Square-integrable pairings and strong involution

Two families of tests recover two different norms. Pairing a bounded operator with the square-integrable unit ball detects its square integrability. Pairing a square-integrable operator with the trace-norm unit ball detects its boundedness. These tests also clarify why a square-integrable factor can repair the discontinuity of adjoint in the strong topology.

Until LQ-04, let $M$ be a von Neumann algebra with a faithful normal semifinite trace $\tau$. No separability, sigma-finiteness or finiteness of $\tau(1)$ is assumed. We allow the zero algebra. Use the bounded trace ideals $\begin{gathered} \mathfrak n_\tau=\{a\in M:\tau(a^*a)<\infty\},\\ \mathfrak m_\tau=\{a\in M:\tau(|a|)<\infty\}. \end{gathered}$ The actual measurable-operator space $L^2=L^2(M,\tau)$, its bounded module actions and its approximation by bounded finite-support operators were constructed in Complete norms on actual measurable operators. In particular, $M\cap L^2=\mathfrak n_\tau$, and $\mathfrak m_\tau\subseteq\mathfrak n_\tau$. Its inner product is linear in the first variable: $\langle x,y\rangle_2=\tau(y^*x).$ The trace Hilbert space proves this formula and the Cauchy–Schwarz inequality. Products of two $L^2$ elements lie in $L^1$, by Hölder's inequality and cyclicity. Products of measurable operators mean the closed products constructed there.

Testing a bounded operator for square integrability

For $x\in M$, put $\begin{gathered} A_x=\sup\{\,|\tau(y^*x)|:\\ y\in\mathfrak m_\tau,\ \|y\|_2\leq1\,\}. \end{gathered}$ Every tested trace exists even before we know that $x\in L^2$: the bounded finite trace ideal is a two-sided star ideal.

Theorem. With extended nonnegative values, $A_x=\tau(x^*x)^{1/2}. \tag{LQ.1}$ Consequently $x\in\mathfrak n_\tau$ if and only if $A_x<\infty$, and then $A_x=\|x\|_2$.

Proof. Let $e$ be any finite-trace projection and put $y=xe$. Its right support is at most $e$, so $y$ lies in the finite-support ideal $\mathfrak m_0\subseteq\mathfrak m_\tau$. Set $t=\|xe\|_2$. The bounded cyclic trace identities in TB-01 give $\begin{gathered} \tau(y^*x)=\tau(ex^*x)\\ =\tau(ex^*xe)=\tau(xex^*)=t^2. \end{gathered}$ For example, cyclically moving the last $e$ in $ex^*xe$ gives $\tau(ex^*x)$. All these products lie in the finite trace ideal. If $t>0$, testing with $y/t$ yields $A_x\geq t$. For $t=0$ the same inequality holds automatically.

By Finite pieces without a countability assumption, the directed net of all finite-trace projections satisfies $e\uparrow1$. Thus $xex^*\uparrow xx^*$, and normality and the trace identity give $\begin{gathered} \sup_e\|xe\|_2^2=\tau(xx^*)\\ =\tau(x^*x). \end{gathered}$ This proves the lower bound in (LQ.1), including the case of an infinite right side. If $\tau(x^*x)<\infty$, Cauchy–Schwarz gives $|\tau(y^*x)|\leq\|y\|_2\|x\|_2,$ which proves the opposite bound. The zero operator and zero algebra cause no exception. $\square$

Testing a square-integrable vector for boundedness

For $x\in L^2$, define $\begin{gathered} B_x=\sup\{\,|\langle x,y\rangle_2|:\\ y\in\mathfrak m_\tau,\ \|y\|_1\leq1\,\},\\ D_x=\sup\{\,\|ax\|_2:\\ a\in\mathfrak n_\tau,\ \|a\|_2\leq1\,\}. \end{gathered}$ The second definition makes sense for every $x\in L^2$, because each $a\in\mathfrak n_\tau$ is bounded and hence acts on $L^2$.

Theorem. The following conditions are equivalent: $x$ is bounded; $B_x<\infty$; $D_x<\infty$. When they hold, $B_x=D_x=\|x\|. \tag{LQ.2}$ Equivalently, either finite testing constant characterizes membership in $\mathfrak n_\tau$ among all $L^2$ vectors.

The upper bounds for bounded vectors. If $x$ is bounded, The trace norm and the dual pairing gives $|\tau(y^*x)|\leq\|y^*\|_1\|x\| =\|y\|_1\|x\|.$ The bounded right module estimate of TI-09 gives $\|ax\|_2\leq\|a\|_2\|x\|.$ Thus both testing constants are at most $\|x\|$.

Finite spectral tests. Write the measurable polar decomposition as $x=uk$, where $k=|x|$ and $u^*u=s(k)$. Its existence and spectral calculus are proved in Positivity, square roots and polar decomposition. For $c>0$, put $e=1_{(c,\infty)}(k).$ The positive spectral integral gives $c^2\tau(e)\leq\tau(k^2)<\infty.$ If $e\ne0$, then $0<\tau(e)<\infty$. Two normalized tests are $\begin{gathered} y_e=\frac{ue}{\tau(e)},\\ a_e=\frac{eu^*}{\sqrt{\tau(e)}}. \end{gathered} \tag{LQ.3}$ Since $e\leq u^*u$, we have $|ue|=e$, and $ueu^*$ is a projection of trace $\tau(e)$. The two supports of $ue$ and $eu^*$ therefore have finite trace. In particular, $\begin{gathered} y_e\in\mathfrak m_0,\quad \|y_e\|_1=1,\\ a_e\in\mathfrak n_\tau,\quad \|a_e\|_2=1. \end{gathered}$ For the second norm identity, $a_e^*a_e=ueu^*/\tau(e)$.

These tests have the exact values $\begin{gathered} \langle x,y_e\rangle_2\\ =\frac{\tau(ek)}{\tau(e)}\geq c,\\ \|a_ex\|_2^2\\ =\frac{\tau(k^2e)}{\tau(e)}\geq c^2. \end{gathered} \tag{LQ.4}$ Here $ek\in L^1$: both $e$ and $k$ are in $L^2$, and their product is integrable by TI-10. Thus the first trace is defined. The identities follow from $e u^*u=e$, spectral commutation, and the closed products $y_e^*x=ek/\tau(e)$ and $a_ex=ek/\sqrt{\tau(e)}$. The second product belongs to $L^2$; its squared modulus is $k^2e/\tau(e)$. No unverified trace of an unbounded product is being used.

If $B_x=C<\infty$, choose $c=C+\varepsilon$ in this construction. A nonzero $e$ would contradict the first inequality in (LQ.4). Hence $1_{(C+\varepsilon,\infty)}(k)=0$ for every $\varepsilon>0$, which implies $k\leq C$ by spectral calculus. Thus $x$ is bounded. If instead $D_x=C<\infty$, the second inequality yields the identical conclusion. This proves both converses independently.

Finally, if $x\ne0$ is bounded and $0<c<\|x\|$, the spectral projection $e$ above is nonzero. Formula (LQ.4) gives $B_x\geq c$ and $D_x\geq c$. Letting $c\uparrow\|x\|$ proves (LQ.2). If $x=0$, all three numbers are zero. $\square$

The adjoints in (LQ.3) have different orientations: $y_e$ is tested through $y_e^*x$, whereas $a_e$ is tested through $a_ex$. The argument therefore applies to arbitrary complex operators, without positivity of $x$.

A square-integrable factor and the strong topology

The sigma-strong topology on $M$ is generated by $p_\varphi(b)=\varphi(b^*b)^{1/2}, \qquad \varphi\in M_*^+.$ The sigma-strong seminorms proves this description and its equivalence with vector seminorms. We first relate it to the square-integrable norm.

Lemma. If $b_i\in\mathfrak n_\tau$, $\sup_i\|b_i\|\leq K<\infty$, and $\|b_i\|_2\to0$, then $b_i\to0$ sigma-strongly in $M$.

Proof. Fix $\varphi\in M_*^+$. The positive trace-density representation supplies $h\in L^1_+$ such that $\varphi(z)=\tau(hz)$. By The complete integrable space, its spectral cutoffs $h_n=h1_{[1/n,n]}(h)$ are bounded, positive and integrable, with $\|h-h_n\|_1\to0$. The trace pairing bound and bounded cyclicity imply $\begin{gathered} 0\leq\varphi(b_i^*b_i)\\ \leq K^2\|h-h_n\|_1\\ {}+\|h_n\|\,\|b_i\|_2^2. \end{gathered} \tag{LQ.5}$ For the last term, cyclicity gives $\begin{gathered} \tau(h_n b_i^*b_i)=\tau(b_i h_n b_i^*),\\ 0\leq b_i h_n b_i^*\leq\|h_n\|b_i b_i^*. \end{gathered}$ Every positive operator in this inequality has finite trace, and $\tau(b_i b_i^*)=\|b_i\|_2^2$. Given a positive error tolerance, first choose $n$ so that the first term of (LQ.5) is small, and then a tail of the original net so that its second term is small. If $K=0$, the conclusion is immediate. It follows that $p_\varphi(b_i)\to0$ for every $\varphi$. $\square$

Theorem. For every fixed $a\in\mathfrak n_\tau$, the map $\begin{gathered} M\longrightarrow M,\\ x\longmapsto ax^* \end{gathered} \tag{LQ.6}$ is sigma-strongly continuous on every operator-norm-bounded subset of $M$.

Proof. First suppose that $x_i\to0$ sigma-strongly and $\sup_i\|x_i\|\leq K$. The element $a^*a$ is bounded, positive and integrable. The bounded trace-density representation therefore gives a positive normal functional $\psi(z)=\tau(a^*az).$ Bounded cyclicity and the definition of sigma-strong convergence give $\begin{gathered} \|ax_i^*\|_2^2=\tau(x_i a^*a x_i^*)\\ =\psi(x_i^*x_i)\longrightarrow0. \end{gathered}$ Moreover, $ax_i^*\in\mathfrak n_\tau$ and $\|ax_i^*\|\leq\|a\|K$. The lemma now proves $ax_i^*\to0$ sigma-strongly in $M$, the required target topology.

For a net $x_i\to x$ in a fixed norm-bounded subset, apply this argument to $x_i-x$, which remains uniformly bounded. Since $a(x_i-x)^*=ax_i^*-ax^*$, the conclusion is continuity at every point of that subset. $\square$

The equality involving $\psi$ already proves convergence in $L^2$. The lemma is the additional step that proves convergence in the algebra itself.

Adjoint is discontinuous in an infinite corner

This section needs no trace. Let $M$ be any von Neumann algebra and let $e\in M$ be an infinite projection. By the definition of infiniteness, there is a partial isometry $v\in eMe$ with $v^*v=e,\qquad vv^*<e.$ Here $e$ is the identity of the corner.

Theorem. Adjoint on the operator-norm unit ball of $eMe$ is not sigma-strongly continuous at zero.

Proof. Put $p=e-vv^*$, a nonzero projection. We have $pv=0$ and $v^*p=0$. For integers $n\geq0$, with $v^0=e$, define $\begin{gathered} w_n=v^np,\\ q_n=w_nw_n^*,\\ z_n=w_n^*. \end{gathered} \tag{LQ.7}$ Because $(v^*)^n v^n=e$, one has $w_n^*w_n=p$; thus $q_n$ is a projection and $\|z_n\|=1$. For $n<m$, $w_n^*w_m=pv^{m-n}p=0,$ so $q_nq_m=0$. The projections $q_n$ are pairwise orthogonal and their finite sums are at most $e$.

For every positive normal functional $\varphi$ on $eMe$, $\sum_{n=0}^N\varphi(q_n)\leq\varphi(e).$ The nonnegative scalar terms consequently tend to zero. Since $\begin{gathered} z_n^*z_n=q_n,\\ z_nz_n^*=p, \end{gathered} \tag{LQ.8}$ the sigma-strong seminorm formula of CP-08 gives $z_n\to0$.

On the other hand, choose a nonzero vector $\xi\in pH$ in a faithful concrete representation of the corner. Its vector functional $\psi(b)=\langle b\xi,\xi\rangle$ is positive and normal, as proved in The concrete predual. Then $p_\psi(z_n^*)^2=\psi(p)=\|\xi\|^2>0$ for every $n$. Hence $z_n^*$ does not converge sigma-strongly to zero. This sequence in the unit ball proves the asserted discontinuity. $\square$

The construction uses the given infinite projection itself. Neither proper infiniteness of $e$ nor a countability assumption on $M$ is needed.

Three tests of the hypotheses

A bounded operator that is not square integrable. Let $M=\ell^\infty(\mathbb N)$ with counting trace, and let $x=1$. The trace is faithful and normal because coordinate suprema commute with each finite coordinate sum; taking the supremum over finite sets gives the full sum. Finite coordinate truncations prove semifiniteness. If $e_N$ projects onto the first $N$ coordinates, then $\begin{gathered} y_N=N^{-1/2}e_N\in\mathfrak m_\tau,\\ \|y_N\|_2=1,\qquad \tau(y_N^*x)=\sqrt N. \end{gathered}$ Thus $A_x=\infty$, in agreement with $\tau(x^*x)=\infty$.

A square-integrable operator that is not bounded. On the same diagonal algebra, put $\tau(b)=\sum_{n\geq1}8^{-n}b_n\qquad(b\geq0).$ This trace is faithful, normal by the same finite-sum argument, and finite with $\tau(1)=1/7$; finiteness implies semifiniteness. Let $x$ be the diagonal operator with entries $2^n$, on its maximal domain $D(x)=\{\xi:\sum_{n\geq1}4^n|\xi_n|^2<\infty\}.$ Finite-coordinate vectors prove domain density. Coordinate convergence of a vector and its image proves closedness, and the diagonal spectral projections give affiliation. The trace of each high spectral tail tends to zero, so The spectral-tail criterion makes $x$ measurable. Its squared norm is $\|x\|_2^2=\sum_{n\geq1}2^{-n}=1.$ For the coordinate projection $q_n$, set $y_n=8^n q_n,\qquad a_n=8^{n/2}q_n.$ Direct evaluation gives $\begin{gathered} \|y_n\|_1=1,\quad \langle x,y_n\rangle_2=2^n,\\ \|a_n\|_2=1,\quad \|a_nx\|_2=2^n. \end{gathered}$ Both testing constants in LQ-02 are infinite although $x$ is a unit vector in $L^2$.

The shift version of the infinite-corner sequence. On $\ell^2(\mathbb N_0)$, let $v\delta_n=\delta_{n+1}$. Then $v^*v=1$, $p=1-vv^*=E_{00}$, and the construction in LQ-04 gives $q_n=E_{nn},\qquad z_n=E_{0n}.$ For each $\xi\in\ell^2$, $\|z_n\xi\|=|\xi_n|\to0$; hence $z_n\to0$ strongly. Since the sequence is uniformly bounded, CP-08 also gives sigma-strong convergence. But $z_n^*\delta_0=\delta_n$ has norm one for every $n$. With the usual trace and $a=p$, one has $az_n^*=0$ for all $n\geq1$. This directly illustrates how the square-integrable factor in LQ-03 changes the continuity behavior.

The corresponding testing and topology statements are Lemmas V.2.20–2.21 and V.2.27–2.28 of M. Takesaki, Theory of Operator Algebras I, printed pp. 323–324 and 326–327.