# not commutative symmetric vs strong symmetric spaces

Let M be a von Neumann algebra with semi-finite normal faithful trace $\tau$, $S(M)$ is space of all measurable operators introduced by I.Segal. For the self-adjoint measurable operator $X\eta M$ consider spectral function $\chi_{[s,\infty)}(X)$, its trace spectral or distribution function $d_X(s)=\tau(\chi_{[s,\infty)}(X))$ and not increasing re-arrangement function or singular value function define as $\mu(t,X)=\inf\{s\geq 0 | d_{|X|}(s)\leq t\}, t\geq 0$. Hardy function is $h(t,X)=\frac{1}{t}\int_0^t \mu(s,|X|) ds$. For the $X, Y\in S(M)$, $Y\prec X$ ( $Y\prec \prec X$) if $\mu(t,Y)\leq \mu(t,X)$ ($h(t,Y)\leq h(t,X)$) for any $t\geq 0$.

The Banach space $E\in S(M)$ with norm $||.||_E$ is called non commutative symmetric if together with $X\in E$ it contains $Y\in S(M)$ when $Y\prec X$; in addition norm inequality $||Y||_E\leq ||X||_E$ holds. The space $E$ is called strongly symmetric if $E$ is symmetric and in addition to $X,Y\in E$, $Y\prec\prec X$ implies $||Y||_E\leq ||X||_E$. Any example of non commutative symmetric and not strongly symmetric space?

-
Please go to help and read "How do I ask a good question?". – Bill Johnson Jun 28 '13 at 22:13
In particular, define your terms: specify which objects live in which spaces; and so forth – Yemon Choi Jun 29 '13 at 0:09
Added all formal definition – alg1oper Jun 29 '13 at 6:33

$$E=\{A\in M: \sup_{t>0}\frac1{\log(t+1)}\int_0^t\mu(s,X)ds<\infty\}.$$ Take $F$ to be weak $L_1$ ideal, that is $$F=\{A\in M: \sup_{t>0}t\mu(t,A)<\infty\}.$$ Clearly, $F\subset E$ and it is known that F is not dense in E. Take H to be the closure of F in E. This answers your question.