Content-Type: multipart/mixed; boundary="-------------0207222231741" This is a multi-part message in MIME format. ---------------0207222231741 Content-Type: text/plain; name="02-319.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="02-319.keywords" eigenvalues, spectral determinants, PT-symmetric oscillators, 1D Schroedinger equations, Bessi and Zinn-Justin conjecture. ---------------0207222231741 Content-Type: application/x-tex; name="ex2.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="ex2.tex" \documentclass[12pt]{amsart} \usepackage{graphicx,amsmath,amssymb} \headheight=6.15pt \textheight=9.25in \textwidth=6.5in \oddsidemargin=0in \evensidemargin=0in \topmargin=-.375in \newcommand{\C}{\mathbb C} \newcommand{\R}{\mathbb R} \newcommand{\Z}{\mathbb Z} \newcommand{\N}{\mathbb N} %Derivatives: % \renewcommand{\d}{\prime} \newcommand{\dd}{{\prime \prime}} \newcommand{\ddd}{{\prime \prime \prime}} \newcommand{\dddd}{{\prime \prime \prime \prime}} \renewcommand{\Re}{{\rm Re}\,} \renewcommand{\Im}{{\rm Im}\,} \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{untheorem}{Theorem} \newtheorem{unlemma}{Lemma} \newtheorem{conjecture}{Conjecture} \newtheorem{claim}{Claim} \newtheorem{problem}{Problem} \newtheorem{definition}{Definition} \newtheorem*{remark}{Remark} \newtheorem*{remarks}{Remarks} \renewcommand{\theuntheorem}{\hspace*{-4pt}} \renewcommand{\theunlemma}{\hspace*{-4pt}} \renewcommand{\theconjecture}{\hspace*{-4pt}} \renewcommand{\theclaim}{\hspace*{-4pt}} \renewcommand{\theproblem}{\hspace*{-4pt}} \renewcommand{\thedefinition}{\hspace*{-4pt}} \renewcommand{\thefootnote}{} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{document} \title[] {The potential $(iz)^m$ generates real eigenvalues only, under symmetric rapid decay conditions} \author[] {K. C. Shin} \address{Department of Mathematics, University of Illinois, Urbana, IL 61801} \date{July 18, 2002} \begin{abstract} We consider the eigenvalue problems $-u^\dd(z)\pm (iz)^m u(z)=\lambda u(z)$, $m\geq 3$, under every rapid decay boundary condition that is symmetric with respect to the imaginary axis in the complex $z$-plane. We prove that the eigenvalues $\lambda$ are all positive real. \end{abstract} \maketitle \begin{center} {\it Preprint.} %{\it To appear in: When the paper has been accepted, the Journal Title goes here} \end{center} \baselineskip = 18pt % \section{Introduction} \label{introduction} % For integers $m\geq 3$ fixed and $1\leq \ell \leq m-1$, we are considering the non-Hermitian eigenvalue problems \begin{equation}\label{H_peq} H_{\ell}u(z):=\left[-\frac{d^2}{dz^2}+(-1)^{\ell} (iz)^m\right]u(z)=\lambda u(z),\quad\text{for some $\lambda\in\C$}, \end{equation} with the boundary condition that \begin{equation}\label{bdcond} \text{$u(z)\rightarrow 0$ exponentially, as $z\rightarrow \infty$ along the two rays}\quad \arg z=-\frac{\pi}{2}\pm \frac{\ell+1}{m+2}\pi. \end{equation} If a non-constant function $u$ along with a complex number $\lambda$ solves (\ref{H_peq}) with the boundary condition (\ref{bdcond}), then we call $u$ an {\it eigenfunction} and $\lambda$ an {\it eigenvalue.} It is known that any solution of (\ref{H_peq}) is entire (analytic on the complex plane). And it is also known that every solution of (\ref{H_peq}) is either decaying to zero or blowing up exponentially as $z$ tends to infinity along any ray $\{z\in \C: \arg z=const.\}$, except along $m+2$ critical rays where the transition between decaying and blowing-up sectors might occur \cite[\S 7.4]{Hille}. Along these $m+2$ critical rays, any non-constant solution is decaying algebraically \cite[\S 7.4]{Hille}. We will explain these asymptotic properties of the solution in Section \ref{properties}. Before we state our main theorem, we first introduce some known facts about the eigenvalues $\lambda$ of $H_{\ell}$, facts due to Sibuya \cite{Sibuya} and Hille \cite{Hille}. \begin{proposition}\label{main2} For $1\leq \ell\leq m-1$, the eigenvalues $\lambda$ of $H_{\ell}$ are the zeros of a certain entire function of order $\frac{1}{2}+\frac{1}{m}\in\left(\frac{1}{2},1\right)$. In particular, the eigenvalues have the following properties. \begin{enumerate} \item[(i)] Eigenvalues are discrete. \item[(ii)] All eigenvalues are simple. \item[(iii)] Infinitely many eigenvalues exist. \end{enumerate} \end{proposition} For our purposes, we will need to examine the proof of this proposition in some details. In Lemmas \ref{equiv} and \ref{equiv1}, we prove that the eigenvalues are zeros of certain entire functions of order $\frac{1}{2}+\frac{1}{m}\in\left(\frac{1}{2},1\right)$. Then the claims (i) and (iii) are consequences of the Hadamard factorization theorem, while the claim (ii) is due to Hille \cite[\S 7.4]{Hille}. In this paper, we will prove the following main theorem regarding the positivity of the eigenvalues. \begin{theorem}\label{main1} The eigenvalues $\lambda$ of $H_{\ell}$ for integers $1\leq \ell\leq m-1$ are all positive real. \end{theorem} The eigenvalues of $H_{\ell}$ are the same as those of $H_{m-\ell}$, as we show in the proof by reflecting $z\mapsto -z$. The case $\ell=1$ of the theorem is due to Dorey, Dunning and Tateo \cite{Dorey}, and we use this in our proof. Also when $m$ is even and $\ell=\frac{m}{2}$, one can see that $H_{\ell}$ is Hermitian on the real line, and hence $\lambda\in \R$. In all other cases Theorem \ref{main1} is new, and provides the first result of its kind for boundary conditions that neither cluster near the negative imaginary axis nor lie on the real axis. We will explain how Theorem \ref{main1} covers {\it every} symmetric rapid decay condition later when we discuss admissible boundary conditions in Section \ref{properties}. For the rest of the Introduction, we will mention brief history and give some background about our main problem and our method of proof, and then in Section \ref{properties}, we will introduce work of Hille \cite[\S 7.4]{Hille} and Sibuya \cite{Sibuya} about some properties of solutions of (\ref{H_peq}). In Section \ref{induction_step}, we establish an induction step on $\ell$, which is the key element in our proof of Theorem \ref{main1}. More precisely, we will prove that for $1\leq \ell\leq\frac{m}{2}-1$, every eigenvalue $\lambda$ of $H_{\ell+1}$ is positive real if all eigenvalues $\sigma$ of $H_{\ell}$ lie in the sector $|\arg \sigma|\leq\frac{2\pi}{m+2}$ in the complex plane. In Section \ref{main_proof}, we will outline a proof of the induction basis $\ell=1$, that is, eigenvalues of $H_1$ lie in the sector $|\arg \lambda|\leq\frac{2\pi}{m+2}$ (in fact $\arg \lambda=0$; see \cite{Dorey} and also \cite{Shin2}). We then prove Theorem \ref{main1} by induction on $\ell$ and the reflection $z\mapsto -z$. In Section \ref{discussion_de}, we discuss some hopes and challanges in extending our method to more general polynomial potentials. Finally, in the last section, we mention some open problems. \subsection*{History and overview of the method} In this subsection, we introduce some earlier work related with our main result, Theorem \ref{main1}. Also, we discuss our method of the proof of Theorem \ref{main1}. The Hamiltonians with the potential $\pm (iz)^m$ have been studied in many physics and mathematics papers, either under a boundary condition on the real axis \cite{CGM,Simon}, $u(\pm \infty+0i)=0$, or under the boundary condition (\ref{bdcond}) with $\ell=1$ \cite{Bender,Dorey,Shin2}. Simon \cite{Simon} and Caliceti et al.\ \cite{CGM} studied the Hamiltonians $-\frac{d^2}{dx^2}+x^2+\beta x^{m}$ on the real line, where $\beta\in \C-\R_-,$ $m=3,4,5,\dots$, and they proved compactness of the resolvent and discreteness of spectrum. Regarding the reality of eigenvalues, Caliceti et al.\ \cite{CGM} showed that eigenvalues for $-\frac{d^2}{dx^2}+x^2+\beta x^{2n+1}$ are real if $\beta$ is small enough. Recently, a conjecture of Bessis and Zinn-Justin has been verified by Dorey et al.\ \cite{Dorey}, (and extended by the author \cite{Shin2}). That is, the eigenvalues $\lambda$ of \begin{equation}\label{zinn} \left[-\frac{d^2}{dz^2}-\alpha(iz)^{3}+\beta z^2\right]u(z)=\lambda u(z),\quad u(\pm\infty+0i)=0\quad\text{for}\quad \alpha\in \R-\{0\},\,\,\beta\in \R, \end{equation} are all positive real. Dorey et al.\ \cite{Dorey} verified for the case $\beta=0$, and later the author extended for the case $\beta\in\R$ \cite{Shin2}. In fact, Dorey et al.\ \cite{Dorey} have proved more. They studied the following eigenvalue problem \begin{equation}\label{bessis} \left[-\frac{d^2}{dz^2}-(iz)^{2M}-\alpha(iz)^{M-1}+\frac{l(l+1)}{z^2}\right]u(z)=\lambda u(z), \end{equation} under the boundary condition (\ref{bdcond}) with $\ell=1$, and $M, \alpha, l$ being all real. They proved that for $M>1,$ $\alpha1,$ $\alpha0$. If for some positive real numbers $\sigma,\, c_1,\, c_2$, we have $M(r,g)\leq c_1 \exp[c_2 r^{\sigma}]$ for all large $r$, then the order of $g$ is finite and less than or equal to $\sigma$. In this paper, we choose $\arg z^\alpha=\alpha \arg z$ for $-\pi<\arg z\leq \pi$ and $\alpha \in \R$. Now we are ready to introduce some existence results and asymptotic estimates of Sibuya \cite{Sibuya}. The existence of an entire solution with a specified asymptotic representation for fixed $\lambda$, is presented as well as an asymptotic expression of the value of the solution at $z=0$ as $\lambda$ tends to infinity. These results are in Theorems 6.1, 7.2, 19.1 and 20.1 of Sibuya's book \cite{Sibuya}. The following is a special case of these theorems that is enough for our argument later. % \begin{proposition}\label{prop} The equation (\ref{rotated}) admits a solution $f(z,\lambda)$ with the following properties. \begin{enumerate} \item[(i)] $f(z,\lambda)$ is an entire function of $(z,\lambda)$. \item[(ii)] $f(z,\lambda)$ and $f^\d(z,\lambda)=\frac{d}{dz}f(z,\lambda)$ admit the following asymptotic expressions. Let $\epsilon>0$. Then \begin{eqnarray} f(z,\lambda)&=&\,\, z^{-\frac{m}{4}}(1+O(z^{-1/2}))\exp\left[-\frac{2}{m+2}z^{\frac{m+2}{2}}\right],\nonumber\\ f^\d(z,\lambda)&=&-z^{\frac{m}{4}}(1+O(z^{-1/2}))\exp\left[-\frac{2}{m+2}z^{\frac{m+2}{2}} \right],\nonumber \end{eqnarray} as $z$ tends to infinity in the sector $|\arg z|\leq \frac{3\pi}{m+2}-\epsilon$, uniformly on each compact set of the complex $\lambda$-plane. \item[(iii)] Properties \textup{(i)} and \textup{(ii)} uniquely determine the solution $f(z,\lambda)$ of (\ref{rotated}). \item[(iv)] For each fixed $\delta>0$, $f$ and $f^\d$ also admit the asymptotic expressions, \begin{eqnarray} f(0,\lambda)&=&\,\,[1+o(1)]\lambda^{-1/4}\exp\left[K\lambda^{\frac{1}{2}+\frac{1}{m}}\right],\label{eq1}\\ f^\d(0,\lambda)&=&-[1+o(1)]\lambda^{1/4}\exp\left[K\lambda^{\frac{1}{2}+\frac{1}{m}}\right],\label{eq2} \end{eqnarray} as $\lambda$ tends to infinity in the sector $|\arg \lambda|\leq \pi-\delta$, where \begin{equation}\label{def_K} K=\int_0^{\infty}\left(\sqrt{1+t^m}-\sqrt{t^m}\right)\, dt. \end{equation} \item[(v)] The entire functions $\lambda\mapsto f(0,\lambda)$ and $\lambda\mapsto f^\d(0,\lambda)$ have orders $\frac{1}{2}+\frac{1}{m}$. \end{enumerate} \end{proposition} \begin{proof} In Sibuya's book \cite{Sibuya}, see Theorem 6.1 for a proof of (i) and (ii); Theorem 7.2 for a proof of (iii); and Theorem 19.1 for a proof of (iv). And (v) is a consequence of (iv) along with Theorem 20.1. Note that properties (i), (ii) and (iv) are summarized on pages 112--113 of Sibuya's book. \end{proof} The next thing we want to introduce is the Stokes multiplier. Let $f(z,\lambda)$ be the function in Proposition \ref{prop}. Note that $f(z,\lambda)$ decays to zero exponentially as $z\rightarrow \infty$ in $S_0$, and blows up in $S_{-1}\cup S_1$. Then one can see that the function \begin{equation}\label{f_kdef} f_k(z,\lambda):=f(\omega^{-k}z,\omega^{-mk}\lambda), \end{equation} which is obtained by rotating $f(z,\omega^{-mk}\lambda)$ in the $z$-variable solves (\ref{rotated}). It is clear that $f_0(z,\lambda)=f(z,\lambda)$, and that $f_k(z,\lambda)$ decays in $S_k$ and blows up in $S_{k-1}\cup S_{k+1}$ since $f(z,\omega^{-mk}\lambda)$ decays in $S_0$. Then since no non-constant solution decays in two consecutive Stokes sectors, $f_{k}$ and $f_{k+1}$ are linearly independent and hence any solution of (\ref{rotated}) can be expressed as a linear combination of these two. Especially, for some coefficients $C_{j,k}(\lambda)$ and $D_{j,k}(\lambda)$, \begin{equation}\label{stokes} f_{j}(z,\lambda)=C_{j,k}(\lambda)f_k(z,\lambda)+D_{j,k}(\lambda)f_{k+1}(z,\lambda),\quad j,\,k\in \Z. \end{equation} These $C_{j,k}(\lambda)$ and $D_{j,k}(\lambda)$ are called {\it the Stokes multipliers of $f_{j}$ with respect to $f_k$ and $f_{k+1}$}. We then see that \begin{equation}\label{CD_def} C_{j,k}(\lambda)=\frac{W_{j,k+1}(\lambda)}{W_{k,k+1}(\lambda)}\quad\text{and}\quad D_{j,k}(\lambda)=-\frac{W_{j,k}(\lambda)}{W_{k,k+1}(\lambda)}, \end{equation} where $W_{j,k}=f_jf_k^\d -f_j^\d f_k$ is the Wronskian of $f_j$ and $f_k$. Since both $f_j$ and $f_k$ are solutions of the same linear equation (\ref{rotated}), we know that the Wronskians are constant functions of $z$. Since $f_k$ and $f_{k+1}$ are linearly independent, $W_{k,k+1}\not=0$ for all $k\in \Z$. In the next lemma, we will show that the Wronskian $W_{k,k+1}(\lambda)$ is constant, which is needed in the proof of our main theorem. \begin{lemma}\label{unit} For each $k\in \Z$, the Wronskian $W_{k,k+1}(\lambda)=-2\omega^{-\frac{m}{4}-k-1}$, which is independent of $\lambda$. \end{lemma} \begin{proof} Since $f_k(z,\,\lambda)=f(\omega^{-k}z,\omega^{-mk}\lambda)$, we get \begin{eqnarray} f_{k+1}(z,\lambda) &=&f(\omega^{-(k+1)}z,\omega^{-m(k+1)}\lambda)\nonumber\\ &=&f_k(\omega^{-1}z,\omega^{-m}\lambda)\nonumber\\ &=&f_k(\omega^{-1}z,\omega^{2}\lambda),\nonumber \end{eqnarray} using $\omega^{m+2}=1$. Also we see that $$f_{k+1}^\d(z,\lambda)=\omega^{-1}f_k^\d(\omega^{-1}z,\omega^{2}\lambda).$$ Thus \begin{equation}\label{kplus1} W_{j+1,k+1}(\lambda)=\omega^{-1}W_{j,k}(\omega^2\lambda). \end{equation} Next we compute \begin{align} W_{-1,0}(\lambda)&=f_{-1}f^\d_0-f_0f^\d_{-1}\nonumber\\ &=f(\omega z,\omega^{-2}\lambda)f^\d(z,\lambda)-f(z,\lambda)\omega f^\d(\omega z,\omega^{-2}\lambda).\nonumber \end{align} Thus as $z$ tends to infinity in $S_0$ for which above asymptotics are valid, we have \begin{align} W_{-1,0}(\lambda)&=-(\omega z)^{-\frac{m}{4}}z^{\frac{m}{4}}(1+O(z^{-\frac{1}{2}}))\exp\left[-\frac{2}{m+2}(\omega z)^{\frac{m+2}{2}}- \frac{2}{m+2} z^{\frac{m+2}{2}}\right]\nonumber\\ &+ z^{-\frac{m}{4}}\omega(\omega z)^{\frac{m}{4}}(1+O(z^{-\frac{1}{2}}))\exp\left[-\frac{2}{m+2}(\omega z)^{\frac{m+2}{2}}- \frac{2}{m+2} z^{\frac{m+2}{2}}\right]\nonumber\\ &=-2\omega^{-\frac{m}{4}}(1+O(z^{-\frac{1}{2}}))\qquad\text{since $\omega^{\frac{m+2}{2}}=-1.$}\nonumber \end{align} Finally we see that \begin{equation}\label{W-10_def} W_{-1,0}(a,\lambda)=-2\omega^{-\frac{m}{4}}, \end{equation} since $W_{j,k}$ is independent of $z$. Thus (\ref{kplus1}) and (\ref{W-10_def}) complete the proof. \end{proof} In the next lemma, we will show that the Wronskians $W_{k,0}(\lambda)$ for $2\leq k\leq m$ have orders $\frac{1}{2}+\frac{1}{m}$, which is needed in establishing the induction step in Theorem \ref{main} and proving the existence of eigenvalues under various boundary conditions. This lemma is due to Sibuya \cite{Sibuya}, but we give a full proof here. \begin{lemma}\label{unit_order} For each $2\leq k \leq m$, the Wronskian $W_{k,0}(\lambda)$ has its order $\frac{1}{2}+\frac{1}{m}\in \left(\frac{1}{2},\,1\right)$. \end{lemma} \begin{proof} First we look at \begin{align} W_{k,0}(\lambda)&=f_{k}(z,\lambda)f^\d_0(z,\lambda)-f_0(z,\lambda)f^\d_{k}(z,\lambda)\nonumber\\ &=f(0,\omega^{2k}\lambda)f^\d(0,\lambda)-f(0,\lambda)\omega^{-k} f^\d(0,\omega^{2k}\lambda),\label{order_w} \end{align} since the Wronskian is independent of $z$ and so we can take $z=0$. We then know that the Wronskian $W_{k,0}(\lambda)$ has order less than or equal to $\frac{1}{2}+\frac{1}{m}$ because by Lemma \ref{prop} (v) each term in the right hand side of (\ref{order_w}) has order $\frac{1}{2}+\frac{1}{m}$. So in order to show $W_{k,0}(\lambda)$ has order equal to $\frac{1}{2}+\frac{1}{m}$, it suffices to show that there exist $c_1>0$ and $c_2>0$ such that $|W_{k,0}(\lambda)|\geq c_1\exp\left[c_2|\lambda|^{\frac{1}{2}+\frac{1}{m}}\right]$ for all large $|\lambda|$ along some ray. Next we examine the right hand side of (\ref{order_w}) along the ray $$\arg \lambda\equiv \theta=\pi-\frac{4\pi}{m+2},$$ by using the asymptotic expressions (\ref{eq1}) and (\ref{eq2}). Notice $\theta\in \left(-\pi+\delta,\pi-\delta\right)$, for all $0<\delta<\frac{\pi}{m+2}$. Recall that the expressions (\ref{eq1}) and (\ref{eq2}) are defined for $\arg \lambda \in (-\pi+\delta,\pi-\delta)$. So in using (\ref{eq1}) and (\ref{eq2}) for $f(0,\omega^{2k}\lambda)$ and $f^\d(0,\omega^{2k}\lambda)$, we will choose \begin{equation}\label{theta_*_sign} \arg (\omega^{2k}\lambda)\equiv \theta_*=\theta+\frac{4k\pi}{m+2}-2\pi,\quad\text{for $2\leq k\leq \frac{m+2}{2}$}, \end{equation} so we have \begin{equation}\label{theta_sign} -\pi<-\frac{m\pi}{m+2}<\theta_*<\theta<\frac{m\pi}{m+2}<\pi,\quad\text{for $2\leq k\leq \frac{m+2}{2}$}. \end{equation} Thus we see \begin{eqnarray} f(0,\omega^{2k}\lambda)&=&\,\,[1+o(1)]|\lambda|^{-\frac{1}{4}}e^{-i\frac{\theta_*}{4}}\exp\left[K|\lambda|^{\frac{1}{2}+\frac{1}{m}}e^{i\frac{m+2}{2m}\theta_*}\right],\nonumber\\ f^\d(0,\omega^{2k}\lambda)&=&-[1+o(1)]|\lambda|^{\frac{1}{4}}e^{i\frac{\theta_*}{4}}\exp\left[K|\lambda|^{\frac{1}{2}+\frac{1}{m}}e^{i\frac{m+2}{2m}\theta_*}\right].\nonumber \end{eqnarray} Hence from this along with (\ref{order_w}), we get \begin{align} W_{k,0}(\lambda)&=-[1+o(1)]|\lambda|^{1/4}e^{i\frac{\theta}{4}}|\lambda|^{-\frac{1}{4}}e^{-i\frac{\theta_*}{4}}\exp\left[K|\lambda|^{\frac{1}{2}+\frac{1}{m}}\left(e^{i\left(\frac{1}{2}+\frac{1}{m}\right)\theta}+e^{i\left(\frac{1}{2}+\frac{1}{m}\right)\theta_*}\right)\right]\nonumber\\ &+[1+o(1)]|\lambda|^{-1/4}e^{-i\frac{\theta}{4}}\omega^{-k}|\lambda|^{\frac{1}{4}}e^{i\frac{\theta_*}{4}}\exp\left[K|\lambda|^{\frac{1}{2}+\frac{1}{m}}\left(e^{i\left(\frac{1}{2}+\frac{1}{m}\right)\theta}+e^{i\left(\frac{1}{2}+\frac{1}{m}\right)\theta_*}\right)\right]\nonumber\\ &=\left[-e^{i\frac{\theta-\theta_*}{4}}+\omega^{-k}e^{i\frac{\theta_*-\theta}{4}} \right][1+o(1)]\exp\left[K|\lambda|^{\frac{1}{2}+\frac{1}{m}}\left(e^{i\left(\frac{1}{2}+\frac{1}{m}\right)\theta}+e^{i\left(\frac{1}{2}+\frac{1}{m}\right)\theta_*}\right)\right]\nonumber\\ &=-2i\omega^{-\frac{k}{2}}[1+o(1)]\exp\left[K|\lambda|^{\frac{1}{2}+\frac{1}{m}}\left(e^{i\left(\frac{1}{2}+\frac{1}{m}\right)\theta}+e^{i\left(\frac{1}{2}+\frac{1}{m}\right)\theta_*}\right)\right],\nonumber \end{align} where the last step is by $\theta-\theta_*=2\pi-\frac{4k\pi}{m+2}.$ So when $\arg \lambda=\pi-\frac{4\pi}{m+2}$, we have $$\Re \left[e^{i\left(\frac{1}{2}+\frac{1}{m}\right)\theta}+e^{i\left(\frac{1}{2}+\frac{1}{m}\right)\theta_*}\right]=\Re \left[\cos\left(\frac{m+2}{2m}\theta\right)+\cos\left(\frac{m+2}{2m}\theta_*\right)\right]>0,$$ where the last step is by (\ref{theta_sign}). So for $2\leq k\leq\frac{m+2}{2}$, since $K>0$, there exists $c_2>0$ such that $$\left|W_{k,0}(\lambda)\right|\geq \exp\left[c_2|\lambda|^{\frac{1}{2}+\frac{1}{m}}\right]\quad\text{for all large $|\lambda|$ on the ray $\arg \lambda=\pi-\frac{4\pi}{m+2}$}.$$ Thus the order of $W_{k,0}(\lambda)$ for $2\leq k\leq \frac{m+2}{2}$ is $\frac{1}{2}+\frac{1}{m}$. Certainly, the Wronskian $W_{k,0}(\lambda)$ for $2\leq k\leq\frac{m+2}{2}$ is blowing up in some other directions as well. If one find $\theta$ and $\theta_*$ satisfying (\ref{theta_sign}), the above argument will show that $W_{k,0}(\lambda)$ is blowing up along $\arg \lambda=\theta.$ Since $f_k(z,\,\lambda)=f(\omega^{-k}z,\omega^{-mk}\lambda)$, we get \begin{eqnarray} f_{k+1}(z,\lambda) &=&f(\omega^{-(k+1)}z,\omega^{-m(k+1)}\lambda)\nonumber\\ &=&f_k(\omega^{-1}z,\omega^{-m}\lambda)\nonumber\\ &=&f_k(\omega^{-1}z,\omega^{2}\lambda),\nonumber \end{eqnarray} using $\omega^{m+2}=1$. Next for $\frac{m+2}{2} 0$ for some $j$ since both $\sigma_j$ and $\overline{\sigma_j}$ are eigenvalues of the $\mathcal{PT}$-symmetric oscillator $H_{2n-1}$. Therefore, from (\ref{D_unit2}) we conclude $\lambda=\overline{\lambda}$, and so $\lambda$ is real. We still need to show positivity of the eigenvalues. The function $v(z)$ solves (\ref{v_eq}) and we know $\lambda\in\R$. Also, one can check that $\overline{v(\overline{z})}$ solves the same equation. Then since the eigenvalues are simple, $v(z)$ and $\overline{v(\overline{z})}$ must be linearly dependent, and hence $v(z)=c\overline{v(\overline{z})}$ for some $c\in \C$. Since $|v(z)|$ and $|v(\overline{z})|$ agree on the real line, we see that $|c|=1$ and so $|v(z)|=|v(\overline{z})|$ for all $z\in \C.$ That is, $|v(x+iy)|$ is even in $y$. From this we have that \begin{equation}\label{real=0} 0=\left.\frac{\partial}{\partial\,y}|v(x+iy)|^2\right|_{y=0}=-2\Im \left(v^\d(x)\overline{v(x)}\right),\quad\text{for all}\quad x\in \R. \end{equation} Let $g(r)=v(r e^{i\theta})$ for $2n\frac{\pi}{m+2}<\theta<(2n+2)\frac{\pi}{m+2}$. Then $g(r)$ satisfies $$e^{-2i\theta}g^\dd(r)+e^{m i\theta}r^mg(r)=\lambda g(r).$$ We then multiply this by $\overline{g(r)}$ and integrate over $r\in [0,\infty)$ to get $$e^{-2i\theta}\int_0^{\infty}g^\dd(r)\overline{g(r)}\, dr+e^{m i\theta}\int_0^{\infty}r^m|g(r)|^2\, dr=\lambda\int_0^{\infty} |g(r)|^2\, dr.$$ Next we integrate the first term by parts and multiply the resulting equation by $e^{i\theta}$ to get $$-e^{-i\theta}g^\d(0)\overline{g(0)}-e^{-i\theta}\int_0^{\infty}|g^\d(r)|^2\, dr+e^{(m+1) i\theta}\int_0^{\infty}r^m|g(r)|^2\, dr=\lambda e^{i\theta}\int_0^{\infty} |g(r)|^2\, dr.$$ The we use $e^{-i\theta}g^\d(0)=v(0)$ and take the imaginary parts to get, for all $\frac{2n\pi}{m+2}<\theta<\frac{(2n+2)\pi}{m+2}$, $$\sin\theta \int_0^{\infty}|g^\d(r)|^2\, dr+\sin[(m+1)\theta]\int_0^{\infty}r^m|g(r)|^2\, dr=\lambda \sin\theta\int_0^{\infty} |g(r)|^2\, dr.$$ We evaluate this at $\theta=\frac{(2n+1)\pi}{m+1}$ to have $\lambda>0$. Note that since $1\leq n\leq\frac{m-2}{4}$, we see that $\frac{2n\pi}{m+2}<\theta=\frac{(2n+1)\pi}{m+1}<\frac{(2n+2)\pi}{m+2}$. \end{proof} \begin{proof}[Proof of Theorem \ref{main} (Case II: $\ell=2n$ is even, with $1\leq n\leq\frac{m-2}{4}$)] Suppose that all the eigenvalues $\tau_j$ of $H_{2n}$ lie in the sector $|\arg \tau|\leq\frac{2\pi}{m+2}$. That is, zeros of the entire function $\tau\mapsto D_{-n-1,n}(\omega\tau)$ lie in the sector $|\arg \tau|\leq\frac{2\pi}{m+2}$. Then we want to show that each eigenvalue $\lambda$ of $H_{2n+1}$ is positive real. First, we examine $D_{-n-1,n}(\tau)$. From Lemma \ref{equiv1} (iii), we know that the zeros of $\tau\mapsto D_{-n-1,n}(\omega\tau)$ are the eigenvalues $\tau_j$ of $H_{2n}$, which lie in the sector $|\arg \tau|\leq \frac{2\pi}{m+2}$, by hypothesis. So we have $$D_{-n-1,n}(\omega\tau)=B_1\prod_{j=1}^{\infty}\left(1-\frac{\tau}{\tau_j}\right),\quad\text{for some $B_1\in \C-\{0\}$,}$$ where $\Im (\omega\tau_j)\geq 0$ for all $j\in \N$. Hence $$D_{-n-1,n}(\lambda)=B_1\prod_{j=1}^{\infty}\left(1-\frac{\omega^{-1}\lambda}{\tau_j}\right).$$ Thus \begin{equation}\label{D_unit4} \left|\frac{D_{-n-1,n}(\lambda)}{D_{-n-1,n}(\overline{\lambda})}\right|=\prod_{j=1}^{\infty}\left|\frac{1-\frac{\omega^{-1}\lambda}{\tau_j}}{1-\frac{\omega^{-1}\overline{\lambda}}{\tau_j}}\right|=\prod_{j=1}^{\infty}\left|\frac{\omega \tau_j -\lambda}{\omega\tau_j -\overline{\lambda}}\right|\leq 1,\quad\text{when $\Im \lambda\geq 0$,} \end{equation} since $\Im(\omega \tau_j)\geq 0$. Next, we suppose that $u(z)$ is an eigenfunction of $H_{2n+1}$ with the eigenvalue $\lambda$. That is, $u(z)$ solves $$-u^\dd(z)+(-1)^{2n+1}(iz)^mu(z)=\lambda u(z),$$ and decays to zero as $z$ tends to infinity along the two rays $\arg z=-\frac{\pi}{2}\pm (2n+2)\frac{\pi}{m+2}.$ We then let $v(z)=u(-iz)$, and so $v(z)$ solves \begin{equation}\label{v_eq1} -v^\dd(z)+(z^m+\lambda) v(z)=0, \end{equation} and decays to zero as $z$ tends to infinity along the two rays $\arg z=\pm (2n+2)\frac{\pi}{m+2}$ that are the center rays of $S_{n+1}$ and $S_{-n-1}$. Then we examine \begin{equation}\label{pdexp1} f_{-n-1}(z,\lambda)=C_{-n-1,n}(\lambda)f_{n}(z,\lambda)+D_{-n-1,n}(\lambda)f_{n+1}(z,\lambda). \end{equation} We see that $C_{-n-1,n}(\lambda)=0$ by Lemma \ref{equiv}, and hence $D_{-n-1,n}(\lambda)\not=0$. So we have \begin{equation}\label{lambda_1} f(\omega^{n+1}z,\omega^{-2n-2}\lambda)=D_{-n-1,n}(\lambda)f(\omega^{-n-1}z,\omega^{2n+2}\lambda). \end{equation} Then we will show that $\left|D_{-n-1,n}(\lambda)\right|\leq 1$ when $\lambda$ is an eigenvalue of $H_{2n+1}$ with $\Im \lambda\geq 0$. Suppose $f(0,\omega^{-2n-2}\lambda)\not=0$. Since $\lambda$ is an eigenvalue, so is $\overline{\lambda}$. Thus we replace $\lambda$ by $\overline{\lambda}$ in (\ref{lambda_1}), and then evaluate the resulting equation at $z=0$ to get $$ f(0,\omega^{-2n-2}\overline{\lambda})=D_{-n-1,n}(\overline{\lambda})f(0,\omega^{2n+2}\overline{\lambda}).$$ Then we take the complex conjugate of this and apply (\ref{eq111}) so that we have $$f(0,\omega^{2n+2}\lambda)=\overline{D_{-n-1,n}(\overline{\lambda})}f(0,\omega^{-2n-2}\lambda).$$Combining this with (\ref{lambda_1}) at $z=0$ gives \begin{equation}\label{D=1} \overline{D_{-n-1,n}(\overline{\lambda})}D_{-n-1,n}(\lambda)=1, \end{equation} since $f(0,\omega^{-2n-2}\lambda)\not=0$. Similarly, when $f^\d(0,\omega^{-2n-2}\lambda)\not=0$, we get (\ref{D=1}) again. The equation (\ref{D=1}) along with the inequality in (\ref{D_unit4}) implies $\left|D_{-n-1,n}(\lambda)\right|\leq 1$ when $\lambda$ is an eigenvalue of $H_{2n+1}$ with $\Im \lambda\geq 0$. So we get from (\ref{lambda_1}) at $z=0$ \begin{equation}\label{D_unit21} 1\geq \left|D_{-n-1,n}(\lambda)\right|=\prod_{j=1}^{\infty}\left|\frac{1-\frac{\omega^{-2n-2}\lambda}{E_j}}{1-\frac{\omega^{2n+2}\lambda}{E_j}}\right|=\prod_{j=1}^{\infty}\left|\frac{\omega^{2n+2}E_j -\lambda}{\omega^{2n+2}E_j -\overline{\lambda}}\right|\geq 1,\quad\text{when $\Im \lambda\geq 0$,} \end{equation} where the last inequality holds since $\Im (\omega^{2n+2}E_j)\leq 0$ if $0\leq\arg (\omega^{2n+2})\leq\pi$ (this is where we use $1\leq n\leq\frac{m-2}{4}$). Then like in the proof for the case $\ell$ odd, we have $\left|D_{-n-1,n}(\lambda)\right|=1,$ which is also obtained when $f(0,\omega^{-2n-2}\lambda)=0$ (and hence $f^\d(0,\omega^{-2n-2}\lambda)\not=0$.) Therefore, we conclude $\lambda=\overline{\lambda}$, and so $\lambda$ is real, like in the proof of the case $\ell$ odd. We still need to show positivity of the eigenvalues $\lambda$. Recall that $\lambda\in\R$, and the function $v(z)=u(-iz)$ solves (\ref{v_eq1}). Let $g(r)=v(r e^{i\theta})$. Then $g(r)$ satisfies $$e^{-2i\theta}g^\dd(r)+e^{m i\theta}r^mg(r)=\lambda g(r).$$ We multiply this by $\overline{g(r)}$ and integrate over $r\in [0,\infty)$ to get $$e^{-2i\theta}\int_0^{\infty}g^\dd(r)\overline{g(r)}\, dr+e^{m i\theta}\int_0^{\infty}r^m|g(r)|^2\, dr=\lambda\int_0^{\infty} |g(r)|^2\, dr.$$ Since $v(z)$ decays exponentially to zero as $z$ tends to infinity in $S_{n+1}\cup S_{-n-1}$, we have integratibility for $(2n+1)\frac{\pi}{m+2}<\theta<(2n+3)\frac{\pi}{m+2}$. Note that since $\lambda\in\R$, the equation (\ref{real=0}) is valid for this case as well. We integrate the first term by parts and multiply the resulting equation by $e^{i\theta}$ to get $$-e^{-i\theta}g^\d(0)\overline{g(0)}-e^{-i\theta}\int_0^{\infty}|g^\d(r)|^2\, dr+e^{(m+1) i\theta}\int_0^{\infty}r^m|g(r)|^2\, dr=\lambda e^{i\theta}\int_0^{\infty} |g(r)|^2\, dr.$$ Then we use $e^{-i\theta}g^\d(0)=v(0)$ and take the imaginary parts with using (\ref{real=0}) to get, for all $\frac{(2n+1)\pi}{m+2}<\theta<\frac{(2n+3)\pi}{m+2}$, $$\sin\theta \int_0^{\infty}|g^\d(r)|^2\, dr+\sin[(m+1)\theta]\int_0^{\infty}r^m|g(r)|^2\, dr=\lambda \sin\theta\int_0^{\infty} |g(r)|^2\, dr.$$ We evaluate this at $\theta=\frac{(2n+1)\pi}{m+1}$ to have $\lambda>0$. Note that since $1\leq n\leq\frac{m-2}{4}$, we see that $\frac{(2n+1)\pi}{m+2}<\theta=\frac{(2n+1)\pi}{m+1}<\frac{(2n+3)\pi}{m+2}$. \end{proof} \section{Proof of Theorem \ref{main1}}\label{main_proof} In proving Theorem \ref{main1}, our induction basis is provided by the following lemma that is due to Dorey et al. \cite{Dorey2} (see also \cite[Theorem 2]{Shin2}). \begin{lemma}\label{H_1_lemma} The eigenvalues $\lambda$ of $H_1$ are all positive real. \end{lemma} Here we will give an outline of the proof. \begin{proof} Let $\lambda$ be an eigenvalue of $H_1$ with the corresponding eigenfunction $u(z)$. Then we set $v(z)=u(-iz)$, and hence $v(z)$ solves $$-v(z)+\left(z^m+\lambda\right) v(z),$$ and $v(z)$ decays in $S_{-1}\cup S_1$. Then we consider \begin{equation}\label{base_eq} f_{-1}(z,\lambda)=C_{-1,0}(\lambda)f_0(z,\lambda)+D_{-1,0}(\lambda)f_1(z,\lambda). \end{equation} So we see that $C_{-1,0}(\lambda)=0$. Moreover, we know that from (\ref{CD_def}) and (\ref{kplus1}), and Lemma \ref{unit}, $$\left|D_{-1,0}(\lambda)\right|=\left|\frac{W_{-1,0}(\lambda)}{W_{0,1}(\lambda)}\right|=1.$$ Next, we use infinite product forms of either (\ref{base_eq}) when $f_{-1}(0,\lambda)\not=0$, or its differentiated form when $f_{-1}^\d(0,\lambda)\not=0$ at $z=0$. Then like in the proof of Theorem \ref{main}, one can show that $\lambda>0$, since the hypothesis in Theorem \ref{main} is needed for showing $\left|D_{-1,0}(\lambda)\right|=1$ only. \end{proof} Now we are ready to prove Theorem \ref{main1}. \begin{proof}[Proof of Theorem \ref{main1}] For integers $1\leq\ell\leq\frac{m}{2}$, the theorem easily follows from induction on $\ell$ along with Theorem \ref{main} and Lemma \ref{H_1_lemma}. So we assume $\frac{m}{2}<\ell\leq m-1$. Suppose $\lambda$ is an eigenvalue of $H_{\ell}$ with the corresponding eigenfunction $u_1(z)$. Then we let $u(z):= u_1(-z)$, and hence $u(z)$ solves $-u^\dd(z)+(-1)^{\ell}(-iz)^mu(z)=\lambda u(z).$ That is, $u(z)$ solves \begin{equation}\nonumber -u^\dd(z)+(-1)^{m-\ell}(iz)^mu(z)=\lambda u(z), \end{equation} since $(-1)^{2\ell}=1$. Next, we examine the boundary condition. It is clear that since $u_1(z)$ decays along the two rays $\arg z=-\frac{\pi}{2}\pm \frac{\ell+1}{m+2}\pi$, $u(z)$ decays along the two rays \begin{eqnarray} \arg z&=&-\pi-\frac{\pi}{2}+ \frac{\ell+1}{m+2}\pi =-\frac{\pi}{2}-\frac{(m-\ell)+1}{m+2}\pi\quad\text{and}\nonumber\\ \arg z&=&\,\,\,\,\,\pi-\frac{\pi}{2}- \frac{\ell+1}{m+2}\pi=-\frac{\pi}{2}+\frac{(m-\ell)+1}{m+2}\pi,\nonumber \end{eqnarray} which is the boundary condition (\ref{bdcond}) with $m-\ell$ in the place of $\ell$. Hence $u(z)$ is an eigenfunction of $H_{m-\ell}$ with the corresponding eigenvalue $\lambda$. Since $\frac{m}{2}<\ell\leq m-1$, we see that $1\leq m-\ell< \frac{m}{2}$. So by induction with help of Theorem \ref{main} and Lemma \ref{H_1_lemma}, we conclude $\lambda>0$. This completes the proof. \end{proof} \begin{remark} {\rm Our method in this paper closely follows the $\ell=1$ method of Dorey et al.\ \cite{Dorey2} and the author \cite{Shin2}. One big difference is that $\ell=1$ implies $\left|D_{-1,0}(\sigma)\right|=1$ for all $\sigma\in\C$, while for $1<\ell< m-1$, the corresponding functions $\sigma\mapsto D_{-n,n}(\omega^{-1}\sigma)$ in (\ref{pdexp}) and $\sigma\mapsto D_{-n-1,n}(\sigma)$ in (\ref{pdexp1}) are entire functions of order $\frac{1}{2}+\frac{1}{m}$. However, when $\lambda$ is an eigenvalue of $H_{\ell+1}$, under some hypothesis on the eigenvalues $\sigma$ of $H_{\ell}$ we are able to show that $\left|D_{-n,n}(\omega^{-1}\lambda)\right|=1$ for $\ell=2n-1$ odd, and $\left|D_{-n-1,n}(\lambda)\right|=1$ for $\ell=2n$ even. This is the new and main idea in proving the induction step, Theorem \ref{main}. } \end{remark} \section{Hopes and challanges in extending to more general potentials}\label{discussion_de} In this section, we mention some hopes and challenges in extending our induction methods to more general polynomial potentials. As mentioned at the end of the previous section, we used the induction on $\ell$ to show that $\left|D_{-n,n}(\omega^{-1}\lambda)\right|=1$ for $\ell=2n-1$ odd, and $\left|D_{-n-1,n}(\lambda)\right|=1$ for $\ell=2n$ even. Of course, if one can find another way of proving these identities it will help us find some new reality results of eigenvalues for more general polynomial potentials. In \cite{Shin2}, the author studied \begin{equation}\label{shin_eq} H_{\ell}^P u(z)=\left[-\frac{d^2}{dz^2}+(-1)^{\ell}(iz)^m+P(iz)\right]u(z)=\lambda u(z), \end{equation} under the boundary condition (\ref{bdcond}) with $\ell=1$, where $P(z)=a_1z^{m-1}+a_2z^{m-2}+\cdots+a_{m-1}z$ is a real polynomial. I proved that if for some $1\leq j\leq\frac{m}{2}$ we have $(j-k)a_k\geq 0$ for all $k$, then the eigenvalues $\lambda$ are all positive real. One might wonder whether or not it is possible for $1<\ell