The p-Relative Distance is a Metric
c 2000 Society for Industrial and Applied Mathematics °
SIAM J. MATRIX ANAL. APPL. Vol. 21, No. 2, pp. 699–702
THE p-RELATIVE DISTANCE IS A METRIC∗ ANDERS BARRLUND†
p Abstract. The conjecture that the p-relative distance, %p (α, α) ˜ = |α − α|/ ˜ p |α|p + |α| ˜ p , is a metric is proved. Key words. relative distance, metric AMS subject classification. 15A45 PII. S0895479898340883
1. Introduction. The p-relative distance between α and α ˜ ∈ C is defined as (1.1)
|α − α ˜| ˜) = p %p (α, α p p |α| + |˜ α|p
for 1 ≤ p ≤ ∞.
For convenience, we define 0/0 := 0. Li [2] presented the conjecture that the p-relative distance is a metric. It is trivial that %p (α, β) ≥ 0, %p (α, β) = 0 if and only if α = β, %p (α, β) = %p (β, α). However, it has been an open question whether (1.2)
%p (α, β) ≤ %p (α, γ) + %p (γ, β).
Li [2] proved (1.2) for α, β, γ ∈ R. Day [1] proved (1.2) for p = ∞. In this paper we prove (1.2) for α, β, γ ∈ C. The paper is outlined as follows: In section 2 we prove a simple lemma which is used to prove (1.2) in section 3. We will let kxk2 denote the Euclidean vector norm. 2. A simple lemma. Lemma 1. Let 1 ≤ p ≤ ∞. If x ≥ 0 and y > 0, then r 1 + xp p (2.1) ≥ min(x, 1/y). 1 + yp If x ≥ y ≥ 1, then also (2.2)
r p
p 1 + xp > x/y. 1 + yp
Proof. If x ≤ 1/y, then xp y p ≤ 1, which implies that 1 + xp ≥ xp y p + xp = (1 + y p )xp , from which (2.1) follows. If x > 1/y, then xp y p > 1, which implies that 1 + y p < xp y p + y p = (1 + xp )y p , from which (2.1) follows. ∗ Received by the editors June 24, 1998; accepted for publication (in revised form) by R. Bhatia September 2, 1998; published electronically January 25, 2000. http://www.siam.org/journals/simax/21-2/34088.html † Department of Computing Science, University of Ume˚ a, S-90187 Ume˚ a, Sweden (abrr@cs. umu.se).
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ANDERS BARRLUND
If u ≥ v ≥ 1, then (uv − 1)(u − v) ≥ 0, which implies that v + u2 v ≥ u + uv 2 and u 1 + u2 ≥ . 1 + v2 v This inequality with u = xp/2 and v = y p/2 gives (2.2). 3. The proof. Theorem 1. Let α, β, γ ∈ C and 1 ≤ p ≤ ∞. Then (3.1)
%p (α, β) ≤ %p (α, γ) + %p (γ, β),
where %p is defined by (1.1). Proof. Without loss of generality, we prove (3.1) in the cases with |α| ≤ |β|. The cases with |β| < |α| are proved by swapping α and β in the rest of the proof. In cases with |γ| ≤ |α| ≤ |β|, we get |α − β| |α − γ| |γ − β| %p (α, β) = p ≤ p + p p p p p p p p |α| + |β| |α| + |β| |α|p + |β|p |α − γ| |γ − β| ≤ p + p = %p (α, γ) + %p (γ, β). p p |α|p + |γ|p |γ|p + |β|p In cases with α = 0, the inequality (3.1) is trivial. Next we consider cases with 0 < |α| ≤ |β| ≤ |γ|. In these cases (2.1) gives p p p β β p 1 + |α/β|p β |α|p + |β|p p (3.2) = p ≥ min(|α/β|, |α/γ|) = , p p p p p α α γ |α| + |γ| 1 + |γ/α| p p p α α p 1 + |β/α|p α |α|p + |β|p p = p ≥ min(|β/α|, |β/γ|) = . p p p p p β β γ |β| + |γ| 1 + |γ/β| We also have (3.3)
β |α − γ| + α |β − γ| = βα − β + βα − α ≥ |α − β|. γ γ γ γ
Combining (3.2) and (3.3) gives (3.1). Finally, we consider cases with 0 < |α| < |γ| < |β|. Let r1 eiθ1 = β/α and r2 eiθ2 = γ/α be the polar decompositions of β/α and γ/α, respectively. Then (3.1) can be rewritten as (3.4) p
1 + r12 − 2r1 cos θ1 p ≤ p 1 + r1p
p p 1 + r22 − 2r2 cos θ2 r12 + r22 − 2r1 r2 cos(θ1 − θ2 ) p p + . p p 1 + r2p r1p + r2p
We now derive some inequalities which can be combined to a proof of (3.4). Since 1 < r2 < r1 and 1 ≤ p ≤ ∞, we have (3.5) p p 1−1/p Z r2 Z r2 q p r2p + r1p − p 1 + r1p 1 1 d p p xp p = x + r1 dx = dx r2 − 1 r2 − 1 1 dx r2 − 1 1 xp + r2p p p 1−1/p Z r1 p r1p + 1 − p r2p + 1 1 xp < dx = . r1 − r2 r 2 xp + 1 r1 − r2
701
THE p-RELATIVE DISTANCE IS A METRIC
p p Combining (3.5) and p r1p + r2p > p 1 + r2p gives q q q q q q p r1p + r2p p 1 + r1p − p 1 + r2p (r2 − 1) > p 1 + r2p p r1p + r2p − p 1 + r1p (r1 − r2 ), which can be rewritten as r1 − 1 r2 − 1 r 1 − r2 p < p + p . p p p p p 1 + r1 1 + r2 r1p + r2p
(3.6)
The second inequality (2.2) in Lemma 1 gives s p p p 1 + r1 (3.7) r1 /r2 p > 1 + r2 and (3.8)
s p
1 1 + r1p = r1p + r2p r2
s p
√ 1√ 1 + r1p > r2 = 1/ r2 . 1 + (r1 /r2 )p r2
From the trigonometric identity sin(x + y) = sin(x) cos(y) + sin(y) cos(x), we get | sin(x + y)| ≤ | sin(x)|| cos(y)| + | sin(y)|| cos(x)| ≤ | sin(x)| + | sin(y)|, which implies that (3.9)
| sin(θ1 /2)| ≤ | sin(θ2 /2)| + | sin((θ2 − θ1 )/2)|.
From the trigonometric identity sin2 (x/2) = (3.10)
1−cos(x) , 2
we get ° ° q ° ° r2 − 1 2 ° ° 1 + r2 − 2r2 cos θ2 = ° √ 2 r2 sin(θ2 /2) °2
and (3.11)
° ° q ° ° r1 − r2 2 2 ° . ° √ r1 + r2 − 2r1 r2 cos(θ1 − θ2 ) = ° 2 r1 r2 sin((θ1 − θ2 )/2) °2
Combining (3.6), (3.7), (3.8), (3.9), (3.10), and (3.11) gives p p q 1 + r22 − 2r2 cos θ2 r12 + r22 − 2r1 r2 cos(θ1 − θ2 ) p p p p 1 + r1 · + p p 1 + r2p r1p + r2p √ √ ° ° ° ° p p ° ° ° ° 1+r1p 1+r1p √ √ ° ° ° ° p p (r1 − r2 ) p (r2 − 1) p p ° ° ° √ ° √ 1+r2 r1 +r2 + = ° p ° ° ° p p p ° 2 √ 1+r1 √r r sin((θ − θ )/2) ° ° 2√ 1+r1 √r sin(θ /2) ° 2 2 ° ° ° ° p 1+rp 1 2 1 2 p p p r +r 2
2
1
2
° ° q ° ° r1 − 1 ° = 1 + r2 − 2r1 cos θ1 . √ ≥° 1 ° 2 r1 sin(θ1 /2) ° 2 That is, we have proved (3.4) and hence also (3.1).
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ANDERS BARRLUND
Acknowledgment. The author would like to thank Ren-Cang Li, who informed him about the paper [1]. REFERENCES [1] D. Day, A New Metric on the Complex Numbers, Sandia Technical Report 98–1754, Sandia National Laboratories, Albuquerque, NM, 1998. [2] R.-C. Li, Relative perturbation theory. I. Eigenvalue and singular value variations, SIAM J. Matrix Anal. Appl., 19 (1998), pp. 956–982.
SIAM J. MATRIX ANAL. APPL. Vol. 21, No. 2, pp. 699–702
THE p-RELATIVE DISTANCE IS A METRIC∗ ANDERS BARRLUND†
p Abstract. The conjecture that the p-relative distance, %p (α, α) ˜ = |α − α|/ ˜ p |α|p + |α| ˜ p , is a metric is proved. Key words. relative distance, metric AMS subject classification. 15A45 PII. S0895479898340883
1. Introduction. The p-relative distance between α and α ˜ ∈ C is defined as (1.1)
|α − α ˜| ˜) = p %p (α, α p p |α| + |˜ α|p
for 1 ≤ p ≤ ∞.
For convenience, we define 0/0 := 0. Li [2] presented the conjecture that the p-relative distance is a metric. It is trivial that %p (α, β) ≥ 0, %p (α, β) = 0 if and only if α = β, %p (α, β) = %p (β, α). However, it has been an open question whether (1.2)
%p (α, β) ≤ %p (α, γ) + %p (γ, β).
Li [2] proved (1.2) for α, β, γ ∈ R. Day [1] proved (1.2) for p = ∞. In this paper we prove (1.2) for α, β, γ ∈ C. The paper is outlined as follows: In section 2 we prove a simple lemma which is used to prove (1.2) in section 3. We will let kxk2 denote the Euclidean vector norm. 2. A simple lemma. Lemma 1. Let 1 ≤ p ≤ ∞. If x ≥ 0 and y > 0, then r 1 + xp p (2.1) ≥ min(x, 1/y). 1 + yp If x ≥ y ≥ 1, then also (2.2)
r p
p 1 + xp > x/y. 1 + yp
Proof. If x ≤ 1/y, then xp y p ≤ 1, which implies that 1 + xp ≥ xp y p + xp = (1 + y p )xp , from which (2.1) follows. If x > 1/y, then xp y p > 1, which implies that 1 + y p < xp y p + y p = (1 + xp )y p , from which (2.1) follows. ∗ Received by the editors June 24, 1998; accepted for publication (in revised form) by R. Bhatia September 2, 1998; published electronically January 25, 2000. http://www.siam.org/journals/simax/21-2/34088.html † Department of Computing Science, University of Ume˚ a, S-90187 Ume˚ a, Sweden (abrr@cs. umu.se).
699
700
ANDERS BARRLUND
If u ≥ v ≥ 1, then (uv − 1)(u − v) ≥ 0, which implies that v + u2 v ≥ u + uv 2 and u 1 + u2 ≥ . 1 + v2 v This inequality with u = xp/2 and v = y p/2 gives (2.2). 3. The proof. Theorem 1. Let α, β, γ ∈ C and 1 ≤ p ≤ ∞. Then (3.1)
%p (α, β) ≤ %p (α, γ) + %p (γ, β),
where %p is defined by (1.1). Proof. Without loss of generality, we prove (3.1) in the cases with |α| ≤ |β|. The cases with |β| < |α| are proved by swapping α and β in the rest of the proof. In cases with |γ| ≤ |α| ≤ |β|, we get |α − β| |α − γ| |γ − β| %p (α, β) = p ≤ p + p p p p p p p p |α| + |β| |α| + |β| |α|p + |β|p |α − γ| |γ − β| ≤ p + p = %p (α, γ) + %p (γ, β). p p |α|p + |γ|p |γ|p + |β|p In cases with α = 0, the inequality (3.1) is trivial. Next we consider cases with 0 < |α| ≤ |β| ≤ |γ|. In these cases (2.1) gives p p p β β p 1 + |α/β|p β |α|p + |β|p p (3.2) = p ≥ min(|α/β|, |α/γ|) = , p p p p p α α γ |α| + |γ| 1 + |γ/α| p p p α α p 1 + |β/α|p α |α|p + |β|p p = p ≥ min(|β/α|, |β/γ|) = . p p p p p β β γ |β| + |γ| 1 + |γ/β| We also have (3.3)
β |α − γ| + α |β − γ| = βα − β + βα − α ≥ |α − β|. γ γ γ γ
Combining (3.2) and (3.3) gives (3.1). Finally, we consider cases with 0 < |α| < |γ| < |β|. Let r1 eiθ1 = β/α and r2 eiθ2 = γ/α be the polar decompositions of β/α and γ/α, respectively. Then (3.1) can be rewritten as (3.4) p
1 + r12 − 2r1 cos θ1 p ≤ p 1 + r1p
p p 1 + r22 − 2r2 cos θ2 r12 + r22 − 2r1 r2 cos(θ1 − θ2 ) p p + . p p 1 + r2p r1p + r2p
We now derive some inequalities which can be combined to a proof of (3.4). Since 1 < r2 < r1 and 1 ≤ p ≤ ∞, we have (3.5) p p 1−1/p Z r2 Z r2 q p r2p + r1p − p 1 + r1p 1 1 d p p xp p = x + r1 dx = dx r2 − 1 r2 − 1 1 dx r2 − 1 1 xp + r2p p p 1−1/p Z r1 p r1p + 1 − p r2p + 1 1 xp < dx = . r1 − r2 r 2 xp + 1 r1 − r2
701
THE p-RELATIVE DISTANCE IS A METRIC
p p Combining (3.5) and p r1p + r2p > p 1 + r2p gives q q q q q q p r1p + r2p p 1 + r1p − p 1 + r2p (r2 − 1) > p 1 + r2p p r1p + r2p − p 1 + r1p (r1 − r2 ), which can be rewritten as r1 − 1 r2 − 1 r 1 − r2 p < p + p . p p p p p 1 + r1 1 + r2 r1p + r2p
(3.6)
The second inequality (2.2) in Lemma 1 gives s p p p 1 + r1 (3.7) r1 /r2 p > 1 + r2 and (3.8)
s p
1 1 + r1p = r1p + r2p r2
s p
√ 1√ 1 + r1p > r2 = 1/ r2 . 1 + (r1 /r2 )p r2
From the trigonometric identity sin(x + y) = sin(x) cos(y) + sin(y) cos(x), we get | sin(x + y)| ≤ | sin(x)|| cos(y)| + | sin(y)|| cos(x)| ≤ | sin(x)| + | sin(y)|, which implies that (3.9)
| sin(θ1 /2)| ≤ | sin(θ2 /2)| + | sin((θ2 − θ1 )/2)|.
From the trigonometric identity sin2 (x/2) = (3.10)
1−cos(x) , 2
we get ° ° q ° ° r2 − 1 2 ° ° 1 + r2 − 2r2 cos θ2 = ° √ 2 r2 sin(θ2 /2) °2
and (3.11)
° ° q ° ° r1 − r2 2 2 ° . ° √ r1 + r2 − 2r1 r2 cos(θ1 − θ2 ) = ° 2 r1 r2 sin((θ1 − θ2 )/2) °2
Combining (3.6), (3.7), (3.8), (3.9), (3.10), and (3.11) gives p p q 1 + r22 − 2r2 cos θ2 r12 + r22 − 2r1 r2 cos(θ1 − θ2 ) p p p p 1 + r1 · + p p 1 + r2p r1p + r2p √ √ ° ° ° ° p p ° ° ° ° 1+r1p 1+r1p √ √ ° ° ° ° p p (r1 − r2 ) p (r2 − 1) p p ° ° ° √ ° √ 1+r2 r1 +r2 + = ° p ° ° ° p p p ° 2 √ 1+r1 √r r sin((θ − θ )/2) ° ° 2√ 1+r1 √r sin(θ /2) ° 2 2 ° ° ° ° p 1+rp 1 2 1 2 p p p r +r 2
2
1
2
° ° q ° ° r1 − 1 ° = 1 + r2 − 2r1 cos θ1 . √ ≥° 1 ° 2 r1 sin(θ1 /2) ° 2 That is, we have proved (3.4) and hence also (3.1).
2
702
ANDERS BARRLUND
Acknowledgment. The author would like to thank Ren-Cang Li, who informed him about the paper [1]. REFERENCES [1] D. Day, A New Metric on the Complex Numbers, Sandia Technical Report 98–1754, Sandia National Laboratories, Albuquerque, NM, 1998. [2] R.-C. Li, Relative perturbation theory. I. Eigenvalue and singular value variations, SIAM J. Matrix Anal. Appl., 19 (1998), pp. 956–982.
