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].
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*}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*}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.
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.
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 \(\lambda \in \spec(N)\text{.}\)
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.
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 \(s(d) = 0\text{.}\)
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.
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 \(UNU^{*} = N'\text{.}\)
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 \(UNU^{*} = N'\text{.}\)