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/Part 7: Operator Algebras/Chapter 40: Examples of Operator Algebras

40.8 $L^{\infty}$

Proposition 40.8.1.label Let $X$ be a compact Hausdorff space and $\mathscr{M}\subset M_{R}(X; \complex)$ be a closed subspace such that:

  1. (A)

    For each $\mu \in \mathscr{M}$ and $\nu \in M_{R}(X; \complex)$ with $\nu \ll \mu$, $\nu \in \mathscr{M}$.

and

\[J: C(X; \complex) \to \mathscr{M}^{*} \quad \dpn{\mu, J(f)}{\mathscr{M}}= \int f d\mu\]

then

  1. (1)

    $J(C(X; \complex))$ is weak*-dense in $\mathscr{M}^{*}$.

  2. (2)

    There exists a unique weak*-continuous involution on $\mathscr{M}^{*}$ such that $J(f^{*}) = J(f)^{*}$ for all $f \in C(X; \complex)$, given by

    \[\dpn{\mu, \phi^*}{\mathscr{M}^*}= \ol{\dpn{\mu, \phi}{\mathscr{M}^*}}\]

  3. (3)

    There exists a unique separately weak*-continuous bilinear map on $\mathscr{M}^{*}$ such that $J(fg) = J(f)J(g)$ for all $f, g \in C(X; \complex)$.

  4. (4)

    $\mathscr{M}^{*}$ equipped with the above involution and product is a commutative unital $C^{*}$-algebra.

Proof. Let $\seqi{\mu}$ be a maximal mutually singular family of Radon measures on $X$. Using Theorem 24.8.2 and the Riesz Representation Theorem, identify

\[\mathscr{M}= [l^{1}(I); L^{1}(\mu_{i}; \complex)] \quad \mathscr{M}^{*} = [l^{\infty}(I); L^{\infty}(\mu_{i}; \complex)]\]

(1): Under the above, $C(X; \complex)$ may be identified as the diagonal

\[\bracsn{f \in C(X; \complex)^I|f_i = f_j \forall i, j \in I}\subset [l^{\infty}(I); L^{\infty}(\mu_{i}; \complex)]\]

By Goldstine’s Theorem, $C(X; \complex)$ is weak*-dense in $C(X; \complex)^{**}$. As a result, $J(C(X; \complex))$ is weak*-dense in $\mathscr{M}^{*}$.

(2): For each $g \in [l^{\infty}(I); L^{\infty}(\mu_{i}; \complex)]$, let $g^{*} = \ol g$. For any $\mu \in [l^{1}(I); L^{1}(\mu_{i}; \complex)]$,

\[\dpn{\mu, g^*}{[l^1(I); L^1(\mu_i; \complex)]}= \dpn{\mu, \ol g}{[l^1(I); L^1(\mu_i; \complex)]}= \ol{\dpn{\mu, g}{[l^1(I); L^1(\mu_i; \complex)]}}\]

so the conjugation map is weak*-continuous.

(3): Let $f, g \in [l^{\infty}(I); L^{\infty}(\mu_{i}; \complex)]$ and $\mu \in [l^{1}(I); L^{1}(\mu_{i}; \complex)]$,

\[\dpn{\mu, fg}{[l^1(I); L^1(\mu_i; \complex)]}= \dpn{f\mu, g}{[l^1(I); L^1(\mu_i; \complex)]}= \dpn{g\mu, f}{[l^1(I); L^1(\mu_i; \complex)]}\]

so the composition map is separately weak*-continuous.

(4): $[l^{\infty}(I); L^{\infty}(\mu_{i}; \complex)]$ is a commutative unital $C^{*}$-algebra.$\square$

Proposition 40.8.2.label Let $(X, \cm, \mu)$ be a localisable measure space, and let $L^{\infty}(X; \complex)$ act on $L^{2}(X; \complex)$ by multiplication, then:

  1. (1)

    The weak* topology on $L^{\infty}(X; \complex)$ is equal to the weak operator topology of $L^{\infty}(X; \complex)$ acting on $L^{2}(X; \complex)$.

  2. (2)

    $L^{\infty}(X; \complex)$ acting on $L^{2}(X; \complex)$ is a von Neumann algebra.

Proof. (2): Let $A \subset B(L^{2}(X; \complex))$ be the von Neumann algebra generated by $L^{\infty}(X; \complex)$ acting on $L^{2}(X; \complex)$. By Theorem 16.3.4, $L^{\infty}(X; \complex)$ is the dual of $L^{1}(X; \complex)$, so (1) and the Banach-Alaoglu Theorem imply that the closed unit ball of $L^{\infty}(X; \complex)$ acting on $L^{2}(X; \complex)$ is weak-operator closed. By the Kaplansky Density Theorem, $\ol{B_{L^\infty(X; \complex)}(0, 1)}= \ol{B_A(0, 1)}$. Therefore $L^{\infty}(X; \complex) = A$.$\square$

Theorem 40.8.3 (Order Structure of $L^{\infty}$).label Let $(X, \cm, \mu)$ be a localisable measure space, then

  1. (1)

    $L^{\infty}(X; \real)$ is order complete.

  2. (2)

    For each $\phi \in L^{1}(X; \real)$ with $\phi \ge 0$ and bounded directed subset $S \subset L^{\infty}(X; \real)$,

    \[\sup_{f \in S}\dpn{\phi, f}{L^1(X; \real)}= \bigg\langle\phi, \sup_{f \in S}f\bigg\rangle_{L^1(X; \real)}\]

Proof. By Proposition 40.8.2, $L^{\infty}(X; \complex)$ acting on $L^{2}(X; \complex)$ is a von Neumann algebra.

(1): Since $L^{\infty}(X; \real)$ is a lattice, it is order complete by Theorem 39.3.1.

(2): By Theorem 39.3.1, $\sup(S) = \sotlim_{f \in S}f = \wotlim_{f \in S}f$. By (1) of Proposition 40.8.2, $\sup_{f \in S}\dpn{\phi, f}{L^1(X; \real)}= \dpn{\phi, \sup_{f \in S}f}{L^1(X; \real)}$.$\square$

Corollary 40.8.4.label Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^{\infty}(X))$ is extremally disconnected.

Proof. By Theorem 40.8.3, $L^{\infty}(X; \real)$ is order complete. By Corollary 38.6.4, $\Omega(L^{\infty}(X))$ is extremally disconnected.$\square$

Lemma 40.8.5.label Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be localisable measure spaces, and $T: L^{\infty}(X; \complex)\to L^{\infty}(Y; \complex)$ such that:

  1. (a)

    $T$ is an isometric isomorphism.

  2. (b)

    For each $f, g \in L^{\infty}(X; \complex)$, $f \ge g$ if and only if $Tf \ge Tg$.

then:

  1. (1)

    $T^{*}(L^{1}(Y; \complex)) \subset L^{1}(X; \complex)$.

  2. (2)

    $T$ is weak*-continuous.

Proof. Let $\phi \in L^{1}(Y; \complex)$ with $\phi \ge 0$, then $T^{*}\phi \in L^{\infty}(X; \complex)^{*}$. For each $\seq{B_n}\subset \cm$ and $B \in \cm$ with $B_{n} \upto B$, $\one_{B} = \sup_{n \in \natp}\one_{B_n}$ as an element of $L^{\infty}(X; \complex)$. Thus (b) and Theorem 40.8.3 imply that

\begin{align*}\sup_{n \in \natp}\dpn{\one_{B_n}, T^*\phi}{L^\infty(X; \complex)}&= \sup_{n \in \natp}\dpn{\phi, T\one_{B_n}}{L^1(Y; \complex)}= \dpn{\phi, T\one_B}{L^1(Y; \complex)}\\&= \dpn{\one_B, T^*\phi}{L^\infty(X; \complex)}\end{align*}

Hence the mapping $B \mapsto \dpn{\one_B, T^*\phi}{L^\infty(X; \complex)}$ is a finite positive measure on $(X, \cm)$, which is absolutely continuous with respect to $\mu$. By the Radon-Nikodym Theorem, there exists $f \in L^{1}(X; \complex)$ with $f \ge 0$ such that $\int_{B} f d\mu = \dpn{\one_{B}, T^*\phi}{L^\infty(X; \complex)}$ for all $B \in \cm$.

Let $g \in L^{\infty}(X; [0, 1])$. By Lemma 26.6.4, there exists simple functions $\seq{g_n}\subset \Sigma(X; [0, 1])$ such that $g_{n} \upto g$ pointwise. In which case, $g = \sup_{n \in \natp}g_{n}$ as an element of $L^{\infty}(X; \real)$, so (b) and Theorem 40.8.3 imply that,

\begin{align*}\dpn{\phi, Tg}{L^1(Y; \complex)}&= \sup_{n \in \natp}\dpn{\phi, Tg_n}{L^1(Y; \complex)}= \sup_{n \in \natp}\dpn{f, g_n}{L^1(X; \complex)}\\&= \dpn{f, g}{L^1(X; \complex)}\end{align*}

By linearity, $\dpn{\phi, Tg}{L^1(Y; \complex)}= \dpn{f, g}{L^1(X; \complex)}$ for all $g \in L^{\infty}(X; \complex)$. Therefore $T^{*}(L^{1}(Y; \complex)) \subset L^{1}(X; \complex)$, and $T$ is weak*-continuous.$\square$

Theorem 40.8.6 (Uniqueness of $L^{\infty}$).label Let $X$ be a compact Hausdorff space, $\mu, \nu: \cb_{X} \to [0, \infty)$ be Radon measures on $X$, and $\Phi: L^{\infty}(\mu; \complex) \to L^{\infty}(\nu; \complex)$ such that:

  1. (a)

    $\Phi$ is a *-isomorphism.

  2. (b)

    $\Phi|_{C(X; \complex)}$ is the identity.

then:

  1. (1)

    $\mu$ and $\nu$ are equivalent.

  2. (2)

    $L^{\infty}(\mu; \complex) = L^{\infty}(\nu; \complex)$.

  3. (3)

    $\Phi$ is the identity map.

Proof, [Theorem 21.4, Zhu93]. As $\Phi$ is a *-isomorphism, $\Phi f \ge \Phi g$ if and only if $f \ge g$ for any $f, g \in L^{\infty}(\mu; \complex)$.

(1): By Lemma 40.8.5, there exists $f \in L^{1}(\mu; \complex)$ such that $\int f g d\mu = \int g d\nu$ for all $g \in C(X; \complex)$. Thus the uniqueness of the Riesz Representation Theorem implies that $\nu = f d\mu$. As the argument is symmetric, the two measures are equivalent.

(3): By Lemma 40.8.5, $\Phi$ is also weak*-continuous. By Goldstine’s Theorem, $C(X; \complex)$ is weak*-dense in $L^{\infty}(\mu; \complex)$ and $L^{\infty}(\nu; \complex)$. As $\Phi$ is the identity on $C(X; \complex)$, $\Phi$ is the identity on $L^{\infty}(\mu; \complex)$.$\square$

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