~\mbox{max} ~( d/2 , 1)$, and $r = ( dp/2 )( p - d/2 )^{-1}$, and if $V_\omega$ satisfies \beq \label{pot} \E \{ \int_{\Lambda_1 (0)} | V_\omega (x)|^{1/pr} \} < \infty , \eeq then the IDS exists almost surely at its points of continuity. The IDS $N(E)$ is monotonic, right continuous, and the points of discontinuity form a countable set. For Anderson-type random potentials satisfying hypotheses (H1)--(H2), condition (\ref{pot}) is satisfied, so that the IDS exists as described above. As a consequence, the DOS $\nu$ measure exists. It is also the vague limit of the finite-volume point measures $\nu_\Lambda$, corresponding to the local Hamiltonians $H_\omega^\Lambda$, as in (\ref{vague0}). These random point measures $\nu_\Lambda$ are constructed from the eigenvalues of $H_\omega^\Lambda$. If $\{ E_j ( \Lambda ; \omega ) \; | \; j \in \Z \}$ is the set of random eigenvalues of $H_\omega^\Lambda$, then we define a random point measure by \beq \label{point1} d \nu_\Lambda ( \lambda ) = \frac{1}{| \Lambda|} \sum_{j \in \Z} \delta ( \lambda - E_j ( \Lambda ; \omega )) ~d \lambda . \eeq For any continuous function $f$ of compact support, we have \beq \label{vague} \lim_{ | \Lambda | \uparrow \infty } \int f (\lambda ) d \nu_\Lambda ( \lambda ) = \int f(\lambda) d \nu ( \lambda) . \eeq Furthermore, for any Borel set $A \subset \R$, we have \beq \label{vague2} \nu (A) \leq \liminf_{| \Lambda | \uparrow \infty } \E \{ \nu_\Lambda (A) \}, \eeq although pointwise convergence in (\ref{vague2}) does not necessarily hold. We now turn to the properties of the DOS measure. \vspace{.1in} \begin{theorem} Under the hypotheses (H1)--(H2), the DOS measure for $H_\omega$ exists and is locally or globally absolutely continuous with respect to Lebesgue measure. For any Borel set $A \subset \R$, as stated in Corollaries 1.2, 1.4, or 1.8, we have \beq \label{holdernu} \nu (A) \leq C |A|^q . \eeq For the cases of Corollary 1.2 and Theorem 1.6, the exponent $q$ satisfies $0 < q < q_1 / (q_1 + 2)$. For the cases of Corollary 1.4 and Theorem 1.7, the exponent $q$ satisfies $0 < q < 1$. The density of the measure belongs to $L^p_{loc} (\R)$, for any $1 \leq p < p_0$, where $p_0 = 1 + \frac{q_1}{2}$ in the cases of Corollary 1.2 and Theorem 1.6, and $p_0 = \infty$, in the case of Corollary 1.4 and Theorem 1.7. \end{theorem} \begin{proof} \noindent 1. We first note that the proofs of Theorems 1.1 and 1.3 can be extended to include arbitrary Borel sets. We unify notation as follows. Let the set $J \subset \R$ denote the interval $\Delta \subset I$ in the proof of Theorem 1.1 for which the Wegner estimate holds, or let it denote the interval $\Delta$ in Theorem 1.3. For an arbitrary Borel set $A \subset \R$, we have \beq \E \{ Tr E_\Lambda (A \cap J ) \} \; \leq \; C_J | A \cap J |^q \; |\Lambda|, \eeq where the notation $|B|$ means the Lebesgue measure of $B$, and the finite positive constant $C_J > 0$ is the constant in the Wegner estimate. The proof of Theorem 1.1 is identical if we replace $\Delta$ by $\Delta \cap J$ throughout. Similarly, the proof of Theorem 1.3 remains the same if we make the same replacement. Hence, if $A$ is such that $| J \cap A | = 0$, we have that $\E \{ Tr E_\Lambda ( J \cap A ) \} = 0$, for all large $\Lambda$. To relate this finite-volume estimate to the DOS measure $\nu$, we use the upper bound (\ref{vague2}). We have for any Borel set $A \subset \R$, \beq \label{holder} \nu (A \cap J ) \; \leq \; \liminf_{|\Lambda| \uparrow \infty} \frac{ \E \{ Tr E_\Lambda ( A \cap J ) \} }{ | \Lambda | } \leq C_J | A \cap J|^q, \eeq for any $q$ in the ranges stated in the theorem. This proves the H\"older bound (\ref{holdernu}). We also conclude from this that for any Borel set $A \subset \R$ of Lebesgue measure zero, the DOS measure $\nu$ satisfies $\nu ( J \cap A ) = 0$, proving the local absolute continuity of the measure. \noindent 2. Let $f_J$ be the local density of the measure $\nu$. By the general theory, the density $f_J \in L_{loc}^1 ( \R)$. Since the IDS $N(E)$ is locally H\"older continuous for any exponent $q >1$, the density is better behaved. For any $\lambda \in \R$, let $F_\lambda$ be the subset of $\R$ defined by $F_\lambda \equiv \{ t \in \R \; | \; f_J (t ) > \lambda \}$. The $\nu$-measure of this set is bounded from below by \beq \label{meas1} \nu (F_\lambda ) = \int_{F_\lambda} f_J (t) ~dt > \lambda | F_\lambda |, \eeq while from the H\"older Continuity of the measure (\ref{holder}), we have \beq \label{meas2} \nu ( F_\lambda ) \leq C_J | F_\lambda |^q, \eeq for any $0 < q < 1$. We conclude from (\ref{meas1}) and (\ref{meas2}) that \beq \label{meas3} | F_\lambda | \leq C_1 | \lambda|^{- \frac{1}{1-q}}. \eeq The $L^p$-norm of the density, for $p \geq 1$, is estimated by \bea \label{dens1} \| f_J \|_{L^p_{loc}}^p & = & \int f_J ( \lambda )^p ~d\lambda \nonumber \\ & \leq & \sum_{n \in \Z} 2^{p(n+1)} ~ | \{ t \; | \; 2^n < f_J(t) < 2^{n+1} \} | \nonumber \\ & \leq & \sum_{n \in \Z} 2^{p(n+1)} | F_{2^n } | \nonumber \\ & \leq & 2^p ~\sum_{n \in \Z} 2^{n ( p - 1/(1-q) ) } . \eea For the cases of Corollary 1.2 and Theorem 1.6, we have $1 < 1/ (1-q) < 1 + \frac{q_1}{2}$, so that the sum in (\ref{dens1}) is finite for any $1 \leq p < 1 + \frac{q_1}{2}$. In the other cases of Corollary 1.4 and Theorem 1.7, we have $1 < 1/(1-q) < \infty$, so the sum in (\ref{dens1}) is finite for any $p \geq 1$. This proves the local result. \end{proof} We remark that we first proved that if the IDS satisfies a H\"older continuity condition with exponent $0 < q < 1$, then the density is in weak-$L^p$, for $p = 1 / (1 - q)$. Let us note that under the best condition of Lipschitz continuity of the IDS, the density $f_J$ is given by the derivative $dN / d \lambda$, and this is in $L^\infty_{loc}$. This corresponds to $q = 1$. On the other hand, if we merely know that the DOS $\nu$ is absolutely continuous with respect to Lebesgue measure, then we are in the $q = 0$ case and the density $f_J$ is in $L^1_{loc}$. \section{Concluding Remarks} \noindent We make some comments about the results. \vspace{.1in} \noindent 1. {\bf Lipschitz Continuity.} It is believed that under hypotheses (H1)--(H2) the IDS is Lipschitz continuous. The above proofs provide for Lipschitz continuity provided there is an improved estimate on the spectral shift function (SSF) (for background on the SSF cf.\ \cite{[BY],[Yafaev]}). In \cite{[CHN]}, the following estimate on the SSF is proved. Suppose that $H_0$ and $H$ are two self-adjoint operators on a separable Hilbert space with $V \equiv H -H_0 \in \mathcal{I}_{1/p}$, for any $p \geq 1$. The Schatten-von Neumann ideals $\mathcal{I}_q$ are the classes of bounded operators $A$ having singular values $\mu_j (A)$ satisfying $\| A \|_q^q \equiv \sum_j \mu_j (A)^q < \infty$, cf.\ \cite{[Simon2]}. Under these conditions, the SSF $\xi ( \lambda ; H , H_0 ) \in L^p ( \R)$, and satisfies the estimate \beq \label{SSF} \| \xi ( \cdot; H , H_0 ) \|_{L^p} \; \leq \; \|V\|_{1/p}^{1/p}. \eeq For Schr\"odinger operators $H_0$ and $H$, the SSF is defined through the resolvents $R_0 (z)$ and $R(z)$. It is not difficult to show that $V_{eff} \equiv R_0(z)^k - R(z)^k \in \mathcal{I}_{1/p }$, for any $p \geq 1$, provided $k > pd/2 + 2$, and provided the perturbation $V = H - H_0$ has bounded support. One can then show that the SSF is bounded by $\|V_{eff} \|_{1/p}^{1/p}$. In the proofs of Theorems 1.1 and 1.3, one considers the expectation with respect to one random variable only. In particular, choosing $\lambda_k$, and for a function $f_\Delta$ with support equal to $\Delta$, one has \bea \label{ssf1} \lefteqn{\int_0^1 h_0 ( \lambda_k ) d \lambda_k \left\{ \frac{\partial}{\partial \lambda_k} Tr f_\Delta (H_\omega^\Lambda) \right\}} && \nonumber \\ & \leq & \|h_0 \|_\infty | Tr \{ f_\Delta (H_\omega^\Lambda)(\lambda_k = 1) - f_\Delta ( H_\omega^\Lambda ) ( \lambda_k = 0) \} | \nonumber \\ & \leq & \|h_0 \|_\infty \left| \int f_\Delta' (t) \xi ( t ; h_1 , h_0 ) ~dt \right| . \eea We use H\"older's inequality to estimate the integral on the right in (\ref{ssf1}). Note that the difference of $ h_1 = H_\omega^\Lambda(\lambda_k = 1)$ and $h_0 = H_\omega^\Lambda ( \lambda_k = 0)$ is proportional to $u( \cdot - k)$. Hence, the SSF for $(h_0, h_1)$ is well-defined and its $L^P$-norm is bounded by $\| V_{eff} \|_{1/p}^{1/p} < \infty$, independent of $|\Lambda |$. From this use of H\"older's inequality, we obtain a fractional power of $|\Delta|$. An improved pointwise bound on the SSF (or the averaged SSF) would eliminate the use of the H\"older inequality and yield Lipschitz continuity of the IDS. This, in turn, implies that the density $dN / d \lambda$ is locally bounded. There are two cases for which Lipschitz continuity of the IDS is known for continuous models in arbitrary dimension. The simplest case is the Anderson-type random potential as in (1.1) when the support of the single-site distribution contains the unit cube. The second case is when $V_\omega$ is a Gaussian process \cite{[FHLM],[HLMW]}. Finally, Tip \cite{[Tip]} proved that the IDS for the Poisson model is absolutely continuous at high energy under a repulsive condition on the single-site potential. \vspace{.1in} \noindent 2. {\bf Landau Hamiltonians.} A nice application of the method of proof of Theorem 1.1 is to the H\"older continuity of the IDS for the random Landau Hamiltonian outside of a discrete set. The unperturbed operator $H_L$ on $L^2 ( \R^2 )$ has the form \begin{equation} \label{landau1} H_L = ( - i \nabla - A)^2, ~~\mbox{where} ~~ A(x_1 , x_2 ) = \frac{B}{2} ( - x_2 , x_1 ), \end{equation} where $B > 0$ is the magnetic field strength. The spectrum is pure point and consists of an increasing sequence of infinitely degenerate, isolated eigenvalues $\{ E_j (B) = (2j + 1)B \; | \; j = 0 , 1, \ldots , \}$. The IDS $N_0 (E)$ for this model is a piecewise constant, monotone increasing function (cf.\ e.\ g.\ \cite{[Nakamura]}). In particular, the IDS $N_0 (E)$ is constant on any open interval of the form $I_n \equiv ( E_n(B) , E_{n+1} (B) )$. Consequently, if ${\tilde \Delta} \subset I_n$, the local estimate (\ref{localtr}) becomes \beq \label{localtr3} Tr E_0^\Lambda({\tilde\Delta})\; \leq \; 2 N_0 (E_+ ) |\tilde\Delta|^{q_1} |\Lambda|, \eeq for any $q_1 \geq 0$. The perturbed family of Landau operators is \begin{equation} \label{landau2} H_\omega = H_L + \lambda V_\omega , \end{equation} where $V_\omega$ is the Anderson-type random perturbation given in (1.1). It is known that $N(E)$ is locally Lipschitz continuous in the following sense. Given an $N > 0$, there is a $B_N >0$ so that for $B > B_N$, the IDS $N(E)$ is Lipschitz continuous on $(0 , (2N+1)B ) \backslash \{ E_j (B) \; | \; j = 0 , 1, \ldots , N \}$ \cite{[CH2],[Wang1]}. Under some additional conditions, Wang \cite{[Wang2]} also proved that $N(E)$ is smooth outside of a given Landau level for sufficiently large magnetic field strength. By applying a slight extension of Theorem 1.1, possible in this case since (\ref{localtr3}) holds for any $q_1 \geq 0$, we are able to prove the following proposition that improves these known results. \vspace{.1in} \begin{prop} The IDS for the random Landau Hamiltonian $H_\omega$, defined in (\ref{landau1})--(\ref{landau2}) is H\"older continuous on each interval $I_n = ( E_n(B) , E_{n+1} (B) )$ with any H\"older exponent $0 < q < 1$. \end{prop} \vspace{.1in} There has been some discussion as to the behavior of the IDS at the Landau energies $E_j (B)$. If the single-site potential $u$ in (1.1) satisfies $u | \Lambda_1 (0) > \epsilon \chi_{\Lambda_1 (0)} > 0$, for some $\epsilon > 0$, then the IDS is locally Lipschitz continuous at all energies \cite{[CH2]}. In \cite{[CHK1]}, we proved a general theorem on the global H\"older continuity of the IDS under assumptions (H1)--(H2), and under the assumption that the magnetic flux through a unit square is $2 \pi$ times a rational number. The proof of this result relies on the methods of the proof of Theorem 1.6. In the absence of the rational flux assumption, we have been able to prove that this IDS is continuous at all energies \cite{[CHK2]}. It is still an open problem to prove the H\"older continuity of the IDS at all energies with no flux condition. \vspace{.1in} \noindent 3. {\bf Regularity.} Very little is known about the regularity of the IDS for the models discussed here. Based on results for lattice models (cf.\ \cite{[BCKP],[CL],[CraigSimon],[PF]}, it is clear that the smoothness of the IDS is related to the smoothness of the probability density $h_0$. 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