Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 3, Article 87, 2005
A MINKOWSKI-TYPE INEQUALITY FOR THE SCHATTEN NORM
MARKUS SIGG
D EPARTMENT OF M ATHEMATICS AND S TATISTICS
U NIVERSITY OF KONSTANZ
G ERMANY
mail@MarkusSigg.de
Received 16 July, 2004; accepted 29 June, 2005
Communicated by F. Hansen
A BSTRACT. Let F be a Schatten p-operator and R, S positive operators. We show that the
1 c
1 c
1 c
inequality |F (R + S) c |p ≤ |F R c |p +|F S c |p for the Schatten p-norm |·|p is true for p ≥ c = 1
and for p ≥ c = 2, conjecture it to be true for p ≥ c ∈ [1, 2], give counterexamples for the other
cases, and present a numerical study for 2 × 2 matrices. Furthermore, we have a look at a
generalisation of the inequality which involves an additional factor σ(c, p).
Key words and phrases: Schatten class, Schatten norm, Norm inequality, Minkowski inequality, Triangle inequality, Powers
of operators, Schatten-Minkowski constant.
2000 Mathematics Subject Classification. 47A30, 47B10.
1. I NTRODUCTION
Let H and K be complex Hilbert spaces and 0 < p ≤ ∞. Following [1], we denote by
cp (H, K) the space of Schatten p-operators T : H −→ K, equipped with the Schatten pnorm or quasi-norm | · |p . Note that [1] deals only with the spaces cp (H) := cp (H, H). The
generalisations cp (H, K) can be found in textbooks like [2] and [3] (there written as Bp (H, K)
and Sp (H, K) respectively).
By L(H) we denote the space of bounded linear operators on H, and by L(H)+ the subset
of positive operators. With |T | := (T ∗ T )1/2 ∈ L(H)+ for T ∈ L(H, K) we have for p < ∞
|T |pp = tr |T |p
for T ∈ cp (H, K), and consequently
|T |pp
for T ∈ cp (H)+ := cp (H) ∩ L(H)+ .
= tr T
p
Applying |T |p = |T ∗ |p for T ∈ cp (H, K), this shows in case of p < ∞
1 2 2
1 2
2 ∗
∗ p2 p
2
F
U
=
F
=
tr
(F
U
F
)
= |F U F ∗ | p
U
p
p
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
This work was supported by the Dr. Helmut Manfred Riedl Foundation.
135-04
2
2
M ARKUS S IGG
for F ∈ cp (H, K) and U ∈ L(H)+ . Because | · |∞ is the usual operator norm,
1 2
2
F U = |F U F ∗ | p
2
p
is also true for p = ∞, with the common convention ∞
:= ∞.
2
Our question, which arose while studying the integration of Schatten operator valued functions in [4], is: For what values of p ∈ (0, ∞] and c ∈ (0, ∞) is the Minkowski-like inequality
1 c
1 c
1 c
c
c
c
F (R + S) ≤ F R + F S (MS)
p
p
p
true for all F ∈ cp (H, K) and R, S ∈ L(H)+ ?
2. T HE C ONJECTURE
Let H, K, p, c, F, R, S be as above.
Theorem 2.1. Inequality (MS) is true for p ≥ c = 1 and for p ≥ c = 2.
Proof. For p ≥ c = 1, the triangle inequality for | · |p shows
|F (R + S)|p = |F R + F S|p ≤ |F R|p + |F S|p .
For p ≥ c = 2, the triangle inequality for | · | p shows
2
2
1 2
1
1 2
F (R + S) 2 = |F (R + S)F ∗ | p ≤ |F RF ∗ | p + |F SF ∗ | p = F R 2 + F S 2 .
2
p
2
2
p
p
Theorem 2.1 suggests the following conjecture.
Conjecture 2.2. Inequality (MS) is true for p ≥ c ∈ [1, 2].
For c ∈ (1, 2) we have at the present time no proof of this conjecture for other than trivial
situations, not even for the special case of 2 × 2 matrices. However, some justification will be
given in Section 4.
3. T HE C ASE p < c AND
THE
C ASE c 6∈ [1, 2]
In this section we will demonstrate, by providing counterexamples, that inequality (MS) is
not necessarily true for other values of (c, p) than those stated in Conjecture 2.2. We will offer
one example for 0 < p < c < ∞, and one for arbitrary p when c < 1 or c > 2, both examples
using 2 × 2 matrices. The power U t for t > 0 of a non-negative matrix U can be calculated
easily with help of the spectral decomposition of U .
Example 3.1. Inequality (MS) is violated for 0 < p < c < ∞ by
1 0
1 0
0 0
F :=
, R :=
, S :=
.
0 1
0 0
0 1
Proof. From U t = U for U ∈ {R, S, R + S} and t ∈ (0, ∞) we get
1
1
1
c
F
U
= |U |p = (tr U p ) p = (tr U ) p ,
p
yielding
1
F R c = 1,
p
J. Inequal. Pure and Appl. Math., 6(3) Art. 87, 2005
1
F S c = 1,
p
1
1
F (R + S) c = 2 p ,
p
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A M INKOWSKI -T YPE I NEQUALITY FOR THE S CHATTEN N ORM
and using p < c,
3
c
1 c
1 c
1 c
p
c
c
c
F R + F S = 2 < 2 = F (R + S) .
p
p
p
The second example makes use of an inequality which is interesting in its own right. Seeming
simple, it is surprisingly fiddly to prove:
Lemma 3.1. For x ∈ (0, 1) ∪ (2, ∞) we have
√ !x √ !x
1
1
3+ 5
3− 5
1+ √
+ 1− √
< 1 + 3x .
2
2
5
5
√
Proof. Setting r := 5, α1 := 1 + 1r , α2 := 1 − 1r , and ω := 3+r
, we have to show
2
α1 ω x + α2 ω −x < 1 + 3x .
The case x ∈ (2, ∞): Set f (x) := α1 ω x , g(x) := α2 ω −x , h(x) := 1 + 3x for x ∈ (0, ∞).
Because α2 > 0 and ω > 1, g is strictly decreasing, thus f (x) + g(x) < f (x) + g(2) for x > 2.
We will show f (x) + g(2) < h(x) for x > 2. Because f (2) + g(2) = h(2), this is done if we
prove f 0 (x) < h0 (x) for x > 2, which is equivalent to α1 ( ω3 )x ln ω < ln 3. This inequality is
true for x = 2. All factors of its left side are positive, and ω < 3, so the left side is strictly
decreasing for x ≥ 2. Hence the inequality is true for x > 2 as well.
ln 3
The case x ∈ (0, 1): After substituting s := ω x and setting δ := ln
, we have to prove the
ω
equivalent inequality
1 1
1
s+ +
s−
< 1 + sδ
s r
s
for s ∈ (1, ω), which can be done by building a sandwich with a suitable polynomial function
inside: Set
1 1
1
s−1
s−
, p(s) := 2 1 +
,
ϕ(s) := s + +
s r
s
ω−1
(s − 1)(s − ω)
(ϕ(3) − p(3))
q(s) :=
(3 − 1)(3 − ω)
for s > 0. The claim is
ϕ(s) < p(s) + q(s) < 1 + sδ
for s ∈ (1, ω). The left inequality is verified by the fact that s · (p(s) + q(s) − ϕ(s)) defines a
polynomial of degree 3 with three zeros {1, ω, 3}, where 1 < ω < 3, and with positive leading
coefficient λ := 21 (ϕ(3) − p(3))/(3 − ω). To prove the second inequality, we inspect
ψ(s) := 1 + sδ − p(s) − q(s)
for s > 0 and get ψ 00 (s) = δ(δ − 1)sδ−2 − 2λ. Because 1 < δ < 2, ψ 00 has a unique zero
1
δ(δ − 1) 2−δ
s0 :=
, 1 < s0 < ω,
2λ
with ψ 00 (s) > 0 for s ∈ (0, s0 ) and ψ 00 (s) < 0 for s ∈ (s0 , ∞). Now ψ(1) = 0, ψ 0 (1) > 0,
and ψ 00 (s) > 0 for s ∈ (1, s0 ) show ψ(s) > 0 for s ∈ (1, s0 ], while ψ(s0 ) > 0, ψ(ω) = 0,
ψ 0 (ω) < 0, and ψ 00 (s) < 0 for s ∈ (s0 , ω) show ψ(s) > 0 for s ∈ [s0 , ω).
Example 3.2. Inequality (MS) is violated for 0 < p ≤ ∞ and c < 1 as well as c > 2 by
0 0
0 0
2 −1
F :=
, R :=
, S :=
.
1 0
0 1
−1 1
J. Inequal. Pure and Appl. Math., 6(3) Art. 87, 2005
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4
M ARKUS S IGG
Proof. Evaluation of the matrix powers for t ∈ (0, ∞) gives
Rt = R, S t =
(R
with r :=
p<∞
√
1
(α1 ω t + α2 ω −t )
2
1
(ω −t − ω t )
r
t
1 1+3
t
+ S) =
t
2
1−3
5, α1 := 1 + 1r , α2 := 1 − 1r , ω :=
3+r
.
2
!
1
(ω −t − ω t )
r
1
(α2 ω t + α1 ω −t )
2
1 − 3t
,
!
1 + 3t
For U ∈ {R, S, R + S} we get in case of
p1 √
p
= ut
|F U t |p = tr (F U 2t F ∗ ) 2
2
with ut being the top left entry of U 2t . Using |F U t |∞ = |F U 2t F ∗ |∞ , the case p = ∞ yields the
same result, thus for all p:
r
1
1
1
(α1 ω 2/c + α2 ω −2/c ),
F R c = 0, F S c =
2
p
p
r
1
1
(1 + 32/c ).
F (R + S) c =
2
p
Substituting 2c by x, we have to prove α1 ω x +α2 ω −x < 1+3x for x ∈ (2, ∞) and for x ∈ (0, 1),
which is the statement of Lemma 3.1.
4. S OME N UMERICAL E VIDENCE
To justify Conjecture 2.2, we present the results of a numerical study performed with 2 × 2
matrices.
From functional calculus it is known: For an operator T ≥ 0 on a complex Hilbert space the
powers T α , T β for α, β ∈ (0, ∞) obey the rule T α T β = T α+β . If T is invertible, then T α can
be defined for α ≤ 0 as well, and T α T β = T α+β is true for all α, β ∈ R.
Before turning to the matrix case, we note the following general lemma.
Lemma 4.1. Let H, K, F be as above and α ∈ (0, ∞).
(a) Let T ∈ L(H)+ . Then F T α = 0 if and only if F T = 0.
(b) Let R, S ∈ L(H)+ . Then F (R + S)α = 0 if and only if F Rα = 0 and F S α = 0.
2
Proof. (a) Suppose F T α = 0. Then |F T α/2 | = |F T α F ∗ | = 0, hence F T α/2 = 0. Repeated
application yields β ∈ (0, 1) with F T β = 0, thus F T = F T β T 1−β = 0.
Now suppose F T = 0. There is nothing to prove in the case of α = 1, so assume α 6= 1.
If T is invertible, then F T α = F T T α−1 = 0. If T is not invertible, then we have 0 ∈ σ(T ),
the spectrum of T . Choose polynomials fn ∈ R[t] for n ∈ N such that fn (x) → xα for
n → ∞ uniformly for x ∈ σ(T ). Then fn (T ) → T α and F fn (T ) → F T α for n → ∞, hence
F fn (T ) = fn (0)F → 0 for n → ∞, thus F T α = 0.
(b) Part (a) shows:
F Rα = 0 ∧ F S α = 0
⇐⇒
FR = 0 ∧ FS = 0
=⇒
F (R + S) = 0
⇐⇒
F (R + S)α = 0.
To prove the missing implication, suppose F (R + S) = 0. Then F RF ∗ + F SF ∗ = 0. Because
F RF ∗ ≥ 0 and F SF ∗ ≥ 0, we get F RF ∗ = 0, thus |F R1/2 |2 = |F RF ∗ | = 0 and F R1/2 = 0.
Applying (a) again gives F R = 0. Symmetry shows F S = 0.
J. Inequal. Pure and Appl. Math., 6(3) Art. 87, 2005
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A M INKOWSKI -T YPE I NEQUALITY FOR THE S CHATTEN N ORM
5
We will also use the following well-known property of 2 × 2 matrices:
Lemma 4.2. A complex 2 × 2 matrix M is positive semidefinite if and only if there exist a, b ∈
[0, ∞) and γ ∈ C with |γ|2 ≤ ab such that
a γ
M=
.
γ b
Lemma 4.1(b) shows that, when checking Conjecture 2.2, one may assume the denominator
to be non-zero, or setting 00 := 0, in
1 c
F (R + S) c p
.
qc,p (F, R, S) := 1 c
1 c
c
c
F R + F S p
p
We are searching for the supremum of qc,p (F, R, S) over all complex 2 × 2 matrices F, R, S
with R, S ≥ 0. For r ∈ [0, ∞) and x ∈ C define r ∧ x := x if |x| ≤ r and r ∧ x := (r/|x|) x
otherwise. Lemma 4.2 shows that R has the structure
α2
|α β| ∧ γ
R=
=: P (α, β, γ)
|α β| ∧ γ
β2
with α, β ∈ R and γ ∈ C, and a corresponding representation is valid for the matrix S.
This means that we have to deal with six complex and four real variables, resulting in a 16dimensional real optimisation problem: For λ = (λ1 , . . . , λ16 ) ∈ R16 we set
λ1 + λ 2 i λ 3 + λ 4 i
Fλ :=
,
λ5 + λ 6 i λ 7 + λ 8 i
Rλ := P (λ9 , λ10 , λ11 + λ12 i),
Sλ := P (λ13 , λ14 , λ15 + λ16 i)
and are asking for
σ(c, p) := sup qc,p (Fλ , Rλ , Sλ ).
λ∈R16
To attack this problem, GNU Octave [5], version 2.1.57, was utilised. It offers a function
for determining the singular values of a matrix, which can be employed for calculating the
Schatten norms. For the optimisation task the implementation [6], version 2002/05/09, with
standard parameters of the Downhill Simplex Method of Nelder and Mead ([7], 10.4) was used.
The results are in perfect agreement with Conjecture 2.2. For visualisation, approximations for
σ(c, p) for c ∈ {1.2, 1.4, 1.6, 1.8, 2.0} have been calculated and plotted with a step size of 0.01
for p, see Figure 4.1.
The apparently smooth shape of p 7→ σ(c, p) for p ≤ c, together with the fact that for each p
a new random starting point λ was used for the Nelder-Mead algorihm, gives some confidence
in the validity of the data.
A closer inspection of some of the calculated numerical values suggests
1
1
σ(2, 1) = 2, σ 32 , 1 = σ 95 , 65 = 2 2 , σ 85 , 65 = σ 2, 32 = 2 3 ,
1
1
σ 45 , 1 = σ 32 , 54 = σ 74 , 75 = σ 2, 85 = 2 4 , σ 65 , 1 = σ 95 , 32 = 2 5 ,
which leads to the idea to look at log2 σ(c, p). It seems there is a linear dependency of log2 σ(c, p)
from c if c ≥ p. This observation will be made precise in the next section.
J. Inequal. Pure and Appl. Math., 6(3) Art. 87, 2005
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6
M ARKUS S IGG
Figure 4.1: Experimental approximations of σ(c, p).
2
1.9
1.8
1.7
1.6
1.5
1.4
1.3
1.2
1.1
1
0.9
c = 1.2
c = 1.4
c = 1.6
c = 1.8
c = 2.0
1
1.2
1.4
1.6
1.8
2
p
5. G ENERALISATION OF (MS)
It is natural to generalise (MS) and to ask for the smallest σ(c, p) ∈ [0, ∞] for c ∈ (0, ∞) and
p ∈ (0, ∞] such that
1 c
1 c
1 c
c
c
c
F (R + S) ≤ σ(c, p) F R + F S p
p
p
for all F ∈ cp (H, K) and R, S ∈ L(H)+ (and for all complex Hilbert spaces H and K). It
is tempting to call σ(c, p) the Schatten-Minkowski constant for (c, p). By choosing F 6= 0 and
setting R to be the identity and S := 0 it can be seen that σ(c, p) ≥ 1. Now Conjecture 2.2 can
be re-phrased using σ(c, p), and, motivated by the numerical results, we add another conjecture:
Conjecture 5.1. (a) For 1 ≤ c ≤ 2 and p ≥ c we have σ(c, p) = 1.
c
(b) For 0 ≤ c ≤ 2 and p ≤ c we have σ(c, p) = 2 p −1 .
Again, the cases c = 1 and c = 2 are not too difficult to prove:
(
(
1
1
for p ≥ 1
(b) σ(2, p) =
Theorem 5.2. (a) σ(1, p) =
1
2
−1
for p ≤ 1
2 p −1
2p
for p ≥ 2
.
for p ≤ 2
Proof. σ(1, p) ≤ 1 for p ≥ 1 and σ(2, p) ≤ 1 for p ≥ 2 is the subject of Theorem 2.1, while
σ(c, p) ≥ 1 is noted above. Example 3.1 tells us that σ(c, p) ≥ 2c/p−1 for 0 < p ≤ c < ∞,
yielding
1
2
and
σ(2, p) ≥ 2 p −1 for p ≤ 2.
σ(1, p) ≥ 2 p −1 for p ≤ 1
Now for the missing ‘≤’ inequalities. For the case c = 1, recall the inequality between the
power means of degrees p ≤ 1 and 1, see e.g. [8], 8.12, which reads
p
1
α + βp p
α+β
≤
or equivalently αp + β p ≤ 21−p (α + β)p
2
2
for α, β ∈ [0, ∞). Together with the quasi-norm inequality of | · |p this gives
|F (R + S)|pp ≤ |F R|pp + |F S|pp ≤ 21−p (|F R|p + |F S|p )p
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A M INKOWSKI -T YPE I NEQUALITY FOR THE S CHATTEN N ORM
7
1
and thus |F (R + S)|p ≤ 2 p −1 (|F R|p + |F S|p ).
For the case c = 2, start with the power means inequality for the degrees p ≤ 2 and 2,
p
1 2
1
p
p
α + β2 2
α + βp p
≤
or equivalently αp + β p ≤ 21− 2 (α2 + β 2 ) 2
2
2
for α, β ∈ [0, ∞). Together with the quasi-norm inequality of | · | p this gives
2
p
1 p
F (R + S) 2 = |F (R + S)F ∗ | p2
p
2
p
2
p
2
p
≤ |F RF ∗ | + |F SF ∗ | p2
2
p2
p
1 p
1 p
1 2
1 2
1−
= F R 2 + F S 2 ≤ 2 2 F R 2 + F S 2 p
p
p
p
and consequently
2
1 2
1 2
1 2
−1
F (R + S) 2 ≤ 2 p
F R 2 + F S 2 .
p
p
p
6. C ONCLUSION
Starting with Conjecture 2.2, which we proved for the cases c = 1 and c = 2 in Theorem 2.1,
a numerical study of 2 × 2 matrices led to the generalised Conjecture 5.1, which we also proved
for c = 1 and c = 2 in Theorem 5.2.
The given proofs make use of the (quasi-) triangle inequality of the Schatten (quasi-) norm.
Another ingredient to Theorem 5.2 is the power means inequality. Presumably, a combination
of these inequalities shall also be central when dealing with the case c 6= 1, 2. However, it is
unclear how to apply the triangle inequality in this situation, because there is no obvious way
to get from F (R + S)1/c to an expression where R and S can be separated.
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[5] J.W. EATON et al, http://www.octave.org.
[6] E. GROSSMANN,
min.html.
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J. Inequal. Pure and Appl. Math., 6(3) Art. 87, 2005
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