\input amstex \documentstyle{amsppt} \magnification=1200 \baselineskip=15 pt \loadbold \TagsOnRight \topmatter \title Bounded eigenfunctions and absolutely continuous spectra for one-dimensional Schr\"odinger operators \endtitle \rightheadtext{Bounded eigenfunctions and a.c.~spectra} \author Barry Simon$^{*}$ \endauthor \leftheadtext{B.~Simon} \affil Division of Physics, Mathematics, and Astronomy \\ California Institute of Technology, 253-37 \\ Pasadena, CA 91125 \endaffil \date March 27, 1995 \enddate \thanks $^{*}$ This material is based upon work supported by the National Science Foundation under Grant No.~DMS-9401491. The Government has certain rights in this material. \endthanks \thanks To be submitted to {\it{Proc.~Amer.~Math.~Soc.}} \endthanks \abstract We provide a short proof of that case of the Gilbert-Pearson theorem that is most often used: That all eigenfunctions bounded implies purely a.c.~spectrum. Two appendices illuminate Weidmann's result that potentials of bounded variation have strictly a.c.~spectrum on a half-axis. \endabstract \endtopmatter \document \flushpar{\bf{\S 1. Introduction and Reduction to $\boldkey m$-functions}} \medpagebreak In this note, I want to consider Schr\"odinger operators and Jacobi matrices on a half-line. Specifically, we'll consider the operator $h$ on $\ell^{2}(\Bbb Z_{+})$ (with $\Bbb Z_{+}=\{1,2,\dots\}$) given by $$\align (hu)(n) &= u(n+1)+u(n-1)+v(n)u(n) \tag 1.1a \\ u(0)&= 0 \tag 1.1b \endalign $$ and the self-adjoint operator on $L^{2}(0,\infty)$ $$\align (Hu)(x) &= -u''(x)+V(x)u(x) \tag 1.2a \\ u(0) &= 0 \tag 1.2b \endalign $$ where we suppose $$ \Gamma(V)\equiv\sup\limits_{x}\, \biggl(\int\limits^{x+1}_{x-1} |V(y)|^{2}\biggr) <\infty. \tag 1.3 $$ For any $E\in\Bbb C$, define two solutions $u_{1}, u_{2}$ of the formal difference (resp.~differential) equation $hu=Eu$ (resp.~$Hu=Eu$) with boundary conditions: $$\xalignat 2 u_{1}(0,E) &= 0 & u_{1}(1,E) & =1 \\ u_{2}(0,E) &= 1 & u_{2}(1,E) & =0 \endxalignat $$ in the discrete case and $$\xalignat 2 u_{1}(0,E) & =0 & u'_{1}(0,E) & =1 \\ u_{2}(0,E) & =1 & u'_{2}(0,E) & =0 \endxalignat $$ in the continuous case. Let $S=\{E\in\Bbb R\mid u_{1}\text{ and $u_{2}$ are bounded on $[0,\infty)$}\}$. Then our purpose here is to prove \proclaim{Theorem 1} On $S$, the spectral measure $\rho$ for $h$ \rom(resp.~$H$\rom) is purely absolutely continuous in the sense that \roster \item"\rom{(i)}" $\rho_{\text{\rom{ac}}}(T)>0$ for any $T\subset S$ with $|T|>0$ \rom(where $|\,\cdot\,|=$ Lebesgue measure\rom) \item"\rom{(ii)}" $\rho_{\text{\rom{sing}}}(S)=0$. \endroster \endproclaim This theorem is not new. In [9,8,11,13], Gilbert, Khan, and Pearson proved a complete characterization of the essential support of $\rho_{\text{\rom{ac}}}$ in terms of mutually subordinate solutions. Their approach has the advantage of not requiring (1.3). Behncke [2] and Stolz [16] have noted that $V$ uniformly $L^{1}_{\text{\rom{loc}}}$ with bounded eigenfunctions allows one to use the Gilbert-Pearson theory. Virtually all applications of [16,12] use the weaker Theorem 1. There seems to be some point in the short proof I'll present here which avoids some of their tricky calculations and which makes the result transparent. In addition, we'll obtain explicit bounds on $m$-functions. I should mention earlier work of Carmona [4] (which is weaker than Theorem 1) and related work of Briet-Mourre [3]. As with Gilbert-Pearson, our proof uses the theory of Weyl $m$-functions. For $E\in\Bbb C_{+}=\{z\mid\text{Im}\, z>0\}$, we can find a unique solution $u_{+}(n,E)$ (resp.~$u_{+}(x,E)$) of (1.1a)/(1.2a) with $u_{+}\in\ell^{2}$ (resp.~$L^2$) at infinity, normalized by $$ u_{+}(0,E)=1. \tag 1.4 $$ Then one defines the $m$ function by $$ m_{+}(E)=u_{+}(1,E) \tag 1.5 $$ in the discrete case and $$ m_{+}(E)=u'_{+}(0,E) \tag 1.6 $$ in the continuous case. By looking at the Wronskian of $u_+$ and $\bar u_{+}$, one gets the well-known formula: $$ \text{Im}\, m_{+}(E)=\text{Im}\, E\sum^{\infty}_{n=1} |u_{+} (n,E)|^{2} \tag 1.7 $$ in the discrete case and $$ \text{Im}\, m_{+}(E)=\text{Im}\, E\int\limits^{\infty}_{0} |u_{+}(x, E)|^{2}\, dx \tag 1.8 $$ in the continuous case. It is known (see [5,15]) that $$ d\rho(E)=\frac{1}{\pi}\,\lim\limits_{\epsilon\downarrow 0}\, \text{Im}\, m_{+}(E+i\epsilon)\, dE. \tag 1.9 $$ It follows [1,7] by the de la Vall\'ee-Poussin theorem that $$ \rho_{\text{\rom{sc}}}\text{ is supported on } \biggr\{E\mid\lim\limits_{\epsilon\downarrow 0}\,\text{Im}\, m_{+}(E+i\epsilon)=\infty\biggl\} $$ and $$ d\rho_{\text{\rom{ac}}}(E)=\frac{1}{\pi}\,\text{Im}\, m_{+} (E+i0)\, dE. $$ Thus, Theorem 1 is an immediate consequence of \proclaim{Theorem 2} If $E\in S$, then $$\alignat2 &\text{\rom{(i)}} &&\qquad \varliminf\,\text{\rom{Im}}\, m_{+} (E+i0)> 0 \tag 1.10 \\ &\text{\rom{(ii)}} &&\qquad \varlimsup\, |m_{+}(E+i0)|<\infty. \tag 1.11 \endalignat $$ \endproclaim \remark{Remark} While the results are stated for the half-line with Dirichlet boundary conditions, Theorem 2 immediately implies the result for any fixed boundary condition and for the whole line. For it is known [1,15] that the essential support $d\rho_{\text{\rom{ac}},\theta}$ for $\theta$ boundary conditions (given by $\sin(\theta)u'(0)+\cos(\theta)u(0)=0$) is $\theta$ independent and that $d\rho_{\text{\rom{sc}},\theta}$ is supported on the set where $m_{+}(E+i0)=-\cot(\theta)$, which cannot happen if (1.10)/(1.11) holds. For the whole line, we can define $S$ via the right half-line condition from which (1.10)/(1.11) and the formula (for the continuous case; the discrete case is similar) $$\align d\rho_{1}(E) &= -\lim\limits_{\epsilon\downarrow 0}\, \frac{1}{\pi}\, \text{Im} \biggl(\frac{1}{m_{+}(E+i\epsilon)+m_{-}(E+i\epsilon)} \biggr)\,dE \\ d\rho_{2}(E) &= \lim\limits_{\epsilon\downarrow 0}\, \frac{1}{\pi}\, \text{Im}\biggl(\frac{1}{m_{+}(E+i0)^{-1}+m_{-}(E+i0)^{-1}}\biggr)\, dE \endalign $$ imply $\rho_{i,\text{\rom{sc}}}(S)=0$. \endremark \medpagebreak It is a pleasure to thank F.~Gesztesy, A.~Kiselev, and G.~Stolz for useful discussions. \newpage \flushpar{\bf{\S 2. The Jacobi Matrix Case}} \medpagebreak In this section, we'll prove Theorem 2 in the discrete case. Define the fundamental or transfer matrix by $$ T(E,n,0)= \pmatrix u_{1}(n+1, E) & u_{2}(n+1, E) \\ u_{1}(n,E) & u_{2}(n,E) \endpmatrix $$ and then $$ T(E,n,m) = T(E,n,0)T(E,m,0)^{-1} \tag 2.1 $$ $T$ is defined so that if $u$ obeys $hu=Eu$, then $\Phi(n)=\binom {u(n+1)}{u(n)}$ obeys $$ \Phi(n)=T(E,n,m)\Phi(m). $$ Constancy of the Wronskian implies $\det\,T=1$ so $\|T^{-1}\|= \|T\|$ and thus by (2.1) $$ C(E)=\sup\limits_{n,m}\,\|T(E,n,m)\|\leq \sup\limits_{n}\, \|T(E,n,0)\|^{2} \tag 2.2 $$ is finite if and only if $E\in S$. We'll prove Theorem 2 in the following explicit form: \proclaim{Theorem 2J} If $E\in S$, then $$\align \varliminf\,\text{\rom{Im}}\, m_{+}(E+i\epsilon) &\geq \frac{1}{4}\, C^{-3} \tag 2.3 \\ \varlimsup\, |m_{+}(E+i\epsilon)| &\leq 4C^{3} \tag 2.4 \endalign $$ where $C(E)$ is given by {\rom{(2.2)}}. \endproclaim \demo{Proof} Let $$ A(E,n)\equiv \pmatrix E-V(n) & -1 \\ 1 & 0 \endpmatrix $$ so $T(E,n,0)=A(E,n)T(E,n-1,0)$. It follows (as a telescoping sum) that $$ T(E+i\epsilon, n,0)=T(E,n,0)+\sum^{n-1}_{j=0}(i\epsilon) T(E,n,j+1)\pmatrix 1 & 0 \\ 0 & 0 \endpmatrix T(E+i\epsilon, j, 0) $$ so by iteration, we get $$ \|T(E+i\epsilon, n,0)\| \leq \sum^{n}_{k=0}\binom{n}{k} C^{k+1} \epsilon^{k}= C(1+C\epsilon)^{n}\leq Ce^{\epsilon Cn} \tag 2.5 $$ By $\|T^{-1}\|=\|T\|$, we see that $$ \biggl\|\binom{u_{+}(E+i\epsilon, n+1)}{u_{+}(E+i\epsilon, n)}\biggr\| \geq C^{-1}e^{-\epsilon Cn} (|m_{+}(E+i\epsilon)|^{2}+1)^{1/2} $$ since $\binom{u_{+}(E+i\epsilon,1)}{u_{+}(E+i\epsilon, 0)} = \binom{m_{+}(E+i\epsilon)}{1}$. Squaring and summing over $n=1,3,\dots$ we see that $$ \sum^{\infty}_{n=1} |u_{+}(E+i\epsilon, n)|^{2} \geq C^{-2} e^{-2\epsilon C} (1-e^{-4\epsilon C})^{-1} (|m_{+}(E+i\epsilon)|^{2} +1). $$ Thus by (1.7) $$ \text{Im}\, m_{+}(E+i\epsilon)\geq \frac{1}{4}\, C^{-3} e^{-2\epsilon C}(4\epsilon C)(1-e^{-4\epsilon C})^{-1} [1+|m_{+}(E+i\epsilon)|^{2}] $$ or $$ \varliminf\,\biggl[\text{Im}\, m_{+}(E+i\epsilon)\big/[1+|m_{+} (E+i\epsilon)|^{2}]\biggr] \geq \frac{1}{4}\, C^{-3} \tag 2.6 $$ Noting that $(1+|m_{+}|^{2})^{-1}\leq 1$, we see that (2.6) immediately implies (2.3). And since $(1+|m_{+}|^{2})/\text{Im}\, m_{+}\geq |m_{+}|$, it also implies (2.4). \qed \enddemo With only minor changes, the theorem extends to the general Jacobi matrix (tridiagonal self-adjoint) matrix: $$ (hu)(n)=a_{n+1} u(n+1)+a_{n}u(n-1)+b_{n}u(n) \tag 2.7 $$ so long as there is $\alpha$ finite with $$ \alpha^{-1}< |a_{n}|<\alpha \tag 2.8 $$ for all $n$. If $d\rho$ is the spectral measure for $u(n)=\delta_{1n}$, then $$ \int\frac{d\rho (E)}{z-E}=m_{+}(z) $$ where $m_{+}(z)$ is defined to be $a^{-1}_{1}u_{+}(1)$ (if $u_+$ is normalized by $u_{+}(0)=1$). (1.7) becomes $$ \text{Im}\, m_{+}(E)=a^{-2}_{1}(\text{Im}\, E) \sum^{\infty}_{n=1} |u_{+}(n,E)|^{2}. $$ It is no longer true that $\|T(E,n,0)^{-1}\|=\|T(E,n,0)\|$ since $\det(T(E,n,0))$ may not be $1$. Rather $\det(T(E,n,0))=\frac{a_1} {a_{n}+1}$ so using (2.8), (2.2) becomes $C(E)\leq \alpha^{2} \sup\limits_{n}\,\mathbreak \|T(E,n,0)\|^{2}$. (2.5) becomes $$ \|T(E+i\epsilon, n,0)\|\leq C(1+C\alpha\epsilon)^{n} \leq Ce^{\epsilon Cn\alpha} $$ and (2.6) becomes $$ \varliminf\, [\text{Im}\,m_{+}(E+i\epsilon)]\big/ [1+a^{2}_{1} |m_{+}(E+i\epsilon)|^{2}] \geq\frac{1}{4}, a^{-2}_{1} C^{-3} \alpha^{-1}. $$ \newpage \flushpar{\bf{\S 3. The Schr\"odinger Case}} \medpagebreak To carry the proof through from the discrete case, we must use (1.3) to bound $u'$ locally by $u$. This is a standard Sobolev-type estimate; we haven't tried to optimize constants. \proclaim{Lemma 3.1} If $u$ obeys $-u''+Vu=Eu$, then $$ |u'(x)|^{2}\leq\biggl[ 4+\frac{3}{4}\,\Gamma(|V-E|)\biggr] \int\limits^{x+1}_{x-1} |u(y)|^{2}\, dy \tag 3.1 $$ where $\Gamma$ is given by {\rom{(1.3)}}. \endproclaim \demo{Proof} By Taylor's theorem with remainder, $$ f'(0)=\frac{1}{2}\,\frac{f(x)-f(-x)}{x}-\frac{1}{2x} \int\limits^{x}_{0} (x-y)[f''(y)+f''(-y)]\,dy. $$ Integrate this from $\frac{1}{2}$ to $1$ to get $$ |f'(0)|\leq\int\limits^{1}_{-1} |f(x)|\,dx +\frac{3}{8}\int \limits^{1}_{-1} |f''(0)|\, dx. $$ Let $f(y)=u(y+x)$ and use $u''=(V-E)u$ and the Schwarz inequality to get (3.1). \qed \enddemo By (3.1), if $E\in S$, $u'$ is also bounded and thus the transfer matrix $T(E,x,y)$ defined by $$ T(E,x,y)\binom{u'(y)}{u(y)}=\binom{u'(x)}{u(x)} $$ is bounded. Let $$ C(E)\equiv\sup\limits_{x,y}\,\|T(E,x,y)\|. $$ \proclaim{Theorem 2S} Let $E\in S$ and define $A(E)=\frac{1}{2} C(E)^{-3}/(9+\frac{3}{2}\Gamma(|E-V|))$. Then $$\align \varliminf\,\text{\rom{Im}}\, m_{+}(E+i\epsilon) &\geq A \\ \varlimsup\,|m_{+}(E+i\epsilon)| &\leq A^{-1}. \endalign $$ \endproclaim \demo{Proof} By mimicking the proof of (2.5), using integrals in place of sums $$ \|T(E+i\epsilon, x,0)\|\leq Ce^{\epsilon C|x|} \tag 3.2 $$ By (3.1) $$ \int\limits^{\infty}_{1} |u'(x)|^{2}\,dx \leq\biggl[ 8+\frac{3}{2}\, \Gamma (|V-E-i\epsilon|)\biggr] \int\limits^{\infty}_{0} |u(y)|^{2}\, dy $$ so if $\beta=1/(9+\frac{3}{2}\Gamma)$, then $$\align \int\limits^{\infty}_{0} |u(y)|^{2}\,dy &\geq \beta\int \limits^{\infty}_{1} [|u(x)|^{2}+|u'(x)|^{2}]\, dx \\ &\geq C^{-2}\beta (1+|m_{+}|^{2})\int\limits^{\infty}_{1} e^{-2\epsilon Cx}\,dx \endalign $$ so by (1.8), $$ \text{Im}\,m_{+}\geq\frac{1}{2}\, C^{-3}\beta e^{-\epsilon C} (1+|m_{+}|^{2}) $$ and the result follows as in the discrete case. \qed \enddemo \vskip 0.3in \flushpar{\bf{Appendix 1: A Discrete Version of Weidmann's Theorem}} \medpagebreak One of the more interesting applications of Theorem 2 is the result of Weidmann [17,18,19] that if $V=V_{1}+V_2$ where $V_{1}\in L^1$ and $V_2$ is of bounded variation with $V_{2}(x)\to 0$ at infinity, then $-\frac{d^2}{dx^2}+V(x)$ has purely a.c.~spectrum on $(0, \infty)$. A key to his argument is a proof that for any $E>0$, solutions are bounded. He does this by noting one can suppose $V_2$ is $C^1$ with $V'_{2}\in L^1$ (by adjusting the breakup) and that if $K(x)=(u')^{2}+ (E-V_{2})u^2$, then $K'(x)=2V_{1}u'u- 2V'_{2}u^{2}\leq C(|V_{1}|+ |V'_{2}|)K(x)$ for $x$ large. Here we'll prove a discrete analog: \proclaim{Theorem A.1} Let $v_n$ be a sequence on $\{1,2,\dots\}$ so that $v_{n}\to 0$ and $$ \sum^{\infty}_{n=1} |v_{n+1}-v_{n}|<\infty. \tag A.1 $$ Then, the operator $h$ of {\rom{(1.1)}} has purely absolutely continuous spectrum on $(-2, 2)$. \endproclaim \remark{Remarks} 1. (A.1) implies $\lim\, v_n$ exist so by adding a constant, it is no loss to suppose $v_{n}\to 0$. 2. If $v_{n}\in\ell^1$, then (A.1) holds so we don't need to consider sums as Weidmann does in the continuous case. \endremark \demo{Proof} Given a solution of $hu=Eu$, let $$ K_{n}=u^{2}_{n+1}+u^{2}_{n}+(v_{n}-E) u_{n}u_{n+1}. $$ Then $$ (K_{n+1}-K_{n}) =(u_{n+2}-u_{n})(u_{n+2}+u_{n}+(v_{n+1}-E)u_{n+1}) +(v_{n}-v_{n+1})u_{n}u_{n+1} $$ so $$|K_{n+1}-K_{n}|\leq |v_{n}-v_{n+1}|\,|u_{n}u_{n+1}|. \tag A.2 $$ Suppose now $E\in (-2,2)$. Then for $n\geq\text{some }N_0$, $2-|v_{n}- E|\geq\delta >0$. For such $n$, $$\align K_{n} &\geq\frac{\delta}{2}\,({u_{n+1}}^{2}+{u_{n}}^{2})+\biggl(1- \frac{\delta}{2}\biggr) (|u_{n+1}|-|u_{n}|)^{2} \\ &\geq\frac{\delta}{2}\, ({u_{n+1}}^{2}+{u_{n}}^{2}) \endalign $$ so (A.2) becomes $$ K_{n+1}\leq \biggl( 1+\frac{2}{\delta}\,|v_{n}-v_{n+1}|\biggr) K_{n} $$ and for all $n\geq N_0$: $$ K_{n}\leq\prod\limits^{\infty}_{m=N_{0}} \biggl(1+\frac{2}{\delta}\, |v_{m}-v_{m+1}|\biggr) K_{N_0}. $$ The product is convergent by (A.1). \qed \enddemo By using the remark at the end of Section 1, Theorem A.1 extends to the operator (2.7) so long as (2.8) holds and $$\alignat2 b_{n} &\to 0, \qquad && \sum^{\infty}_{n=1} |b_{n+1}-b_{n}|<\infty \\ a_{n} &\to 1, \qquad && \sum^{\infty}_{n=1} |a_{n+1}-a_{n}|<\infty. \endalignat $$ We merely define $K_n$ by $$ K_{n}=a_{n+1}u^{2}_{n+1}+a_{n}u^{2}_{n}+(b_{n}-E)e_{n}u_{n+1}. $$ This is related to results of [6]. \vskip 0.3in \flushpar{\bf{Appendix 2: Eigenfunctions for Weidmann's Theorem}} \medpagebreak We want to further elucidate Weidmann's theorem by showing how to actually find the asymptotics of the eigenfunctions. We'll suppose $V(x)=V_{1}(x)+V_{2}(x)$ with $V_{1}\in L^1$ and $V_{2}$ a $C^1$ function with $V'_{2}\in L^1$ and $V_{2}\to 0$ at infinity. We claim: \proclaim{Theorem B.1} Fix $E=k^{2}>0$ with $k>0$. Then every solution of $(-\frac{d^2}{dx^2}+V(x))u=Eu$ is bounded; indeed, there exist $a,b$ so that $$\align |u(x)-au_{+}(x)-bu(x)| &\to 0 \\ |u'(x)-iaku_{+}(x)+ibku_{-}(x)| &\to 0 \endalign $$ where $$ u_{\pm}(x)=\exp\biggl(\pm i\int\limits^{x}_{x_0} \sqrt{k^{2}- V_{2}(x)}\, dx\biggr) \tag B.1 $$ where $x_0$ is chosen so large that $V_{2}(x)x_0$. \endproclaim \remark{Remarks} 1. Since $(k^{2}-V_{2}(x))^{-1/4}\to k^{-1/2}$, we could use the {\eightpoint{WKB}} form instead of (B.1), but the form (B.1) is what enters naturally. 2. This theorem and proof can be regarded as specializations of arguments in Hinton-Shaw [10]. \endremark \demo{Proof} Define $u_\pm$ by (B.1). Note that $u_\pm$ are $C^2$ and $$ -u''_{\pm}+(V(x)-E)u_{\pm}=F_{\pm}u_{\pm} \tag B.2a $$ where $$ F_{\pm}(x)=V_{1}(x)\pm\frac{i}{2}\,V'_{2}(x) (k^{2}-V_{2}(x))^{-1/2} \tag B.2b $$ is in $L^1$ near infinity. Let $W(x)$ be the Wronskian of $u_+$ and $u_-$. Clearly, $W(x)=2ik +o(1)$. Define $a(x), b(x)$ by the equations (variation of parameters) $$\align u(x) &= a(x)u_{+}(x)+b(x)u_{-}(x) \\ u'(x) &= a(x)u'_{+}(x)+b(x)u'_{-}(x). \endalign $$ A straightforward and standard calculation (see prob.~98 on pg.~395 of [14]) shows that $a,b$ obey the equations $$ \binom{a(x)}{b(x)}' = M(x)\binom{a(x)}{b(x)} $$ where $$ M(x)=W(x)^{-1} \pmatrix -F_{+} & -u^{2}_{-}F_{-} \\ F_{+}u^{2}_{+} & F_{-} \endpmatrix. $$ Since this is in $L^1$, standard arguments show that $\lim\limits _{k\to\infty}\binom{a(x)}{b(x)}=\binom{a}{b}$ exists. \qed \enddemo If, moreover, $V$ obeys (1.3) (a mild restriction), this and Theorem 1 implies that $\sigma_{\text{\rom{ac}}}(H)=[0,\infty)$, $\sigma_{\text{\rom{sing}}}\cap (0,\infty)=\emptyset$. \vskip 0.3in \Refs \widestnumber\key{19} \ref\key 1 \by N.~Aronszajn \paper On a problem of Weyl in the theory of Sturm-Liouville equations \jour Am.~J.~Math. \vol 79 \yr 1957 \pages 597--610 \endref \ref\key 2 \by H. 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