数理解析研究所講究録
1187 巻 2001 年 110-117

110

Absolute Norms on$\mathbb{C}^{n}$

新潟大学理学部
九州工業大学工学部

斎藤吉助 (Kichi-Suke Saito)
加藤幹雄 (Mikio Kato)

岡山県立大学情報工学部

高橋泰嗣 (Yasuji Takahashi)

there corresponds
It is known that for every absolute normalized norm ona unique convex function on $[0,1]$ satisfying certain suitable conditions (see BonsallDuncan [1], also [2] . Recently the authors [3] extended this result to the n-dimensionalcase. In this note we shall present a brief introduction of our result. We first recall the2-dimensional case, and then we treat 3-and -dimensional cases, where we focus ourdiscussion on the 3-dimensional case which will illustrate the -dimensional situation.
We shall also present a characterization of the strict convexity of these norms, which
extends our previous result in [4].
$\mathbb{C}^{2}$$)$

$n$

$n$

A norm$||\cdot||$(1)on$\mathbb{C}^{n}$is called absolute if$||(|_{X_{1}}|, \cdots, |X_{n}|)||=||(x_{1,n}\ldots, x)||$$\forall(X_{1,)}\ldots Xn)\in \mathbb{C}^{n}$,and is called normalized if(2)$||(1,0, \cdots, 0)||=||(0,1,0, \cdots, 0)||=\cdots=||(0, \cdots, 0,1)||=1$ .

The$\ell_{p}$-norms$||\cdot||_{p}$are such examples:$(|x_{1}|p+\cdots+|X|^{p}n)^{1}/p$

if $1\leq p<\infty$ ,$\max(|x_{1}|, \cdots, |x_{n}|)$if $p=\infty$ .

$||(x_{1,2,n}x\cdots, X)||_{p}=\{$

Let$AN_{n}$be the set of all absolute normalized norms on$\mathbb{C}^{n}$.

We recall the 2-dimensional case. For any $||\cdot||\in AN_{2}$ let(3)$\psi(t)=||(1-t, t)||$Then$\psi$(4)$(0\leq t\leq 1)$.

is convex continuous on $[0,1]$ and$\psi(0)=\psi(1)=1$ ,$\max\{1-t, t\}\leq\psi(t)\leq 1$ .

denote the set of all convex continuous functions on $[0,1]$ which satisfies (4).
Let
Then the converse is valid.
$\Psi_{2}$111

Theorem 1 (Bonsall-Duncan [1]). The sets
respondence under (3). That is, for any
let$AN_{2}$and$\Psi_{2}$are in $one_{-}to$ -one cor-$\psi\in\Psi_{2}$(5)$(|z|+|w|) \psi(\frac{|w|}{|z|+|w|})$

if $(z, w)\neq(0,0)$ ,$0$

if $(z, w)=(0,0)$ .

$||(z, w)||_{\psi}=\{$

Then$||\cdot||_{\psi}\in N_{2}$and$\psi(t)=||(1-t, t)||_{\psi}$

1. Absolute Norms on$(0\leq t\leq 1)$.

$\mathbb{C}^{3}$

Lemma 2. Let $||\cdot||\in AN_{3}$ . Then$||(0, y, z)||$$\leq$$||(x, y, Z)||$,$||(x, 0, Z)||$$\leq$$||(x, y, Z)||$,$||(x, y, \mathrm{o})||$$\leq$$||(x, y, z)||$.

In particular$||\cdot||_{\infty}\leq||\cdot||\leq||\cdot||_{1}$Proof. For any$(x, y, z)\in \mathbb{C}^{3}$$||(x, y, \mathrm{o})||$.

, we have$=$$\frac{1}{2}||(x, y, z)+(x, y, -Z)||$$\leq$$\frac{1}{2}(||(x, y, z)||+||(x, y, -z)||)=||(x, y, \mathcal{Z})||$.

Similarly we have the other inequalities.

Lemma 3.

Let $||\cdot||\in AN_{3}$ .
$||(x_{1,y_{1}}, z_{1})||\leq||(x_{2,y_{2}}, z_{2})||$ .

If$|x_{1}|\leq|x_{2}|,$and$|y_{1}|\leq|y_{2}|$Let$\triangle_{3}=\{(s, t) :0\leq s+t\leq 1, s, t\geq 0\}$.

For any $||\cdot||\in AN_{3}$ , we put
(6)

Then

$\psi(s, t)=||(1-s-t, s, t)||$$\psi$is a continuous convex on$\triangle_{3}$for$(s, t)\in\triangle_{3}$and satisfies that$\psi(0,0)=\psi(1,0)=\psi(0,1)=1$ .

.

$|z_{1}|\leq|Z_{2}|$, then

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Further, by Lemma 2, we have$\psi(s, t)$$\psi(s, t)$$=$$||(1-s-t, S, t)||\geq||(0, s, t)||$$=$$(s+t)||(0,$$=$$||(1-s-t, S, t)||\geq||(1-s-t, \mathrm{o}, t)||$$=$$(1-S)||(1- \frac{t}{1-s},$ $0,$ $\frac{t}{1-s})||=(1-s)\psi(0,$ $\frac{t}{1-s})$ ,$\frac{s}{s+t},$$\frac{t}{s+t})||=(s+t)\psi(\frac{s}{s+t},$ $\frac{t}{s+t})$,and similarly.

$\psi(_{S}, t)\geq(1-t)\psi(\frac{s}{1-t},$ $0)$

denote the family of all continuous convex functions onLet
following conditions:$\Psi_{3}$(7)$\psi(\mathrm{o}, \mathrm{o})$$=$$\psi(1,0)=\psi(0,1)=1$ ,(8)$\psi(s, t)$$\geq$$(s+t) \psi(\frac{s}{s+t},$(9)$\psi(s, t)$$\geq$$(1-s)\psi(0,$ $\frac{t}{1-s})$ ,(10)$\psi$$(s, t)$$\geq$$(1-t) \psi(\frac{s}{1-t},$ $0)$$\frac{t}{s+t})$.

Lemma 4. Let$\psi\in\Psi_{3}$$\triangle_{3}$satisfying the,.

. Then$1 \geq\psi(s, t)\geq\psi_{\infty}(s, t)\geq\frac{1}{3}$for all

Remark 5. We consider the following function$(s, t)\in\triangle_{3}$$\psi$

on$\triangle_{3}$.

:$\psi(s, t)=\max\{1-2s, 1-2t, 2s+2t-1\}$ .

Then is continuous convex on. Then sincethat is in, we have$\psi$$\psi$$\Psi_{3}$$\triangle_{3}$and satisfies (7), but not in. Indeed, suppose$\psi(\frac{s}{s+t}, \frac{t}{s+t})=\psi(0, \frac{t}{1-s})=\psi(\frac{s}{1-t}, 0)=1$$\triangle_{3}$$\psi(s, t)\geq\max\{s+t, 1-s, 1-t\}\geq\frac{2}{3}$by (8) $-(10)$ , whereas$\Psi_{3}$$\psi(\frac{1}{3}, \frac{1}{3})=\frac{1}{3}$, which is a contradiction.

for all $(s, t)\in$113

Theorem 6. For any $||\cdot||\in AN_{3f}$ put(11)Then(12)for$\psi(s, t)=||(1-s-t, s, t)||$$\psi\in\Psi_{3}$. Conversely, for any$\psi\in\Psi_{3}$$(s, t)\in\triangle_{3}$.

define$(|x|+|y|+|Z|) \psi(\frac{|y|}{|x|+|y|+|z|},$ $\frac{|z|}{|x|+|y|+|z|})$if $(x, y, z)\neq(0,0,0)$ ,$0$

if$||(x, y, z)||_{\psi}=\{$

Then$||\cdot||_{\psi}$

is in$AN_{3}$

and$||\cdot||_{\psi}$$(x, y, z)=(\mathrm{O}, 0, \mathrm{o})$.

satisfies (11).

We see the converse assertion. Let
the triangular inequality$\psi\in\Psi_{3}$, and define$||\cdot||_{\psi}$by (12). We only see$||(_{X_{1}}, y1, z1)+(X_{2y_{2},)|},z2|_{\psi}\leq||(_{X_{1,y}}1, z1)||_{\psi}+||(X2, y2, z_{2})||\psi\cdot$

To do this, we show that, if $0\leq p\leq a,$ $0\leq q\leq b$ and $0\leq r\leq c$ , then$||(p, q, r)||_{\psi}\leq||(a, b, c)||\psi$,that is,(13)$(p+q+r) \psi(\frac{q}{p+q+r},\frac{r}{p+q+r})\leq(a+b+c)\psi(\frac{b}{a+b+c},$ $\frac{c}{a+b+c})$ .

At first, we show that, if $0\leq p<a$ , then(14)$(p+q+r) \psi(\frac{q}{p+q+r},\frac{r}{p+q+r})\leq(a+q+r)\psi(\frac{q}{a+q+r},$ $\frac{r}{a+q+r})$ .

Take any $(s, t)\in\triangle_{3}$ such that $0<s+t<1$ . We consider the line segment. For any real number, we putsuch that 1in$s’=\lambda s$$(s’,t’)$. Then. Since
is in
and
$[(s, t), ( \frac{s}{s+t}, \frac{t}{s+t})]$$\lambda$$\triangle_{3}$$t’=\lambda t$$< \lambda\leq\frac{1}{s+t}$$[(s, t), ( \frac{s}{s+t}, \frac{t}{s+t})]$$(s’, t’)=( \lambda S, \lambda t)=\frac{(_{S}+t)(\lambda-1)}{1-s-t}(\frac{s}{s+t}$$\frac{t}{s+t})+\frac{1-\lambda(_{S+}t)}{1-s-t}(S, t)$,we have$\psi(_{S’}, t’)\leq\frac{(_{S}+t)(\lambda-1)}{1-s-t}\psi(\frac{s}{s+t},$$\frac{t}{s+t})+\frac{1-\lambda(_{S+}t)}{1-s-t}\psi(s, t)$

by the convexity of . Therefore we have$\psi$$\frac{\psi(s,t)}{s}-\frac{\psi(S^{J},t’)}{s’}$$\frac{t}{s+t})+\frac{1-\lambda(_{S+}t)}{1-s-t}\psi(s, t)\}$$\geq$$\frac{\psi(s,t)}{s}-\frac{1}{\lambda s}\{\frac{(_{S}+t)(\lambda-1)}{1-s-t}\psi(\frac{s}{s+t},$$=$$\frac{\lambda-1}{\lambda s(1-s-t)}\{\psi(s, t)-(S+t)\psi(\frac{s}{s+t},$ $\frac{t}{s+t})\}\geq 0$.

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Thus,

$\frac{\psi(s,t)}{s}\geq\frac{\psi(_{S’’},t)}{s},$. Put$s’= \frac{q}{p+q+r},$

respectively. Since$t’= \frac{r}{p+q+r},$$\frac{s’}{s}=\frac{t’}{t}=\frac{a+q+r}{p+q+r}>1$$s= \frac{q}{a+q+r})$and$t= \frac{r}{a+q+r}$, we have$\frac{\psi(\frac{q}{p+q+r},\frac{r}{p+q+r})}{\frac{q}{p+q+r}}\leq\frac{\psi(\frac{q}{a+q+r},\frac{r}{a+q+r})}{\frac{q}{a+q+r}}$.

This implies (14). Repeating a similar discussion, we obtain (13).
Now, let. Then$(x_{1,y_{1}}, z_{1}),$(15)$(x_{2,y_{2}}, z_{2})\in \mathbb{C}^{3}$$||(x_{1}, y_{1}, z1)+(x2, y2, \bigwedge_{2})\sim||_{\psi}$$=$$||(_{X_{1}+x_{2,y+Z}}1y_{21},+z2)||_{\psi}$$=$$||(|x_{1}+x_{2}|, |y1+y2|, |_{Z}1+Z_{2}|)||\psi$$\leq$$||(|x_{1}|+|x2|,$ $|y1|+|y2|,$ $|_{Z}1|+|_{Z_{2}|)}||_{\psi}$$=$$K \psi(\frac{|y_{1}|+|y_{2}|}{K},$$\frac{|z_{1}|+|_{Z_{2}|}}{K})$$(K:=|x_{1}|+|X_{2}|+|y_{1}|+|y_{2}|+|_{Z_{1}}|+|z_{2}|)$$\leq$$K \{\frac{|x_{1}|+|y_{1}|+|Z_{1}|}{K}\psi(\frac{|y_{1}|}{|x_{1}|+|y_{1}|+|Z_{1}|},$ $\frac{|z_{1}|}{|x_{1}|+|y_{1}|+|Z_{1}|})$$+ \frac{|x_{2}|+|y_{2}|+|Z_{2}|}{K}\psi(\frac{|y_{2}|}{|x_{2}|+|y_{2}|+|Z_{2}|},$$=$$||(x_{1}, y_{1}, z_{1})||\psi+||(X2, y2, z_{2})||\psi$

2. Absolute Norms onLemma 7.

$\cdot$$\frac{|z_{2}|}{|x_{2}|+|y_{2}|+|Z_{2}|})\}$.

$\mathbb{C}^{n}$

Let $||\cdot||\in AN_{n}$ . Then$(B_{1})$$||(0, X_{2}, X_{3}, \cdots, xn)||$$\leq$$||(x_{1,n}\ldots, x)||$,$(B_{2})$$||(x_{1},0, x_{3}, \cdots, x)n||$$\leq$$||(x_{1}, \cdots, x_{n})||$,$||(x_{1,n}\ldots, x)||$.

:$(B_{n})$$||(x_{1}, x_{2}, \cdots, X_{n-}1,0)||$$\leq$

In particular,$||\cdot||_{\infty}\leq||\cdot||\leq||\cdot||_{1}$.

115

Now let.

$\triangle_{n}=\{(_{S_{1},S_{2,n-1}}\cdot 2\cdot, S) : s_{1}+s_{2}+\cdots+sn-1\leq 1, s_{i}\geq 0(\forall i)\}$Take any $||\cdot||\in AN_{n}$ and let(16)$\psi(s)=||(1-s_{1^{-S_{2}}} -.

Then$\psi$.

.

-S_{n-1}, S_{1}, \cdots, S_{n}-1)||$is continuous and convex on$(A_{0})$$\triangle_{n}$for $s=(s_{1}, \cdots , s_{n-1})\in\triangle_{n}$, and$\psi(1,0, \cdots, 0)=\cdots=\psi(0, \cdots, 0,1)=1$ ,$\psi(\mathrm{o}, \cdots, \mathrm{o})$$=$$(A_{1})$$\psi(s_{1}, \cdots, S_{n}-1)$$\geq$$(s_{1}+ \cdots+s_{n}-1)\psi(\frac{s_{1}}{S_{1}+\cdots+S_{n}}, \cdots, \frac{s_{n.-1}}{S_{1}+\cdot\cdot+Sn-1})$$(A_{2})$$\psi(s_{1,n-1}\ldots, S)$$\geq$$(1-s_{1}) \psi(\mathrm{o}, \frac{s_{2}}{1-s_{1}}, \cdots, \frac{s_{n-1}}{1-s_{1}})$$(A_{n})$$\psi(s_{1,n-1}\ldots, S)$$\geq$$(1-S_{n-1}) \psi(\frac{s_{1}}{1-s_{n-1}}, \cdots, \frac{s_{n-2}}{1-s_{n-2}},0)$

Let..,,$(A_{n})$,be the family of all continuous convex functions on. The functions corresponding to -norms are$\Psi_{n}$.

$\triangle_{n}$.

satisfying$(A_{0}),$ $(A_{1})$,$\ell_{p}$$((1- \sum_{i=1}^{n}-1si)p+s_{1}^{p}+\cdots+s_{n-1})^{1/p}p$

if $1\leq p<\infty$ ,$\max(1-\sum_{i}^{n}=1s-1s_{i},1, \cdots, s_{n-}1)$

if $p=\infty$ .

$\psi_{p}(_{S_{1},s}2, \cdots, S-1)n=\{$Lemma 8. Let$\psi\in\Psi_{n}$. Then$\frac{1}{n}\leq\psi_{\infty}(_{S_{1},S_{2},\cdots,S}n-1)\leq\psi(s_{1}, S_{2}, \cdots, Sn-1)\leq 1$

for all$(s_{1}, s_{2}, \cdots, S_{n}-1)\in\triangle_{n}$.

Theorem 9. The sets $AN_{n}$ andare in one-to-one correspondence with (16).
That is, for any $||\cdot||\in AN_{n}$ the function defined by$\Psi_{n}$$\psi$(16) $\psi(s_{1}, \cdots, s_{n}-1)=||(1-s_{1} -. . . -S_{n-1}, S_{1}, \cdots, S_{n}-1)||$is in$\Psi_{n}$. Conversely, for any$\psi\in\Psi_{n}$for$(s_{1}, \cdots, s_{n-1})\in\triangle_{n}$define/$(|X_{1}|+ \cdots+|X|n)\psi(\frac{|x_{2}|}{|x_{1}|+\cdots+|x_{n}|},$$||(X1, \cdots, Xn)||_{\psi}=$$\iota 0$

Then$||\cdot||_{\psi}\in AN_{n}$and$||\cdot||_{\psi}$

satisfies (16).

$\cdots,$$\frac{|x_{n}|}{|x_{1}|+\cdots+|x_{n}|})$

if$(_{X_{1},\cdots,x_{n}})\neq(0, \cdots, 0)$,if$(x_{1}, \cdots, x_{n})=(0, \cdots, 0)$.

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3. Strict Convexity of Absolute Norms on$\mathbb{C}^{n}$$y\in X(x\neq y,$ $||x||=$A Banach space $X$ is called strictly convex if for all$||y||=1)$ we haveis called strictly convex if for all. A function
on$t\in\triangle_{n}(s\neq t)$ we have $\psi(\frac{1}{2}(s+t))<\frac{1}{2}(\psi(s)+\psi(t))$ .
by $|x|=(|x_{1}|, \cdots, |x_{n}|)$ . We say that
, we defineFor$|x|\leq|y|$ if
for $1\leq j\leq n$ . Further, we say that $|x|<|y|$ if $|x|\leq|y|$ and$|x_{j}|<|y_{j}|$ for some
. Then we have the following lemma.
$x,$$|| \frac{x+y}{2}||<1$$\triangle_{n}$$\psi$$s,$$x=(x_{1}, \cdots, x_{n})\in \mathbb{C}^{n}$$|x|$$|x_{j}|\leq|y_{j}|$$j$. Let $x=(x_{1}, \cdots, x_{n}),$Lemma 10 ([3]). Let.
(i) $If|x|\leq|y|$ , then$|x|<|y|$, thenandstrictly
convexis(ii) If$y=(y_{1}, \cdots, y_{n})\in \mathbb{C}^{n}$$\psi\in\Psi_{n}$. Then$||x||_{\psi}\leq||y||_{\psi}$$||x||_{\psi}<||y||_{\psi}$$\psi$

Theorem 11 ([3]) Let.
strictly convex on$\psi\in\Psi_{n}$. Then$||\cdot||_{\psi}$.

is strictly convex if and only if$\psi$

is$\triangle_{n}$is strictly convex. Suppose that is not strictly
We see the proof in brief. Let$t=(t_{1}, t_{2}, \cdots, t_{n-1})\in\triangle_{n}(s\neq t)$
convex. Then there exist $s=$such that $\psi(\frac{1}{2}(s+t))=\frac{1}{2}(\psi(s)+\psi(t))$ . Put $x=(1-s_{1} -. .. -s_{n-1}, s_{1}, \cdots, s_{n}-1)$
and $y=$ $(1 -t_{1} -. .. -t_{n-1}, t_{1,n-}\ldots, t1)$ . Then a direct calculation shows that$||x+y||_{\psi}=||x||_{\psi}+||y||_{\psi}$ . Since
is strictly convex, and are colinear, that is,there exists a $k>0$ such that $x=ky$ . This implies that $k=1$ , and so $x=y$ . Hence
we have $s=t$ , a contradiction.
. Take any $x=(X_{1,-1}\ldots, X_{n})$ ,Conversely, suppose that is strictly convex on$x\neq y$ , such that $||x||=||y||=1$ . Put$\psi$$||\cdot||_{\psi}$$(s_{1}, s_{2}, \cdots , s_{n-1}),$$x$$||\cdot||_{\psi}$$y$$\triangle_{n}$$\psi$$y=(y_{1}, \cdots, y_{n-1})\in \mathbb{C}^{n},$$s=( \frac{|x_{2}|}{|x_{1}|+\cdots+|x_{n}|},$

Then,

$s,$

$t\in\triangle_{n}$$\cdots,$$\frac{|x_{n}|}{|x_{1}|+\cdots+|x_{n}|})$

and$t=( \frac{|y_{2}|}{|y_{1}|+\cdots+|y_{n}|}$,$\cdot$..,$\frac{|y_{n}|}{|y_{1}|+\cdots+|y_{n}|})$.

. If $s\neq t$ , in the same way as (15), we have$||X+y||_{\psi<}||x||_{\psi}+||y||_{\psi=}2$by Lemma 10 since is strictly convex. In case of $s=t$ , we have $|x_{j}|=|y_{j}|(1\leq j\leq$$n)$ . So, there exists a positive number. Sincesuch that$x\neq y$ , there exists a, whence $0<\theta_{j_{0}}<2\pi$ . Put $c=|1+e^{i\theta_{j_{0}}}|/2$ .
such that
Then $0<c<1$ , and we have$\psi$$\theta_{j}(0\leq\theta_{j}<2\pi)$$x_{j_{0}}\neq y_{j_{0}}$$j_{0}$$||x+y||_{\psi}$$x_{j}=e^{i\theta_{j}}y_{j}$$=$$||(|_{X_{1}}+y_{1}|, \cdots, |x_{n}+y_{n}|)||_{\psi}$$=$$||(|1+e^{i}|\theta 1|y1|, \cdots, |1+e^{i}\theta_{j}0||y_{j_{0}}|, \cdots, |1+e^{i}|\theta n|yn|)||_{\psi}$$\leq$$||(2|y_{1}|, \cdots, 2_{C}|y_{j}0|, \cdots, 2|yn|)||_{\psi}$$=$$2||(|y_{1}|, \cdots, C|yj_{0}|, \cdots, |yn|)||_{\psi}$$<$$2||(|y_{1}|, \cdots f|y_{j}\mathrm{o}|, \cdots, |yn|)||\psi=2$

by Lemma 10, as is desired.

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References[1] F. F. Bonsall and J. Duncan, Numerical Ranges II, London Math. Soc. LNS No.
10, 1973.

[2] K.-S. Saito, M. Kato and Y. Takahashi, Von Neumann-Jordan constant of absolutenormalized norms on , J. Math. Anal. Appl. 244 (2000), 515-532.
$\mathbb{C}^{2}$[3] K.-S. Saito, M. Kato and Y. Takahashi, Absolute norms on
Math. Anal. Appl.

$\mathbb{C}^{n}$, to appear in J.

[4] Y. Takahashi, M. Kato and K.-S. Saito, Strict convexity of absolute norms on
and direct sums of Banach spaces, to appear in J. Inequal. Appl.

$\mathbb{C}^{2}$