数理解析研究所講究録903 巻 1995 年 148-149

148

Two remarks on the subadditivity inequalities
in von Neumann algebras 1$by$

O. E. TIKHONOVTaking into account a developed theory of operator monotone and operator convex functions (see e.g. [1-3]) it appears interesting to study operator
subadditive functions. We do this within the context of von Neumann algebras-matrices.
though the main result is essentially a statelllent on:
In what follows we suppose that $kf$ is a von Neumann algebra and.
ofsubsetsboundedis a Borel measurable function bounded onwlth respect to [ or briefly M. We say tluat is operatorof positive operators
subadditive if $\phi(a+b)\leq\phi(a)+\phi(b)$ for every pair$M$from .
are MEXAMPLES. It is easy to see that the following functions on
subadditive for any $M$ :$2\cross 2$$\phi$$[0, \infty)arrow \mathbb{R}$$[0, \infty)$$subadd\dot{\iota}tive$$\phi$$\Lambda/$$a,$$b$$[0, \infty)$;1)$\emptyset(t)=1/(\alpha t+1)(\alpha\in \mathbb{R}^{+})$ ;2).
for some3) an arbitrary function satisfying$\phi(0)=0$
fol.
Thesatisfy
Subadditive real functions used in analysis often
lowing theorem shows ) the class of operator subadditive functions satisfying
this condition uses to be very small.
be a von Neumann algebra and let there exist a function
THEOREM 1. Let
such that $\phi(0)=0,$ is $M$ -subadditive, and is not of the form:. Then $M$ is commutative.
with$M$
Proof. Let be nonconnnutative. Suppose is $M$ -subadditive and $\phi(0)=$.
for some. We will show that has to be of the form
Since $M$ is noncolmnutative, it is easy to check that there exist two equivalent
exists a nonzero
and mutually orthogonal nonzero projections in $M$ , i.e.,$q=vv^{*}$ are$p=v^{*}v$
andpartial isolnetry $v\in M$ such that the projections
andsuch thatmutually orthogonal (see e.g. [4]). Take positive realsconsider the pair of operators:$\phi(t)=\alpha t+\beta(\alpha\in \mathbb{R}, \beta\in \mathbb{R}^{+})$$\mathrm{t}\mathrm{l}$$\alpha\in \mathrm{R}^{+}$$\alpha\leq\emptyset(t)\leq 2\alpha$$\phi$$\mathrm{a}\mathrm{t}$$\lambda/I$$\phi$$[0, \infty)arrow \mathbb{R}$$\phi$$\phi$$\alpha\in \mathbb{R}$$\emptyset(t)=\alpha t$$\phi$$0$$\alpha\in \mathbb{R}$$\emptyset(t)=\alpha t$$\phi$$\mathrm{t}1_{1}\mathrm{e}\mathrm{r}\mathrm{e}$$\epsilon\leq\delta$$\delta$$\epsilon,$$a=\epsilon p+\sqrt{\epsilon(\delta-\epsilon)}v+\sqrt{\epsilon(\delta-\epsilon)}v^{*}+(\delta-\epsilon)q$,$b=\epsilon_{P^{-}}\sqrt{\epsilon(\delta-\epsilon)}v-\sqrt{\epsilon(\delta-\epsilon)}v+*(\delta-\epsilon)q$.

and a straightObserve that and are positive scalar multiples of
forward computation shows that $\phi(a+b)=\phi(2\epsilon)p+\phi(2(\delta-\epsilon))q,$ $\phi(a)=$the inequality $\phi(a+b)\leq$, after.
.
$\phi(a)+\phi(b)$ by
right we obtainfrom the left alld
. As the only restriction imposed on and 6 isHence,is a constant, say , on $(0, \infty)$ . Thus,, it follows that.
on
$b$

$a$

$(\phi(\delta)/\mathit{5})a,$$\mathrm{I}$$\phi(b)=(\phi(\mathit{5})/\delta)b$$l\wedge^{\gamma}11\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}$)$\mathrm{r}\mathrm{o}\mathrm{j}\mathrm{e}\mathrm{C}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}$$1\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}_{\mathrm{P}^{1}}\mathrm{y}\mathrm{i}\mathrm{n}\mathrm{g}$$\phi(2\epsilon)p\leq(2\epsilon\phi(\delta)/\delta)p$$\mathrm{t}\mathrm{l}\mathrm{l}\mathrm{e}$$p$$\phi(2\epsilon)/2\epsilon\leq\phi(\delta)/\delta$$0<\epsilon\leq\delta$$\emptyset(t)/t$$\epsilon$$\alpha$$[0, \infty)$

1Supported by Russian Foundation for Basic Research, grant 93-011-16099.

$\emptyset(t)=\alpha t$

149

By similar arguments, we can also prove the following.
THEOREM 2. Let be a semifinite normal faithful trace on a noncommutative von Neumann algebra M. Let
satisfy $\phi(0)=0$ and$\tau$$\phi:[0, \infty)arrow \mathbb{R}$$(*)$$\tau(\phi(a+b))\leq\tau(\phi(a))+\tau(\phi(b))$

such that both sides of thefor every pair a, of positive operators from.
inequality are well-defined. Then is concave on
$M$

$b$

$[0, \infty)$$\phi$

This theorem may be viewed as a converse to the well-known assertion that,
under certain restrictions, the inequality $(*)$ holds if is supposed to be awith $\phi(0)=0$ (see [5], [6], [7]).
concave function onACKNOWLEDGEMENT. The author would like to thank P. G. Ovchinnikov
for a fruitful conversation.
$\phi$$[0, \infty)$

REFERENCES

1. K. L\"owner, \"Uber monotone MatriXfunktio.nen, Math. Z. 38 (1934), 177-216.

2. J. Bendat and S. Sherman, Monotone and convex operator functions,
Trans. Amer. Math. Soc. 79 (1955), 58-71.
3. F. Hansen and G. K. Pedersen, Jensen’s inequality for operators andL\"owner’s theorem, Math. Ann. 258 (1982), 229-241.
4. M. Takesaki, Theory of operator algebras I (Springcr-Verlag, 1979).
5. T. Fack and H. Kosaki, Generalized -numbers of -measurable operators,
Pacific J. Math. 123 (1982), 257-262.
6. L. G. Brown and H. Kosaki, Jensen’s inequality in semi-finite von Neumann algebras, J. Oper. Theory 23 (1990), 3-19.
7. O. E. Tikhonov, Convex functions and inequalities for traces, in: Konstr.
Teor. Funktsli i Funktsional. Anal. 6 (Kazan Univ., 1987), 77-82 (in Russian).
$\mathrm{s}$alldResearch Institute ofKazan University,
Universitetskaya 17,Kazan, Tatarstan,420008, Russia.
-mail: tikhonov@niimm.kazan.su$\mathrm{M}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{I}11\mathrm{a}\mathrm{t}\mathrm{i}_{\mathrm{C}}\mathrm{s}$$\mathrm{E}$$\mathrm{M}\mathrm{c}\mathrm{c}11\mathrm{a}\mathrm{n}\mathrm{i}_{\mathrm{C}\mathrm{S}}$$\tau$