\input amstex \documentstyle{amsppt} \TagsOnRight \NoBlackBoxes \topmatter \title Operators with Singular Continuous Spectrum: \\ I. General Operators \endtitle \rightheadtext{Singular Continuous Spectrum: General Operators} \author B.~Simon$^{*}$ \endauthor \leftheadtext{B.~Simon} \thanks ${*}$ This material is based upon work supported by the National Science Foundation under Grant No. DMS-9101715. The Government has certain rights in this material. \endthanks \thanks To appear in {\it Ann.~Math.} \endthanks \affil Division of Physics, Mathematics, and Astronomy \\ California Institute of Technology 253-37 \\ Pasadena, CA 91125 \endaffil \endtopmatter \document \bigpagebreak \flushpar {\bf \S 0. Introduction} The Baire category theorem implies that the family, $\Cal F$, of dense sets $G_\delta$ in a fixed metric space, $X$, is a candidate for generic sets since it is closed under countable intersections; and if $X$ is perfect (has no isolated point), then $A\in\Cal F$ has uncountable intersections with any open ball in $X$. There is a long tradition of soft arguments to prove that certain surprising sets are generic. For example in $\Cal C[0, 1]$, a generic function is nowhere differentiable. Closer to our concern here, Zamfirescu [20] has proven that a generic monotone function has purely singular continuous derivative, and Halmos [7]-Rohlin [14] have proven that a generic ergodic process is weak mixing but not mixing. We will say a set $S\subset X$ is Baire typical if it is a dense $G_\delta$ and a set $S\subset X$ is Baire null if its complement is Baire typical. Our goal is to look at generic sets of self-adjoint operators and show that their spectrum is quite often purely singular continuous. Here are three of our results that give the flavor of what we will prove in \S 3 and \S 4. Consider the sequence space, $[-a, a]^{\Bbb Z}$, of sequences $v_n$ with $|v_{n}|\leq a$. Given any such $v$, we can define a Jacobi matrix $J(v)$ as the tridiagonal matrix with $J_{n, n\pm 1}=1$ and $J_{n,n}=v_n$. View $J$ as a self-adjoint operator on $\ell^{2}(\Bbb Z)$. It is known (e.g.~[4,17,16]) that if one puts a product of normalized Lebesgue measures on $[-a, a]^{\Bbb Z}$ (i.e., the $v_n$ are independent random variables each uniformly distributed in $[-a, a]$) then, $J(v)$ is a.e.~an operator with spectrum $[a-2, a+2]$ and the spectrum there is pure point. So our first result is somewhat surprising. \proclaim{Theorem 1} View $[-a, a]^{\Bbb Z}$ in the product topology. Then $\{v\mid J(v)\text{\rom{ has spectrum }}\mathbreak [-a-2, a+2] \text{\rom{ and the spectrum is purely singular continuous}}\}$ is Baire typical. \endproclaim We also have some results if $\Bbb Z$ is replaced by $\Bbb Z^{\nu}$ and the Jacobi matrix by the multidimensional discrete Schr\"odinger operator. One might think that the weakness of the topology and the one dimension are critical. They are not, as our second result shows. For $V\in\Cal C(\Bbb R^{\nu})$, let $S(V)$ be the Schr\"odinger operator $-\Delta +V$ on $L^{2}(\Bbb R^{\nu})$. \proclaim{Theorem 2} Let $\Cal C_{\infty}(\Bbb R^{\nu})$ be the continuous functions vanishing at infinity in $\|\cdot\|_{\infty}$. Then $$\{V\mid S(V)\text{\rom{ has purely singular continuous spectrum on }} (0,\infty)\}$$ is Baire typical. \endproclaim Note that for $V\in\Cal C_{\infty}(\Bbb R)$, the essential spectrum, $\text{spec}_{\text{\rom ess}}(S(V))=[0, \infty)$, so Theorem 2 says that generically, the singular continuous spectrum, $\text {spec}_{\text{\rom sc}}(S(V))=[0, \infty)$, the absolutely continuous spectrum, $\text{spec}_{\text{\rom ac}}=\emptyset$, and the pure point spectrum, $\text{spec}_{\text{\rom pp}}(S(V))\subset (-\infty, 0]$. For the discrete one-dimensional (Jacobi matrix) case, we will be able to say something about decay. For example when $\nu=1$, a generic $v\in\ell^{p}$ ($20$. Suppose there exist eigenvectors $\eta_n$ of $A_n$: $$A_{n}\eta_{n}=\lambda_{n}\eta_{n}$$ with $\|\eta_{n}\|=1$, $\lambda_{n}\in K$ and $|\langle\eta_{n}, \varphi\rangle|\geq\epsilon$. Then $A$ has an eigenvector $\eta$ with $$A\eta = \lambda\eta$$ with $\lambda\in K$, $\|\eta\|=1$ and $|\langle\eta, \varphi\rangle)\geq \epsilon$. \endproclaim \demo{Proof} $K$ is compact and $\{\psi\in\Cal H\mid \|\psi\|\leq 1\}$ is compact in the weak topology. So we can pass to a subsequence and suppose $\eta_{n}\to\eta_{\infty}$ weakly and $\lambda_{n}\to\infty$. We will show that $\eta_{\infty}\in D(A)$ and $A\eta_{\infty}=\lambda \eta_{\infty}$. Since $|(\eta_{\infty}, \varphi)|\geq\epsilon$, we have $\eta_{\infty}\neq 0$ and so $\eta=\eta_{\infty}/\|\eta_{\infty}\|$ is the required vector. Let $\psi\in\Cal H$ be arbitrary. Then \align \left((A+i)^{-1}\eta_{\infty}, \psi\right)&=\left(\eta_{\infty}, (A-i)^{-1}\psi\right) \\ &= \lim_{n}\left(\eta_{n}, (A_{n}-i)^{-1}\psi\right) \tag 1.1 \\ &=\lim_{n}\left((A_{n}+i)^{-1}\eta_{n}, \psi\right) \\ &=\lim_{n}\left((\lambda_{n}+i)^{-1}\eta_{n}, \psi\right) \\ &=\left((\lambda+i)^{-1}\eta_{\infty}, \psi\right). \endalign It follows that $\eta_{\infty}=(\lambda+i)$, $(A+i)^{-1}\eta_{\infty} \in D(A_{\infty})$ and $A\eta_{\infty}=\lambda\eta_{\infty}$. (1.1) holds because $(A_{n}-i)^{-1}\psi$ converges to $(A-i)^{-1}\psi$ in norm and $\eta_{n}\to\eta_{\infty}$ weakly with $\|\eta_{n}\|\leq 1$. \qed \enddemo \demo{Proof of Theorem \rom{1.1}} Fix $K\subset\Bbb R$ compact, $\epsilon>0$ and $\varphi\in\Cal H$. Then Lemma 1.4 implies that $$\multline Q(K, \varphi, \epsilon)= \\ \{A\in X\mid\exists\eta\in D(A)\text{ with }\|\eta\|=1, |\langle\varphi, \eta\rangle|\geq \epsilon,\, A\eta=\lambda\eta \text{ for some } \lambda\in K\} \endmultline$$ is a closed subset of $X$. Fix $\{\varphi_{l}\}^{\infty}_{l=1}$ an orthonormal basis of $\Cal H$. For $n, l, m\in\Bbb Z_+$, let $$Q_{n,l,m}=Q(C\cap [-n, n], \varphi_{l}, m^{-1}).$$ Then $$\cup Q_{n,l,m}=\{A\mid A\text{ has an eigenvalue in }C\}$$ is an $F_\sigma$, so its complement is a $G_\delta$ as claimed. \qed \enddemo \proclaim{Lemma 1.5} Let $(a, b)$ be a fixed open interval in $\Bbb R^{n}$ and let $d\mu$ be a measure on $\Bbb R$. Then $\mu$ is purely singular on $(a, b)$ if and only if for each $n>2$, there exists $\epsilon_{n}>0$ and $f_n$ obeying \roster \item"\rom{(1)}" $0\leq f_{n}\leq 1$, \item"\rom{(2)}" $f_n$ is supported in $(a-\epsilon_{n}, b+\epsilon_{n})$, \item"\rom{(3)}" $\int\limits^{\infty}_{-\infty} f_{n}(s)ds<2^{-n}$, \item"\rom{(4)}" $\mu\left(\chi_{[a-\epsilon_{n}, b+\epsilon_{n}]} -f_{n}\right)<2^{-n}$, \item"\rom{(5)}" $\epsilon_{n}<2^{-n}$. \endroster \endproclaim \demo{Proof} Suppose such $\epsilon_n$ and $f_n$ exist. Let $$C_{n}=\left\{x\mid f_{n}(x)>\frac12\right\}.$$ Then (with $|\cdot|=$ Lebesgue measure): $$\gathered |C_{n}|<2^{-n+1} \\ \mu([a-\epsilon_{n}, b+\epsilon_{n}]\backslash C_{n})<2^{-n+1} \endgathered$$ and $$C_{n}\subset [a-\epsilon_{n}, b+\epsilon_{n}].$$ It follows that $$C=\bigcap_{m}\,\bigcup^{\infty}_{n=m} C_n$$ obeys $|C|=0$ and $\mu([a, b]\backslash C)=0$. Conversely, suppose that $\mu$ is purely singular continuous on $(a, b)$. Find $C$ in $(a, b)$ so $|C|=0$ and $\mu((a, b)\backslash C)=0$. By adding $a$ and/or $b$ to $C$, we can suppose $C\subset [a, b]$ and $\mu([a, b]\backslash C)=0$. Since $\lim\limits_{\epsilon\downarrow 0} \mu([a-\epsilon, a))=0$ and $\lim\limits_{\epsilon\downarrow 0} \mu((b, b+\epsilon])=0$, we can choose $\epsilon_{n}<2^{-n}$ so that $$\mu([a-\epsilon_{n}, a))+\mu((b, b+\epsilon_{n}])<2^{-n-1}.$$ By regularity of measures, we can find $K_{n}\subset C\subset U_{n} \subset (a-\epsilon_{n}, b+\epsilon_{n})$ so that $|U_{n}|<2^{-n}, \mu([a, b]\backslash K_{n})<2^{-n-1}$. By Urysohn's lemma, find $f$ continuous with $0\leq f\leq 1$, $f\equiv 1$ on $K_n$ and $\text{supp} f\subset U_n$. Then $$\int\limits^{\infty}_{-\infty} f_{n}(s)ds\leq|U_{n}|<2^{-n}$$ while $$\mu\left(\chi_{[a-\epsilon_{n}, b+\epsilon_{n}]}-f_{n}\right) \leq \mu([a-\epsilon_{n}, a))+\mu([a, b]\backslash K_{n})+ \mu((b, b+\epsilon_{n}])<2^{-n}$$ as required. \qed \enddemo \demo{Proof of Theorem \rom{1.2}} Let $\varphi\in\Cal H$, $a, b\in\Bbb R$ and $$Q(\varphi, a, b)=\{A\mid d\mu^{A}_{\varphi}\text{ is purely singular on } (a, b)\}.$$ By Lemma 1.5 $$Q(\varphi, a, b)=\bigcap^{\infty}_{n=2}\,\bigcup_{(f,\epsilon)\in B_{n}} \Cal A_{n}(f, \epsilon; \varphi)$$ where $B_n$ is the set of pairs $(f, \epsilon)$ obeying (1--3; 5) of Lemma 1.5 and $$\Cal A_{n}(f, \epsilon; \varphi)=\{A\mid (\varphi, [\chi_{[a-\epsilon, b+\epsilon]}(A)-f(A)]\varphi)<2^{-n}\}.$$ We claim each $A$ is open, equivalently that $$\Cal A^{c}_{n}(f, \epsilon; \varphi)=\{A\mid (\varphi, \chi_{[a- \epsilon, b+\epsilon]}(A)-f(A)\varphi)\geq2^{-n}\}$$ is closed. For let $A_{l}\in\Cal A^{c}_{n}$ converge to $A$ in strong resolvent sense. Then \linebreak $\lim(\varphi, f(A_{l})\varphi) =(\varphi, f(A)\varphi)$ (see, e.g., [12]). Let $h_m$ be continuous functions with $h_{m}\downarrow\chi_{[a-\epsilon, b+\epsilon]}$ monotonically. Then $h_{m}(A)\to h_{m}(A)$ strongly, so \align (\varphi, \chi_{[a-\epsilon, b+\epsilon]}(A)\varphi) &=\inf_{m}(\varphi, h_{m}(A)\varphi) \\ &=\inf_{m}[\lim_{n}(\varphi, h_{m}({}_{l})\varphi)] \\ &\geq \overline{\lim_{n}}(\varphi, \chi_{[a-\epsilon, b+\epsilon]}(A_{l})\varphi) \endalign so the claim is proven. Any open set $U$ is a countable union of open intervals $I_{n}=(a_{n}, b_{n})$. Let $\varphi_{l}$ be an orthonormal basis for $\Cal H$. Then the set that the theorem asserts is a $G_\delta$ is just $$\bigcap^{\infty}_{n=1}\,\bigcap^{\infty}_{l=1} Q(\varphi_{l}, a_{n}, b_{n})$$ which is indeed therefore a $G_\delta$. \qed \enddemo The following is an expression of the well-known fact of lower semicontinuity of the spectrum under strong limits. \proclaim{Lemma 1.6} If $A_{n}\to A$ in strong resolvent sense and $(a, b)\cap\text{\rom{spec}}(A_{n})=\emptyset$, then $(a, b)\cap \text{\rom{spec}}(A)=\emptyset$. \endproclaim \demo{Proof} Let $f$ be the function $f(x)=\text{dist}(x, \Bbb R \backslash(a, b))$. Then $(a, b)\cap\text{spec}(B)=\emptyset$ if and only if $f(B)=0$. By the continuity of the functional calculus of $A_{n}\to A$ in strong resolvent sense, then $f(A)=\text{s-lim}\, f(A_{n})=0$ if $(a, b)\cap\text{spec}(A_{n})=\emptyset$. \qed \enddemo \demo{Proof of Theorem \rom{1.3}} Let $\lambda_n$ be a countable dense set in $K$. Then $$\{A\mid K\subset\text{spec}(A)\}=\bigcap_{n}\{A\mid\lambda_{n}\in \text{spec}(A)\}$$ so we need only consider the cases where $K=\{\lambda\}$. But $$\{A\mid\lambda\notin\text{spec}(A)\}=\bigcup^{\infty}_{n=1} \left\{A\mid\left(\lambda - \frac1n ,\, \lambda + \frac 1n\right) \cap\text{spec}(A)=\emptyset\right\}$$ is an $F_\sigma$ by Lemma 1.6. Thus, its complement is a $G_\delta$. \qed \enddemo \bigpagebreak \flushpar {\bf \S 2. Welcome to Wonderland} The main point in the way to generate generic singular spectrum is \proclaim{Theorem 2.1} Let $X$ be a regular metric space of self-adjoint operators. Suppose that for some interval $(a, b)$, we have that \roster \item"\rom{(i)}" $\{A\mid A\text{\rom{ has purely continuous spectrum on$(a, b)$}}\}$ is dense in $X$. \item"\rom{(ii)}" $\{A\mid A\text{\rom{ has purely singular spectrum on$(a, b)$}}\}$ is dense in $X$. \item"\rom{(iii)}" $\{A\mid A\text{\rom{ has$(a, b)$in its spectrum}}\}$ is dense in $X$. \endroster Then $\{A\mid (a, b)\subset\text{\rom{spec}}_{\text{\rom{sc}}}(A), (a, b)\cap\text{\rom{spec}}_{\text{\rom{pp}}}(A)=\emptyset, (a, b)\cap\text{\rom{spec}}_{\text{\rom{ac}}}(A)=\emptyset\}$ is a dense $G_\delta$. \endproclaim \demo{Proof} Because $(a, b)$ is an $F_\delta$, each of the sets in (i)--(iii) is a $G_\delta$ by Theorems 1.1--3. (For example, the set in (i) is the intersection of the same sets for $[a+\frac1n, b-\frac1n]$.) Thus, by hypothesis they are dense $G_\delta$'s. By the Baire category theorem, their intersection is a dense $G_\delta$. \qed \enddemo \remark{Remarks} 1. We pick an interval for definiteness. In many cases, one can say things about other sets. 2. We pick the same set $(a, b)$ for convenience. In some examples later, we will take $(a, b)=\Bbb R$ in (ii), but replace $(a, b)$ by a closed set in (i). \endremark Here is a spectacular corollary, which we call the Wonderland Theorem: \proclaim{The Wonderland Theorem} Let $X$ be a regular metric space of operators. Suppose \roster \item"\rom{(a)}" $\{A\mid A\text{\rom{ has purely absolutely continuous spectrum}}\}$ is dense in $X$; \item"\rom{(b)}" $\{A\mid A\text{\rom{ has purely point spectrum}}\}$ is dense in $X$. \endroster Then Baire typically, $A$ has only singular continuous spectrum. \endproclaim \demo{Proof} Strictly speaking, this is not a corollary of the theorem but of its proof, since we do not specify the spectrum. By Theorem 1.1 and (a) $$\{A\mid A\text{ has purely continuous spectrum}\}$$ is Baire typical. Similarly, by Theorem 1.2 and (b) $$\{A\mid A\text{ has purely singular spectrum}\}$$ is Baire typical. So their intersection is Baire typical. \qed \enddemo \bigpagebreak \flushpar {\bf \S 3. General Operators} We apply the theory to general self-adjoint operators first. Throughout, $\Cal H$ is a fixed separable Hilbert space. \proclaim{Theorem 3.1} Fix $a>0$. Let $X=\{A\mid A\text {\rom{ is self-adjoint}},\, \|A\|\leq a\}$ which is a complete metrizable space in the strong topology. Then $$\{A\mid\text{\rom{spec}}(A)=[-a, a]; A\text{\rom{ has purely singular continuous spectrum}}\}$$ is Baire typical. \endproclaim \remark{Remark} For example, if $\varphi_n$ is an orthonormal basis $$\rho (A, A')=\sum^{\infty}_{n=1} \min (2^{-n}, \|(A-A') \varphi_{n}\|)$$ is a metric. \endremark \demo{Proof} This will use the Wonderland Theorem. By the Weyl-von Neumann theorem, the operators with point spectrum are norm dense, but there is a simpler argument since we only need strong density. Since the same argument is needed for dense absolutely continuous spectrum, we give it. Pick an orthonormal basis $\{\varphi_{n}\}^{\infty}_{n=-\infty}$ (this way of counting will be convenient) and let $P_N$ be the projection onto $\{\varphi_{n}\}_{|n|\leq N}$ so that $P_{N}\to 1$ strongly. Let $\alpha_n$ be a counting of the rationals in $[-a, a]$ and let $B$ be the diagonal operator $B\varphi_{n}=\alpha_{n}\varphi_{n}$. Then $$P_{N}AP_{N}+(1-P_{N})B(1-P_{N})\overset s\to\rightarrow A.$$ The operator on the left has spectrum $[-a, a]$ and it is pure point. So we have two of the three hypotheses of the Wonderland Theorem. To prove that absolutely continuous spectrum operators are dense, we need only prove that an operator $A$ with point spectrum and $\|A\| \leq a-\epsilon$ can be approximated since we have just proven such operators are dense. Let $\{ \varphi_{n}\}$ be the eigenvectors of $A$ (say, $A\varphi_{n}=\alpha_{n}\varphi_{n}$) and let $A_{N}=P_{N}AP_{N}$. Fix a sequence $\delta_N$ with $0<\delta_{N}<\frac{\epsilon}{2}$ and $\delta_{N}\to 0$. Let $B_N$ be defined by $$B_{N}\varphi_{n}=\delta_{N}(\varphi_{n+(2N+1)}+\varphi_{n-(2N+1}) +\beta_{n}\varphi_n$$ where $\beta_{n}=\alpha_j$ for the unique $j$ with $n\equiv j$ mod $(2N-1)$. Then $\|B_{N}\|\leq a$ since $\delta_{N}\leq\frac{\epsilon}{2}$ and $B_{N}\to A$ strongly as $N\to\infty$. Each $B_{N}$ is a direct sum of $N+1$ operators of the form $$\alpha_{n}\Bbb I + \delta_{N}J$$ where $J$ is the tridiagonal operator with zeros on diagonal and 1 on the two principal off diagonals. $J$ has absolutely continuous spectrum and thus so does $\alpha_{n}\Bbb I + \delta_{N}J$ and $B_N$. \qed \enddemo Surprisingly, the strong topology is only relevant to be sure that the spectrum is $[-a, a]$: \proclaim{Theorem 3.2} Fix $a1$. If $A$ has no absolutely continuous spectrum, one can take $p=1$. \endremark \demo{Proof} By the Baire category theorem, it suffices to prove the set with (i), (ii) separately are given by dense $G_\delta$'s. By Theorem 1.2, the set of operators $B$ with $\text{spec}_{\text{\rom{ac}}} (A+B)$ empty is a $G_\delta$, and by the Weyl-von Neumann theorem, it is dense so (i) yields a dense $G_\delta$. By Weyl-von Neumann and a simple additional argument, given $\epsilon$, we can find $B_0$ with $\|B_{0}\|_{2}<\frac{\epsilon}{2}$ so $A_{0}\equiv A+B_0$ has simple pure point spectrum. Let $\varphi$ be a cyclic vector for $A_0$ and let $P_0$ be the projection onto $\{\alpha \varphi\mid\alpha\in\Bbb C\}$. By a theorem of [3], $A_{0}+\lambda P_{0}$ has no eigenvalues in $\text{spec}(A_{0})$ for Baire typical $\lambda$ so we can find $|\lambda_{0}|<\frac{\epsilon}2$ so that $A_{0}+ \lambda_{0} P_{0}$ has no eigenvalues on $\text{spec}_{\text{\rom{ess}}} (A)$. Take $B=B_{0}+\lambda_{0} P_{0}$ so $\|B\|_{2}<\epsilon$. This proves the density of the set in (ii). It is a $G_\delta$ by Theorem 1.1. \qed \enddemo \bigpagebreak \flushpar {\bf \S 4. Jacobi Matrices and Schr\"odinger Operators} We will begin with the Jacobi matrix case and prove Theorem 1 of the introduction. \proclaim{Theorem 4.1} Fix $a>0$. Let $X$ be the set of Jacobi matrices on $\ell^{2}(\Bbb Z)$: $$Au_{n}=u_{n+1}+u_{n-1}+x_{n}u_{n}$$ where $x_n$ is an arbitrary sequence with $|x_{n}|\leq a$. Put the topology of pointwise convergence on $\{x_{n}\}$ \rom(so $X$ is a compact metrizable space\rom). Then $$\{A\in X\mid\text{\rom{spec}}(A)=[-a-2, a+2],\, \text{\rom{spec}}(A)\text {\rom{ is purely singular continuous}}\}$$ is Baire typical. \endproclaim \demo{Proof} We use the Wonderland Theorem. Let $d\mu$ be the product of Lebesgue measures $(2a)^{-1}dx_{n}$ so $\text{supp}(d\mu)=[-a, a]^{\Bbb Z}$. Let $D=\{A\in X\mid\text{spec}(A)=[-a-2, a+2],\, \text{spec}(A)\text{ is pure point}\}$. Then $\mu(X\backslash D)=0$ by Anderson localization (see, e.g., [14]). $D$ is dense by the support result. Given any $x_n$, let \align x^{j}_{n}&=x_{n} \quad |n|\leq j \\ &=\text{chosen to be periodic of period } 2j+1\ \quad \text{if } n>|j|. \endalign Thus, $x^{(j)}\to x$ and the Jacobi matrix associated to $x^{(j)}$ has purely absolutely continuous spectrum. \qed \enddemo \remark{Remark} We do not need the full proof of Anderson localization; it suffices that the Jacobi matrices associated to Lebesgue typical sequences have no a.c.~ spectrum and this is easier to prove. For random Jacobi matrices in higher dimension, it is believed that there is sometimes a.c.~spectrum, but that is not so for the generic matrix. Let $\Bbb Z^\nu$ have the norms $|n|=\sum\limits^{\nu}_{j=1}|n_{j}|$ and $\|n\|=\sup\limits_{j}|n_{j}|$. \endremark \proclaim{Theorem 4.2} Fix $a>0$. Let $X$ be the set of Jacobi matrices on $\ell^{2}(\Bbb Z^{\nu})$ $$Au_{n}=\sum_{|j|=1}u_{n+j}+x_{n}u_n$$ where $x$ is an arbitrary multisequence with $|x_{n}|\leq n$. Put the topology of pointwise convergence on $\{x_{n}\}$. Then $$\{A\in X\mid\text{\rom{spec}}(A)=[-a-2\nu, a+2\nu];\, \text{\rom{spec}} (A)\text{\rom{ is purely singular continuous}}\}$$ is Baire typical. \endproclaim We need a lemma which shows how loose'' generic really is: \proclaim{Lemma 4.3} In the setup of Theorem {\rom{4.2}}, suppose that there is a single operator $A_{0}\in X$, $\text{\rom{spec}}_{\text{\rom {ac}}}(A_{0})=\emptyset$. Then, $\text{\rom{spec}}_{\text{\rom{ac}}} (A)=\emptyset$ for a dense set of $A$ in $X$. \endproclaim \demo{Proof} Let $x^{(0)}_{n}$ be the multisequence defining $A_0$. Given $B\in X$ with multisequence $x_n$, define $A_j$ by the multisequence $x^{(j)}_n$ where \alignat2 x^{(j)}_{n}&=x_{n} &&\quad |n|\leq j \\ &=x^{(0)}_{n}&&\quad |n|> j. \endalignat Since $x^{(j)}_{n}\to x_{n}$ pointwise, $A_{j}\to B$. But $A_{j}\to A_0$ is finite rank, so $\text{spec}_{\text{\rom{ac}}}(A_{j})=\text{spec}_ {\text{\rom{ac}}}(A_{0})=\emptyset$. \qed \enddemo \demo{Proof of Theorem \rom{4.2}} We use the Wonderland Theorem. For any rational $q\in [-a, a]$, the set of potentials $x_n$ equal to $q$ if $|n|\geq j$ for some $j$ is dense. Such a potential yields an operator $A$ with $[q-2\nu, q+2\nu]\in\text{spec}(A)$, so generically $\operatornamewithlimits{\cup}\limits_{q} [q-2\nu, q+2\nu]$ is in ] $\text{spec}(A)$. As in the proof of Theorem 4.1, the periodic multisequences are dense and each yields an operator $A$ with no point spectrum, so the operators with no point spectrum are dense. By the lemma, we need only find the operator $A$ in our space with no a.c.~spectrum. Let $\{y_{i}\}_{i\in\Bbb Z}$ be a specific sequence in $[\frac{-a}\nu, \frac{a}\nu]^{\Bbb Z}$ whose one-dimensional Jacobi matrix $J_0$ has only dense point spectrum in $[-2-\frac{a}\nu, 2+ \frac{a}\nu]$. Let $$x_{n}=y_{n_1}+\cdots + y_{n_\nu}$$ so the corresponding $A$ has the form $$J\otimes 1\otimes\cdots\otimes 1+1\otimes J\otimes\cdots\otimes 1+\cdots +1\otimes 1\otimes\cdots\otimes J$$ in $\ell^{2}(\Bbb Z^{\nu})=\ell^{2}(\Bbb Z)\otimes\ell^{2}(\Bbb Z)\otimes\cdots\otimes\ell^{2}(\Bbb Z)$. Then $\text{spec}(A)$ is also pure point. \qed \enddemo \proclaim{Theorem 4.4} Let $c_0$ be the sequences $\{x_{n}\}_{n\in \Bbb Z}$ with $|x_{n}|\to 0$. For $x\in\ell^p$ or in $c_0$, let $J(x)$ be the corresponding Jacobi matrix on $\ell^{2}(\Bbb Z)$. Then $\text{\rom{spec}}_{\text{\rom{ess}}}(J(x))=[-2, 2]$ and $$\{x\mid (x)\text{\rom{ has purely singular continuous spectrum on }} [-2, 2]\}$$ is Baire typical in $c_0$ and in each $\ell^{p}$ $(p>2)$ when these spaces are given the norm topology. \endproclaim \demo{Proof} Since $x_{n}\to 0$ at $\pm\infty$, the diagonal matrix is compact and $\text{spec}_{\text{\rom{ess}}}(J(x))=\text{spec}_{\text {\rom{ess}}}(J(x=0))=[-2, 2]$. Thus, it suffices to find dense sets with no point spectrum in $[-2, 2]$ and with no a.c.~spectrum in $[-2, 2]$. If $x$ has finite support, then any solution of $J(x)u=\lambda u$ with $\lambda\in (-2, 2)$ must be a plane wave outside a finite set and so is not in $\ell^2$. Since the sequences, $x$, of compact support are dense, we have the required density of operators without point spectrum. As in the proof of the last theorem, we need only find one $x$ in our space with no a.c.~spectrum. In [15], Simon showed that if $a_n$ is a typical random sequence, independent and uniformly distributed in $[-1, 1]$, then $x_{n}=(|n|+1)^{-\beta}a_n$ yields a $J(x)$ with pure point spectrum so long as $\beta<\frac12$. This yields the required examples in $\ell^p$ or $c_0$. \qed \enddemo \remark{Remarks} 1. One could instead look at sequences $x_n$ with $\sup\left|(1+|n|)^{\beta}x_{n}\right|<\infty$ in the obvious norm and get the result so long as $\beta<\frac12$. 2. For $p=1$, or $\beta >1$ (in the language of Remark 1), $J(x)$ has lots of a.c.~spectrum, so the result requires some slow falloff hypothesis. It is likely the result remains true for $1<\beta\leq\frac12$ and $1R$ for some $R$. Then $\text{spec}_{\text {\rom{ac}}}(-\Delta+W)=\emptyset$ by using Dirichlet decoupling as in Deift-Simon [2]. Any $W_{0}\in\Cal C_{\infty}(\Bbb R^{\nu})$ is a limit of functions equal to $V$ outside of some ball, so we get the required density. Thus we need only find one $V$\!. To find the required $V$, we choose $V$ spherically symmetric and given by a typical potential in the analysis of Kotani-Ushiroya [10]. These go to zero at infinity and are known to have $\text{spec}(-\frac{d^2}{dx^2}+ V(r))$ pure point. Each partial wave Hamiltonian $-\frac{d^2}{dr^2} + \frac{c}{r^2}+V(r)$ also has no a.c.~spectrum by trace class theory, so $-\Delta+V$ has no a.c.~spectrum. \qed \enddemo \remark{Remark} By looking carefully at [10], the result extends to $L^{p}(\Bbb R^{\nu})$, $p>2n$. \endremark Here is a typical example for random Schr\"odinger operators. \proclaim{Theorem 4.6} For $v\in [-a, a]^{\Bbb Z^{\nu}}$, define $V$ on $\Bbb R^\nu$ by $$V(x)=v(i(x))$$ where $i(x)$ is defined by  i(x)=j \quad \text{\rom{if }} j_{\alpha}\leq x_{\alpha}2$, we have pure point spectrum for Lebesgue typical$\theta$and purely s.c.~spectrum for Baire typical$\theta$. 3. 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