Orthogonally additive polynomials on non-commutative $L^p$-spaces

Abstract

Let ${{\mathscr {M}}}$ be a von Neumann algebra with a normal semifinite faithful trace $\tau$. We prove that every continuous m-homogeneous polynomial P from $L^p({{\mathscr {M}}},\tau )$, with $0<p<\infty$, into each topological linear space X with the property that $P(x+y)=P(x)+P(y)$ whenever x and y are mutually orthogonal positive elements of $L^p({{\mathscr {M}}},\tau )$ can be represented in the form $P(x)=\varPhi (x^m)$$(x\in L^p({{\mathscr {M}}},\tau ))$ for some continuous linear map $\varPhi :L^{p/m}({{\mathscr {M}}},\tau )\rightarrow X$.

1 Introduction

In [16], the author succeeded in providing a useful representation of the orthogonally additive homogeneous polynomials on the spaces $L^p([0,1])$ and $\ell ^p$ with $1\le p<\infty$. In [12] (see also [6]), the authors obtained a similar representation for the space C(K), for a compact Hausdorff space K. These results were generalized to Banach lattices [4] and Riesz spaces  [9]. Further, the problem of representing the orthogonally additive homogeneous polynomials has been also considered in the context of Banach function algebras [1, 19] and non-commutative Banach algebras [2, 3, 11]. Notably, [11] can be thought of as the natural non-commutative analogue of the representation of orthogonally additive polynomials on C(K)-spaces, and the purpose to this paper is to extend the results of [16] on the representation of orthogonally additive homogeneous polynomials on $L^p$-spaces to the non-commutative $L^p$-spaces.

The non-commutative $L^p$-spaces that we consider are those associated with a von Neumann algebra ${{\mathscr {M}}}$ equipped with a normal semifinite faithful trace $\tau$. From now on, $S({{\mathscr {M}}},\tau )$ stands for the linear span of the positive elements x of ${{\mathscr {M}}}$ such that $\tau \bigl ({{\,\mathrm{supp}\,}}(x)\bigr )<\infty$; here ${{\,\mathrm{supp}\,}}(x)$ stands for the support of x. Then $S({{\mathscr {M}}},\tau )$ is a $*$-subalgebra of ${{\mathscr {M}}}$ with the property that $\vert x\vert ^p\in S({{\mathscr {M}}},\tau )$ for each $x\in S({{\mathscr {M}}},\tau )$ and each $0<p<\infty$. For $0<p<\infty$, we define $\Vert \cdot \Vert _p:S({{\mathscr {M}}},\tau )\rightarrow {\mathbb {R}}$ by $\Vert x\Vert _p=\tau (\vert x\vert ^p)^{1/p}$$(x\in S({{\mathscr {M}}},\tau ))$. Then $\Vert \cdot \Vert _p$ is a norm or a p-norm according to $1\le p<\infty$ or $0<p<1$, and the space $L^p({{\mathscr {M}}},\tau )$ can be defined as the completion of $S({{\mathscr {M}}},\tau )$ with respect to $\Vert \cdot \Vert _p$. Nevertheless, for our purposes here, it is important to realize the elements of $L^p({{\mathscr {M}}},\tau )$ as measurable operators. Specifically, the set $L^0({{\mathscr {M}}},\tau )$ of measurable closed densely defined operators affiliated to ${{\mathscr {M}}}$ is a topological $*$-algebra with respect to the strong sum, the strong product, the adjoint operation, and the topology of the convergence in measure. The algebra ${{\mathscr {M}}}$ is a dense $*$-subalgebra of $L^0({{\mathscr {M}}},\tau )$, the trace $\tau$ extends to the positive cone of $L^0({{\mathscr {M}}},\tau )$ in a natural way, and we can define

$$\begin{aligned}\begin{gathered} \Vert x\Vert _p=\tau \bigl (\vert x\vert ^p\bigr )^{1/p}\quad (x\in L^0({{\mathscr {M}}},\tau )), \\ L^p({{\mathscr {M}}},\tau )=\bigl \{x\in L^0({{\mathscr {M}}},\tau ) : \Vert x\Vert _p<\infty \bigr \}. \end{gathered}\end{aligned}$$

Also we set $L^\infty ({{\mathscr {M}}},\tau )={{\mathscr {M}}}$ (with $\Vert \cdot \Vert _\infty :=\Vert \cdot \Vert$, the operator norm). Operators $x,y\in L^0({{\mathscr {M}}},\tau )$ are mutually orthogonal, written $x\perp y$, if $xy^*=y^*x=0$. This condition is equivalent to requiring that x and y have mutually orthogonal left, and right, supports. Further, for $x,y\in L^p({{\mathscr {M}}},\tau )$ with $0<p<\infty$, the condition $x\perp y$ implies that $\Vert x+y\Vert _p^p=\Vert x\Vert _p^p+\Vert y\Vert _p^p$, and conversely, if $\Vert x\pm y\Vert _p^p=\Vert x\Vert _p^p+\Vert y\Vert _p^p$ and $p\ne 2$, then $x\perp y$ (see [14, Fact 1.3]). The orthogonal additivity considered in [16] for the spaces $L^p([0,1])$ and $\ell ^p$ can of course equally well be considered for the space $L^p({{\mathscr {M}}},\tau )$. Let P be a map from $L^p({{\mathscr {M}}},\tau )$ into a linear space X. Then P is:

• (i) orthogonally additive on a subset ${\mathscr {S}}$ of $L^p({{\mathscr {M}}},\tau )$ if

  $$\begin{aligned} x,y \in {\mathscr {S}}, \, x\perp y=0 \ \Rightarrow \ P(x+y)=P(x)+P(y); \end{aligned}$$

• (ii) an m-homogeneous polynomial if there exists an m-linear map $\varphi$ from $L^p({{\mathscr {M}}},\tau )^m$ into X such that

  $$\begin{aligned} P(x)=\varphi \left( x,\dotsc ,x \right) \quad (x\in L^p({{\mathscr {M}}},\tau )). \end{aligned}$$

  Here and subsequently, $m\in {\mathbb {N}}$ is fixed with $m\ge 2$ and the superscript m stands for the m-fold Cartesian product. Such a map is unique if it is required to be symmetric. Further, in the case where X is a topological linear space, the polynomial P is continuous if and only if the symmetric m-linear map $\varphi$ associated with P is continuous.

Given a continuous linear map $\varPhi :L^{p/m}({{\mathscr {M}}},\tau )\rightarrow X$, where X is an arbitrary topological linear space, the map $P_\varPhi :L^p({{\mathscr {M}}},\tau )\rightarrow X$ defined by

$$\begin{aligned} P_\varPhi (x)=\varPhi (x^m)\quad (x\in L^p({{\mathscr {M}}},\tau )) \end{aligned}$$

is a natural example of a continuous m-homogeneous polynomial which is orthogonally additive on $L^p({{\mathscr {M}}},\tau )_{\mathrm{sa}}$ (Theorem 4), and we will prove that every continuous m-homogeneous polynomial which is orthogonally additive on $L^p({{\mathscr {M}}},\tau )_{\mathrm{sa}}$ is actually of this special form (Theorem 5). Here and subsequently, the subscripts “sa” and $+$ are used to denote the self-adjoint and the positive parts of a given subset of $L^0({{\mathscr {M}}},\tau )$, respectively.

We require a few remarks about the setting of our present work. Throughout the paper we are concerned with m-homogeneous polynomials on the space $L^p({{\mathscr {M}}},\tau )$ with $0<p$, and thus one might wish to consider polynomials with values in the space $L^q({{\mathscr {M}}},\tau )$, especially with $q\le p$. Further, in the case case where $p/m<1$ and the von Neumann algebra ${{\mathscr {M}}}$ has no minimal projections, there are no non-zero continuous linear functionals on $L^{p/m}({{\mathscr {M}}},\tau )$; since one should like to have non-trivial “orthogonally additive” polynomials on $L^p({{\mathscr {M}}},\tau )$, some weakening of the normability must be allowed to the range space (see Corollary 2). For these reasons, throughout the paper, X will be a (complex and Hausdorff) topological linear space. In the case where the von Neumann algebra ${{\mathscr {M}}}$ is commutative, the prototypical polynomials $P_\varPhi$ mentioned above are easily seen to be orthogonally additive on the whole domain. In contrast, we will point out in Propositions 1 and 3 that this is not the case for the von Neumann algebra ${{\mathscr {B}}}(H)$ of all bounded operators on a Hilbert space H whenever $\dim H\ge 2$.

We assume a basic knowledge of $C^*$-algebras and von Neumann algebras, tracial non-commutative $L^p$-spaces, and polynomials on topological linear spaces. For the relevant background material concerning these topics, see [5, 7, 8, 10, 13, 17, 18].

2 $C^*$-algebras and von Neumann algebras

Our approach to the problem of representing the orthogonally additive homogeneous polynomials on the non-commutative $L^p$-spaces relies on the representation of those polynomials on the von Neumann algebras.

Recall that two elements x and y of a $C^*$-algebra ${\mathscr {A}}$ are mutually orthogonal if $xy^*=y^*x=0$, in which case the identity $\Vert x+y\Vert =\max \{\Vert x\Vert ,\Vert y\Vert \}$ holds. The reader should be aware that we have chosen the standard definition of orthogonality in the setting of non-commutative $L^p$-spaces. This definition is slightly different from the one used in [11], which is the standard one in the setting of Banach algebras. In [11] the orthogonality of two elements x and y is defined by the relation $xy=yx=0$, and, further, the orthogonally additive polynomials on the self-adjoint part of a $C^*$-algebra are automatically orthogonally additive on the whole algebra. The important point to note here is that both the definitions of orthogonality agree on the self-adjoint part of the $C^*$-algebra. Thus, for a polynomial on a $C^*$-algebra, the property of being orthogonally additive on the self-adjoint part according to our definition is the same as being orthogonally additive according to [11]. Nevertheless, in contrast to [11], there are no non-zero orthogonally additive polynomials from the von Neumann algebra ${{\mathscr {B}}}(H)$ into any topological Banach space according to our definition (Proposition 1).

Suppose that ${\mathscr {A}}$ is a linear space with an involution $*$. Recall that for a linear functional $\varPhi :{\mathscr {A}}\rightarrow {\mathbb {C}}$, the map $\varPhi ^*:{\mathscr {A}}\rightarrow {\mathbb {C}}$ defined by $\varPhi ^*(x)=\overline{\varPhi (x^{*})}$$(x\in {\mathscr {A}})$ is a linear functional, and $\varPhi$ is said to be hermitian if $\varPhi ^*=\varPhi$. Similarly, for an m-homogeneous polynomial $P:{\mathscr {A}}\rightarrow {\mathbb {C}}$, the map $P^*:{\mathscr {A}}\rightarrow {\mathbb {C}}$ defined by $P^*(x)=\overline{P(x^*)}$$(x\in {\mathscr {A}})$ is an m-homogeneous polynomial, and we call P hermitian if $P^*=P$.

Lemma 1

Let X and Y be linear spaces, and let $P:X\rightarrow Y$ be an m-homogeneous polynomial. Suppose that P vanishes on a convex set $C\subset X$. Then P vanishes on the linear span of C.

Proof

Set $x_1,x_2,x_3,x_4\in C$. Let $\eta :Y\rightarrow {\mathbb {C}}$ be a linear functional, and define $f:{\mathbb {C}}^4\rightarrow {\mathbb {C}}$ by

$$\begin{aligned} f(\alpha _1,\alpha _2,\alpha _3,\alpha _4)= \eta \bigl (P(\alpha _1x_1+\alpha _2 x_2+\alpha _3x_3+\alpha x_4)\bigr )\quad (\alpha _1,\alpha _2,\alpha _3,\alpha _4\in {\mathbb {C}}). \end{aligned}$$

Then f is a complex polynomial function in four complex variables that vanishes on the set

$$\begin{aligned} \bigl \{(\rho _1,\rho _2,\rho _3,\rho _4)\in {\mathbb {R}}^4 : 0\le \rho _1,\rho _2,\rho _3,\rho _4, \ \rho _1+\rho _2+\rho _3+\rho _4=1\bigr \}. \end{aligned}$$

This implies that f is identically equal to 0 on ${\mathbb {C}}^4$, and, in particular,

$$\begin{aligned} \eta \bigl (P(\rho _1x_1-\rho _2x_2+i\rho _3x_3-i\rho _4x_4)\bigr )=f(\rho _1,-\rho _2,i\rho _3,-i\rho _4)=0 \end{aligned}$$

for all $\rho _1,\rho _2,\rho _3,\rho _4\ge 0$. Since this identity holds for each linear functional $\eta$, it may be concluded that $P(\rho _1x_1-\rho _2x_2+i\rho _3x_3-i\rho _4x_4)=0$ for all $\rho _1,\rho _2,\rho _3,\rho _4\ge 0$. Thus P vanishes on the set

$$\begin{aligned} \bigl \{ \rho _1x_1-\rho _2x_2+i\rho _3x_3-i\rho _4x_4 : \rho _j\ge 0, \ x_j\in C \ (j=1,2,3,4) \bigr \}, \end{aligned}$$

which is exactly the linear span of the set C. $\square$

Theorem 1

Let ${\mathscr {A}}$ be a $C^*$-algebra, let X be a topological linear space, and let $\varPhi :{\mathscr {A}}\rightarrow X$ be a continuous linear map. Then:

• (i) the map $P_\varPhi :{\mathscr {A}}\rightarrow X$ defined by $P_\varPhi (x)=\varPhi (x^m)$$(x\in {\mathscr {A}})$ is a continuous m-homogeneous polynomial which is orthogonally additive on ${\mathscr {A}}_{\mathrm{sa}}$;

• (ii) the polynomial $P_\varPhi$ is uniquely specified by the map $\varPhi$.

Suppose, further, that X is a q-normed space, $0<q\le 1$. Then:

• (iii) $2^{-1/q}\Vert \varPhi \Vert \le \Vert P_\varPhi \Vert \le \Vert \varPhi \Vert$.

Moreover, in the case where $X={\mathbb {C}}$,

• (iv) the functional $\varPhi$ is hermitian if and only if the polynomial $P_\varPhi$ is hermitian, in which case $\Vert P_\varPhi \Vert =\Vert \varPhi \Vert$.

Proof

(i) It is clear that the map $P_\varPhi$ is continuous and that $P_\varPhi$ is the m-homogeneous polynomial associated with the symmetric m-linear map $\varphi :{\mathscr {A}}^m\rightarrow X$ defined by

$$\begin{aligned} \varphi (x_1,\ldots ,x_m)= \frac{1}{m!}\sum _{\sigma \in \mathfrak {S}_m} \varPhi \left( x_{\sigma (1)} \cdots x_{\sigma (m)} \right) \quad (x_1,\ldots ,x_m\in {\mathscr {A}}); \end{aligned}$$

here and subsequently, we write $\mathfrak {S}_m$ for the symmetric group of order m.

Suppose that $x,y\in {\mathscr {A}}_{\mathrm{sa}}$ are such that $x\perp y$. Then $xy=yx=0$, and so $(x+y)^m=x^m+y^m$, which gives

$$\begin{aligned} P_\varPhi (x+y)=\varPhi \bigl ((x+y)^m\bigr )= \varPhi \bigl (x^m+y^m\bigr )=\varPhi \bigl (x^m\bigr )+\varPhi \bigl (y^m\bigr )= P_\varPhi (x)+P_\varPhi (y). \end{aligned}$$

(ii) Assume that $\Psi :{\mathscr {A}}\rightarrow X$ is a linear map with the property that $P_\Psi =P_\varPhi$. If $x\in {\mathscr {A}}_+$, then

$$\begin{aligned} \varPhi (x)=\varPhi \bigl ((x^{1/m})^{m}\bigr )=P(x^{1/m})=\Psi \bigl ((x^{1/m})^{m}\bigr )=\Psi (x). \end{aligned}$$

By linearity we also get $\Psi (x) = \varPhi (x)$ for each $x \in {\mathscr {A}}$.

(iii) Next, assume that X is a q-normed space. For each $x\in {\mathscr {A}}$, we have

$$\begin{aligned} \Vert P_\varPhi (x)\Vert =\Vert \varPhi (x^m)\Vert \le \Vert \varPhi \Vert \Vert x^m\Vert \le \Vert \varPhi \Vert \Vert x\Vert ^m, \end{aligned}$$

which implies that $\Vert P_\varPhi \Vert \le \Vert \varPhi \Vert$. Now take $x\in {\mathscr {A}}$, and let $\omega \in {\mathbb {C}}$ with $\omega ^m=-1$. Then $x=\mathfrak {R}x+i\mathfrak {I}x$, where

$$\begin{aligned} \mathfrak {R}x=\frac{1}{2}(x^*+x), \, \mathfrak {I}x=\frac{i}{2}(x^*-x)\in {\mathscr {A}}_{\mathrm{sa}}, \end{aligned}$$

and, further, $\Vert \mathfrak {R}x\Vert ,\Vert \mathfrak {I}x\Vert \le \Vert x\Vert$. Moreover, $\mathfrak {R}x=x_1-x_2$ and $\mathfrak {I}x=x_3-x_4$, where $x_1,x_2,x_3,x_4\in {\mathscr {A}}_+$, $x_1\perp x_2$, and $x_3\perp x_4$. Since $x_1\perp x_2$ and $x_3\perp x_4$, it follows that $x_1^{1/m}\perp x_2^{1/m}$ and $x_3^{1/m}\perp x_4^{1/m}$. Consequently,

$$\begin{aligned} {\begin{matrix} \Vert \mathfrak {R}x\Vert &{}=\max \bigl \{\Vert x_1\Vert ,\Vert x_2\Vert \bigr \},\\ \Vert \mathfrak {I}x\Vert &{}=\max \bigl \{\Vert x_3\Vert ,\Vert x_4\Vert \bigr \}, \end{matrix}} \end{aligned}$$

(1)

and

$$\begin{aligned} {\begin{matrix} {\big \Vert } x_1^{1/m}+\omega x_2^{1/m} {\big \Vert }&{}= \max \bigl \{ {\big \Vert } x_1^{1/m}{\big \Vert },{\big \Vert } x_2^{1/m} {\big \Vert } \bigr \},\\ {\big \Vert } x_3^{1/m}+\omega x_4^{1/m} {\big \Vert }&{}= \max \bigl \{ {\big \Vert }x_3^{1/m}{\big \Vert },{\big \Vert } x_4^{1/m} {\big \Vert } \bigr \}. \end{matrix}} \end{aligned}$$

(2)

Since

$$\begin{aligned} \big \Vert x_1^{1/m}\big \Vert = \Vert x_1\Vert ^{1/m}, \ \big \Vert x_2^{1/m}\big \Vert = \Vert x_2\Vert ^{1/m}, \ \big \Vert x_3^{1/m}\big \Vert = \Vert x_3\Vert ^{1/m}, \ \big \Vert x_4^{1/m}\big \Vert = \Vert x_4\Vert ^{1/m}, \end{aligned}$$

it follows, from (1) and (2), that

$$\begin{aligned} {\begin{matrix} {\big \Vert } x_1^{1/m}+\omega x_2^{1/m}{\big \Vert }^m &{}= \max \bigl \{\Vert x_1\Vert ,\Vert x_2\Vert \bigr \}=\Vert {\mathfrak {R}} x\Vert ,\\ {\big \Vert } x_3^{1/m}+\omega x_4^{1/m}{\big \Vert }^m &{}= \max \bigl \{\Vert x_3\Vert ,\Vert x_4\Vert \bigr \}=\Vert {\mathfrak {I}} x\Vert . \end{matrix}} \end{aligned}$$

(3)

On the other hand, we have

$$\begin{aligned} \bigl (x_1^{1/m}+\omega x_2^{1/m}\bigr )^m=x_1-x_2=\mathfrak {R}x,\quad \bigl (x_3^{1/m}+\omega x_4^{1/m}\bigr )^m=x_3-x_4=\mathfrak {I}x, \end{aligned}$$

and so

$$\begin{aligned} {\begin{matrix} \varPhi (x) &{}= \varPhi (\mathfrak {R}x)+i\varPhi (\mathfrak {I}x)= \varPhi \bigl (\bigl (x_1^{1/m}+\omega x_2^{1/m}\bigr )^m\bigr )+ i\varPhi \bigl (\bigl (x_3^{1/m}+\omega x_4^{1/m}\bigr )^m\bigr )\\ &{}= P_\varPhi \bigl (x_1^{1/m}+\omega x_2^{1/m}\bigr )+ iP_\varPhi \bigl (x_3^{1/m}+\omega x_4^{1/m}\bigr ). \end{matrix}} \end{aligned}$$

Hence, by (3),

$$\begin{aligned} \Vert \varPhi (x)\Vert ^q\le & {} \big \Vert P_\varPhi \bigl (x_1^{1/m}+\omega x_2^{1/m}\bigr )\big \Vert ^q+ \big \Vert P_\varPhi \bigl (x_3^{1/m}+\omega x_4^{1/m}\bigr )\big \Vert ^q\\\le & {} \Vert P_\varPhi \Vert ^q\big \Vert x_1^{1/m}+\omega x_2^{1/m}\big \Vert ^{mq}+ \Vert P_\varPhi \Vert ^q\big \Vert x_3^{1/m}+\omega x_4^{1/m}\big \Vert ^{mq}\\= & {} \Vert P_\varPhi \Vert ^q\left( \Vert \mathfrak {R}x\Vert ^q+\Vert \mathfrak {I}x\Vert ^q\right) \\\le & {} \Vert P_\varPhi \Vert ^q 2\Vert x\Vert ^q. \end{aligned}$$

This clearly forces $\Vert \varPhi \Vert \le 2^{1/q}\Vert P_\varPhi \Vert$, as claimed.

(iv) It is straightforward to check that $P_\varPhi ^*=P_{\varPhi ^*}$. Consequently, if $\varPhi$ is hermitian, then $P_\varPhi ^*=P_{\varPhi ^*}=P_\varPhi$ so that $P_\varPhi$ is hermitian. Conversely, if $P_\varPhi$ is hermitian, then $P_{\varPhi ^*}=P_\varPhi ^*=P_\varPhi$ and (ii) implies that $\varPhi ^*=\varPhi$. Finally, assume that $\varPhi$ is a hermitian functional. For the calculation of $\Vert P_\varPhi \Vert$ it suffices to check that $\Vert \varPhi \Vert \le \Vert P_\varPhi \Vert$. For this purpose, let $\varepsilon \in {\mathbb {R}}^+$, and choose $x\in {\mathscr {A}}$ such that $\Vert x\Vert =1$ and $\Vert \varPhi \Vert -\varepsilon <\vert \varPhi (x)\vert$. We take $\alpha \in {\mathbb {C}}$ with $\vert \alpha \vert =1$ and $\vert \varPhi (x)\vert =\alpha \varPhi (x)$, so that

$$\begin{aligned} \Vert \varPhi \Vert -\varepsilon <\vert \varPhi (x)\vert =\varPhi (\alpha x)= \overline{\varPhi (\alpha x)}=\varPhi \bigl ((\alpha x)^*\bigr ). \end{aligned}$$

Note that $\Vert \mathfrak {R}(\alpha x)\Vert \le 1$ and $\Vert \varPhi \Vert -\varepsilon <\varPhi (\mathfrak {R}(\alpha x))$. Now we consider the decomposition $\mathfrak {R}(\alpha x)=x_1-x_2$ with $x_1,x_2\in {\mathscr {A}}_+$ and $x_1\perp x_2$ and take $\omega \in {\mathbb {C}}$ with $\omega ^m=-1$. As in (3), we see that $\big \Vert x_1^{1/m}+\omega x_2^{1/m}\big \Vert =\Vert \mathfrak {R}(\alpha x)\Vert ^{1/m}\le 1$. Moreover, we have

$$\begin{aligned} P_\varPhi \bigl ( x_1^{1/m}+\omega x_2^{1/m}\bigr )= \varPhi \bigl (\bigl ( x_1^{1/m}+\omega x_2^{1/m}\bigr )^m\bigr )= \varPhi (\mathfrak {R}(\alpha x)), \end{aligned}$$

which gives $\Vert \varPhi \Vert -\varepsilon <\Vert P_\varPhi \Vert$. $\square$

Lemma 2

Let ${\mathscr {A}}$ be a $C^*$-algebra, let ${\mathscr {R}}$ be a $*$-subalgebra of ${\mathscr {A}}$, let X be a topological linear space, and let $\varPhi :{\mathscr {R}}\rightarrow X$ be a linear map. Suppose that the polynomial $P:{\mathscr {R}}\rightarrow X$ defined by $P(x)=\varPhi (x^m)$$(x\in {\mathscr {R}})$ is continuous and that ${\mathscr {R}}$ satisfies the following conditions:

• (i) $\vert x\vert \in {\mathscr {R}}$ for each $x\in {\mathscr {R}}_{\mathrm{sa}}$;

• (ii) $x^{1/m}\in {\mathscr {R}}$ for each $x\in {\mathscr {R}}_+$.

Then $\varPhi$ is continuous.

Proof

Let U be a neighbourhood of 0 in X. Let V be a balanced neighbourhood of 0 in X with $V+V+V+V\subset U$. The set $P^{-1}(V)$ is a neighbourhood of 0 in ${\mathscr {R}}$, which implies that there exists $r\in {\mathbb {R}}^+$ such that $P(x)\in V$ whenever $x\in {\mathscr {R}}$ and $\Vert x\Vert <r$. Take $x\in {\mathscr {R}}$ with $\Vert x\Vert <r^m$. Since ${\mathscr {R}}$ is a $*$-subalgebra of ${\mathscr {A}}$, we see that $\mathfrak {R}x,\mathfrak {I}x\in {\mathscr {R}}_{\mathrm{sa}}$. We write $\mathfrak {R}x=x_1-x_2$ and $\mathfrak {I}x=x_3-x_4$, as in the proof of Theorem 1, where, on account of the condition (i),

$$\begin{aligned} x_1&=\tfrac{1}{2}\bigl (\vert \mathfrak {R}x\vert +\mathfrak {R}x\bigr )\in {\mathscr {R}}_+,&x_2&=\tfrac{1}{2}\bigl (\vert \mathfrak {R}x\vert -\mathfrak {R}x\bigr ) \in {\mathscr {R}}_+,\\ x_3&=\tfrac{1}{2}\bigl (\vert \mathfrak {I}x\vert +\mathfrak {I}x\bigr )\in {\mathscr {R}}_+,&x_4&=\tfrac{1}{2}\bigl (\vert \mathfrak {I}x\vert -\mathfrak {I}x\bigr ) \in {\mathscr {R}}_+. \end{aligned}$$

For each $j\in \{1,2,3,4\}$, condition (ii) gives $x_j^{1/m}\in {\mathscr {R}}$, and, further, we have $\Vert x_{j}^{1/m}\big \Vert =\Vert x_j\Vert ^{1/m}\le \Vert x\Vert ^{1/m}<r$. Hence

$$\begin{aligned} \varPhi (x)&= \varPhi \bigl ( \bigl (x_{1}^{1/m}\bigr )^m-\bigl (x_{2}^{1/m}\bigr )^m+i\bigl (x_{3}^{1/m}\bigr )^m -i\bigl (x_{4}^{1/m}\bigr )^m\bigr ) \\&= \varPhi \bigl (\bigl (x_{1}^{1/m}\bigr )^m\bigr )- \varPhi \bigl (\bigl (x_{2}^{1/m}\bigr )^m\bigr )+ i\varPhi \bigl (\bigl (x_{3}^{1/m}\bigr )^m\bigr )- i\varPhi \bigl (\bigl (x_{4}^{1/m}\bigr )^m\bigr ) \\&= P\bigl (x_{1}^{1/m}\bigr )-P\bigl (x_{2}^{1/m}\bigr )+iP\bigl (x_{3}^{1/m}\bigr ) -iP\bigl (x_{4}^{1/m}\bigr )\in V+V+V+V\subset U, \end{aligned}$$

which establishes the continuity of $\varPhi$. $\square$

Theorem 2

Let ${\mathscr {A}}$ be a $C^*$-algebra, let X be a locally convex space, and let $P:{\mathscr {A}}\rightarrow X$ be a continuous m-homogeneous polynomial. Then the following conditions are equivalent:

• (i) there exists a continuous linear map $\varPhi :{\mathscr {A}} \rightarrow X$ such that $P(x)=\varPhi (x^m)$$(x\in {\mathscr {A}})$;

• (ii) the polynomial P is orthogonally additive on ${\mathscr {A}}_{\mathrm{sa}}$;

• (iii) the polynomial P is orthogonally additive on ${\mathscr {A}}_+$.

If the conditions are satisfied, then the map $\varPhi$ is unique.

Proof

Theorem 1 gives (i)$\Rightarrow$(ii), and obviously (ii)$\Rightarrow$(iii). The task is now to prove that (iii)$\Rightarrow$(i).

Suppose that (iii) holds. For each continuous linear functional $\eta :X\rightarrow {\mathbb {C}}$, set $P_\eta =\eta \circ P$. Then $P_\eta$ is a complex-valued continuous m-homogeneous polynomial. We claim that $P_\eta$ is orthogonally additive on ${\mathscr {A}}_{\mathrm{sa}}$. Take $x,y\in {\mathscr {A}}_{\mathrm{sa}}$ with $x\perp y$. Then we can write $x=x_{+}-x_{-}$ and $y=y_{+}-y_{-}$ with $x_{+},x_{-},y_{+},y_{-}\in {\mathscr {A}}_{+}$ mutually orthogonal. Define $f:{\mathbb {C}}^2\rightarrow {\mathbb {C}}$ by

$$\begin{aligned} f(\alpha ,\beta )\!= \!P_\eta (x_++\alpha x_-\!+\!y_++\beta y_-)\!-\! P_\eta (x_+\!+\!\alpha x_-)\!-\!P_\eta (y_+\!+\!\beta y_-)\quad (\alpha ,\beta \in {\mathbb {C}}^2). \end{aligned}$$

Then f is a complex polynomial function in two complex variables. If $\alpha ,\beta \in {\mathbb {R}}^+$, then $x_++\alpha x_-,y_++\beta y_-\in {\mathscr {A}}_+$ are mutually orthogonal, and so, by hypothesis, $P(x_++\alpha x_-+y_++\beta y_-)=P(x_++\alpha x_-)+P(y_++\beta y_-)$. This shows that $f(\alpha ,\beta )=0$. Since f vanishes on ${\mathbb {R}}^{+}\times {\mathbb {R}}^{+}$, it follows that f vanishes on ${\mathbb {C}}^2$, which, in particular, implies

$$\begin{aligned} P_\eta (x+y)-P_\eta (x)-P_\eta (y)=f(-1,-1)=0. \end{aligned}$$

Having proved that $P_\eta$ is orthogonally additive on ${\mathscr {A}}_{\mathrm{sa}}$ we can apply [11, Theorem 2.8] to obtain a unique continuous linear functional $\varPhi _\eta$ on ${\mathscr {A}}$ such that

$$\begin{aligned} \eta \bigl (P(x)\bigr )=\varPhi _\eta (x^m)\quad (x\in {\mathscr {A}}). \end{aligned}$$

(4)

Each $x\in {\mathscr {A}}$ can be written in the form $x_1^m+\cdots +x_k^m$ for suitable $x_1,\ldots ,x_k\in {\mathscr {A}}$, and we define

$$\begin{aligned} \varPhi (x)=\sum _{j=1}^kP(x_j). \end{aligned}$$

Our next goal is to show that $\varPhi$ is well-defined. Suppose that $x_1,\ldots ,x_k\in {\mathscr {A}}$ are such that $x_1^m+\cdots +x_k^m=0$. For each continuous linear functional $\eta$ on X, (4) gives

$$\begin{aligned} \eta \left( \sum _{j=1}^kP(x_j)\right) = \sum _{j=1}^k\eta \bigl (P(x_j)\bigr )= \sum _{j=1}^k\varPhi _\eta (x_j^m)= \varPhi _\eta \left( \sum _{j=1}^k x_j^m\right) = 0. \end{aligned}$$

Since X is locally convex, we conclude that $\sum _{j=1}^kP(x_j)=0$.

It is a simple matter to check that $\varPhi$ is linear and, by definition, $P(x)=\varPhi (x^m)$$(x\in {\mathscr {A}})$. The continuity of $\varPhi$ then follows from Lemma 2.

The uniqueness of the map $\varPhi$ follows from Theorem 1(ii). $\square$

The assumption that the space X be locally convex can be removed by requiring that the $C^*$-algebra ${\mathscr {A}}$ be sufficiently rich in projections. The real rank zero is the most important existence of projections property in the theory of $C^*$-algebras. We refer the reader to [5, Section V.3.2] and [7, Section V.7] for the basic properties and examples of $C^*$-algebras of real rank zero. This class of $C^*$-algebras contains the von Neumann algebras and the $C^*$-algebras ${\mathscr {K}}(H)$ of all compact operators on any Hilbert space H. Let us remark that every $C^*$-algebra of real rank zero has an approximate unit of projections (but not necessarily increasing).

Theorem 3

Let ${\mathscr {A}}$ be a $C^*$-algebra of real rank zero, let X be a topological linear space, and let $P:{\mathscr {A}}\rightarrow X$ be a continuous m-homogeneous polynomial. Suppose that ${\mathscr {A}}$ has an increasing approximate unit of projections. Then the following conditions are equivalent:

• (i) there exists a continuous linear map $\varPhi :{\mathscr {A}} \rightarrow X$ such that $P(x)=\varPhi (x^m)$$(x\in {\mathscr {A}})$;

• (ii) the polynomial P is orthogonally additive on ${\mathscr {A}}_{\mathrm{sa}}$;

• (iii) the polynomial P is orthogonally additive on ${\mathscr {A}}_+$.

If the conditions are satisfied, then the map $\varPhi$ is unique.

Proof

Theorem 1 gives (i)$\Rightarrow$(ii), and it is clear that (ii)$\Rightarrow$(iii). We will henceforth prove that (iii)$\Rightarrow$(i).

We first note that such a map $\varPhi$ is necessarily unique, because of Theorem 1(ii).

Suppose that (iii) holds and that ${\mathscr {A}}$ is unital. Let $\varphi :{\mathscr {A}}^m\rightarrow X$ be the symmetric m-linear map associated with P and define $\varPhi :{\mathscr {A}}\rightarrow X$ by

$$\begin{aligned} \varPhi (x)=\varphi (x,1,\ldots ,1)\quad (x\in {\mathscr {A}}). \end{aligned}$$

Let $Q:{\mathscr {A}}\rightarrow X$ be the m-homogeneous polynomial defined by

$$\begin{aligned} Q(x)=\varPhi (x^m)\quad (x\in {\mathscr {A}}). \end{aligned}$$

We will prove that $P=Q$. On account of Lemma 1, it suffices to show that $P(x)=Q(x)$ for each $x\in {\mathscr {A}}_{\mathrm{sa}}$.

First, consider the case where $x\in {\mathscr {A}}_{\mathrm{sa}}$ has finite spectrum, say $\{\rho _1,\ldots ,\rho _k\}\subset {\mathbb {R}}$. This implies that x can be written in the form

$$\begin{aligned} x=\sum _{j=1}^k\rho _j e_j, \end{aligned}$$

where $e_1,\ldots ,e_k\in {\mathscr {A}}$ are mutually orthogonal projections (specifically, the projection $e_j$ is defined by using the continuous functional calculus for x by $e_j=\chi _{\{\rho _j\}}(x)$ for each $j\in \{1,\ldots ,k\}$). We also set $e_0=1-(e_1+\cdots +e_k)$, so that the projections $e_0,e_1,\ldots ,e_k$ are mutually orthogonal, and $\rho _0=0$. We claim that if $j_1,\ldots ,j_m\in \{0,\ldots ,k\}$ and $j_l\ne j_{l'}$ for some $l,l'\in \{1,\ldots ,m\}$, then

$$\begin{aligned} \varphi (e_{j_1},\ldots ,e_{j_m})=0. \end{aligned}$$

(5)

Let $\varLambda _1=\bigr \{n\in \{1,\ldots ,m\} : j_n=j_l\bigr \}$ and $\varLambda _2=\bigr \{n\in \{1,\ldots ,m\} : j_n\ne j_l\bigr \}$. For each $\alpha _1,\ldots ,\alpha _m\in {\mathbb {R}}^+$, the elements $\sum _{n\in \varLambda _1}\alpha _{n}e_{j_n}$ and $\sum _{n\in \varLambda _2}\alpha _{n}e_{j_n}$ are positive and mutually orthogonal, so that the orthogonal additivity of P on ${\mathscr {A}}_+$ gives

$$\begin{aligned} P\left( \sum _{n=1}^m\alpha _ne_{j_n} \right) = P\left( \sum _{n\in \varLambda _1}\alpha _n e_{j_n}\right) + P\left( \sum _{n\in \varLambda _2}\alpha _n e_{j_n}\right) . \end{aligned}$$

This implies that, for each linear functional $\eta :X\rightarrow {\mathbb {C}}$, the function $f:{\mathbb {C}}^m\rightarrow {\mathbb {C}}$ defined by

$$\begin{aligned} f(\alpha _1,\ldots ,\alpha _m)= \eta \left( P\left( \sum _{n=1}^m\alpha _n e_{j_n}\right) - P\left( \sum _{n\in \varLambda _1}\alpha _n e_{j_n}\right) - P\left( \sum _{n\in \varLambda _2}\alpha _n e_{j_n}\right) \right) , \end{aligned}$$

for all $\alpha _1,\ldots ,\alpha _m\in {\mathbb {C}}$, is a complex polynomial function in m complex variables vanishing in $\bigl ({\mathbb {R}}^+\bigr )^m$. Therefore f vanishes on ${\mathbb {C}}^m$. Moreover, we observe that the coefficient of the monomial $\alpha _1\cdots \alpha _m$ is given by $n!\eta \bigl (\varphi (e_{j_1},\ldots ,e_{j_m})\bigr )$, because both $\varLambda _1$ and $\varLambda _2$ are different from $\{1,\ldots ,m\}$. We thus get

$$\begin{aligned}n!\eta \bigl (\varphi (e_{j_1},\ldots ,e_{j_m})\bigr )=0 .\end{aligned}$$

Since this identity holds for each linear functional $\eta$, our claim follows. Property (5) now leads to

$$\begin{aligned} {\begin{matrix} P(x) &{} = \varphi \left( \sum _{j=1}^k\rho _{j} e_j,\ldots ,\sum _{j=1}^k\rho _{j}e_j\right) = \sum _{j_1, \ldots , j_m =1}^k\rho _{j_{1}}\cdots \rho _{j_{m}}\varphi \left( e_{j_1},\ldots , e_{j_m}\right) \\ &{}= \sum _{j=1}^{k} \rho _{j}^m\varphi \left( e_j,\ldots ,e_j\right) \end{matrix}} \end{aligned}$$

and

$$\begin{aligned} {\begin{matrix} Q(x) &{}=\! \varphi \left( \!\left( \!\sum _{j=0}^k\rho _j e_j \!\right) ^m,\sum _{j=0}^k e_j,\ldots , \sum _{j=0}^k e_j\!\right) \!=\! \varphi \left( \!\sum _{j=0}^k \rho _{j}^m e_j , \sum _{j=0}^k e_j,\ldots ,\sum _{j=0}^k e_j \!\right) \\ &{}= \sum _{j_1,\ldots , j_m =0}^k \rho _{j_1}^m \varphi \left( e_{j_1},\ldots ,e_{j_m}\right) = \sum _{j=1}^k \rho _{j}^m \varphi \left( e_j , \ldots ,e_j \right) . \end{matrix}} \end{aligned}$$

We thus get $P(x)=Q(x)$.

Now suppose that $x\in {\mathscr {A}}_{\mathrm{sa}}$ is an arbitrary element. Since ${\mathscr {A}}$ has real rank zero, it follows that there exists a sequence $(x_n)$ in ${\mathscr {A}}_{\mathrm{sa}}$ such that each $x_n$ has finite spectrum and $\lim x_n=x$. On account of the above case, we have $P(x_n)=Q(x_n)$ ($n\in {\mathbb {N}}$), and the continuity of both P and Q now yields $P(x)=\lim P(x_n)=\lim Q(x_n)=Q(x)$, as required.

We are now in a position to prove the non-unital case. By hypothesis, there exists an increasing approximate unit of projections $(e_\lambda )_{\lambda \in \varLambda }$. For each $\lambda \in \varLambda$, set ${\mathscr {A}}_\lambda =e_\lambda {\mathscr {A}}e_\lambda$. Then ${\mathscr {A}}_\lambda$ is a unital $C^*$-algebra (with identity $e_\lambda$) and has real rank zero (because ${\mathscr {A}}_\lambda$ is a hereditary $C^*$-subalgebra of ${\mathscr {A}}$). From what has previously been proved, it follows that there exists a unique continuous linear map $\varPhi _\lambda :{\mathscr {A}}_\lambda \rightarrow X$ such that

$$\begin{aligned} P(x)=\varPhi _\lambda (x^m)\quad (x\in {\mathscr {A}}_\lambda ). \end{aligned}$$

(6)

Define

$$\begin{aligned} {\mathscr {R}}=\bigcup _{\lambda \in \varLambda }{\mathscr {A}}_\lambda \end{aligned}$$

and, for each $x\in {\mathscr {R}}$, set

$$\begin{aligned} \varPhi (x)=\varPhi _\lambda (x), \end{aligned}$$

where $\lambda \in \varLambda$ is such that $x\in {\mathscr {A}}_\lambda$. We will show that $\varPhi$ is well-defined. Suppose $\lambda ,\mu \in \varLambda$ are such that $x\in {\mathscr {A}}_\lambda \cap {\mathscr {A}}_\mu$. Then there exists $\nu \in \varLambda$ with $\lambda ,\mu \le \nu$. Since the net $(e_\lambda )_{\lambda \in \varLambda }$ is increasing, we see that $e_\lambda ,e_\mu \le e_\nu$ and therefore ${\mathscr {A}}_\lambda ,{\mathscr {A}}_\mu \subset {\mathscr {A}}_\nu$. The uniqueness of the representation of P on both ${\mathscr {A}}_\lambda$ and ${\mathscr {A}}_\mu$ implies that $\varPhi _\nu \mid _{{\mathscr {A}}_\lambda }=\varPhi _\lambda$ and $\varPhi _\nu \mid _{{\mathscr {A}}_\mu }=\varPhi _\mu$, which implies that $\varPhi _\lambda (x)=\varPhi _\nu (x)=\varPhi _\mu (x)$. We now show that ${\mathscr {R}}$ is a $*$-subalgebra of ${\mathscr {A}}$ and that $\varPhi$ is linear. Take $x,y\in {\mathscr {R}}$ and $\alpha ,\beta \in {\mathbb {C}}$. We take $\lambda ,\mu \in \varLambda$ such that $x\in {\mathscr {A}}_{\lambda }$ and $y\in {\mathscr {A}}_\mu$. Then $x^*\in {\mathscr {A}}_\lambda \subset {\mathscr {R}}$. Now set $\nu \in \varLambda$ with $\lambda ,\mu \le \nu$. Hence $x,y\in {\mathscr {A}}_\nu$, so that $\alpha x+\beta y,xy\in {\mathscr {A}}_\nu \subset {\mathscr {R}}$, which shows that ${\mathscr {R}}$ is a subalgebra of ${\mathscr {A}}$. Further, we have

$$\begin{aligned} \varPhi (\alpha x+\beta y)=\varPhi _\nu (\alpha x+\beta y)= \alpha \varPhi _\nu (x)+\beta \varPhi _\nu (y)=\alpha \varPhi (x)+\beta \varPhi (y), \end{aligned}$$

which shows that $\varPhi$ is linear.

From (6) we deduce that $P(x)=\varPhi (x^m)$ for each $x\in {\mathscr {R}}$.

Our next goal is to show that ${\mathscr {R}}$ satisfies the conditions of Lemma 2. If $x\in {\mathscr {R}}_{\mathrm{sa}}$ ($x\in {\mathscr {R}}_+$), then there exists $\lambda \in \varLambda$ such that $x\in \bigl ({\mathscr {A}}_\lambda \bigr )_{\mathrm{sa}}$ ($x\in \bigl ({\mathscr {A}}_\lambda \bigr )_{+}$, respectively) and therefore $\vert x\vert \in {\mathscr {A}}_\lambda \subset {\mathscr {R}}$ ($x^{1/m}\in {\mathscr {A}}_\lambda \subset {\mathscr {R}}$, respectively). Since the polynomial $P\mid _{\mathscr {R}}$ is continuous, Lemma 2 shows that the map $\varPhi$ is continuous.

Since $(e_\lambda )_{\lambda \in \varLambda }$ is an approximate unit, it follows that ${\mathscr {R}}$ is dense in ${\mathscr {A}}$, and hence that the map $\varPhi$ extends uniquely to a continuous linear map from ${\mathscr {A}}$ into the completion of X. By abuse of notation we continue to write $\varPhi$ for this extension. Since both P and $\varPhi$ are continuous, it may be concluded that $P(x)=\varPhi (x^m)$ for each $x\in {\mathscr {A}}$. We next prove that the image of $\varPhi$ is actually contained in X. Of course, it suffices to show that $\varPhi$ takes ${\mathscr {A}}_+$ into X. If $x\in {\mathscr {A}}_+$, then

$$\begin{aligned} \varPhi (x)= \varPhi \bigl (\bigl (x^{1/m}\bigr )^m\bigr )= P\bigl (x^{1/m}\bigr )\in X, \end{aligned}$$

as required. $\square$

Since every von Neumann algebra is unital and has real rank zero, Theorem 3 applies in this setting and gives the following.

Corollary 1

Let ${{\mathscr {M}}}$ be a von Neumann algebra, let X be a topological linear space, and let $P:{{\mathscr {M}}}\rightarrow X$ be a continuous m-homogeneous polynomial. Then the following conditions are equivalent:

• (i) there exists a continuous linear map $\varPhi :{{\mathscr {M}}} \rightarrow X$ such that $P(x)=\varPhi (x^m)$$(x\in {{\mathscr {M}}})$;

• (ii) the polynomial P is orthogonally additive on ${{\mathscr {M}}}_{\mathrm{sa}}$;

• (iii) the polynomial P is orthogonally additive on ${{\mathscr {M}}}_+$.

If the conditions are satisfied, then the map $\varPhi$ is unique.

Proposition 1

Let H be a Hilbert space with $\dim H \ge 2$, let X be a topological linear space, and let $P:{\mathscr {B}}(H) \rightarrow X$ be a continuous m-homogeneous polynomial. Suppose that P is orthogonally additive in ${\mathscr {B}}(H)$. Then $P=0$.

Proof

For each unitary $v\in {\mathscr {B}}(H)$, the map $P_v:{\mathscr {B}}(H)\rightarrow X$ defined by

$$\begin{aligned} P_v(x)=P(vx)\quad (x\in {\mathscr {B}}(H)) \end{aligned}$$

is easily seen to be a continuous m-homogeneous polynomial that is orthogonally additive on ${\mathscr {B}}(H)$. In particular, $P_v$ is orthogonally additive on ${\mathscr {B}}(H)_{\mathrm{sa}}$, and Corollary 1 then gives a unique continuous linear map $\varPhi _v:{\mathscr {B}}(H)\rightarrow X$ such that

$$\begin{aligned} P(vx)=\varPhi _v(x^m)\quad (x\in {\mathscr {B}}(H)). \end{aligned}$$

We claim that, if $e,e'\in {\mathscr {B}}(H)$ are equivalent projections with $e\perp e'$, then $P(e)=P(e')=0$. Let $u\in {\mathscr {B}}(H)$ be a partial isometry such that $u^*u=e$ and $uu^*=e'$. Then

$$\begin{aligned} \left\| u^2 \right\| ^4= \left\| (u^2)^*u^2 \right\| ^2= \left\| \left( (u^2)^*u^2\right) ^2\right\| = \left\| u^{*}ee^{\prime } eu \right\| =0, \end{aligned}$$

which gives $u^2=0$. From this we see that $u\perp u^*$, and therefore

$$\begin{aligned} P(vu+vu^*)= P_v(u+u^*)=P_v(u)+P_v(u^*)=\varPhi _v\left( u^m\right) +\varPhi _v\left( (u^*)^m\right) =0. \end{aligned}$$

(7)

We now take $\omega \in {\mathbb {C}}$ with $\omega ^m=-1$, and define

$$\begin{aligned} {\begin{matrix} v &{} = 1+u+u^*-e-e',\\ v_\omega &{} = 1+\omega u+u^*-e-e'. \end{matrix}} \end{aligned}$$

It is immediately seen that both v and $v_\omega$ are unitary, and so applying (7) (and using the orthogonal additivity of P and that $e\perp e'$), we see that

$$\begin{aligned} 0&=P(vu+vu^*)=P(e+e')=P(e)+P(e'),\\ 0&=P(v_\omega u+v_\omega u^*)=P(e+\omega e')=P(e)+P(\omega e')=P(e)-P(e'). \end{aligned}$$

By comparing both identities, we conclude that $P(e)=P(e')=0$, as claimed.

Our next objective is to prove that $P(e)=0$ for each projection $e\in {\mathscr {B}}(H)$. Suppose that $e\in {\mathscr {B}}(H)$ is a rank-one projection. Since $\dim H\ge 2$, it follows that there exists an equivalent projection $e'$ such that $e'\perp e$. Then it follows from the above claim that $P(e)=0$. Let $e\in {\mathscr {B}}(H)$ be a finite projection. Then there exist mutually orthogonal projections $e_1,\ldots ,e_n$ such that $e_1+\cdots +e_n=e$. Using the preceding observation and the orthogonal additivity of P we get $P(e)=P(e_1)+\cdots +P(e_n)=0$. We now assume that $e\in {\mathscr {B}}(H)$ is an infinite projection. Then there exist mutually orthogonal, equivalent projections $e_1$ and $e_2$ such that $e_1+e_2=e$. By the claim, we have $P(e)=P(e_1)+P(e_2)=0$.

We finally proceed to show that $P=0$. By Lemma 1, it suffices to show that $P(x)=0$ for each $x\in {\mathscr {B}}(H)_+$. Suppose that $x\in {\mathscr {B}}(H)_+$ can be written in the form $x=\sum _{j=1}^k\rho _j e_j$, where $e_1,\ldots ,e_k\in {\mathscr {B}}(H)$ are mutually orthogonal projections and $\rho _1,\ldots ,\rho _k\in {\mathbb {R}}^+$. Then we have $P(x)=\sum _{j=1}^k{\rho _j}^mP(e_j)=0$. Now let $x\in {\mathscr {B}}(H)_{+}$ be an arbitrary element. From the spectral decomposition we deduce that there exists a sequence $(x_n)$ in ${\mathscr {B}}(H)_{+}$ such that each $x_n$ is a positive linear combination of mutually orthogonal projections and $\lim x_n=x$. On account of the preceding observation, $P(x_n)=0$ ($n\in {\mathbb {N}}$), and the continuity of P implies that $P(x)=\lim P(x_n)=0$, as required. $\square$

3 Non-commutative $L^p$-spaces

Before giving the next results we make the following preliminary remarks.

A fundamental fact for us is the behaviour of the product of $L^0({{\mathscr {M}}},\tau )$ when restricted to the $L^p$-spaces. Specifically, if $0<p,q,r\le \infty$ are such that $\tfrac{1}{p}+\tfrac{1}{q}=\tfrac{1}{r}$, then the Hölder inequality states that

$$\begin{aligned} x\in L^p({{\mathscr {M}}},\tau ), \ y\in L^q({{\mathscr {M}}},\tau ) \ \Rightarrow \ xy\in L^r({{\mathscr {M}}},\tau ) \text { and } \Vert xy\Vert _r\le \Vert x\Vert _p\Vert y\Vert _q. \end{aligned}$$

(8)

Suppose that $x,y\in L^p({{\mathscr {M}}},\tau )_+$, $0<p<\infty$, are mutually orthogonal and that $\omega \in {\mathbb {C}}$ with $\vert \omega \vert =1$. Then it is immediately seen that $\vert x+\omega y\vert =x+y$, and it follows, by considering the spectral resolutions of x, y, and $x+y$, that $(x+y)^p=x^p+y^p$. Hence

$$\begin{aligned} \Vert x+\omega y\Vert _p^p=\Vert x\Vert _p^p+\Vert y\Vert _p^p. \end{aligned}$$

(9)

Each $x\in L^{p}({{\mathscr {M}}},\tau )$ can be written in the form

$$\begin{aligned} \begin{aligned} x=x_1-x_2+i(x_3-x_4), \text { with }&x_1,x_2,x_3,x_4\in L^{p}({{\mathscr {M}}},\tau )_+, \\&x_1\perp x_2, \ x_3\perp x_4,\\&\Vert x_1\Vert _p^p+\Vert x_2\Vert _p^p=\Vert x_1-x_2\Vert _p^p\le \Vert x\Vert _p^p,\\&\Vert x_3\Vert _p^p+\Vert x_4\Vert _p^p=\Vert x_3-x_4\Vert _p^p\le \Vert x\Vert _p^p. \end{aligned} \end{aligned}$$

(10)

Indeed, first we write $x=\mathfrak {R}x+i\mathfrak {I}x$, where

$$\begin{aligned} \mathfrak {R}x=\frac{1}{2}(x^*+x), \, \mathfrak {I}x=\frac{i}{2}(x^*-x)\in L^{p}({{\mathscr {M}}},\tau )_{\mathrm{sa}}, \end{aligned}$$

and, since $\Vert x^*\Vert _{p}=\Vert x\Vert _{p}$, it follows that $\Vert \mathfrak {R}x\Vert _{p},\Vert \mathfrak {I}x\Vert _{p}\le \Vert x\Vert _{p}$. Further, we take the positive operators

$$\begin{aligned} x_1 \!=\!\tfrac{1}{2}\left( \vert \mathfrak {R}x\vert +\mathfrak {R}x \right) ,\ x_2 \!=\!\tfrac{1}{2} \left( \vert \mathfrak {R}x\vert -\mathfrak {R}x \right) ,\ x_3 \!=\!\tfrac{1}{2} \left( \vert \mathfrak {I}x\vert +\mathfrak {I}x \right) ,\ x_4 \!=\!\tfrac{1}{2} \left( \vert \mathfrak {I}x\vert -\mathfrak {I}x \right) . \end{aligned}$$

Then $x_1,x_2,x_3,x_4\in L^p({{\mathscr {M}}},\tau )$, $\mathfrak {R}x=x_1-x_2$ with $x_1\perp x_2$, so that (9) gives

$$\begin{aligned} \Vert \mathfrak {R}x\Vert _p^p=\Vert x_1\Vert _p^p+\Vert x_2\Vert _p^p, \end{aligned}$$

and $\mathfrak {I}x= x_3-x_4$ with $x_3\perp x_4$, so that (9) gives

$$\begin{aligned} \Vert \mathfrak {I}x\Vert _p^p=\Vert x_3\Vert _p^p+\Vert x_4\Vert _p^p. \end{aligned}$$

Theorem 4

Let ${{\mathscr {M}}}$ be a von Neumann algebra with a normal semifinite faithful trace $\tau$, let X be a topological linear space, and let $\varPhi :L^{p/m}({{\mathscr {M}}},\tau )\rightarrow X$ be a continuous linear map with $0<p<\infty$. Then:

• (i) the map $P_\varPhi :L^p({{\mathscr {M}}},\tau )\rightarrow X$ defined by $P_\varPhi (x)=\varPhi (x^m)$$(x\in L^p({{\mathscr {M}}},\tau ))$ is a continuous m-homogeneous polynomial which is orthogonally additive on $L^p({{\mathscr {M}}},\tau )_{\mathrm{sa}}$;

• (ii) the polynomial $P_\varPhi$ is uniquely specified by the map $\varPhi$.

Suppose, further, that X is a q-normed space, $0<q\le 1$. Then:

• (iii) $2^{-1/q}\Vert \varPhi \Vert \le \Vert P_\varPhi \Vert \le \Vert \varPhi \Vert$.

Moreover, in the case where $X={\mathbb {C}}$,

• (iv) the functional $\varPhi$ is hermitian if and only if the polynomial $P_\varPhi$ is hermitian, in which case $\Vert P_\varPhi \Vert =\Vert \varPhi \Vert$.

Proof

The proof of this result is similar to that establishing Theorem 1.

(i) It follows immediately from (8) that, for each $x_1,\ldots ,x_m\in L^p({{\mathscr {M}}},\tau )$,

$$\begin{aligned} x_1\cdots x_m\in L^{p/m}({{\mathscr {M}}},\tau ) \text { and } \Vert x_1\cdots x_m\Vert _{p/m}\le \Vert x_1\Vert _p\cdots \Vert x_m\Vert _p. \end{aligned}$$

(11)

On the one hand, this clearly implies that the map $P_\varPhi$ is well-defined, on the other hand, the map $x\mapsto x^m$ from $L^p({{\mathscr {M}}},\tau )$ into $L^{p/m}({{\mathscr {M}}},\tau )$ is continuous, and so $P_\varPhi$ is continuous. Further, $P_\varPhi$ is the m-homogeneous polynomial associated with the symmetric m-linear map $\varphi :L^p({{\mathscr {M}}},\tau )^m\rightarrow X$ defined by

$$\begin{aligned} \varphi (x_1,\ldots ,x_m)= \frac{1}{m!}\sum _{\sigma \in \mathfrak {S}_m} \varPhi \left( x_{\sigma (1)} \cdots x_{\sigma (m)} \right) \quad (x_1,\ldots ,x_m\in L^p({{\mathscr {M}}},\tau )). \end{aligned}$$

Suppose that $x,y\in L^p({{\mathscr {M}}},\tau )_{\mathrm{sa}}$ are such that $x\perp y$. Then $xy=yx=0$, and so $(x+y)^m=x^m+y^m$, which gives

$$\begin{aligned} P_\varPhi (x+y)=\varPhi \bigl ((x+y)^m\bigr )= \varPhi \bigl (x^m+y^m\bigr )=\varPhi \bigl (x^m\bigr )+\varPhi \bigl (y^m\bigr )= P_\varPhi (x)+P_\varPhi (y). \end{aligned}$$

(ii) Suppose that $\Psi :L^{p/m}({{\mathscr {M}}},\tau )\rightarrow X$ is a linear map such that $P_\Psi =P_\varPhi$. For each $x\in L^{p/m}({{\mathscr {M}}},\tau )_+$, we have $x^{1/m}\in L^{p}({{\mathscr {M}}},\tau )$ and

$$\begin{aligned} \varPhi (x)= \varPhi \bigl (\bigl (x^{1/m}\bigr )^m\bigr )= P\bigl (x^{1/m}\bigr )= \Psi \bigl (\bigl (x^{1/m}\bigr )^m\bigr )= \Psi (x). \end{aligned}$$

By linearity we obtain $\varPhi =\Psi$.

(iii) Next, assume that X is a q-normed space. For each $x\in L^p({{\mathscr {M}}},\tau )$, by (11), we have

$$\begin{aligned} \Vert P_\varPhi (x)\Vert =\Vert \varPhi (x^m)\Vert \le \Vert \varPhi \Vert \Vert x^m\Vert _{p/m}\le \Vert \varPhi \Vert \Vert x\Vert _p^m, \end{aligned}$$

which clearly implies that $\Vert P_\varPhi \Vert \le \Vert \varPhi \Vert$. Now take $x\in L^{p/m}({{\mathscr {M}}},\tau )$, and take $\omega \in {\mathbb {C}}$ with $\omega ^m=-1$. Write

$$\begin{aligned} x= \mathfrak {R}x+i\mathfrak {I}x= x_1-x_2+i(x_3-x_4) \end{aligned}$$

as in (10) (with p / m instead of p). Since $x_1\perp x_2$ and $x_3\perp x_4$, it follows that $x_1^{1/m}\perp x_2^{1/m}$ and $x_3^{1/m}\perp x_4^{1/m}$, so that (9) gives

$$\begin{aligned} {\begin{matrix} \Vert \mathfrak {R}x\Vert _{p/m}^{p/m} &{}= \Vert x_1\Vert _{p/m}^{p/m}+\Vert x_2\Vert _{p/m}^{p/m},\\ \Vert \mathfrak {I}x\Vert _{p/m}^{p/m} &{}= \Vert x_3\Vert _{p/m}^{p/m}+\Vert x_4\Vert _{p/m}^{p/m}, \end{matrix}} \end{aligned}$$

(12)

and

$$\begin{aligned} {\begin{matrix} {\big \Vert } x_1^{1/m}+\omega x_2^{1/m} {\big \Vert }_p^p &{} = {\big \Vert } x_1^{1/m}{\big \Vert }_p^p+{\big \Vert }x_2^{1/m} {\big \Vert }_p^p,\\ {\big \Vert } x_3^{1/m}+\omega x_4^{1/m} {\big \Vert }_p^p &{} = {\big \Vert } x_3^{1/m}{\big \Vert }_p^p+{\big \Vert } x_4^{1/m} {\big \Vert }_p^p. \end{matrix}} \end{aligned}$$

(13)

Further, we have $x_1^{1/m},x_2^{1/m},x_3^{1/m},x_4^{1/m}\in L^p({{\mathscr {M}}},\tau )$ and

$$\begin{aligned} \bigl \Vert x_1^{1/m}\big \Vert _p\!= \!\Vert x_1\Vert _{p/m}^{1/m}, \ \big \Vert x_2^{1/m}\big \Vert _p\!\!=\! \Vert x_2\Vert _{p/m}^{1/m}, \ \big \Vert x_3^{1/m}\big \Vert _p\!\!=\! \Vert x_3\Vert _{p/m}^{1/m}, \ \big \Vert x_4^{1/m}\big \Vert _p\!\!=\! \Vert x_4\Vert _{p/m}^{1/m}, \end{aligned}$$

so that (12) and (13) give

$$\begin{aligned} {\begin{matrix} {\big \Vert } x_1^{1/m}+\omega x_2^{1/m}{\big \Vert }_p^p &{}= \Vert {\mathfrak {R}} x\Vert _{p/m}^{p/m},\\ {\big \Vert } x_3^{1/m}+\omega x_4^{1/m}{\big \Vert }_p^p &{}= \Vert {\mathfrak {I}} x\Vert _{p/m}^{p/m}. \end{matrix}} \end{aligned}$$

(14)

On the other hand, we have

$$\begin{aligned} \left( x_1^{1/m}+\omega x_2^{1/m}\right) ^m=x_1-x_2=\mathfrak {R}x,\quad \left( x_3^{1/m}+\omega x_4^{1/m}\right) ^m=x_3-x_4=\mathfrak {I}x, \end{aligned}$$

whence

$$\begin{aligned} \varPhi (x)&= \varPhi (\mathfrak {R}x)+i\varPhi (\mathfrak {I}x)= \varPhi \bigl (\bigl (x_1^{1/m}+\omega x_2^{1/m}\bigr )^m\bigr )+ i\varPhi \bigl (\bigl (x_3^{1/m}+\omega x_4^{1/m}\bigr )^m\bigr ) \\&= P_\varPhi \bigl (x_1^{1/m}+\omega x_2^{1/m}\bigr )+ iP_\varPhi \bigl (x_3^{1/m}+\omega x_4^{1/m}\bigr ). \end{aligned}$$

Hence, by (14),

$$\begin{aligned} \Vert \varPhi (x)\Vert ^q&\le \big \Vert P_\varPhi \bigl (x_1^{1/m}+\omega x_2^{1/m}\bigr )\big \Vert ^q+ \big \Vert P_\varPhi \bigl (x_3^{1/m}+\omega x_4^{1/m}\bigr )\big \Vert ^q\\&\le \Vert P_\varPhi \Vert ^q\big \Vert x_1^{1/m}+\omega x_2^{1/m}\big \Vert _p^{mq}+ \Vert P_\varPhi \Vert ^q\big \Vert x_3^{1/m}+\omega x_4^{1/m}\big \Vert _p^{mq}\\&= \Vert P_\varPhi \Vert ^q\left( \Vert \mathfrak {R}x\Vert ^q+\Vert \mathfrak {I}x\Vert ^q\right) \\&\le \Vert P_\varPhi \Vert ^q 2\Vert x\Vert ^q. \end{aligned}$$

This clearly forces $\Vert \varPhi \Vert \le 2^{1/q}\Vert P_\varPhi \Vert$, as claimed.

(iv) It is straightforward to check that $P_\varPhi ^*=P_{\varPhi ^*}$. From this deduce that $\varPhi$ is hermitian if and only if $P_\varPhi$ is hermitian as in the proof of Theorem 1(iv). Suppose that $\varPhi$ is a hermitian functional. By direct calculation, we see that $P_\varPhi$ is hermitian, and it remains to prove that $\Vert P_\varPhi \Vert =\Vert \varPhi \Vert$. We only need to show that $\Vert \varPhi \Vert \le \Vert P_\varPhi \Vert$. To this end, let $\varepsilon \in {\mathbb {R}}^+$, and choose $x\in L^{p/m}({{\mathscr {M}}},\tau )$ such that $\Vert x\Vert _{p/m}=1$ and $\Vert \varPhi \Vert -\varepsilon <\vert \varPhi (x)\vert$. We take $\alpha \in {\mathbb {C}}$ with $\vert \alpha \vert =1$ and $\vert \varPhi (x)\vert =\alpha \varPhi (x)$, so that

$$\begin{aligned} \Vert \varPhi \Vert -\varepsilon <\vert \varPhi (x)\vert =\varPhi (\alpha x)= \overline{\varPhi (\alpha x)}=\varPhi \bigl ((\alpha x)^*\bigr ). \end{aligned}$$

We see that $\mathfrak {R}(\alpha x)\in L^{p/m}({{\mathscr {M}}},\tau )_{\mathrm{sa}}$, $\Vert \mathfrak {R}(\alpha x)\Vert _{p/m}\le 1$, and $\Vert \varPhi \Vert -\varepsilon <\varPhi (\mathfrak {R}(\alpha x))$. Now we consider the decomposition $\mathfrak {R}(\alpha x)=x_1-x_2$ as in (10) (with p / m instead of p), and take $\omega \in {\mathbb {C}}$ with $\omega ^m=-1$. As in (14), we see that $\big \Vert x_1^{1/m}+\omega x_2^{1/m}\big \Vert =\Vert \mathfrak {R}(\alpha x)\Vert ^{1/m}\le 1$. Moreover, we have

$$\begin{aligned} P_\varPhi \bigl ( x_1^{1/m}+\omega x_2^{1/m}\bigr )= \varPhi \bigl (\bigl ( x_1^{1/m}+\omega x_2^{1/m}\bigr )^m\bigr )= \varPhi (\mathfrak {R}(\alpha x)), \end{aligned}$$

and so $\Vert \varPhi \Vert -\varepsilon <\Vert P_\varPhi \Vert$. $\square$

Theorem 5

Let ${{\mathscr {M}}}$ be a von Neumann algebra with a normal semifinite faithful trace $\tau$, let X be a topological linear space, and let $P:L^p({{\mathscr {M}}},\tau )\rightarrow X$ be a continuous m-homogeneous polynomial with $0<p<\infty$. Then the following conditions are equivalent:

• (i) there exists a continuous linear map $\varPhi :L^{p/m}({{\mathscr {M}}},\tau )\rightarrow X$ such that $P(x)=\varPhi (x^m)$$(x\in L^p({{\mathscr {M}}},\tau ))$;

• (ii) the polynomial P is orthogonally additive on $L^p({{\mathscr {M}}},\tau )_{\mathrm{sa}}$;

• (iii) the polynomial P is orthogonally additive on $S({{\mathscr {M}}},\tau )_+$.

If the conditions are satisfied, then the map $\varPhi$ is unique.

Proof

Theorem 4 shows that (i)$\Rightarrow$(ii), and it is obvious that (ii)$\Rightarrow$(iii). We proceed to prove that (iii)$\Rightarrow$(i).

Suppose that (iii) holds. Let $e\in {\mathscr {M}}$ be a projection such that $\tau (e)<\infty$, and consider the von Neumann algebra ${{\mathscr {M}}}_e=e{{\mathscr {M}}}e$. We claim that ${\mathscr {M}}_e\subset S({{\mathscr {M}}},\tau )$ and that there exists a unique continuous linear map $\varPhi _e:{\mathscr {M}}_e\rightarrow X$ such that

$$\begin{aligned} P(x)=\varPhi _e(x^m)\quad (x\in {\mathscr {M}}_e). \end{aligned}$$

(15)

Set $x\in {\mathscr {M}}_e$, and write $x=(x_1-x_2)+i(x_3-x_4)$ with $x_1,x_2,x_3,x_4\in {\mathscr {M}}_{e \, +}$. Then ${{\,\mathrm{supp}\,}}(x_j)\le e$ and therefore $\tau \bigl ({{\,\mathrm{supp}\,}}(x_j)\bigr )\le \tau (e)<\infty$$(j\in \{1,2,3,4\})$. This shows that $x_j\in S({\mathscr {M}},\tau )$$(j\in \{1,2,3,4\})$, whence $x\in S({\mathscr {M}},\tau )$. Our next goal is to show that the restriction $P\mid _{{\mathscr {M}}_e}$ is continuous (with respect to the norm that ${\mathscr {M}}_e$ inherits as a closed subspace of ${\mathscr {M}}$). Let $x\in {\mathscr {M}}_e$, and let $U\subset X$ be a neighbourhood of P(x). Since P is continuous, the set $P^{-1}(U)$ is a neighbourhood of x in $L^p({\mathscr {M}},\tau )$, which implies that there exists $r\in {\mathbb {R}}^+$ such that $P(y)\in U$ whenever $y\in L^p({\mathscr {M}},\tau )$ and $\Vert y- x\Vert _p<r$. If $y\in {\mathscr {M}}_e$ is such that $\Vert y- x\Vert <r/\Vert e\Vert _p$, then, from (8), we obtain

$$\begin{aligned} \Vert y-x\Vert _p=\Vert e(y-x)\Vert _p\le \Vert e\Vert _p\Vert y- x\Vert <r \end{aligned}$$

and therefore $P(y)\in U$. Hence $P\mid _{{\mathscr {M}}_e}$ is continuous. Since, by hypothesis, the polynomial $P\mid _{{\mathscr {M}}_e}$ is orthogonally additive on ${\mathscr {M}}_{e \, +}$, Corollary 1 states that there exists a unique continuous linear map $\varPhi _e:{\mathscr {M}}_e\rightarrow X$ such that (15) holds.

For each $x\in S({\mathscr {M}},\tau )$, define

$$\begin{aligned} \varPhi (x)=\varPhi _e(x), \end{aligned}$$

where $e\in {\mathscr {M}}$ is any projection such that

$$\begin{aligned} ex=xe=x\quad \text {and}\quad \tau (e)<\infty . \end{aligned}$$

(16)

We will show that $\varPhi$ is well-defined. For this purpose we first check that, if $x\in S({\mathscr {M}},\tau )$, then there exists a projection e such that (16) holds. Indeed, we write $x=\sum _{j=1}^k\alpha _j x_j$ with $\alpha _1,\ldots ,\alpha _k\in {\mathbb {C}}$ and $x_1,\ldots ,x_k\in S({\mathscr {M}},\tau )_+$, and define $e={{\,\mathrm{supp}\,}}(x_1)\vee \cdots \vee {{\,\mathrm{supp}\,}}(x_k)$. Then $ex=xe=x$ and $\tau (e)\le \sum _{j=1}^k\tau \bigl ({{\,\mathrm{supp}\,}}(x_j)\bigr )<\infty$, as required. Suppose that $x\in S({\mathscr {M}},\tau )$ and that $e_1,e_2\in {\mathscr {M}}$ are projections satisfying (16). Then the projection $e=e_1\vee e_2$ satisfies (16) and ${\mathscr {M}}_{e_1},{\mathscr {M}}_{e_2}\subset {\mathscr {M}}_e$. The uniqueness of the representation (15) on both ${\mathscr {M}}_{e_1}$ and ${\mathscr {M}}_{e_2}$ gives $\varPhi _{e}\mid _{{\mathscr {M}}_{e_1}}=\varPhi _{e_1}$ and $\varPhi _{e}\mid _{{\mathscr {M}}_{e_2}}=\varPhi _{e_2}$, which implies that $\varPhi _{e_1}(x)=\varPhi _{e}(x)=\varPhi _{e_2}(x)$.

We now show that $\varPhi$ is linear. Take $x_1,x_2\in S({\mathscr {M}},\tau )$ and $\alpha ,\beta \in {\mathbb {C}}$. Let $e_1,e_2\in {\mathscr {M}}$ be projections such that $e_jx_j=x_je_j=x_j$ and $\tau (e_j)<\infty$$(j\in \{1,2\})$. Then the projection $e=e_1\vee e_2$ satisfies

$$\begin{aligned}\begin{gathered} ex_j=x_je=x_j\quad (j\in \{1,2\}), \\ e(\alpha x_1+\beta x_2)=(\alpha x_1+\beta x_2)e=\alpha x_1+\beta x_2, \end{gathered}\end{aligned}$$

and

$$\begin{aligned} \tau (e)\le \tau (e_1)+\tau (e_2)<\infty . \end{aligned}$$

Thus

$$\begin{aligned} \varPhi (x_j)=\varPhi _e(x_j)\quad (j\in \{1,2\}) \end{aligned}$$

and

$$\begin{aligned} \varPhi (\alpha x_1+\beta x_2)= \varPhi _e(\alpha x_1+\beta x_2)= \alpha \varPhi _e(x_1)+\beta \varPhi _e(x_2)= \alpha \varPhi (x_1)+\beta \varPhi (x_2). \end{aligned}$$

We see from the definition of $\varPhi$ that

$$\begin{aligned} P(x)=\varPhi (x^m) \quad (x\in S({\mathscr {M}},\tau )). \end{aligned}$$

(17)

Our next concern will be the continuity of $\varPhi$ with respect to the norm $\Vert \cdot \Vert _{p/m}$. Let U be a neighbourhood of 0 in X. Let V be a balanced neighbourhood of 0 in X with $V+V+V+V\subset U$. The set $P^{-1}(V)$ is a neighbourhood of 0 in $L^p({\mathscr {M}},\tau )$, which implies that there exists $r\in {\mathbb {R}}^+$ such that $P(x)\in V$ whenever $x\in L^p({\mathscr {M}},\tau )$ and $\Vert x\Vert _p<r$. Take $x\in S({\mathscr {M}},\tau )$ with $\Vert x\Vert _{p/m}<r^m$, and write $x=(x_1-x_2)+i(x_3-x_4)$ as in (10) (with p / m instead of p). Then it is immediate to check that actually $x_1,x_2,x_3,x_4\in S({\mathscr {M}},\tau )_+$ and, further, $\Vert x_j\Vert _{p/m}\le \Vert x\Vert _{p/m}$$(j\in \{1,2,3,4\})$. For each $j\in \{1,2,3,4\}$, we have

$$\begin{aligned} \big \Vert x_{j}^{1/m}\big \Vert _p&= \tau \bigl (x_{j}^{p/m}\bigr )^{1/p}= \bigl (\tau \bigl (x_{j}^{p/m}\bigr )^{m/p}\bigr )^{1/m}= {\Vert x_j\Vert _{p/m}}^{1/m} \\&\le \Vert x\Vert _{p/m}^{1/m}<r, \end{aligned}$$

whence

$$\begin{aligned} \varPhi (x)&= \varPhi \left( \bigl (x_{1}^{1/m}\bigr )^m-\bigl (x_{2}^{1/m}\bigr )^m+i\bigl (x_{3}^{1/m}\bigr )^m-i\bigl (x_{4}^{1/m}\bigr )^m\right) \\&= \varPhi \left( \bigl (x_{1}^{1/m}\bigr )^m\right) - \varPhi \left( \bigl (x_{2}^{1/m}\bigr )^m\right) + i\varPhi \left( \bigl (x_{3}^{1/m}\bigr )^m\right) - i\varPhi \left( \bigl (x_{4}^{1/m}\bigr )^m\right) \\&= P\bigl (x_{1}^{1/m}\bigr )-P\bigl (x_{2}^{1/m}\bigr ) \\&\quad {}+iP\bigl (x_{3}^{1/m}\bigr )-iP\bigl (x_{4}^{1/m}\bigr )\in V+V+V+V\subset U, \end{aligned}$$

which establishes the continuity of $\varPhi$. Since $S({\mathscr {M}},\tau )$ is dense in $L^{p/m}({\mathscr {M}},\tau )$, the map $\varPhi$ extends uniquely to a continuous linear map from $L^{p/m}({\mathscr {M}},\tau )$ into the completion of X. By abuse of notation we continue to write $\varPhi$ for this extension. Since both P and $\varPhi$ are continuous, (17) gives $P(x)=\varPhi (x^m)$ for each $x\in L^p({\mathscr {M}})$. The task is now to show that the image of $\varPhi$ is actually contained in X. Of course, it suffices to show that $\varPhi$ takes $L^{p/m}({\mathscr {M}},\tau )_+$ into X. Let $x\in L^{p/m}({\mathscr {M}},\tau )_+$. Then $x^{1/m}\in L^{p}({\mathscr {M}},\tau )_+$ and

$$\begin{aligned} \varPhi (x)= \varPhi \bigl (\bigl (x^{1/m}\bigr )^m\bigr )= P\bigl (x^{1/m}\bigr )\in X, \end{aligned}$$

as required.

The uniqueness of the map $\varPhi$ is given by Theorem 4(ii). $\square$

Let us note that the space of all continuous m-homogeneous polynomials from $L^p({\mathscr {M}},\tau )$ into any topological linear space X which are orthogonally additive on $S({\mathscr {M}},\tau )_+$ is sufficiently rich in the case where $p/m\ge 1$, because of the existence of continuous linear functionals on $L^{p/m}({\mathscr {M}},\tau )$. However, some restriction on the space X must be imposed when we consider the case $p/m<1$ and the von Neumann algebra ${\mathscr {M}}$ has no minimal projections, because in this case the dual of $L^{p/m}({\mathscr {M}},\tau )$ is trivial ([15]). In fact, there are no non-zero continuous linear maps from $L^p({\mathscr {M}},\tau )$ into any q-normed space X with $q>p$. We think that this property is probably well-known, but we have not been able to find any reference, so that we next present a proof of this result for completeness.

Proposition 2

Let ${\mathscr {M}}$ be a von Neumann algebra with a normal semifinite faithful trace $\tau$ and with no minimal projections, let X be a q-normed space, $0<q\le 1$, and let $\varPhi :L^p({\mathscr {M}},\tau )\rightarrow X$ be a continuous linear map with $0<p<q$. Then $\varPhi =0$.

Proof

The proof will be divided in a number of steps.

Our first step is to show that for each projection $e_0\in {\mathscr {M}}$ with $\tau (e_0)<\infty$ and each $0\le \rho \le \tau (e_0)$, there exists a projection $e\in {\mathscr {M}}$ such that $e\le e_0$ and $\tau (e)=\rho$. Set

$$\begin{aligned} {\mathscr {P}}_1=\bigl \{e\in {\mathscr {M}} : e\text { is a projection, } e\le e_0, \ \tau (e)\ge \rho \bigr \}. \end{aligned}$$

Note that $e_0\in {\mathscr {P}}_1$, so that ${\mathscr {P}}_1$ is non-empty. Let ${\mathscr {C}}$ be a chain in ${\mathscr {P}}_1$, and let $e'=\wedge _{e\in {\mathscr {C}}}e$. Then $e'$ is a projection and $e'\le e_0$. For each $e\in {\mathscr {C}}$, since $\tau (e_0)<\infty$, it follows that $\tau (e_0)-\tau (e)=\tau (e_0-e)$. From the normality of $\tau$ we now deduce that

$$\begin{aligned} \tau (e_0)-\inf _{e\in {\mathscr {C}}}\tau (e)&= \sup _{e\in {\mathscr {C}}}\bigl (\tau (e_0)-\tau (e)\bigr ) = \sup _{e\in {\mathscr {C}}}\tau (e_0-e) \\&= \tau \left( \vee _{e\in {\mathscr {C}}}(e_0-e)\right) = \tau (e_0-e'). \end{aligned}$$

Hence $\tau (e')=\inf _{e\in {\mathscr {C}}}\tau (e)\ge \rho$, which shows that $e'$ is a lower bound of ${\mathscr {C}}$, and so, by Zorn’s lemma, ${\mathscr {P}}_1$ has a minimal element, say $e_1$. We now consider the set

$$\begin{aligned} {\mathscr {P}}_2=\bigl \{e\in {\mathscr {M}} : e\text { is a projection, } e\le e_1, \ \tau (e)\le \rho \bigr \}. \end{aligned}$$

Note that $0\in {\mathscr {P}}_2$, so that ${\mathscr {P}}_2$ is non-empty. Let ${\mathscr {C}}$ be a chain in ${\mathscr {P}}_2$, and let $e'=\vee _{e\in {\mathscr {C}}}e$. Then $e'\le e_1$, and the normality of $\tau$ yields

$$\begin{aligned} \tau (e')=\sup _{e\in {\mathscr {C}}}\tau (e)\le \rho . \end{aligned}$$

This implies that $e'$ is an upper bound of ${\mathscr {C}}$, and so, by Zorn’s lemma, ${\mathscr {P}}_2$ has a maximal element, say $e_2$. Assume towards a contradiction that $e_1\ne e_2$. Since, by hypothesis, ${\mathscr {M}}$ has no minimal projections, it follows that there exists a non-zero projection $e<e_1-e_2$. Since $e\perp e_2$, we see that $e_2+e$ is a projection. Further, we have $e_2<e_2+e<e_1$. The maximality of $e_2$ implies that $\tau (e_2+e)>\rho$, which implies that $e_2+e\in {\mathscr {P}}_1$, contradicting the minimality of $e_1$. Thus $e_1=e_2$, and this clearly implies that $\tau (e_1)=\tau (e_2)=\rho$.

Our next goal is to show that $\varPhi (e_0)=0$ for each projection $e_0$ with $\tau (e_0)<\infty$. From the previous step, it follows that there exists a projection $e\le e_0$ with $\tau (e)=\frac{1}{2}\tau (e_0)$. Set $e'=e_0-e$. Then $\tau (e')=\frac{1}{2}\tau (e_0)$. Further,

$$\begin{aligned} \Vert \varPhi (e_0)\Vert ^q= \Vert \varPhi (e)+\varPhi (e')\Vert ^q\le \Vert \varPhi (e)\Vert ^q+\Vert \varPhi (e')\Vert ^q, \end{aligned}$$

and therefore either $\Vert \varPhi (e)\Vert ^q\ge \frac{1}{2}\Vert \varPhi (e_0)\Vert ^q$ or $\Vert \varPhi (e')\Vert ^q\ge \frac{1}{2}\Vert \varPhi (e_0)\Vert ^q$. We define $e_1$ to be any of the projections $e,e'$ for which the inequality holds. We thus get $e_1\le e_0$, $\tau (e_1)=\frac{1}{2}\tau (e_0)$, and $\Vert \varPhi (e_1)\Vert \ge 2^{-1/q}\Vert \varPhi (e_0)\Vert$. By repeating the process, we get a decreasing sequence of projections $(e_n)$ such that

$$\begin{aligned} \tau (e_n)=2^{-n}\tau (e_0)\quad \text {and}\quad \Vert \varPhi (e_n)\Vert \ge 2^{-n/q}\Vert \varPhi (e_0)\Vert \quad (n\in {\mathbb {N}}). \end{aligned}$$

Then

$$\begin{aligned} \big \Vert 2^{n/q}e_n\big \Vert _p=2^{n/q}\tau (e_n)^{1/p}=2^{n(1/q-1/p)}\tau (e_0)^{1/p}, \end{aligned}$$

which converges to zero, because $p<q$. Since $\varPhi$ is continuous and $\Vert \varPhi (e_0)\Vert \le \big \Vert \varPhi \bigl (2^{n/q}e_n\bigr )\big \Vert _p$$(n\in {\mathbb {N}})$, it may be concluded that $\varPhi (e_0)=0$, as claimed.

Our next concern is to show that $\varPhi$ vanishes on $S({\mathscr {M}},\tau )$. Of course, it suffices to show that $\varPhi$ vanishes on $S({\mathscr {M}},\tau )_+$. Take $x\in S({\mathscr {M}},\tau )_+$, and let $e={{\,\mathrm{supp}\,}}(x)$, so that $\tau (e)<\infty$. The spectral decomposition implies that there exists a sequence $(x_n)$ in ${\mathscr {M}}_+$ such that $\lim x_n=x$ with respect to the operator norm and each $x_n$ is of the form $x_n=\sum _{j=1}^k\rho _j e_j$, where $\rho _1,\ldots ,\rho _k\in {\mathbb {R}}^+$ and $e_1,\ldots ,e_k\in {\mathscr {M}}$ are mutually orthogonal projections with $e_je=ee_j=e_j$$(j\in \{1,\ldots ,k\})$. From the previous step, we conclude that $\varPhi (x_n)=0$$(n\in {\mathbb {N}})$. Further, from (8) we deduce that

$$\begin{aligned} \Vert x-x_n\Vert _p=\Vert e(x-x_n)\Vert _p\le \Vert e\Vert _p\Vert x-x_n\Vert \rightarrow 0, \end{aligned}$$

and the continuity of $\varPhi$ implies that $\varPhi (x)=0$, as required.

Finally, since $S({\mathscr {M}},\tau )$ is dense in $L^p({\mathscr {M}},\tau )$ and $\varPhi$ is continuous, it may be concluded that $\varPhi =0$. $\square$

Corollary 2

Let ${\mathscr {M}}$ be a von Neumann algebra with a normal semifinite faithful trace $\tau$ and with no minimal projections, let X be a q-normed space, $0<q\le 1$, and let $P:L^p({\mathscr {M}},\tau )\rightarrow X$ be a continuous m-homogeneous polynomial with $0<p/m<q$. Suppose that P is orthogonally additive on $S({\mathscr {M}},\tau )_+$. Then $P=0$.

Proof

This is a straightforward consequence of Theorem 5 and Proposition 2. $\square$

We now turn our attention to the complex-valued polynomials. In this setting the representation given in Theorem 5 has a particularly significant integral form, because of the well-known representation of the dual of the $L^p$-spaces. The trace gives rise to a distinguished contractive positive linear functional on $L^1({\mathscr {M}},\tau )$, still denoted by $\tau$. By (8), if $\frac{1}{p}+\frac{1}{q}=1$, for each $\zeta \in L^q({\mathscr {M}},\tau )$, the formula

$$\begin{aligned} \varPhi _\zeta (x)=\tau (\zeta x)\quad (x\in L^p({\mathscr {M}},\tau )) \end{aligned}$$

(18)

defines a continuous linear functional on $L^p({\mathscr {M}},\tau )$. Further, in the case where $1\le p<\infty$, the map $\zeta \mapsto \varPhi _\zeta$ is an isometric isomorphism from $L^q({\mathscr {M}},\tau )$ onto the dual space of $L^p({\mathscr {M}},\tau )$. It is immediate to see that $\varPhi _\zeta ^*=\varPhi _{\zeta ^*}$, so that $\varPhi _\zeta$ is hermitian if and only if $\zeta$ is self-adjoint.

Corollary 3

Let ${\mathscr {M}}$ be a von Neumann algebra with a normal semifinite faithful trace $\tau$, and let $P:L^p({\mathscr {M}},\tau )\rightarrow {\mathbb {C}}$ be a continuous m-homogeneous polynomial with $m\le p<\infty$. Then the following conditions are equivalent:

• (i) there exists $\zeta \in L^{r}({\mathscr {M}},\tau )$ such that $P(x)=\tau (\zeta x^m)$$(x\in L^p({\mathscr {M}},\tau ))$, where $r=p/(p-m)$ (with the convention that $p/0=\infty$);

• (ii) the polynomial P is orthogonally additive on $L^p({\mathscr {M}},\tau )_{\mathrm{sa}}$;

• (iii) the polynomial P is orthogonally additive on $S({\mathscr {M}},\tau )_+$.

If the conditions are satisfied, then $\zeta$ is unique and $\Vert P\Vert \le \Vert \zeta \Vert _{r}\le 2\Vert P\Vert$; moreover, if P is hermitian, then $\zeta$ is self-adjoint and $\Vert \zeta \Vert _{r}=\Vert P\Vert$.

Proof

This follows from Theorems 4 and 5. $\square$

Let H be a Hilbert space. We denote by ${{\,\mathrm{Tr}\,}}$ the usual trace on the von Neumann algebra ${\mathscr {B}}(H)$. Then $L^p({\mathscr {B}}(H),{{\,\mathrm{Tr}\,}})$, with $0<p<\infty$, is the Schatten class $S^p(H)$. In the case where $0<p<q$, we have $S^p(H)\subset S^q(H)\subset {\mathscr {K}}(H)$ and $\Vert x\Vert \le \Vert x\Vert _q\le \Vert x\Vert _p$$(x\in S^p(H))$. It is clear that $S({\mathscr {B}}(H),{{\,\mathrm{Tr}\,}})={\mathscr {F}}(H)$, the two-sided ideal of ${\mathscr {B}}(H)$ consisting of the finite-rank operators. Thus, the following result is an immediate consequence of Corollary 3.

Corollary 4

Let H be a Hilbert space, and let $P:S^p(H)\rightarrow {\mathbb {C}}$ be a continuous m-homogeneous polynomial with $m< p<\infty$. Then the following conditions are equivalent:

• (i) there exists $\zeta \in S^{r}(H)$ such that $P(x)={{\,\mathrm{Tr}\,}}(\zeta x^m)$$(x\in S^p(H))$, where $r=p/(p-m)$;

• (ii) the polynomial P is orthogonally additive on $S^p(H)_{\mathrm{sa}}$;

• (iii) the polynomial P is orthogonally additive on ${\mathscr {F}}(H)_+$.

If the conditions are satisfied, then $\zeta$ is unique and $\Vert P\Vert \le \Vert \zeta \Vert _{r}\le 2\Vert P\Vert$; moreover, if P is hermitian, then $\zeta$ is self-adjoint and $\Vert \zeta \Vert _{r}=\Vert P\Vert$.

Corollary 5

Let H be a Hilbert space, and let $P:{\mathscr {K}}(H)\rightarrow {\mathbb {C}}$ be a continuous m-homogeneous polynomial. Then the following conditions are equivalent:

• (i) there exists $\zeta \in S^1(H)$ such that $P(x)={{\,\mathrm{Tr}\,}}(\zeta x^m)$$(x\in {\mathscr {K}}(H))$;

• (ii) the polynomial P is orthogonally additive on ${\mathscr {K}}(H)_{\mathrm{sa}}$;

• (iii) the polynomial P is orthogonally additive on ${\mathscr {F}}(H)_+$.

If the conditions are satisfied, then $\zeta$ is unique and $\Vert P\Vert \le \Vert \zeta \Vert _{1}\le 2\Vert P\Vert$; moreover, if P is hermitian, then $\zeta$ is self-adjoint and $\Vert \zeta \Vert _{1}=\Vert P\Vert$.

Proof

In order to prove the equivalence of the conditions we are reduced to prove that (iii)$\Rightarrow$(i). Suppose that (iii) holds. Let $x,y\in {\mathscr {K}}(H)_+$ such that $x\perp y$. From the spectral decomposition of both x and y we deduce that there exist sequences $(x_n)$ and $(y_n)$ in ${\mathscr {F}}(H)_+$ such that $\lim x_n=x$, $\lim y_n=y$, and $x_m\perp y_n$$(m,n\in {\mathbb {N}})$. Then

$$\begin{aligned} P(x+y)=\lim P(x_n+y_n)=\lim \bigl (P(x_n)+P(y_n)\bigr )=P(x)+P(y). \end{aligned}$$

This shows that P is orthogonally additive on ${\mathscr {K}}(H)_+$. Since the $C^*$-algebra ${\mathscr {K}}(H)$ has real rank zero and the net consisting of all finite-rank projections is an increasing approximate unit, Theorem 3 applies and gives a continuous linear functional $\varPhi$ on ${\mathscr {K}}(H)$ such that $P(x)=\varPhi (x^m)$$(x\in {\mathscr {K}}(H))$. It is well-known that the map $\zeta \mapsto \varPhi _\zeta$, as defined in (18), gives an isometric isomorphism from $S^1(H)$ onto the dual of ${\mathscr {K}}(H)$, so that there exists $\zeta \in S^1(H)$ such that $\varPhi (x)={{\,\mathrm{Tr}\,}}(\zeta x)$$(x\in {\mathscr {K}}(H))$ and $\Vert \zeta \Vert _1=\Vert \varPhi \Vert$. Thus we obtain (i). The additional properties of the result follow from Theorem 1. $\square$

Corollary 6

Let H be a Hilbert space, and let $P:S^p(H)\rightarrow {\mathbb {C}}$ be a continuous m-homogeneous polynomial with $0<p\le m$. Then the following conditions are equivalent:

• (i) there exists $\zeta \in {\mathscr {B}}(H)$ such that $P(x)={{\,\mathrm{Tr}\,}}(\zeta x^m)$$(x\in S^p(H))$;

• (ii) the polynomial P is orthogonally additive on $S^p(H)_{\mathrm{sa}}$;

• (iii) the polynomial P is orthogonally additive on ${\mathscr {F}}(H)_+$.

If the conditions are satisfied, then $\zeta$ is unique and $\Vert P\Vert \le \Vert \zeta \Vert \le 2\Vert P\Vert$; moreover, if P is hermitian, then $\zeta$ is self-adjoint and $\Vert \zeta \Vert =\Vert P\Vert$.

Proof

By Theorems 4 and 5, it suffices to show that the map $\zeta \mapsto \varPhi _\zeta$, as defined in (18), gives isometric isomorphism from ${\mathscr {B}}(H)$ onto the dual of $S^{p/m}(H)$. This is probably well-known, but we are not aware of any reference. Consequently, it may be helpful to include a proof of this fact. If $\zeta \in {\mathscr {B}}(H)$ and $x\in S^{p/m}(H)$, then, by (8), $\zeta x\in S^{p/m}(H)$, so that $\zeta x\in S^1(H)$ and

$$\begin{aligned} \big \Vert {{\,\mathrm{Tr}\,}}(\zeta x)\big \Vert \le \Vert \zeta x\Vert _1\le \Vert \zeta \Vert \Vert x\Vert _{1}\le \Vert \zeta \Vert \Vert x\Vert _{p/m}, \end{aligned}$$

which shows that $\varPhi _\zeta$ is a continuous linear functional on $S^{p/m}(H)$ with $\Vert \varPhi _\zeta \Vert \le \Vert \zeta \Vert$. Conversely, assume that $\varPhi$ is a continuous linear functional on $S^{p/m}(H)$. For each $\xi ,\eta \in H$, let $\xi \otimes \eta \in {\mathscr {F}}(H)$ defined by

$$\begin{aligned} \bigl (\xi \otimes \eta \bigr )(\psi )=\langle \psi \vert \eta \rangle \xi \quad (\psi \in H), \end{aligned}$$

and define $\varphi :H\times H\rightarrow {\mathbb {C}}$ by

$$\begin{aligned} \varphi (\xi ,\eta )=\varPhi (\xi \otimes \eta )\quad (\xi ,\eta \in H). \end{aligned}$$

It is easily checked that $\varphi$ is a continuous sesquilinear functional with $\Vert \varphi \Vert \le \Vert \varPhi \Vert$. Therefore there exists $\zeta \in {\mathscr {B}}(H)$ such that $\langle \zeta (\xi )\vert \eta \rangle =\varphi (\xi ,\eta )$ for all $\xi ,\eta \in H$ and $\Vert \zeta \Vert \le \Vert \varPhi \Vert$. The former condition implies that

$$\begin{aligned} \varPhi _\zeta (\xi \otimes \eta )= {{\,\mathrm{Tr}\,}}(\zeta \xi \otimes \eta )= \langle \zeta (\xi )\vert \eta \rangle = \varphi (\xi ,\eta )= \varPhi (\xi \otimes \eta ) \end{aligned}$$

for all $\xi ,\eta \in H$, which gives $\varPhi _\zeta (x)=\varPhi (x)$ for each $x\in {\mathscr {F}}(H)$. Since ${\mathscr {F}}(H)$ is dense in $S^{p/m}(H)$, it follows that $\varPhi _\zeta =\varPhi$. Further, we have $\Vert \zeta \Vert \le \Vert \varPhi \Vert =\Vert \varPhi _\zeta \Vert \le \Vert \zeta \Vert$. Finally, it is immediate to see that $\varPhi _\zeta ^*=\varPhi _{\zeta ^*}$, so that $\varPhi _\zeta$ is hermitian if and only if $\zeta$ is self-adjoint. $\square$

Proposition 3

Let H be a Hilbert space with $\dim H \ge 2$, let X be a topological linear space, and let $P:S^p(H) \rightarrow X$ be a continuous m-homogeneous polynomial with $0<p<\infty$. Suppose that P is orthogonally additive on $S^p(H)$. Then $P=0$.

Proof

Since ${\mathscr {F}}(H)$ is dense in $S^p(H)$ and P is continuous, it suffices to prove that P vanishes on ${\mathscr {F}}(H)$. On account of Lemma 1, we are also reduced to prove that P vanishes on ${\mathscr {F}}(H)_{\mathrm{sa}}$. We continue to use the notation $\xi \otimes \eta$ which was introduced in the proof of Corollary 6.

Let $x\in {\mathscr {F}}(H)_{\mathrm{sa}}$. Then $x=\sum _{j=1}^k\alpha _j\xi _j\otimes \xi _j$, where $k\ge 2$, $\alpha _1,\ldots ,\alpha _k\in {\mathbb {R}}$, and $\{\xi _1,\ldots ,\xi _k\}$ is an orthonormal subset of H. It is clear that the subalgebra ${\mathscr {M}}$ of ${{\mathscr {B}}}(H)$ generated by $\bigl \{\xi _i\otimes \xi _j : i,j\in \{1,\ldots ,k\}\bigr \}$ is contained in ${{\mathscr {F}}}(H)$ and it is $*$-isomorphic to the von Neumann algebra ${{\mathscr {B}}}(K)$, where K is the linear span of the set $\{\xi _1,\ldots ,\xi _k\}$. By Proposition 1, $P\mid _{{\mathscr {M}}}=0$, and therefore $P(x)=0$. We thus get $P\mid _{{{\mathscr {F}}}(H)_{\mathrm{sa}}}=0$, as required. $\square$