44.4 Noncommutative $L^{p}$ Spaces

Proof. Since $\tau$ is faithful and normal, $\tau(p_{n}) \upto 1$ and Theorem 43.3.1 imply that $p_{n} \upto 1_{M}$ in the strong operator topology.

($\le$): Let $E: \cb_{[0, \infty)}\to M$ be the resolution of the identity for $x$. For each $\alpha > 0$,

By the Monotone Convergence Theorem,

($\ge$): Assume without loss of generality that $\tau(x) < \infty$. In which case, $\int_{[0, \infty)}|\lambda^{1/2}|^{2} E_{\wh 1_M, \wh 1_M}(d\lambda) < \infty$, and $\wh 1_{M} \in D(x^{1/2})$. Let $J: L^{2}(M, \tau) \to L^{2}(M, \tau)$ be the canonical conjugation operator, then for each $p \in \text{Proj}(M)$ with $pH \subset D(x)$, $JpJ\hat 1_{M} = Jp \hat 1_{M} = p \hat 1_{M}$. By Theorem 44.2.8, $JpJ \in M'$, so

and $\limsup_{n \to \infty}\normn{x^{1/2}p_n \wh 1_M}_{L^2(M, \tau)}^{2} \le \tau(x)$.$\square$

Proof. (1): By Lemma 44.3.4, there exist non-decreasing sequences $\seq{p_n}, \seq{q_n}\subset \text{Proj}(M)$ such that:

1. (i)

   $p_{n}L^{2}(M, \tau) \subset D(x)$ and $q_{n}L^{2}(M, \tau) \subset D(y)$ for all $n \in \natp$

2. (ii)

   $\tau(p_{n}) \upto 1$ and $\tau(q_{n}) \upto 1$ as $n \to \infty$.

For each $n \in \natp$, let $r_{n} = p_{n} \wedge q_{n}$, then $r_{n}L^{2}(M, \tau) \subset D(x) \cap D(y) \subset D(x + y)$. Moreover, $\seq{r_n}\subset \text{Proj}(M)$ is non-decreasing, with $\tau(r_{n}) \upto 1$ as $n \to \infty$. As such, by Lemma 44.4.2,

(3): Let $x = u|x|$ be the polar decomposition of $x$, and $E: \cb_{[0, \infty)}\to B(L^{2}(M, \tau))$ be the resolution of the identity for $|x|^{2}$. By Proposition 44.3.2, $u \in U(M)$ and $E(B) \in M$ for all $B \in \cb_{[0, \infty)}$. For each $n \in \natp$, let $p_{n} = E([0, n])$, then by Lemma 44.4.2,

For each $n \in \natp$, let $q_{n} = up_{n}u^{*} \in M$ be the projection onto $up_{n}L^{2}(M, \tau)$. Then, $\tau(q_{n}) = \tau(up_{n}u^{*}) = \tau(p_{n} u^{*}u) = \tau(p_{n})$, so $\tau(q_{n}) \upto 1$ as $n \to \infty$. Therefore Lemma 44.4.2 implies that

$\square$

Proof, [Theorem 7.3.2, AP]. (Well-Defined): Let $\xi \in L^{2}(M, \tau)$ and $u \in U(M') = U(R(M))$. By Theorem 44.2.8, there exists $v \in U(M)$ such that $u = R_{v}$. In which case, for each $x \in M$,

so $L_{\xi}^{0}$ is affiliated with $M$. By Lemma 44.3.7, $L_{\xi}$ is also affiliated with $M$.

(Isometric): Let $\xi \in L^{2}(M, \tau)$ and $E: \cb_{[0, \infty)}\to M$ be the resolution of the identity for $|L_{\xi}|$. By Definition 43.8.12, $D(|L_{\xi}|) = D(L_{\xi})$. In particular, $\wh 1_{M} \in D(L_{\xi})$, so

(Surjective): Let $T \in \mathcal{L}^{2}(M, \tau)$ and $\xi = T \wh 1_{M}$, then for each $x \in M$,

so $L_{\xi} \subset T$. By Proposition 44.3.3, $L_{\xi} = T$.$\square$

Proof. (2) $\Rightarrow$ (1): Let $\alpha = \norm{\text{Id}}_{L^\infty([0, \infty), \cb_{[0, \infty)}, E_{\wh 1_M, \wh 1_M}; \complex)}$, then $\tau(E((\alpha, \infty])) = 0$, and as $\tau$ is faithful, $E((\alpha, \infty]) = 0$. As such, $x = (\text{Id}\cdot \one_{[0, \alpha]})(x)$, so $x \in M$ with $\norm{x}_{M} \le \alpha$.$\square$

Proof. (1): For each $z \in \td M_{+}$, let $E(z, \cdot): \cb_{[0, \infty)}\to M$ be the resolution of the identity for $z$.

Let $x \in L^{p}(M, \tau)_{+}$. For every $n \in \natp$, let $p_{n} = E(x, [0, n])$, then $E(xp_{n}, d\lambda) = E(x, \one_{[0, n]}d\lambda)$ and $E(x(1 - p_{n}), d\lambda) = E(x, \one_{(n, \infty)}d\lambda)$. By the Dominated Convergence Theorem,

As $\seq{xp_n}\subset M_{+}$, $M_{+}$ is dense in $L^{p}(M, \tau)_{+}$. By Theorem 43.7.2, positive simple elements of $M$ are dense in $L^{p}(M, \tau)_{+}$.

(2): By (1) applied to $0 \vee x$ and $-(0 \wedge x)$.$\square$

Proof. Assume without loss of generality that $p, q \in (1, \infty)$.

(Simple): Suppose that $x$ and $y$ are simple and positive. Write $x = \sum_{j = 1}^{m} \lambda_{j} p_{j}$ and $y = \sum_{k = 1}^{n} \mu_{k} q_{k}$, where $\bracsn{p_j}_{1}^{m}, \bracsn{q_k}_{1}^{n} \subset \text{Proj}(M)$ are families of mutually orthogonal projections in $M$, with $1_{M} = \sum_{j = 1}^{m} p_{j} = \sum_{k = 1}^{n} q_{k}$.

Let $u, v \in M$ with $\norm{u}_{M}, \norm{v}_{M} \le 1$, then by the classical Hölder’s Inequality,

(Bounded): By Theorem 43.7.2, there exists positive simple operators $\seq{x_n}, \seq{y_n}\subset M$ such that $x_{n} \to x$ and $y_{n} \to y$ in $M$ as $n \to \infty$. By Definition 44.4.6, $\norm{z}_{L^p(M, \tau)}\le \norm{z}_{M}$ for all $z \in M$.

Let $u, v \in M$ with $\norm{u}_{M}, \norm{v}_{M} \le 1$, then by the simple case,

(Arbitrary): Supopse that $x \in L^{p}(M, \tau)$ and $y \in L^{q}(M, \tau)$ are arbitrary. Since $\norm{x}_{L^p(M, \tau)}= \norm{|x|}_{L^p(M, \tau)}$ and $\norm{y}_{L^q(M, \tau)}= \norm{|y|}_{L^q(M, \tau)}$, assume without loss of generality that $x$ and $y$ are positive.

By Lemma 44.3.4, there exists non-decreasing sequences $\seq{p_n}, \seq{q_n}\subset \text{Proj}(M)$ such that:

1. (i)

   $p_{n}L^{2}(M, \tau) \subset D(x)$ and $q_{n}L^{2}(M, \tau) \subset D(y)$ for all $n \in \nat$.

2. (ii)

   $xp_{n}, yq_{n} \in M$ for all $n \in \nat$.

3. (iii)

   $\tau(p_{n}), \tau(q_{n}) \upto 1$ as $n \to \infty$.

By Lemma 44.3.6, there exists a non-decreasing sequence $\seq{r_n}\subset \text{Proj}(M)$ such that:

1. (iv)

   $r_{n} L^{2}(M, \tau) \subset D(xy) \subset D(|xy|)$ and $yr_{n}L^{2}(M, \tau) \subset p_{n}L^{2}(M, \tau)$ for all $n \in \natp$.

2. (v)

   $r_{n} \le p_{n} \wedge q_{n}$ for all $n \in \natp$.

3. (vi)

   $\tau(r_{n}) \upto 1$ as $n \to \infty$.

Let $xy = u|xy|$ be the polar decomposition of $xy$, then by Lemma 44.4.2 and the bounded case,

$\square$

Proof. By Lemma 44.4.8, the linear extension theorem, and Hölder’s Inequailty.$\square$