Let $M$ be a semifinite von Neumann algebra with a trace $\tau$.Let $S(M)$ be the algebra of all affiliated operators measurable with respect to $M$. The $L^p$ norm on $M$ is given by \begin{equation} \a\^p_{L^p(M)}=\tau(a^p),\quad a\in S(M). \end{equation} Now, there is a notion of the generalised singular values for $a\in S(M)$ \begin{equation} \mu_t(a)=\inf\{s\geq 0\colon \tau(E_{(s,+\infty)}(A)\leq t\}, \end{equation} where $E_{(s,+\infty)}(A)$ is the spectral projection of the modulus of $A$. It has been shown In F. Thierry and H. Kosaki. Generalized snumbers of τmeasurable operators. Pacific J. Math., 123(2):269–300, 1986. Corollary 2.8 \begin{equation} \a\_{L^p(M)} =\left(\tau(a^p\right)^{\frac1p}=\\ \left( \int\limits^{+\infty}_0 \mu_t(a)^p\,dt \right)^{\frac1p}\quad 0<p<+\infty. \end{equation} Now, we are almost ready to formulate the question. Let $M$ be an arbitrary von Neumann algebra with a normal faithful weight $\varphi_0$. Let $L^1(M)$ be the Haagerup $L^1$space. To the best of my understanding, $L^1(M)$ is the space of all operators $a$ which are RadonNykodime derivative of a weight $\varphi$ with respect to a chosen and fixed weight $\varphi_0$. Then its $L^1$norm $\a\_{L^1(M)}=\varphi(1)$. Now, regarding the $\mu_t(a)$ for an operator $a$ from general von Neumann algebra. The crossproduct $R(M,\sigma^{\varphi_0})$ of $M$ with respect to the modular automorphism group $\sigma^{\varphi_0}$ has always a semifinite trace $\tau$. Then using the definition for $\mu_t(a)$ above it has been shown in http://www.sciencedirect.com/science/article/pii/0022123684900557 Lemma B that \begin{equation} \mu_t(a)=\frac1{t}\a\_{L^1(M)}, \end{equation} here $\mu_t(a)$ is computed with respect to the trace on $R(M,\sigma^{\varphi_0})$. Then one should get \begin{equation} \a\_{L^1(M)}=\int\limits^{+\infty}_0 \mu_t(a)dt=\a\_{L^1(M)}\int\limits^{+\infty}_0\frac{dt}{t}. \end{equation} Since the integral is divergent, we get absurd. I have to be wrong somewhere. Could anyone help me understand my mistake? I might suggest that the different traces are used. Thanks a lot.
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2$\begingroup$ In your formulas μ_t is taken with respect to two different traces: the trace τ on M and the newly constructed trace on the Takesaki dual R(M,σ). The latter trace equals ∞ when restricted to M, consistent with your last formula. This explains the apparent discrepancy. $\endgroup$ – Dmitri Pavlov Jun 11 '16 at 16:16

$\begingroup$ @DmitriPavlov Do I get it right then that $\tau(a^p)=\int\limits^{+\infty}_0 \mu_t(a)^p\,dt$ for the trace $\tau$ on the crossedproduct $R(M,\sigma)$? What reference one might give up to a ""statement x on page y in work z"? $\endgroup$ – Rauan Akylzhanov Jun 11 '16 at 22:40

$\begingroup$ The canonical trace τ on the Takesaki dual R(M,σ) is a faithful semifinite trace, hence the cited result by Fack and Kosaki is applicable. The relevant integral is infinite, thus the element a does not belong to L^1(R(M,σ),τ). $\endgroup$ – Dmitri Pavlov Jun 12 '16 at 10:25

$\begingroup$ @DmitriPavlov: Let ${\mathrm Tr}$ be the trace on Haagerup's $L^1(M)$ and $\tau$ be the canonical trace on the crossedproduct $R(M,\sigma)$. Is the cited result by Fack and Kosaki still holds for $Tr$? In other words, does this formula hold true $\a\_{L^1(M)}=\int\limits^{+\infty}_0 \mu_t(a)\,dt$, where $\mu_t(a)$ is taken with respect to the trace ${\mathrm Tr}$ on $L^1(M)$? $\endgroup$ – Rauan Akylzhanov Jun 27 '16 at 14:38

$\begingroup$ I don't think I understand your question: in order to apply the FackKosaki theorem you need a trace on a von Neumann algebra. Tr is not such a trace because L^1(M) is not a von Neumann algebra. On the other hand, τ is such a trace and the FackKosaki theorem correctly computes the answer for the case of τ, i.e., ∞. $\endgroup$ – Dmitri Pavlov Jun 28 '16 at 6:40