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Some eigenvalue inequalities

Let $\sH$ be a separable (infinite-dimensional) Hilbert space, and denote by $\sL(\sH)$ the algebra of bounded operators on $\sH$. Let $\sK = \sK(\sH)$ be the ideal of compact operators on $\sH$. Each $T \in \sK$ has a polar decomposition $T = U\vert T\vert$, where $\vert T\vert = (T^*T)^{1/2} \in \sK$, and $U \in \sL(\sH)$ is a partial isometry. This factorization is unique if we require that $U = 0$ on $\ker\vert T\vert$, since $U$ must map the range of $\vert T\vert$ isometrically onto the range of $T$.

The spectral theorem yields an orthonormal family $\{\psi_k\}$ in $\sH$, such that

\begin{displaymath}
\vert T\vert = \sum_{k\geq 0} s_k(T)  \ketbra{\psi_k}{\psi_k}, \qquad
T = \sum_{k\geq 0} s_k(T)  \ketbra{U\psi_k}{\psi_k}.
\end{displaymath}

(If $\vert T\vert$ is invertible, this is an orthonormal basis for $\sH$. Otherwise, we can adjoin an orthonormal basis for $\ker\vert T\vert$ to the family $\{\psi_k\}$.) Since $\phi_k := U\psi_k$ gives another orthonormal family, any $T \in \sK$ has an expansion of the form
\begin{displaymath}
T = \sum_{k\geq 0} s_k(T)  \ketbra{\phi_k}{\psi_k},
\end{displaymath} (16)

for some pair of orthonormal families $\{\phi_k\}$, $\{\psi_k\}$.

If $V_1$, $V_2$ are unitary operators on $\sH$, we can then write

\begin{displaymath}
V_1 T V_2 = \sum_{k\geq 0} s_k(T)  \ketbra{V_1\phi_k}{V_2^*\psi_k},
\end{displaymath}

and conclude that $s_k(V_1 T V_2) = s_k(T)$ for each $k$, and hence that

\begin{displaymath}
\sg_N(V_1 T V_2) = \sg_N(T).
\end{displaymath}

Therefore, any norm $\snorm{T}$ that is built from the sequence $\set{s_k(T) : k \in \bN}$ is unitarily invariant, that is, $\snorm{V_1 T V_2} = \snorm{T}$ for $V_1$, $V_2$ unitary.


\begin{example}
If $\Vert T\Vert$ is the usual operator norm on $\sK$, then
\be...
...subset \sL^r \subset \sL^p \subset \sK$ for
$1 < r < p < \infty$.
\end{example}

Soon, we shall introduce a ``Dixmier trace class'' $\sL^{1+}(\sH)$, with yet another norm built from singular values, such that $\sL^1 \subset \sL^{1+} \subset \sL^p$ for $p > 1$.

Much is known about the singular values of compact operators. For instance, the following relation holds, for $T \in \sK$:

\begin{displaymath}
s_k(T) = \inf\set{ \Vert T(1 - P)\Vert : P = P^2 = P^*, \dim P(\sH) \leq k}.
\end{displaymath} (17)

This comes from a well-known minimax principle: see [RS], for instance. The infimum is indeed attained at the projector $Q$ of rank $k$ whose range is $Q(\sH) := \spn\{\psi_0,\dots,\psi_{k-1}\}$, when $T$ is given by ([*]), since $T(1 - Q) = T - TQ = \sum_{j\geq k} s_j(T) \ketbra{\phi_j}{\psi_j}$ is an operator with norm $\Vert T - TQ\Vert = s_k(T)$.


\begin{lem}
If $T \in \sK$, then
\begin{subequations}
\begin{equation}
\sg_N(T) ...
...r(PAP) : P = P^2 = P^*, \rank P = N}.
\end{equation}\end{subequations}\end{lem}


\begin{proof}
If $P$ is a projector of finite rank $N$, then
$(TP)^*(TP) = PT^*...
...en
follows that $\Tr(QAQ) = \Tr(AQ) = \Vert AQ\Vert _1 = \sg_N(A)$.
\end{proof}


\begin{cor}
Each $\sg_N$ is a norm on $\sK$:
$\sg_N(S + T) \leq \sg_N(S) + \sg_N(T)$ for $S, T \in \sK$.
\end{cor}


\begin{proof}
This follows from $\Vert SP + TP\Vert _1 \leq \Vert SP\Vert _1 + \Vert TP\Vert _1$
for $P = P^2 = P^*$, $\rank P = N$.
\end{proof}


\begin{lem}
If $T \in \sK$, then
\begin{displaymath}
\sg_N(T)
= \inf\set{\Vert R...
...,\Vert S\Vert : R,S \in \sK \text{ with } R + S = T}.
\end{displaymath}\end{lem}


\begin{proof}
If $T = U\vert T\vert$, then $\vert T\vert = U^* T$ (by the detai...
...) - N s_N(T)$, while
$\Vert\Tilde{S}\Vert = s_N(T)$ by inspection.
\end{proof}

The triangle inequality in Corollary [*] is not good enough for our needs: our goal is get an additive functional, rather than just a subadditive one. The next step is to extract from ([*]) a sort of ``wrong-way triangle inequality'', at least for positive compact operators.


\begin{lem}
If $A \geq 0$, $B \geq 0$ are positive compact operators, and if
$M...
...splaymath}
\sg_{M+N}(A + B) \geq \sg_M(A) + \sg_N(B).
\end{displaymath}\end{lem}


\begin{proof}
{}From \eqref{eq:sigmaN-norm-pos} we obtain
$\sg_M(A) = \sup\set{\...
...
of $\Tr$, hence the restriction to the case of positive operators.)
\end{proof}


\begin{cor}
If $A,B \in \sK$ with $A \geq 0$, $B \geq 0$, then
\begin{displayma...
...\sg_N(A) + \sg_N(B) \leq \sg_{2N}(A + B).
\eqno{\qed}
\end{displaymath}\end{cor}

We see that the functional $A \mapsto \sg_N(A)/\log N$ is not far from being additive functional on the positive cone $\sK_+$. But to get a truly additive functional, we must try to take the limit $N \to \infty$, and here things become more interesting.


next up previous contents
Next: Dixmier traces Up: Spectral Growth and Dixmier Previous: Logarithmic divergence of spectra   Contents
Pawel Witkowski 2006-03-14