, x + tu Kn x for all t > 0. Consequently, for any h ∈ H, we have
h, (fsoc)0(x + tu) − (fsoc)0(x) h ≥ 0.
Note that (fsoc)0(x) is continuously differentiable. The last inequality implies that 0 ≤h, (fsoc)00(x)(u, h) = hu, (fsoc)00(x)(h, h)i.
By the self-duality of K and the arbitrariness of u in K, this means that (fsoc)00(x)(h, h) ∈ K. Together with Proposition 2.25(a), it follows that f is SOC-convex.
Example 2.13. The following functions are SOC-monotone.
(a) The function f (t) = tr is SOC-monotone on [0, ∞) if and only if 0 ≤ r ≤ 1.
(b) The function f (t) = −t−r is SOC-monotone on (0, ∞) if and only if 0 ≤ r ≤ 1.
(c) The function f (t) = ln(t) is SOC-monotone on (0, ∞).
(d) The function f (t) = − cot(t) is SOC-monotone on (0, π).
(e) The function f (t) = c+tt with c ≥ 0 is SOC-monotone on (−∞, c) and (c, ∞).
(f ) The function f (t) = c−tt with c ≥ 0 is SOC-monotone on (−∞, c) and (c, ∞).
Example 2.14. The following functions are SOC-convex.
(a) The function f (t) = tr with r ≥ 0 is SOC-convex on [0, ∞) if and only if r ∈ [1, 2].
Particularly, f (t) = t2 is SOC-convex on IR.
(b) The function f (t) = t−r with r > 0 is SOC-convex on (0, ∞) if and only if r ∈ [0, 1].
(c) The function f (t) = tr with r ≥ 0 is SOC-concave if and only if r ∈ [0, 1].
(d) The entropy function f (t) = t ln t is SOC-convex on [0, ∞).
(e) The logarithmic function f (t) = − ln t is SOC-convex on (0, ∞).
(f ) The function f (t) = t−σt with σ ≥ 0 is SOC-convex on (σ, ∞).
(g) The function f (t) = −t+σt with σ ≥ 0 is SOC-convex on (−σ, ∞).
(h) The function f (t) = 1−tt2 is SOC-convex on (−1, 1).
Next we illustrate the applications of the SOC-monotonicity and SOC-convexity of certain functions in establishing some important inequalities. For example, by the SOC-monotonicity of −t−r and tr with r ∈ [0, 1], one can get the order-reversing inequality and the L¨owner-Heinz inequality, and by the SOC-monotonicity and SOC-concavity of
−t−1, one may obtain the general harmonic-arithmetic mean inequality.
Proposition 2.29. For any x, y ∈ H and 0 ≤ r ≤ 1, the following inequalities hold:
(a) y−r Kn x−r if x Kn y Kn 0;
(b) xrKn yr if x Kn y Kn 0;
(c) [βx−1+ (1 − β)y−1]−1 K βx + (1 − β)y for any x, y Kn 0 and β ∈ (0, 1).
From the second inequality of Proposition 2.29, we particularly have the following result which generalizes [63, Eq.(3.9)], and is often used when analyzing the properties of the generalized Fischer-Burmeister (FB) SOC complementarity function φp(x, y) :=
(|x|p+ |y|p)1/p− (x + y). To know more about this function φp, please refer to [121].
Proposition 2.30. For any x, y ∈ H, let z(x, y) := (|x|p+ |y|p)1/p for any p > 1. Then, z(x, y) Kn |x| Kn x and z(x, y) Kn |y| Kn y.
The SOC-convexity can also be used to establish some matrix inequalities. From (2.51) we see that, when H reduces to the n-dimensional Euclidean space Rn, the differ-ential operator (fsoc)0(x) becomes the following n × n symmetric matrix:
b1(x) c1(x)xTe
c1(x)xe a0(x)I + (b1(x) − a0(x))xexTe
where a0(x), b1(x) and c1(x) are same as before, and I is an identity matrix. Thus, from Proposition 2.28, we have the following result which is hard to get by direct calculation.
Proposition 2.31. If f ∈ C2(J ) is SOC-convex on the open interval J , then for any x, y ∈ S with x Kn y,
b1(x) c1(x)xTe
c1(x)xe a0(x)I + (b1(x) − a0(x))xexTe
b1(y) c1(y)xTe
c1(y)xe a0(y)I + (b1(y) − a0(y))xexTe
. Particularly, when f (t) = t2 (t ∈ R), this conclusion reduces to the following implication
x Kn y =⇒ x0 xTe xe x0I
y0 yTe ye y0I
.
As mentioned earlier, with certain SOC-monotone and SOC-convex functions, one can easily establish some determinant inequalities. Below is a stronger version of Proposition 1.8(b).
Proposition 2.32. For any x, y ∈ K and any real number p ≥ 1, it holds that pp
det(x + y) ≥ 22p−2 pp
det(x) +pp
det(y) .
Proof. In light of Example 2.12(b), we see that f (t) = t1/p is SOC-concave on [0, ∞), which says
x + y 2
1/p
Kn x1/p+ y1/p
2 .
This together with the fact that det(x) ≥ det(y) whenever x Kn y Kn 0 implies
2−2pdet √p
x + y = det r x + yp 2
!
≥ det
√p
x +√p y 2
≥ det(√p
x) + det(√p y)
4 ,
where det(x + y) ≥ det(x) + det(y) for x, y ∈ K is used for the last inequality. In addition, by the definition of det(x), it is clear that det (√p
x) = pdet(x). Thus, from thep last equation, we obtain the desired inequality. The proof is complete.
Comparing Example 2.13 with Example 2.14, we observe that there are some rela-tions between SOC-monotone and SOC-convex funcrela-tions. For example, f (t) = t ln t and f (t) = − ln t are SOC-convex on (0, ∞), and its derivative functions are SOC-monotone on (0, ∞). This is similar to the case for matrix-convex and matrix-monotone func-tions. However, it is worthwhile to point out that they can not inherit all relations between matrix-convex and matrix-monotone functions, since the class of continuous SOC-monotone (SOC-convex) functions coincides with the class of continuous monotone (convex) functions of order 2 only, and there exist gaps between matrix-monotone (matrix-convex) functions of different orders (see [69, 113]). Then, a question occurs to us: which relations for matrix-convex and matrix-monotone functions still hold for SOC-convex and SOC-monotone functions.
Lemma 2.10. Assume that f : J → IR is three times differentiable on the open interval J . Then, f is a non-constant SOC-monotone function if and only if f0 is strictly positive and (f0)−1/2 is concave.
Proof. “⇐”. Clearly, f is a non-constant function. Also, by [68, Proposition 2.2], we have f (t2) − f (t1)
t2− t1
≤p
f0(t2)f0(t1), ∀t2, t1 ∈ J.
This, by the strict positivity of f0 and Proposition 2.21, shows that f is SOC-monotone.
“⇒”. The result is direct by [58, Theorem III] and Proposition 2.23.
Using Lemma 2.10, we may verify that SOC-monotone and SOC-convex functions inherit the following relation of matrix-monotone and matrix-convex functions.
Proposition 2.33. If f : J → IR is a continuous SOC-monotone function, then the function g(t) =Rt
af (s)ds with some a ∈ J is SOC-convex.
Proof. It suffices to consider the case where f is a non-constant SOC-monotone function.
Due to Proposition 2.22, we may assume f ∈ C3(J ). By Lemma 2.10, f0(t) > 0 for all t ∈ J and (f0)−1/2 is concave. Since g ∈ C4(J ) and g00(t) = f0(t) > 0 for all t ∈ J , in order to prove that g is SOC-convex, we only need to argue
g00(t)g(4)(t)
48 ≥ g(3)(t)2
36 ⇐⇒ f0(t)f(3)(t)
48 ≥ [f00(t)]2
36 , ∀t ∈ J. (2.71) Since (f0)−1/2 is concave, its second-order derivative is nonpositive. From this, we have
1
32(f00(t))2 ≤ 1
48f0(t)f(3)(t), ∀t ∈ J, which implies the inequality (2.71). The proof is complete.
Similar to matrix-monotone and matrix-convex functions, the converse of Proposition 2.33 does not hold. For example, f (t) = 1−tt2 on (−1, 1) is SOC-convex by Example 2.14(g), but its derivative g0(t) = (1−t)1 2 − 1 is not SOC-monotone by Proposition 2.21.
As a consequence of Proposition 2.33 and Proposition 2.28, we have the following result.
Proposition 2.34. Let f ∈ C2(J ). If f0 is SOC-monotone, then f is SOC-convex. This is equivalent to saying that for any x, y ∈ S with x Kn y,
(f0)soc(x) Kn (f0)soc(y) =⇒ (fsoc)0(x) − (fsoc)0(y) 0.
From [21, Theorem V. 2.5], we know that a continuous function f mapping [0, ∞) into itself is matrix-monotone if and only if it is matrix-concave. However, for such f we cannot prove that f is SOC-concave when it is SOC-monotone, but f is SOC-concave under a little stronger condition than SOC-monotonicity, i.e., the matrix-monotonicity of order 4.
Proposition 2.35. Let f : [0, ∞) → [0, ∞) be continuous. If f is matrix-monotone of order 4, then f is SOC-concave.
Proof. By [107, Theorem 2.1], if f is continuous and matrix-monotone of order 2n, then f is matrix-concave of order n. This together with Proposition 2.27 gives the result. Note that Proposition 2.35 verifies Conjecture 2.2 partially and also can be viewed as the converse of Proposition 2.8. From [21], we know that the functions in Example 2.13(a)-(c) are all matrix-monotone, and so they are SOC-concave by Proposition 2.35(b).
In addition, using Proposition 2.35(b) and noting that −t−1 (t > 0) is SOC-monotone and SOC-concave on (0, ∞), we readily have the following consequence.
Proposition 2.36. Let f : (0, ∞) → (0, ∞) be continuous. If f is matrix-monotone of order 4, then the function g(t) = f (t)1 is SOC-convex.
Proposition 2.37. Let f be a continuous real function on the interval [0, α). If f is SOC-convex, then the indefinite integral of f (t)t is also SOC-convex.
Proof. The result follows directly by [114, Proposition 2.7] and Proposition 2.27. For a continuous real function f on the interval [0, α), [21, Theorem V. 2.9] states that the following two conditions are equivalent:
(i) f is matrix-convex and f (0) ≤ 0;
(ii) The function g(t) = f (t)t is matrix-monotone on (0, α).
At the end of this section, let us look back to Conjecture 2.1. By looking into Example 2.13(a)-(c) and (f)-(g), we find that these functions are continuous, nondecreasing and concave. Then, one naturally asks whether such functions are SOC-monotone or not, which recalls Conjecture 2.1(b). The following counterexample shows that Conjecture 2.1(b) does not hold generally. To the contrast, Conjecture 2.1(a) remains open.
Example 2.15. Let f : (0, ∞) → IR be f (t) = −t ln t + t, if t ∈ (0, 1).
1, if t ∈ [1, +∞). Then, the function f (t) is not SOC-monotone.
Solution. This function is continuously differentiable, nondecreasing and concave on (0, +∞). However, letting t1 = 0.1 and t2 = 3,
f0(t1)f0(t2) − f (t1) − f (t2) t1− t2
2
= − −t1ln t1+ t1− 1 t1− t2
2
= −0.0533.
By Proposition 2.21, we know that the function f is not SOC-monotone.
Chapter 3
Algorithmic Applications
In this Chapter, we will see details about how the characterizations established in Chap-ter 2 be applied in real algorithms. In particular, the SOC-convexity are often involved in the solution methods of convex SOCPs; for example, the proximal-like methods. We present three types of proximal-like algorithms, and refer the readers to [115, 116, 118]
for their numerical performance.