are well-defined (note this is true only when n = 2) and coincide with the definition (4.7), (4.8). Then, the following inequalities
x ∧ y
K2 H(x, y)
K2 G(x, y)
K2 L(x, y)
K2 A(x, y)
K2 Q(x, y)
K2 x ∨ y hold as well.
By applying Proposition 1.1(a), we immediately obtain one trace inequality for SOC mean.
Proposition 4.5. Let A(x, y), H(x, y), G(x, y) and L(x, y) be defined as in (4.3)-(4.4), (4.7)-(4.8), respectively. For any x Kn 0, y Kn 0, there holds
tr(x ∧ y) ≤ tr(H(x, y)) ≤ tr(G(x, y)) ≤ tr(L(x, y)) ≤ tr(A(x, y)) ≤ tr(x ∨ y).
In 1995, Ando [3] showed the singular value version of Young inequality that sj(AB) ≤ sj Ap
p + Bq q
for all 1 ≤ j ≤ n, (4.9) where A and B are positive definite matrices. Note that both positive semidefinite cone and second-order cone belong to symmetric cones [61]. It is natural to ask whether there is a similar version in the setting of second-order cone. First, in view of the classical Young inequality, one may conjecture that the Young inequality in the SOC setting is in form of
x ◦ y Kn xp p +yq
q .
However, this inequality does not hold in general (a counterexample is presented later).
Here “◦” is the Jordan product associated with second-order cone. Next, according to Ando’s inequality (4.9), we naively make another conjecture that the eigenvalue version of Young inequality in the SOC setting may look like
λj(x ◦ y) ≤ λj xp p +yq
q
, j = 1, 2. (4.10)
Although we believe it is true, it is very complicated to prove the inequality directly due to the algebraic structure of xpp + xqq. Eventually, we seek another variant and establish the SOC trace version of Young inequality. Accordingly, we further deduce the SOC trace versions of H¨older and Minkowski inequalities.
As mentioned earlier, one may conjecture that the Young inequality in the SOC setting is in form of
x ◦ y Kn xp p +yq
q .
However, this inequality does not hold in general. For example, taking p = 3, q = 32, x = (18,18, 0), and y = (18, 0,18), we obtain x3 = (1281 ,1281 , 0), y32 = (161, 0,161). Hence,
x ◦ y = 1 64, 1
64, 1 64
and x3 3 +y32
3 2
= 17 384, 1
384, 16 384
, which says
x3 3 + y32
3 2
− x ◦ y = 11 384, −5
384, 10 384
∈ K/ n.
In view of this and motivated by the Ando’s singular value version of Young inequality as in (4.9), we turn to derive the eigenvalue version of Young inequality in the setting of second-order cone. But, we do not succeed in achieving such type inequality. Instead, we consider the SOC trace version of the Young inequality.
Proposition 4.6. (Young inequality-Type I) For any x, y ∈ Kn, there holds tr(x ◦ y) ≤ tr xp
p + yq q
,
where 1 < p, q < ∞ and 1 p+ 1
q = 1.
Proof. First, we note x◦y = (x1y1+ hx2, y2i, x1y2+ y1x2) and denote xp p +yq
q := (w1, w2) where
w1 = λ1(x)p+ λ2(x)p
2p +λ1(y)q+ λ2(y)q
2q ,
w2 = λ2(x)p− λ1(x)p 2p
x2
kx2k +λ2(y)q− λ1(y)q 2q
y2 ky2k. Then, the desired result follows by
tr(x ◦ y) ≤ λ1(x)λ1(y) + λ2(x)λ2(y)
≤ λ1(x)p
p + λ1(y)q q
+ λ2(x)p
p + λ2(y)q q
= tr xp p +yq
q
,
where the last inequality is due to the Young inequality on real number setting. Remark 4.1. When p = q = 2, the Young inequality in Proposition 4.6 reduces to
2hx, yi = tr(x ◦ y) ≤ tr x2 2 + y2
2
= kxk2+ kyk2,
which is equivalent to 0 ≤ kx − yk2. As a matter of fact, for any x, y ∈ IRn, the inequality (x − y)2 Kn 0 always holds, which implies 2x ◦ y Kn x2+ y2. Therefore, by Proposition 1.1(a), we obtain tr(x ◦ y) ≤ tr
x2
2 + y22
as well.
We note that the classical Young inequality can be extended to nonnegative real numbers, that is,
|ab| = |a| · |b| ≤ |a|p p +|b|q
q , ∀a, b ∈ IR.
This motivates us to consider further generalization of the SOC trace version of Young inequality as in Proposition 4.6. However, |x|◦|y| and |x◦y| are unequal in general; and no relation between them. To see this, taking x = (√
2, 1, 1) ∈ K3 and y = (√
2, 1, −1) ∈ K3, yields x ◦ y = (2, 2√
2, 0) /∈ K3. In addition, it implies
|x| ◦ |y| = (2, 2√
2, 0) Kn (2√
2, 2, 0) = |x ◦ y|.
On the other hand, let x = (0, 1, 0), y = (0, 1, 1), which give |x| = (1, 0, 0), |y| = (√
2, 0, 0). However, we see that
|x ◦ y| = (1, 0, 0) Kn (√
2, 0, 0) = |x| ◦ |y|.
From these two examples, it also indicates that there is no relationship between tr(|x|◦|y|) and tr(|x ◦ y|). In other words, there are two possible extensions of Proposition 4.6:
tr(|x| ◦ |y|) ≤ tr |x|p
p + |y|q q
or tr(|x ◦ y|) ≤ tr |x|p
p +|y|q q
. Fortunately, these two types of generalizations are both true.
Proposition 4.7. (Young inequality-Type II) For any x, y ∈ IRn, there holds tr(|x| ◦ |y|) ≤ tr |x|p
p + |y|q q
,
where 1 < p, q < ∞ and 1 p+ 1
q = 1.
Proof. Following the proof of Proposition 4.6, we have tr(|x| ◦ |y|)
≤ λ1(|x|)λ1(|y|) + λ2(|x|)λ2(|y|)
= min
i {|λi(x)|} min
i {|λi(y)|} + max
i {|λi(x)|} max
i {|λi(y)|}
≤ (mini{|λi(x)|})p
p +(mini{|λi(y)|})q
q + (maxi{|λi(x)|})p
p +(maxi{|λi(y)|})q q
= |λ1(x)|p
p + |λ2(x)|p p
+ |λ1(y)|q
q + |λ2(y)|q q
= tr |x|p
p +|y|q q
,
where the last inequality holds by the Young inequality on real number setting. We point out that Proposition 4.7 is more general than Proposition 4.6 because it is true for all x, y ∈ IRn, not necessary restricted to x, y ∈ Kn. For real numbers, it is clear that ab ≤ |a| · |b|. It is natural to ask whether tr(x ◦ y) is less than tr(|x| ◦ |y|) or not.
Before establishing the relationship, we need the following technical lemma.
Lemma 4.3. For 0 Kn u Kn x and 0 Kn v Kn y, there holds 0 ≤ tr(u ◦ v) ≤ tr(x ◦ y).
Proof. Suppose 0 Kn u Kn x and 0 Kn v Kn y, we have tr(x ◦ y) − tr(u ◦ v)
= tr(x ◦ y − u ◦ v)
= tr(x ◦ y − x ◦ v + x ◦ v − u ◦ v)
= tr(x ◦ (y − v) + (x − u) ◦ v)
= tr(x ◦ (y − v)) + tr((x − u) ◦ v)
≥ 0,
where the inequality holds by Property 1.3(d).
Proposition 4.8. For any x, y ∈ IRn, there holds tr(x ◦ y) ≤ tr (|x| ◦ |y|) . Proof. For any x ∈ IRn, it can be expressed by x = [x]++ [x]−, and then
tr(x ◦ y) = tr(([x]++ [x]−) ◦ y)
= tr([x]+◦ y) + tr((−[x]−) ◦ (−y))
≤ tr([x]+◦ |y|) + tr((−[x]−) ◦ |y|)
= tr(([x]+− [x]−) ◦ |y|)
= tr(|x| ◦ |y|), where the inequality holds by Lemma 4.3.
There is some interpretation from geometric view for Proposition 4.8. More specifi-cally, by the definition of trace in second-order cone, we notice
tr(x ◦ y) = 2hx, yi = 2kxk · kyk cos θ
where θ is the angle between the vectors x and y. According to the definition of absolute value associated with second-order cone, we know the equality in Proposition 4.8 holds whenever x, y ∈ Kn or x, y ∈ −Kn. Otherwise, it can be observed that the angle between
|x| and |y| is smaller than the angle between x and y since the vector x, |x| and the axis of second-order cone are in a hyperplane.
Proposition 4.9. For any x, y ∈ IRn, the following inequalities hold.
(a) tr((x + y)2) ≤ tr((|x| + |y|)2), i.e., kx + yk ≤ k|x| + |y|k.
(b) tr((x − y)2) ≥ tr((|x| − |y|)2), i.e., kx − yk ≥ k|x| − |y|k.
Proof. (a) From Proposition 4.8, we have
tr (x + y)2 = tr x2+ 2x ◦ y + y2 ≤ tr |x|2 + 2|x| ◦ |y| + |y|2 = tr (|x| + |y|)2 .
This is equivalent to kx + yk2 ≤ k|x| + |y|k2, which implies kx + yk ≤ k|x| + |y|k.
(b) The proof is similar to part(a).
In contrast to Proposition 4.8, applying Proposition 1.1(a), it is clear that tr(x ◦ y) ≤ tr (|x ◦ y|) because x ◦ y Kn |x ◦ y|. In view of this, we try to achieve another extension as below.
Proposition 4.10. (Young inequality-Type III) For any x, y ∈ IRn, there holds
tr(|x ◦ y|) ≤ tr |x|p
p + |y|q q
,
where 1 < p, q < ∞ and 1 p+ 1
q = 1.
Proof. For analysis needs, we write x = (x1, x2) ∈ IR × IRn−1 and y = (y1, y2) ∈ IR × IRn−1. Note that if x ◦ y ∈ Kn∪ (−Kn), the desired inequality holds immediately by Proposition 4.7 and Proposition 4.8. Thus, it suffices to show the inequality holds for x ◦ y /∈ Kn∪ (−Kn). In fact, we only need to show the inequality for the case of x1 ≥ 0 and y1 ≥ 0. The other cases can be derived by suitable changing variable like
|x ◦ y| = | − (x ◦ y)| = |(−x) ◦ y| = |x ◦ (−y)| = |(−x) ◦ (−y)|.
To proceed, we first claim the following inequality
2kx1y2+ y1x2k ≤ |λ1(x)λ1(y)| + |λ2(x)λ2(y)|, (4.11) which is also equivalent to 4kx1y2 + y1x2k2 ≤ (|λ1(x)λ1(y)| + |λ2(x)λ2(y)|)2. Indeed, we observe that
4kx1y2+ y1x2k2 = 4 x21ky2k2+ y12kx2k2+ 2x1y1hx2, y2i . On the other hand,
(|λ1(x)λ1(y)| + |λ2(x)λ2(y)|)2
= [λ1(x)λ1(y)]2+ [λ2(x)λ2(y)]2 + 2 |λ1(x)λ1(y)λ2(x)λ2(y)|
= 2(x1y1+ kx2kky2k)2+ 2(x1ky2k + y1kx2k)2+ 2
x21− kx2k2
y12− ky2k2
= 2 x21y12+ kx2k2ky2k2+ x21ky2k2+ y12kx2k2 + 8x1y1kx2kky2k + 2
x21− kx2k2
y12− ky2k2 .
Therefore, we conclude that (4.11) is satisfied by checking (|λ1(x)λ1(y)| + |λ2(x)λ2(y)|)2− 4kx1y2 + y1x2k2
= 2 x21y12+ kx2k2ky2k2+ x21ky2k2+ y12kx2k2 + 8x1y1kx2kky2k + 2
x21− kx2k2
y12− ky2k2
− 4 x21ky2k2+ y12kx2k2+ 2x1y1hx2, y2i
= 2 x21y12+ kx2k2ky2k2− x21ky2k2− y12kx2k2 + 8x1y1(kx2kky2k − hx2, y2i) +2
x21− kx2k2
y12− ky2k2
= 2 x21 − kx2k2
y21− ky2k2 + 2
x21− kx2k2
y21 − ky2k2 +8x1y1(kx2kky2k − hx2, y2i)
≥ 0,
where the last inequality is due to the Cauchy-Schwarz Inequality.
Suppose that x ◦ y /∈ Kn∪ (−Kn). From the simple calculation, we have
|x ◦ y| =
kx1y2 + y1x2k, x1y1+ hx2, y2i
kx1y2+ y1x2k(x1y2+ y1x2)
, which says tr(|x ◦ y|) = 2kx1y2+ y1x2k. Using inequality (4.11), we obtain
tr(|x ◦ y|) ≤ |λ1(x)λ1(y)| + |λ2(x)λ2(y)|
≤ |λ1(x)|p
p + |λ1(y)|q q
+ |λ2(x)|p
p + |λ2(y)|q q
= tr |x|p
p +|y|q q
,
where the last inequality holds by the classical Young inequality on real number setting.
There also exist some trace versions of Young inequality in the setting of Euclidean Jordan algebra, please see [13, Theorem 23] and [78, Theorem 3.5-3.6]. Using the SOC trace versions of Young inequality, we can derive the SOC trace versions of H¨older in-equality as below.
Proposition 4.11. (H¨older inequality-Type I) For any x, y ∈ IRn, there holds tr(|x| ◦ |y|) ≤ [tr(|x|p)]1p · [tr(|x|q)]1q ,
where 1 < p, q < ∞ and 1 p+ 1
q = 1.
Proof. Let α = [tr(|x|p)]1p and β = [tr(|x|q)]1q. By Proposition 4.7, we have
tr |x|
α ◦ |y|
β
≤ tr ||x|α|p
p +||y|β|q q
!
= 1
ptr |x|p αp
+1
qtr |y|q βq
= 1 p +1
q = 1.
Therefore, we conclude that
tr(|x| ◦ |y|) ≤ α · β = [tr(|x|p)]1p · [tr(|x|q)]1q because α, β > 0.
Proposition 4.12. (H¨older inequality-Type II) For any x, y ∈ IRn, there holds tr(|x ◦ y|) ≤ [tr(|x|p)]1p · [tr(|x|q)]1q ,
where 1 < p, q < ∞ and 1 p+ 1
q = 1.
Proof. The proof is similar to Proposition 4.11 by using Proposition 4.10.
Remark 4.2. When p = q = 2, both inequalities in Proposition 4.11 and Proposition 4.12 deduce
2hx, yi
= tr(|x ◦ y|) ≤tr(|x|2)12
·tr(|x|2)12
= 2kxk · kyk, which is equivalent to the Cauchy-Schwarz inequality in IRn.
Next, we present the SOC trace version of Minkowski inequality.
Proposition 4.13. (Minkowski inequality) For any x = (x1, x2) ∈ IR × IRn−1 and y = (y1, y2) ∈ IR × IRn−1, and p > 1, there holds
[tr(|x + y|p)]1p ≤ [tr(|x|p)]1p + [tr(|y|p)]p1 .
Proof. We partition the proof into three parts. Let q > 1 and 1p + 1q = 1.
(i) For x + y ∈ Kn, we have |x + y| = x + y, then we have
tr(|x + y|p) = tr(|x + y| ◦ |x + y|p−1) = tr((x + y) ◦ |x + y|p−1)
= tr(x ◦ |x + y|p−1) + tr(y ◦ |x + y|p−1)
≤ [tr(|x|p)]1p ·tr(|x + y|(p−1)q)1q
+ [tr(|y|p)]1p ·tr(|x + y|(p−1)q)1q
=
[tr(|x|p)]p1 + [tr(|y|p)]p1
· [tr(|x + y|p)]1q , which implies [tr(|x + y|p)]1p ≤ [tr(|x|p)]1p + [tr(|y|p)]1p.
(ii) For x + y ∈ −Kn, we have |x + y| = −x − y, then we have tr(|x + y|p) = tr((−x) ◦ |x + y|p−1) + tr((−y) ◦ |x + y|p−1)
≤ [tr(|x|p)]1p ·tr(|x + y|(p−1)q)1q
+ [tr(|y|p)]1p ·tr(|x + y|(p−1)q)1q
=
[tr(|x|p)]p1 + [tr(|y|p)]p1
· [tr(|x + y|p)]1q ,
which also implies [tr(|x + y|p)]1p ≤ [tr(|x|p)]1p + [tr(|y|p)]1p.
(iii) For x + y /∈ Kn∪ (−Kn), we note that λ1(x + y) < 0 and λ2(x + y) > 0, which says,
|λ1(x + y)| = |x1+ y1− kx2+ y2k| = kx2+ y2k − x1− y1 ≤ kx2k + ky2k − x1− y1,
|λ2(x + y)| = |x1+ y1+ kx2+ y2k| = x1+ y1+ kx2+ y2k ≤ x1+ y1+ kx2k + ky2k.
This yields
[tr(|x + y|p)]1p = [|λ1(x + y)|p+ |λ2(x + y)|p]p1
≤ [(kx2k + ky2k − x1− y1)p+ (kx2k + ky2k + x1+ y1)p]1p
= [(−λ1(x) − λ1(y))p+ (λ2(x) + λ2(y))p]1p
= [|λ1(x) + λ1(y)|p+ |λ2(x) + λ2(y)|p]p1
≤ [|λ1(x)|p+ |λ2(x)|p]1p + [|λ1(y)|p+ |λ2(y)|p]p1
= [tr(|x|p)]1p + [tr(|y|p)]p1 ,
where the last inequality holds by the classical Minkowski inequality on real number setting.
Remark 4.3. We elaborate more about Proposition 4.13. We can define a norm ||| · |||p on IRn by
|||x|||p := [tr(|x|p)]1p,
and hence it induces a distance d(x, y) = |||x − y|||p on IRn. In particular, this norm will deduce the Euclidean-norm when p = 2, and the inequality reduces to the triangular inequality. In addition, this norm is similar to Schatten p-norm, which arise when ap-plying the p-norm to the vector of singular values of a matrix. For more details, please refer to [21].
According to the arguments in Proposition 4.13, if we wish to establish the SOC trace version of Minkowski inequality in general case without any restriction, the crucial key is verifying the SOC triangular inequality
|x + y| Kn |x| + |y|.
Unfortunately, this inequality does not hold. To see this, checking x = (√
2, 1, −1) and y = (−√
2, −1, 0) will lead to a counterexample. More specifically, x ∈ Kn, y ∈ −Kn, and x + y = (0, 0, −1) /∈ Kn∪ (−Kn), which says |x + y| = (1, 0, 0) and |x| + |y| = x + (−y) = (2√
2, 2, −1). Hence,
|x| + |y| − |x + y| = (2√
2 − 1, 2, −1) /∈ Kn∪ (−Kn).
Moreover, we have
λ1(|x + y|) = 1 > 2√ 2 −√
5 = λ1(|x| + |y|), λ2(|x + y|) = 1 < 2√
2 +√
5 = λ2(|x| + |y|).
Nonetheless, we build another SOC trace version of triangular inequality as below.
Proposition 4.14. (Triangular inequality) For any x = (x1, x2) ∈ IR × IRn−1 and y = (y1, y2) ∈ IR × IRn−1, there holds
tr(|x + y|) ≤ tr(|x|) + tr(|y|).
Proof. In order to complete the proof, we discuss three cases.
(i) If x + y ∈ Kn, then |x + y| = x + y Kn |x| + |y|, and hence tr(|x + y|) ≤ tr(|x|) + tr(|y|) by Proposition 1.1(a).
(ii) If x + y ∈ −Kn, then |x + y| = −x − y Kn |x| + |y|, and hence tr(|x + y|) ≤ tr(|x|) + tr(|y|).
(iii) Suppose x + y /∈ Kn∪ (−Kn), we have |x + y| =
kx2+ y2k,kxx1+y1
2+y2k(x2 + y2) from simple calculation, then
tr(|x + y|) = 2kx2+ y2k.
If one of x, y is in Kn (for convenience, we let x ∈ Kn), we have two subcases: y ∈ −Kn and y /∈ Kn∪ (−Kn). For y ∈ −Kn, we have |y| = −y and −y1 ≥ ky2k, and hence
tr(|x| + |y|) = tr(x − y) = 2(x1− y1) ≥ 2(kx2k + ky2k) ≥ 2kx2+ y2k = tr(|x + y|).
For y /∈ Kn∪ (−Kn), we have |y| =
ky2k,kyy1
2ky2
, and hence
tr(|x| + |y|) = 2(x1 + ky2k) ≥ 2(kx2k + ky2k) ≥ 2kx2+ y2k = tr(|x + y|).
If one of x, y is in −Kn, then the argument is similar. To complete the proof, it remains to show the inequality holds for x, y /∈ Kn∪ (−Kn). Indeed, in this case, we have
tr(|x| + |y|) = 2(kx2k + ky2k) ≥ 2kx2+ y2k = tr(|x + y|).
Hence, we complete the proof.
To close this section, we comment a few words about the aforementioned inequalities.
In real analysis, Young inequality is the main tool to derive the H¨older inequality, and then the Minkowski inequality can be derived by applying H¨older inequality. Tao et al.
[143] establish a trace p-norm in the setting of Euclidean Jordan algebra. In particular, they directly show the trace version of Minkowski inequality, see [143, Theorem 4.1].
As an application of trace versions of Young inequality, we use the approach which follows the same idea as in real analysis to derive the trace versions of H¨older inequality.
Furthermore, the SOC trace version of Minkowski inequality is also deduced. On the other hand, the trace version of Triangular inequality holds for any Euclidean Jordan algebra, see [96, Proposition 4.3] and [143, Corollary 3.1]. In the setting of second-order cone, we prove the inequality by discussing three cases directly.
min 𝑎, 𝑏
𝐻 𝑎, 𝑏 𝐺 𝑎, 𝑏 𝐿 𝑎, 𝑏 𝐴 𝑎, 𝑏
m𝑎𝑥 𝑎, 𝑏
𝐻 𝑎, 𝑏 𝐺 𝑎, 𝑏 𝐴 𝑎, 𝑏
𝜆 1
2 𝜆 𝑏
𝑎
𝑀 𝑎, 𝑏
𝜈 1
2 𝜈 1
𝜆 1
Figure 4.1: Relationship between means defined on real number.