Arveson's Theorem Revisited

Section4Arveson's Theorem Revisited

In this section we apply the results concerning restricted diagonalization to prove a few key facts which will yield a reformulation and extension of Arveson's theorem (Theorem 4.3). Our first result in this direction is Theorem 4.2 which characterizes the condition $(d_n) \in \Lim\big(\spec(N)\big)$ in terms of restricted diagonalization. In order to prove Theorem 4.2, we use a straightforward geometric lemma which serves a similar purpose as Lemma 1 [2].

Lemma4.1

Suppose $\lambda_1,\ldots,\lambda_m \in \mathbb{C}$ are distinct and $x = \sum_{j=1}^m c_j \lambda_j$ is a convex combination, and $L$ is a line separating $\lambda_k$ from the remaining $\lambda_j\text{.}$ If $x$ lies on a line parallel to $L$ separating $\lambda_k$ from $L\text{,}$ then

$\begin{equation*} \sum_{\substack{j=1 \\ j\not=k}}^m c_j \le \frac{\abs{x - \lambda_k}}{\dist(\lambda_k, L)}. \end{equation*}$

Proof

Relabel the $\lambda_j$ if necessary so that $k=1\text{.}$ By applying a rotation, translation and scaling (which preserve proportional distances), we may suppose that $\lambda_1 = 1$ and $L = -a + i\mathbb{R}$ for some $a \ge 0$ so that the real part $\Re(x) = 0\text{.}$ Note that $-a \ge \max_{j \ge 2} \{ \Re(\lambda_j)\}\text{.}$ Since $0 \in [-a,1]$ we may write

\begin{equation*} t (-a) + (1-t) 1 = 0, \quad\text{for}\quad t = \frac{1}{1 + a} \end{equation*}

Now

\begin{equation*} 0 = \Re(x) = \sum_{j=1}^m c_j \Re(\lambda_j) \le \Bigg( \sum_{j=2}^m c_j \Bigg) \max_{j \ge 2} \{\Re(\lambda_j)\} + c_1 \lambda_1 \le \Bigg( \sum_{j=2}^m c_j \Bigg) (-a) + c_1 1. \end{equation*}

Since we have two convex combinations of $-a, 1$ and the latter is closer to $1$ than the former, the convexity coefficients satisfy

\begin{equation*} \sum_{j = 2}^m c_j \le t = \frac{1}{1+a} = \frac{\dist(\Re(x),\lambda_1)}{\dist(\lambda_1,L)} \le \frac{\abs{x-\lambda_1}}{\dist(\lambda_1,L)}. \qedhere \end{equation*}

Theorem4.2

Let $N$ be a normal operator with finite spectrum and diagonal $(d_n)\text{,}$ and let $X$ be the vertices of the convex hull of its essential spectrum. Then $(d_n) \in \Lim(X)$ if and only if $\essspec(N) = X$ and $N$ is diagonalizable by a unitary which is a Hilbert–Schmidt perturbation of the identity.

Proof

We first reduce to the case when $\spec(N) = \essspec(N)\text{.}$ Since $N$ is a normal operator with finite spectrum, by the spectral theorem there is a finite rank perturbation $N'$ of $N$ for which $N'$ is normal and $\spec(N') = \essspec(N') = \essspec(N)\text{.}$ In particular, if $P_{\lambda}$ are the spectral projections of $N$ onto $\{\lambda\}\text{,}$ and $\lambda' \in \essspec(N)$ is a distinguished element, then we can choose

\begin{equation*} N' := \lambda' P + \sum_{\lambda \in \essspec(N)} \lambda P_{\lambda}, \quad\text{where}\quad P = \sum_{\lambda \notin \essspec(N)} P_{\lambda}. \end{equation*}

Since $N'-N$ is finite rank, the diagonals of $N'$ and $N$ differ by an absolutely summable sequence, so $(d_n) \in \Lim(X)$ if and only if the diagonal of $N'$ is in $\Lim(X)\text{.}$ Moreover, the spectral projections of $N$ and $N'$ differ from one another by finite projections. Therefore, the spectral projections of $N$ each differ from a diagonal projection by a Hilbert–Schmidt operator if and only if the same holds true for $N'\text{.}$ By Theorem 3.4, $N$ is diagonalizable by a unitary which is a Hilbert–Schmidt perturbation of the identity if and only if $N'$ is as well. Therefore, by the above reduction, it suffices to prove the theorem with the added assumption that $\spec(N) = \essspec(N)\text{.}$

(Proof of $\Rightarrow$) Enumerate the elements of $\essspec(N) = \spec(N)$ as $\lambda_1,\ldots,\lambda_m\text{.}$ Let $P_j$ denote the spectral projection corresponding to the eigenvalue $\lambda_j\text{,}$ so that $N = \sum_{j=1}^m \lambda_j P_j\text{.}$ Let $\{e_n\}_{n=1}^{\infty}$ denote the orthonormal basis corresponding to the diagonal $(d_n)\text{.}$ Suppose $(d_n) \in \Lim(X)\text{,}$ and so there exist $x_n \in X$ for which $(d_n - x_n) \in \ell^1\text{.}$ Let $\Lambda_k := \{ n \in \mathbb{N} \mid x_n = \lambda_k\}$ be the index set where the sequence $(x_n)$ takes the value $\lambda_k \in X\text{.}$

The projections $P_j$ sum to the identity, so for each $n \in \mathbb{N}\text{,}$ $\sum_{j=1}^m \angles{P_j e_n,e_n} = 1$ and therefore

\begin{equation*} d_n = \sangles{Ne_n, e_n} = \sum_{j=1}^m \sangles{P_j e_n, e_n} \lambda_j \end{equation*}

is a convex combination of the spectrum.

For $\lambda_k \in X\text{,}$ let $L_k$ be a line separating $\lambda_k$ from the remaining elements of $\essspec(N)\text{.}$ Such a line $L_k$ exists because $\lambda_k$ is an extreme point of the convex hull of $\essspec(N)\text{,}$ and this is a finite set. Since $(d_n) \in \Lim (X)$ we know that $(d_n - \lambda_k)_{n \in \Lambda_k}$ is absolutely summable for every $k\text{.}$ Therefore, for all but finitely many indices $n \in \Lambda_k\text{,}$ the diagonal entry $d_n$ lies on a line parallel to $L_k$ separating $\lambda_k$ from $L_k$ and hence also $\essspec(N) \setminus \{ \lambda_k \}\text{.}$

By Lemma 4.1, for these indices $n \in \Lambda_k\text{,}$

\begin{equation} \sum_{\substack{j=1 \\ j\not=k}}^m \sangles{P_j e_n, e_n} \le \frac{\abs{d_n - \lambda_k}}{\dist(\lambda_k,L_k)}.\label{eq-P_j-in-ell-1}\tag{4.1} \end{equation}

Since this inequality holds for all but finitely many $n \in \Lambda_k\text{,}$ and $\dist(\lambda_k,L_k)$ is independent of $n \in \Lambda_k\text{,}$ and $(d_n - \lambda_k)_{n \in \Lambda_k}$ is absolutely summable, (4.1) proves $\big(\langle P_j e_n, e_n \rangle\big)_{n \in \Lambda_k}$ lies in $\Lim (\{0\}) = \ell^1$ when $j \not= k\text{.}$ If $\lambda_j \in \essspec(N) \setminus X\text{,}$ by letting $\lambda_k$ run through $X\text{,}$ we find $\big(\langle P_j e_n, e_n \rangle\big)_{n \in \mathbb{N}}$ is absolutely summable since $\bigcup_{\lambda_k \in X} \Lambda_k = \mathbb{N}\text{.}$ This implies $P_j$ is trace-class and hence a finite projection, contradicting the fact that $\lambda_j \in \essspec(N)\text{.}$ Therefore $X = \essspec(N)\text{.}$

Now consider $\lambda_j \in X = \essspec(N)\text{.}$ In analogy with the previous paragraph, using the fact that $\big(\langle P_j e_n, e_n \rangle\big)_{n \in \Lambda_k} \in \ell^1$ when $j \not= k$ and letting $\lambda_k$ run through $X \setminus \lambda_j\text{,}$ we find $\big(\langle P_j e_n, e_n \rangle\big)_{n \notin \Lambda_j} \in \ell^1\text{.}$ Finally, for $n \in \Lambda_j\text{,}$

\begin{equation*} 1 - \sangles{P_j e_n, e_n} = \sum_{\substack{k=1 \\ k\not=j}}^m \sangles{P_k e_n, e_n}, \end{equation*}

and hence $\big(1- \sangles{P_j e_n, e_n}\big)_{n \in \Lambda_k}$ is a finite sum of absolutely summable sequences, and is therefore absolutely summable. Thus $\big(\sangles{P_j e_n, e_n}\big)_{n \in \Lambda_k} \in \Lim(\{1\})\text{,}$ so $\big(\sangles{P_j e_n,e_n}\big) \in \Lim(\{0,1\})\text{.}$ Therefore, by Corollary 3.1, $P_j$ differs from a diagonal projection by a Hilbert–Schmidt operator. Since this is true of all the spectral projections of $N\text{,}$ we may apply Theorem 3.4 to conclude that $N$ is a diagonalizable by a Hilbert–Schmidt perturbation of the identity.

(Proof of $\Leftarrow$) This implication is a direct corollary of Theorem 3.8. To see this, suppose $\essspec(N) = X$ and $N$ is diagonalizable by a unitary $U$ which is a Hilbert–Schmidt perturbation of the identity. Thus $UNU^{*} = \diag(x_n)$ for some sequence $x_n \in \essspec(N) = X\text{.}$ Then by Theorem 3.8, $E(N-UNU^{*})$ is trace-class. That is, $\trace \big(E(N-UNU^{*})\big) = \sum_{n=1}^{\infty} (d_n-x_n)$ is an absolutely summable series, so $(d_n) \in \Lim(X)\text{.}$

We now establish our generalized operator-theoretic reformulation of Arveson's Theorem 1.5 by means of Theorem 3.8. After the proof we will explain how to derive Theorem 1.5 from Theorem 4.3.

Theorem4.3

Let $N$ be a normal operator with finite spectrum. If $N$ is diagonalizable by a unitary which is a Hilbert–Schmidt perturbation of the identity, then there is a diagonal operator $N'$ with $\spec(N') \subseteq \spec(N)$ for which $E(N-N')$ is trace-class, and for any such $N'\text{,}$ $\trace\big(E(N-N')\big) \in K_{\spec(N)}\text{.}$ In particular,

$\begin{equation} \trace\big(E(N-N')\big) = \sum_{\lambda \in \spec(N)} [P_{\lambda}:Q_{\lambda}] \lambda,\label{eq-trace-in-K-spec-N}\tag{4.2} \end{equation}$

where $P_{\lambda},Q_{\lambda}$ are the spectral projections onto $\{\lambda\}$ of $N,N'$ respectively. Moreover, $P_{\lambda}-Q_{\lambda}$ is Hilbert–Schmidt for each $\lambda \in \spec(N)\text{.}$

Proof

Suppose $N$ is normal operator with finite spectrum which is diagonalizable by a unitary $U$ that is a Hilbert–Schmidt perturbation of the identity. Then by Theorem 3.8, $E(UNU^{*}-N)$ is trace-class with trace zero. Moreover, $\spec(UNU^{*}) = \spec(N)\text{,}$ thereby proving that an $N'$ as in the statement exists.

Now, let $N'$ be any diagonal operator with $\spec(N') \subseteq \spec(N)$ for which $E(N-N')$ is trace-class. Since $N'$ and $UNU^{*}$ are diagonal, we find

\begin{equation} UNU^{*}-N' = E(UNU^{*}-N') = E(UNU^{*}-N)+E(N-N')\label{eq-diagonal-split}\tag{4.3} \end{equation}

is trace-class, diagonal, and has finite spectrum contained in the set of differences $\spec(N)-\spec(N)\text{.}$ Together, these conditions imply this operator is finite rank. Moreover, the (diagonal) spectral projections of $UNU^{*}, N'\text{,}$ which we denote $R_{\lambda},Q_{\lambda}\text{,}$ respectively for $\lambda \in \spec(N)\text{,}$ each differ by a finite rank operator. Here we allow for the case $Q_{\lambda} = 0$ when $\lambda \in \spec(N) \setminus \spec(N')\text{.}$ This guarantees

\begin{equation*} [R_{\lambda}:Q_{\lambda}] = \trace(R_{\lambda}-Q_{\lambda}), \end{equation*}

using, for example, Proposition 2.4; however, this formula for essential codimension holds whenever the difference of the projections is trace-class and is widely known (see for instance [4] Theorem 4.1, [3] Theorem 3, or [9] Corollary 3.3).

Therefore,

\begin{equation} \trace(UNU^{*}-N') = \trace \left( \sum_{\lambda \in \spec(N)} (\lambda R_{\lambda} - \lambda Q_{\lambda}) \right) = \sum_{\lambda \in \spec(N)} [R_{\lambda} : Q_{\lambda}] \lambda.\label{eq-trace-ess-codim-formula}\tag{4.4} \end{equation}

Moreover, we can replace $R_{\lambda}$ with $P_{\lambda}$ in the right-most side of the above display. Indeed, since $U$ conjugates $P_{\lambda}, R_{\lambda}\text{,}$ $[P_{\lambda} : R_{\lambda}] = 0$ by Proposition 2.3, and furthermore $[P_{\lambda} : Q_{\lambda}] = [P_{\lambda} : R_{\lambda}] + [R_{\lambda} : Q_{\lambda}]$ by Proposition 2.2(iii).

Finally, since $\trace\big(E(UNU^{*}-N)\big) = 0\text{,}$ using (4.3) and (4.4) we find that

\begin{equation*} \trace\big(E(N-N')\big) = \trace(UNU^{*}-N') = \sum_{\lambda \in \spec(N)} [P_{\lambda}:Q_{\lambda}] \lambda. \qedhere \end{equation*}

We now illustrate how our results may be used to provide a new proof of Arveson's theorem.

Proof

Let $X = \{\lambda_1,\ldots,\lambda_m\}$ and $d = (d_1,d_2,\ldots)$ be as in Theorem 1.5. That is, $X$ is the set of vertices of a convex polygon in $\mathbb{C}\text{,}$ and $d$ satisfies

\begin{equation*} \sum_{n=1}^{\infty} \abs{f(d_n)} \lt \infty, \end{equation*}

where $f(z) = (z-\lambda_1)(z-\lambda_2)\cdots(z-\lambda_m)\text{.}$ As we remarked after Theorem 1.5, this summability condition is equivalent to $d \in \Lim (X)$ by Proposition 2 of [2]. Now suppose $d$ is the diagonal of an operator $N \in \mathcal{N}(X)$ (i.e., $N$ is normal with $\spec(N) = \essspec(N) = X$). Then by Theorem 4.2, $N$ is diagonalizable by a unitary $U = I+K$ with $K$ Hilbert–Schmidt. Therefore, we may apply Theorem 4.3 to conclude that $\trace\big(E(N-N')\big) \in K_{\spec(N)} = K_X$ for some diagonal operator $N'$ with $\spec(N') \subseteq \spec(N)$ and $E(N-N')$ is trace-class. Finally, equation (3.2) of Proposition 3.5 establishes

\begin{equation*} \sum_{n=1}^{\infty} (d_n - x_n) = \trace\big(E(N-N')\big) \in K_X \end{equation*}

where $(x_n)$ is the diagonal of $N'\text{,}$ so $x_n \in \spec(N') = \spec(N) = X\text{.}$ Hence $s(d) = 0\text{.}$

Remark4.4

In [6], Bownik and Jasper completely characterized the diagonals of selfadjoint operators with finite spectrum. A few of the results we have presented herein are generalizations of Theorem 4.1 of [6], which consists of some necessary conditions for a sequence to be the diagonal of a finite spectrum selfadjoint operator. In particular, the statement $(d_n) \in \Lim(X)$ implies $X = \essspec(N)$ of our Theorem 4.2 is an extension to finite spectrum normal operators of their corresponding result [6] Theorem 4.1(ii) for selfadjoint operators. Similarly, our formula (4.2) of Theorem 4.3 generalizes [6] Theorem 4.1(iii).

We conclude with another perspective on the trace $\trace\big(E(N-N')\big)\text{.}$ Our next corollary shows that when the $\mathbb{Z}$ -module $K_{\spec(N)}$ has full rank (i.e., $\rank K_{\spec(N)}$ is one less than the number of elements in the spectrum), this trace is zero if and only if $N'$ is a diagonalization of $N$ by a unitary $U = I + K$ with $K$ Hilbert–Schmidt.

Corollary4.5

Suppose $N$ is a normal operator with $\spec(N) = \{\lambda_1,\ldots,\lambda_m\}$ such that $\lambda_1-\lambda_2,\ldots,\lambda_1-\lambda_m$ are linearly independent in the $\mathbb{Z}$ -module $K_{\spec(N)}\text{.}$ Suppose further that $N$ is diagonalizable by a unitary which is a Hilbert–Schmidt perturbation of the identity. If $N'$ is a diagonal operator such that $E(N-N')$ is trace-class and $\trace\big(E(N-N')\big) = 0\text{,}$ then there is a unitary $U = I+K$ with $K$ Hilbert–Schmidt such that $UNU^{*} = N'\text{.}$

Proof

By Theorem 4.3, the differences $P_k - Q_k$ are Hilbert–Schmidt and

\begin{equation*} 0 = \trace\big(E(N-N')\big) = \sum_{k=1}^m [P_k:Q_k]\lambda_k \end{equation*}

Since $\sum_{k=1}^m [P_k:Q_k] = 0\text{,}$ we have $[P_1:Q_1] = - \sum_{k=2}^m [P_k:Q_k]$ and so we may rearrange the equality above to

\begin{equation*} 0 = \sum_{k=2}^m [P_k:Q_k](\lambda_1 - \lambda_k). \end{equation*}

Since $\lambda_1-\lambda_2,\ldots,\lambda_1-\lambda_m$ are linearly independent in $K_{\spec(N)}\text{,}$ we conclude that the coefficients $[P_k:Q_k] = 0$ for $2 \le k \le m\text{.}$ In turn, this implies $[P_1:Q_1] = 0\text{.}$ Therefore, by Lemma 3.2, there is a unitary $U = I+K$ with $K$ Hilbert–Schmidt conjugating each $P_k$ to $Q_k\text{.}$ Thus $UNU^{*} = N'\text{.}$