text:
\begin{document}

\title{Singular spectrum for radial trees}

\author{Jonathan Breuer}
\address{Jonathan Breuer, Mathematics 253-37, California Institute of Technology, Pasadena, CA 91125, USA}
\email{jbreuer@caltech.edu}

\author{Rupert L. Frank}
\address{Rupert L. Frank, Department of Mathematics,
Princeton University, Washington Road, Princeton, NJ 08544, USA}
\email{rlfrank@math.princeton.edu}

\begin{abstract}
We prove several results showing that absolutely continuous spectrum for the Laplacian on
radial trees is a rare event. In particular, we show that metric trees with unbounded edges
have purely singular spectrum and that generically (in the sense of Baire) radial trees have purely
singular continuous spectrum.
\end{abstract}

\date{}

\maketitle

\sloppy

\section{Introduction and main results}

This paper deals with the absence of absolutely continuous spectrum for the Laplacian on radial trees.
In particular, our purpose is to demonstrate that the existence of absolutely continuous spectrum imposes
rather stringent restrictions on the structure of the tree, so that generally, the occurrence of absolutely
continuous spectrum is a rather exceptional event.
While our primary concern are metric trees, let us demonstrate the above first in the discrete setting.

Let $\Gamma_d$ be a radial discrete tree, that is, a rooted,
radially symmetric tree graph. Denote the root by $O$ and, for any
vertex $x$, the branching number of $x$ (that is, the number of forward
nearest neighbors of $x$) by $b(x)$. The symmetry implies that
$b(x)$ is a function of the distance of $x$ to the root. We only consider infinite trees. Therefore,
there exists a sequence of natural numbers, $\{b_n\}_{n=1}^\infty$,
such that if $n=d(O,x)$ then $b(x)=b_n$.

We have

\begin{theorem} \label{discrete}
Assume $\{b_n\}_{n=1}^\infty$ is bounded. Let $\Delta$ be the
discrete Laplacian on $\Gamma_d$. Then, if $\Delta$ has
nonempty absolutely continuous spectrum, the sequence
$\{b_n\}_{n=1}^\infty$ is eventually periodic. That is, there exists
$N \in \bbN$ such that $\{b_n\}_{n=N}^\infty$ is a (one sided)
periodic sequence.
\end{theorem}
\begin{remark*}
Different authors use different definitions for $\Delta$. The
theorem holds both for

\beq \no (\Delta f)(x)=\sum_{y \sim x}f(y) \eeq (AKA the adjacency
matrix), and for
\beq \no (\Delta f)(x)= \left\{
\begin{array}{cc} \left(b(x)+1 \right)f(x)-\sum_{y \sim x}f(y) & x \neq O \\
\left(b(x)\right)f(x)- \sum_{y \sim x}f(y) & x=O \end{array} \right.
\eeq where $y \sim x$ means $y$ is a nearest neighbor of $x$.
\end{remark*}

Regarding radial \emph{metric} trees (see, e.g., \cite{solomyak}
for an introduction to the concept of metric trees), a moment's
reflection shows that such a result can not hold since the edge lengths
are now continuous parameters. We can, however, show that unbounded edge lengths does rule out absolutely
continuous spectrum.

Explicitly, let $\Gamma_c$ be a radial metric tree with root $O$. As
before, let $b(x)$ be the branching number of a vertex $x$. We assume $b(O)=1$
and $b(x)>1$ for any other vertex. If $x$ is a vertex with $n+1$
vertices on the unique geodesic connecting it to $O$ (including the
endpoints), we denote $d(O,x)=t_n$ and $b(x)=b_n$. The parameters $b_n$ and $t_n$
are well defined because of the radial symmetry. We shall assume
that
\begin{enumerate}
\item $\inf_n \left(t_{n+1}-t_n\right)>0$.
\end{enumerate}

\begin{remark*}
In the metric tree literature, the trees we consider here are
usually called \emph{regular trees} \cite{naim-solom,solomyak}. Since this paper also
has a theorem about discrete trees, where regularity usually means that every vertex has
the same number of neighbors, we chose the term \emph{radial} for
both settings as a unifying compromise.
\end{remark*}

Let $-\Delta$ be the operator in $L^2(\Gamma_c)$
defined through the quadratic form \beq \no \int_{\Gamma_c} |u'(x)|^2\,dx
\eeq for functions $u \in H^1_0(\Gamma_c)$, the Sobolev space of
functions continuous along the edges and satisfying $\int_{\Gamma_c}
\left(|u'(x)|^2+|u(x)|^2 \right)\,dx  <\infty$, $u(O)=0$. Functions
in the operator domain of $-\Delta$ satisfy Dirichlet boundary conditions at
the root and Kirchhoff boundary conditions at the vertices.

We shall prove

\begin{theorem}\label{continuous}
Under assumption (1) above, if
\beq \no
\limsup_{n \rightarrow \infty}(t_{n+1}-t_n)=\infty \,,
\eeq
then the spectrum of $-\Delta$ is purely singular, in the
sense that any spectral measure for $-\Delta$ is supported on a set of Lebesgue measure zero.
\end{theorem}

Put differently, this theorem says that \emph{any} subsequence of unbounded edges destroys absolutely continuous spectrum,
no matter what happens between these unbounded edges.
The trees described in Theorem \ref{continuous} are sparse in the sense that their branchings become sparse as the
distance from the root increases (at least along some sequence). The fact that sparse potentials for one-dimensional
Schr\"odinger operators may lead to singular spectral measures was discovered by Pearson \cite{pearson} thirty years
ago and has been explored extensively since then (see \cite{last-review} and references therein). The paper
\cite{remling-disc}, whose main result is one of the principal ingredients in our analysis, has a remarkably general
result (Corollary 1.5 there) in this area.

A restricted family of discrete sparse trees was
studied in \cite{breuer-sparse, breuer-mol}, where it was shown that by controlling the various parameters defining
the tree, it is possible to control also the `degree of continuity' of the spectral measures. Theorem \ref{continuous}
above is much softer, but is considerably more general. It is also, to the best of our knowledge, the first theorem
of this type to be proven for metric trees. We find it remarkable that such a small class of radial metric trees
has absolutely continuous spectrum. The next two theorems show that pure point spectrum does not occur too often as well.

\begin{theorem}\label{simonstolz}
Assume that the sequence $\{b_n\}$ is bounded. Then, if
\beq \label{sparseness}
\limsup_{n \rightarrow \infty}\frac{t_{n+1}-t_n}{n^{2n}}>0 \,,
\eeq
the spectrum of $-\Delta$ coincides with $[0,\infty)$ and is purely singular continuous.
\end{theorem}

For any $\varepsilon>0$, $C>0$ we consider the set $\mathcal{T}^{\varepsilon,C}$ of radial trees whose
defining parameter sequences, $\{t_n,b_n\}$, satisfy
\begin{enumerate}
\item[$(1)_\epsilon$] $\inf \left(t_{n+1}-t_n\right)\geq\varepsilon,\qquad t_1 \geq \varepsilon$.
\item[$(2)_C$] $\sup b_n \leq C$.
\end{enumerate}
Moreover, we allow $\{t_n,b_n\}$ to be a finite (possibly empty) sequence. This means that we also consider trees which have only a finite number of vertices and which contain half-lines.

Identifying sequences $\{t_n,b_n\}$ with measures $\sum_n \beta_n \delta_{t_n}$
(with $\beta_n=\frac{\sqrt{b_n}+1}{\sqrt{b_n}-1}$)
we can consider
$\mathcal{T}^{\varepsilon,C}$ as a (compact) metric space with convergence being induced from weak convergence of measures.
This convergence is natural for us since, as we shall see,
convergence of the trees implies strong resolvent convergence
of the corresponding Laplacians (up to a natural unitary transformation taking the
different Hilbert spaces into account).

{F}rom Theorem \ref{simonstolz} we shall deduce the somewhat surprising

\begin{theorem}\label{wonderland}
In the space $\mathcal{T}^{\varepsilon,C}$, with the topology of weak convergence for the corresponding measures,
the set of trees whose spectrum on $[0,\infty)$ is purely singular continuous, is a dense $G_\delta$ set.
\end{theorem}

Recall that a $G_\delta$ set is a countable intersection of open sets.

As with other works dealing with radial trees (see e.g.\ \cite{breuer-sparse, breuer-mol, carlson, naim-solom, solomyak}),
the radial symmetry reduces the analysis to a one-dimensional
problem. Thus, the exclusion of eigenvalues in Theorem \ref{simonstolz} follows by a Simon-Stolz type argument
\cite{simon-stolz} and Theorem~\ref{wonderland} follows from Theorem \ref{simonstolz} with the help of Simon's
Wonderland Theorem~\cite{simon-wonderland} applied to the corresponding families of one-dimensional operators.

The new ingredient which enters in the proofs of Theorems
\ref{discrete} and \ref{continuous} is a recent theorem of Remling's \cite{remling-disc, remling-cont}
following work by Breimesser and Pearson \cite{bp1, bp2}. Remling's Theorem leads to various
explicit restrictions on one-dimensional discrete and continuous Schr\"odinger
operators with absolutely continuous spectrum. The structural restrictions on trees with
absolutely continuous spectrum are a consequence of these restrictions.
In particular, Theorem \ref{discrete} is an immediate corollary of Theorem 1.1 of \cite{remling-disc},
given Theorem 2.4 of \cite{breuer-sparse}.

In contrast with the discrete case, Theorem \ref{continuous} is not
an immediate consequence of the results of \cite{remling-disc} or
its continuous counterpart \cite{remling-cont}. The difficulty lies
in the fact that the objects appearing in the direct sum
decomposition of $-\Delta$ (see, e.g., \cite{carlson, naim-solom})
are not `standard' Schr\"odinger operators, but rather
Sturm-Liouville operators on weighted $L^2$ spaces with rather singular weights.

The better part of the rest of this paper is devoted to demonstrating the
applicability of Remling's Theorem to these operators. A crucial point in the
analysis is the proof that for a whole-line potential that is reflectionless
(see Section 4.1 for the definition) on a set of positive Lebesgue
measure, the part of the potential lying to the left of $0$ uniquely
determines the part lying to the right of $0$. In the context of
Jacobi matrices and Schr\"odinger operators with measure valued
potentials this is, indeed, a simple realization relying on
classical results. In our case, however, this seems to be a new
result (in particular, see Proposition \ref{unique}). In order to prove this, we have made use of the
Kre\u\i n formula for the difference of the resolvents of two
different self-adjoint extensions of a closed, densely defined,
symmetric operator, as it appears in \cite{posilicano} (see Section 3).
We are not aware of any previous application of this formula in
the spectral theory of Schr\"odinger operators on radial trees, and we believe that it may be useful also beyond the
context of the present paper.

Examples of trees, and more general graphs, for which the graph Laplacian has singular spectrum,
have been constructed before \cite{breuer-sparse, breuer-mol, hislop-post, teplyaev, graph-lap, teplyaev1}.
However, to the best of our knowledge, the theorems above are the first of their kind in terms of the generality
in which they hold.
 In particular, we do not know of another Wonderland-type theorem for trees. Interestingly enough, it is not clear
how to formulate an interesting analogue of Theorem \ref{wonderland} for discrete trees.
The reason is that in order to exclude eigenvalues one needs a `free operator' which approximates the tree.
The natural free operator in the discrete case is the discrete Laplacian which has spectrum in [-2,2]. Since,
in general, discrete trees might have a significant portion of their spectrum outside [-2,2], the exclusion of
eigenvalues there does not have the implications of Theorem \ref{simonstolz}.

The rest of this paper is structured as follows. Section 2 describes the reduction of the above theorems to theorems for
one-dimensional Schr\"odinger operators with point interactions. Section 3 proves a resolvent formula and
a uniqueness result for such operators. Section 4 completes the proof of the theorems. As noted above, Theorem
\ref{discrete} is a direct consequence of Theorem 1.1 in \cite{remling-disc}. Thus, no additional discussion will be
devoted to its proof.

\textbf{Acknowledgments}.
We are grateful to Barry Simon for useful discussions.
RF appreciates the warm hospitality of Caltech, where part of this work has been done, and acknowledges support through DAAD grant D/06/49117.

\section{Reduction to the one-dimensional case}\label{sec:reduction}

Using the radial symmetry of the tree we shall deduce our main theorems from results about one-dimensional operators. In this section we describe this reduction and state the corresponding theorems in the one-dimensional context.

As in the previous section, let $\Gamma_c$ be a radial metric tree associated with parameters $\{(t_n,b_n)\}_{n=1}^\infty$, which are assumed to satisfy (1). We put $(t_0,b_0):=(0,0)$. For any integer $k\geq 0$, we introduce the self-adjoint operator $A_k^+$ in $L^2(t_k,\infty)$ defined by $(A_k^+ f)(r)=-f''(r)$ for $r\in \cup_{n=k}^\infty (t_n,t_{n+1})$ with domain consisting of all functions
$f\in H^2\left(\cup_{j=k}^\infty (t_n,t_{n+1})\right)$ satisfying $f(t_k)=0$ and
\beq \label{boundaries}
f(t_n+)=\sqrt{b_n}f(t_n-), \qquad f'(t_n+)=\frac{1}{\sqrt{b_n}}f'(t_n-)
\eeq
for $n>k$. These operators appear naturally in the direct sum decomposition of the Laplacian \cite{carlson,naim-solom,solomyak}. Indeed, one has

\begin{proposition}[\cite{solomyak}]\label{decomposition}
$-\Delta$ is unitarily equivalent to the direct sum \beq
\label{directsum} -\Delta \cong A^+_0\oplus \sum_{k=1}^\infty \oplus
\left( A_k^+ \otimes I_{\bbC^{b_1\cdots b_{k-1}(b_k-1)}} \right) .\eeq
\end{proposition}

It follows from this proposition that our main results, Theorems \ref{continuous}, \ref{simonstolz} and~\ref{wonderland},
will be proved if we can show the corresponding results for any of the operators $A^+_k$, $k\geq 0$.
Since we consider general sequences $\{(t_n,b_n)\}_{n=1}^\infty$ we may, without loss of generality, restrict our
attention to $A^+_0$. To simplify notation we denote this operator from now on by $A^+$.
Moreover, when studying these operators we need no longer assume that the $b_n$'s are integer-valued. All we need is
\begin{enumerate}
\item[(2)] $\inf_{n\geq 1} b_n >1$.
\end{enumerate}
Using boundary conditions \eqref{boundaries} and integrating by parts one easily finds that
\begin{equation}
 \label{eq:apos}
(f,A^+ f) = \int_0^\infty |f'|^2 \,dt
\end{equation}
for $f\in\dom A^+$. In particular, $A^+$ is a non-negative operator.

We now state our results concerning the operators $A^+$ which will imply Theorems~\ref{continuous}, \ref{simonstolz} and~\ref{wonderland}.

\begin{theorem} \label{delta-operator}
Let $\{(t_n,b_n)\}_{n=1}^\infty$ satisfy assumptions (1) and (2) and let $A^+$ be the associated operator. Then, if $\limsup_{n \rightarrow \infty} (t_{n+1}-t_n)=\infty$, the absolutely continuous spectrum of the operator $A^+$ is empty.
\end{theorem}

One easily sees that if $\limsup_{n \rightarrow \infty} (t_{n+1}-t_n)=\infty$, then the spectrum of $A^+$ coincides with the interval $[0,\infty)$. According to Theorem \ref{delta-operator} this spectrum might have a singular continuous and a pure point component. However, if a subsequence of the differences $t_{n+1}-t_n$ grows sufficiently fast, we can rule out the existence of eigenvalues following an argument of Simon and Stolz \cite{simon-stolz} and we obtain

\begin{theorem}\label{no-ev}
Let $\{(t_n,b_n)\}_{n=1}^\infty$ with $\sup_n b_n<\infty$ and let $A^+$ be the associated operator. Then, if
$ \limsup_{n \rightarrow \infty} n^{-2n} (t_{n+1}-t_n)>0$, the spectrum of $A^+$ coincides with $[0,\infty)$ and is purely singular continuous.
\end{theorem}

Our next result states that singular continuous spectrum is indeed the `generic' situation.
In order to define what we mean by `generic' we will introduce a natural topology on the sequences
$\{(t_n,b_n)\}_{n=1}^\infty$ as above.
It will be convenient to identify such a sequence with a measure $\mu = \sum \beta_n \delta_{t_n}$ on $\R_+$ where $\beta_n =(\sqrt{b_n}+1)/(\sqrt{b_n}-1)$.
For later use we will consider at once the case of measures on the whole line.

For any $\varepsilon>0$ we denote by $\mathcal{M}_{\rm a}^{\varepsilon}$ the set of all non-negative atomic measures $\mu$ on $\R$ of the form $\mu=\sum_{n\in J} {\beta}_n \delta_{{t}_n}$ where $ \beta_n\in [1,\infty)$ and where $ t_n$ are real numbers satisfying $|{t}_{n}-{t}_{m}| \geq \varepsilon$ for all $n\neq m$. The index set $J$ may be finite, infinite or empty.  Moreover, we denote by $\mathcal{M}_{\rm a}^{\varepsilon,+}$ the subsets consisting of all $\mu\in \mathcal{M}_{\rm a}^{\varepsilon}$ with $\supp\mu\subset [\epsilon,\infty)$, and we put
$$
\mathcal{M}_{\rm a}^{0} := \bigcup_{\varepsilon>0} \mathcal{M}_{\rm a}^{\varepsilon} \,,
\qquad
\mathcal{M}_{\rm a}^{0,+} := \bigcup_{\varepsilon>0} \mathcal{M}_{\rm a}^{\varepsilon,+} \,.
$$
Finally, for $C\geq 2$ let $\mathcal{M}_{\rm a}^{\varepsilon,C,+}$ be the subset consisting of all $\mu\in \mathcal{M}_{\rm a}^{\varepsilon,+}$ with $1+C^{-1} \leq \beta_n \leq C$ for all $n\in J$.

With any measure $\mu= \sum_{n\in J} {\beta}_n \delta_{{t}_n} \in \mathcal{M}_{\rm a}^{0,+}$ we associate an operator $A^+_\mu$ in $L^2(\R_+)$ acting as $A^+_\mu f = -f''$ in $\R_+\setminus\supp\mu$ on functions satisfying $f(0)=0$ and \eqref{boundaries} for all $n\in J$ where $\beta_n = (\sqrt{b_n} +1)/(\sqrt{b_n}-1)$ with $b_n\in (1,\infty]$. (For $b_n=\infty$, \eqref{boundaries} is interpreted as $f(t_n-)=0$ and $f'(t_n+)=0$.) If $J$ is infinite and all $b_n$'s are finite, this is precisely the operator $A^+$ defined above.

The one-dimensional analog of Theorem \ref{wonderland} is

\begin{theorem}\label{wonderland1d}
In the space $\mathcal{M}_{\rm a}^{\varepsilon, C,+}$ with the topology of weak convergence,
the set of $\mu$'s for which the spectrum of $A_\mu^+$ is purely singular continuous is a dense $G_\delta$ set.
\end{theorem}

\begin{remark*}
 The proof will show that the same is true if we restrict the $b_n$'s to be integers. This is what we need when we deduce Theorem \ref{wonderland}. Moreover, to deduce Theorem \ref{wonderland} we also use that a countable intersection of dense $G_{\delta}$'s is a dense $G_\delta$ by Baire's Category Theorem.
\end{remark*}

\section{The resolvent and the m-function}

\subsection{A resolvent formula}

In this subsection we derive a convenient expression for the resolvent of the operator $A^+=A^+_\mu$ for $\mu\in
\mathcal{M}_{\rm a}^{0,+}$. We write
\begin{equation}\label{eq:mu}
\mu = \sum_{n\in J} \frac{\sqrt{b_n}+1}{\sqrt{b_n}-1} \ \delta_{t_n}\,,
\qquad
0 < t_1 < t_2 < \ldots \,,
\qquad
1< b_n\leq\infty \,,
\end{equation}
where $J$ is either of the form $\{1,2,\ldots,\#\supp\mu\}$ if there is a finite number of atoms, or $J=\N$ if there are infinitely many atoms. Let $A^+_0 :=-d^2/dt^2$ be the Dirichlet Laplacian in $L^2(\R_+)$ and recall that its resolvent $(A^+_0-z)^{-1}$, $z\in\C\setminus[0,\infty)$, has integral kernel
$$
g_z(t,u) := \frac{i}{2k} \left( e^{ik|t-u|} -e^{ik(t+u)} \right)\,,
\qquad z=k^2,\, \im k>0\,.
$$
Put $\mathfrak H := \ell(J,\C^2)$ and  let $\gamma$ be the trace operator from $L^2(\R_+)$ to $\mathfrak H$ with domain $\dom\gamma=\dom A^+_0$, that is,
$$
(\gamma f)_n := \begin{pmatrix} f(t_n) \\ f'(t_n) \end{pmatrix} \,.
$$
For any $z\in\C\setminus[0,\infty)$ we define an operator $T(z)$ in $\mathfrak H$ by
\begin{align*}
T(z)_{nm} :=
\begin{pmatrix}
    \frac{1}{2ik} \left( e^{ik|t_n-t_m|} -e^{ik(t_n+t_m)} \right) &
    \frac{1}{2} \left( \sigma_{mn} e^{ik|t_n-t_m|} -e^{ik(t_n+t_m)} \right) \\
    \frac{1}{2} \left( \sigma_{nm} e^{ik|t_n-t_m|} -e^{ik(t_n+t_m)} \right) &
    -\frac{ik}{2} \left( e^{ik|t_n-t_m|} + e^{ik(t_n+t_m)} \right)
  \end{pmatrix}\,
\end{align*}
where $\sigma_{mn}:=\sgn(t_m-t_n)$,
with the convention $\sgn(0)=0$. Finally, we define the multiplication operator $B$ in $\mathfrak H$ by
$$
B_{nm} :=  \delta_{nm} \frac12 \frac{\sqrt {b_n} +1}{\sqrt {b_n}-1}
\begin{pmatrix}
0 & 1 \\ 1 & 0
\end{pmatrix} \,.
$$
The following expression for the resolvent of the operator $A^+=A^+_\mu$ will be very useful for us.

\begin{lemma}\label{res}
For any $z\in\C\setminus[0,\infty)$ one has
\begin{equation}\label{eq:res}
(A^+-z)^{-1} = (A^+_0-z)^{-1}  + \left( \gamma (A^+_0-\overline z)^{-1} \right)^* \left( T(z) + B \right)^{-1}  \gamma (A^+_0-z)^{-1} \,.
\end{equation}
\end{lemma}

\begin{proof}
Obviously, $T(\overline z)=T(z)^*$. It is straightforward to check that
$$
\gamma_\pm \left( \gamma (A^+_0-\overline z)^{-1} \right)^* = -T(z) \pm \frac 12 J\,,
\qquad J_{nm}:=\delta_{nm}
\begin{pmatrix}
0 & 1 \\ -1 & 0
\end{pmatrix} \,.
$$
Therefore, the resolvent formula implies that
$$
T(z) - T(\zeta) = (\zeta -z) \gamma_{\pm}\left(A^+_0 -\zeta\right)^{-1} \left( \gamma \left(A^+_0 -\overline z\right)^{-1} \right)^* \,
$$
and so
$$
T(z) - T(\zeta) = (\zeta -z) \gamma\left(A^+_0 -\zeta\right)^{-1} \left( \gamma \left(A^+_0 -\overline z\right)^{-1} \right)^* \,.
$$
Hence, by the abstract result of Posilicano \cite{posilicano} there exists a self-adjoint operator $G$, say, with $(G-z)^{-1}$ given by the RHS of \eqref{eq:res}. We need to prove that $G=A^+$. For any function $f\in L^2(\R_+) \cap H^2(\R_+\setminus\{t_n\}_{n\in J})$ we introduce
$$
(\gamma_\pm f)_n := \begin{pmatrix} f(t_n\pm) \\ f'(t_n\pm) \end{pmatrix} \,.
$$
Now assume that $f\in\dom \,G$. From \eqref{eq:res} one sees that $f\in H^2(\R_+\setminus\{t_n\}_{n\in J})$ and  $Gf=-f''$. Let $a_\pm :=\gamma_\pm f$ and $c:=\gamma (A^+_0-z)^{-1} (G-z)f$. (Note that $(A^+_0-z)^{-1} (G-z)f$ and its derivative are continuous.) Applying \eqref{eq:res} to $(G-z)f$ we learn that
\begin{equation}\label{eq:bc}
a_\pm = c +  (-T(z) \pm \tfrac 12 J ) (T(z)+B)^{-1} c = (B \pm \tfrac 12 J ) (T(z)+B)^{-1} c   \,.
\end{equation}
Decomposing $J=\hat J \dot\cup \check J$ where $\hat J := \{n\in J : \, b_n<\infty \}$ and accordingly $\mathfrak H=\hat {\mathfrak H} \oplus \check{\mathfrak H}$ and $a_\pm = \hat a_\pm + \check a_\pm$, we see from \eqref{eq:bc} after eliminating $c$ that
$
\hat a_+ = (B + \frac 12 J )  (B - \frac 12 J )^{-1} \hat a_- \,.
$
(Note that the inverse $(B - \frac 12 J )^{-1}$ is well-defined on $\hat{\mathfrak H}$.) Calculating the product of the two matrices we find jump condition \eqref{boundaries} for $f$. For $n\in\check J$  \eqref{eq:bc} says that the second component of $a_{+,n}$ and the first component of $a_{-,n}$ are zero, which again are the claimed boundary conditions for $f$.
\end{proof}

Our first application of the resolvent formula is to prove that weak convergence of measures implies strong resolvent convergence of the associated operators.

\begin{proposition}\label{strongresconv}
Let $\{\mu^{(j)}\}_{j=1}^\infty \subset \mathcal{M}_{\rm a}^{\varepsilon,+}$ for some $\epsilon>0$ and assume that $\mu^{(j)} \to \mu$ weakly. Then $A^+_{\mu^{(j)}} \to A^+_{\mu}$ in strong resolvent sense.
\end{proposition}

\begin{proof}
Let $f\in L^2(0,\infty)$ with compact support. The assertion will follow if we can prove that for some sufficiently large $\kappa>0$ one has
$$
\left(f, (A^+_{\mu^{(j)}}+\kappa^2)^{-1}f\right) \to \left(f, (A^+_{\mu}+\kappa^2)^{-1}f\right) \,.
$$
(Here we use that weak resolvent convergence is the same as strong resolvent convergence and that the operators $A^+_{\mu^{(j)}}$ and $A^+_{\mu}$ are all non-negative, so that it suffices to verify the convergence at a single point $-\kappa^2$ of the resolvent set.)
We introduce the operators $T^{(j)}(-\kappa^2)$, $B^{(j)}$ and $\gamma^{(j)}$ in the obvious way and write $a^{(j)} := \gamma^{(j)} (A^+_0 +\kappa^2)^{-1} f$ and $a := \gamma (A^+_0 +\kappa^2)^{-1} f$. In view of the resolvent formula \eqref{eq:res} we need to show that
\begin{equation}\label{eq:convt}
\left( a^{(j)}, \left( T^{(j)}(-\kappa^2) + B^{(j)}\right)^{-1}  a^{(j)} \right)
\to \left( a, \left( T(-\kappa^2) + B \right)^{-1}  a \right) \,.
\end{equation}
We shall assume that $J^{(j)}=J=\N$ for any $j$, the other cases being similar.
We claim that
\begin{equation}\label{eq:conva}
a^{(j)} \to a
\qquad \text{in} \ \ell^2(\N,\C^2) \,.
\end{equation}
Indeed, the weak convergence implies that $t_n^{(j)} \to t_n$ and hence, since $(A^+_0 +\kappa^2)^{-1} f$ is $C^1$, that $a^{(j)}_n \to a_n$ for each $n$. Moreover, for all $t_n^{(j)}\geq \sup\supp f:=M$ one has $a^{(j)}_n = c e^{-\kappa t_n^{(j)}} (1,- \kappa )$ with a constant $c$ depending on $f$ but not on $j$ or $n$. Since $t_n^{(j)} \geq \epsilon n$, it follows that $|a^{(j)}_n|^2 \leq |c|^2 (1+\kappa^2) e^{- 2\kappa \epsilon n}$ for
$n\geq M/\varepsilon$ and similarly for $a$. From this one easily deduces \eqref{eq:conva}.

In the proof of Proposition \ref{unique} below we show that for sufficiently large $\kappa$, the operators $\left( T^{(j)}(-\kappa^2) + B^{(j)}\right)^{-1}$ are uniformly bounded in $j$. Moreover, the weak convergence of the measures implies that $T^{(j)}(-\kappa^2) + B^{(j)} \to T(\kappa^2) + B$ strongly. With the help of the resolvent identity one deduces that $\left(T^{(j)}(-\kappa^2) + B^{(j)}\right)^{-1} \to \left( T(\kappa^2) + B\right)^{-1}$ strongly. This, together with \eqref{eq:conva}, implies \eqref{eq:convt}.
\end{proof}

Our second application of \eqref{eq:res} will be to derive an expression for the m-function. Let us recall the definition.
For later purposes we consider whole-line operators.
As in Section \ref{sec:reduction} we can associate to each measure $\mu\in \mathcal{M}_{\rm a}^{0}$ on $\R$ a whole line operator $A=A_\mu$ acting as the Laplacian away from $\supp\mu$ on functions satisfying `jump conditions' \eqref{boundaries} for $t_n\in\supp\mu$ and $(\sqrt{b_n}+1)/(\sqrt{b_n}-1)=\mu(\{t_n\})$ (with the same modification as before if $b_n=\infty$).
This defines a self-adjoint, non-negative operator in $L^2(\R)$.
Therefore, for any $z\in\C\setminus[0,\infty)$ there exist functions $f_\pm(z;\cdot)$ solving $-f''=zf$ in
$\R\setminus\supp\mu$, satisfying `jump conditions' \eqref{boundaries} and lying in $L^2$ at $\pm\infty$.
(For example, choose $f_+(z;t)=\left((A-z)^{-1}g\right)(t)$ where $g$ is supported near $-\infty$ and $t$ is to
the right of $\supp g$, and continue $f$ to the left.)
Since $f_\pm$ is defined uniquely only up to a multiplicative constant, it is natural to consider
$$
m_\pm (z;t)=\pm \frac{f'_\pm(z;t)}{f_\pm(z;t)} \,,
$$
the m-functions of $A$. Note that if $\mu(\{0\})=0$ and $t\geq 0$, then $m_\pm(z;t)$ depends only on the restriction of $\mu$ to $\R_+$, and therefore we will also speak of the m-function of $A^+$.

The promised formula is

\begin{corollary}\label{mmatrix}
 Let $\mu\in \mathcal{M}_{\rm a}^{0,+}$ as in \eqref{eq:mu}. Then for all $z=k^2\in\C\setminus[0,\infty)$, $\im k>0$,
\begin{equation}\label{eq:mmatrix}
m_+(k^2;0)=ik + \sum_{n,m} e^{ik(t_n+t_m)}
\begin{pmatrix} 1\\ ik \end{pmatrix} ^T
(T(k^2)+B)^{-1}_{n,m}
\begin{pmatrix} 1\\ ik \end{pmatrix} \,.
\end{equation}
Here $(T(k^2)+B)^{-1}_{n,m}$ is the $(n,m)$-entry (a $2\times 2$-matrix) of the operator $(T(k^2)+B)^{-1}$.
\end{corollary}

\begin{proof}
Since
$$
m_+(z;0) = \frac{\partial^2}{\partial t \partial u} (A^+-z)^{-1} (t,u) |_{(t,u)=(0,0)} \,,
$$
this follows from \eqref{eq:res}.
\end{proof}

\subsection{Uniqueness}\label{sec:unique}

Our goal in this subsection is to prove that the m-function of $A^+$ uniquely determines the measure $\mu$.

\begin{proposition}\label{unique}
Let $\mu,\tilde\mu \in \mathcal{M}_{\rm a}^{0,+}$ with corresponding operators $A^+=A^+_\mu$, $\tilde A^+=A^+_{\tilde\mu}$. Assume that the corresponding $m$-functions satisfy $m_+(z;t)=\tilde m_+(z;t)$ for some $0\leq t<\min\{\inf\supp\mu,\inf\supp\tilde\mu\}$ and all $z\in\C^+$. Then $\mu=\tilde\mu$.
\end{proposition}

This is an analog of the famous Borg-Marchenko result in the
Schr\"odinger case.
It has been generalized to perturbations by measures in \cite{ben-amor}, but the result seems to be new for perturbations by boundary conditions \eqref{boundaries}. Our proof below relies on the expression \eqref{eq:mmatrix} from which we will derive

\begin{lemma}\label{asymp}
Let $0\neq \mu \in \mathcal{M}_{\rm a}^{0,+}$ and put $t_1:=\inf\supp\mu$ and $\mu(\{t_1\}) =: (\sqrt{b_1}+1)/(\sqrt{b_n}-1)$. Then for large, real $\kappa$,
\begin{equation}\label{eq:asymp}
m_+(-\kappa^2;0) + \kappa = - \frac{b_1 -1}{b_1 +1} \ 2 \kappa e^{-2\kappa t_1} (1+o(1)) \,.
\end{equation}
\end{lemma}

Of course, if $b_1=\infty$ then $\frac{b_1 -1}{b_1 +1}=1$. Accepting Lemma \ref{asymp} for the moment, we turn to the

\begin{proof}[Proof of Proposition \ref{unique}]
By translation invariance we may assume that $t=0$. Hence by Lemma \ref{asymp} either both $\mu$ and $\tilde\mu$ are zero, or else they are both not and then $t_1:=\inf\supp\mu = \inf\supp\tilde\mu$ and $\beta_1:= \mu(\{t_1\}) = \tilde\mu(\{t_1\})$. Now choose $t_1<s<\min\{\inf\supp(\mu-\beta_1\delta_{t_1}), \inf\supp(\tilde\mu-\beta_1\delta_{t_1}) \}$. Solving the equation $-f_+''=zf_+$ explicitly on the interval $[0,s]$ we can write $m_+(z;s)$ as a fractional linear function of $m_+(z;0)$ with coefficients depending only on $ s, z, r_1, \beta_1$. Hence $m_+(z;s)=\tilde m_+(z;s)$ for all $z$. Now iterate.
\end{proof}

\begin{proof}[Proof of Lemma \ref{asymp}]
We shall use the expression for $m(-\kappa^2;0)$ from Corollary \ref{mmatrix}. In order to calculate the asymptotics as $\kappa\to\infty$ we decompose
$$
T(-\kappa^2)=T^0(-\kappa^2) + T^R(-\kappa^2) \,,
\qquad
T^0(-\kappa^2)_{nm} : = \delta_{nm}
\begin{pmatrix}
 -\frac{1}{2\kappa} & 0 \\
 0 & \frac{\kappa}{2}
  \end{pmatrix}\,.
$$
One easily estimates for all large $\kappa$
$$
\| T^R(-\kappa^2)_{n,m} \|_{\C^2\to\C^2} \leq
\left\{
\begin{array}{ll}
\const \kappa e^{- 2 \kappa t_n} & \qquad \text{if} \ n=m\,, \\
\const \kappa e^{- \kappa |t_n-t_m|} & \qquad \text{if} \ n\neq m\,.
\end{array}
\right.
$$
Hence by a matrix-valued version of Schur's lemma
$$
\| T^R(-\kappa^2) \|_{\mathfrak H \to\mathfrak H} \leq
\sup_{m} \sum_n \|T^R(-\kappa^2)_{n,m} \|_{\C^2\to\C^2} \leq
\const \kappa \left( e^{- 2 \kappa t_1} + e^{- \kappa \varepsilon} \right) \,,
$$
where we used that $\varepsilon:= \inf_{n\neq m} |t_n-t_m|>0$ and hence $|t_n-t_m|\geq\varepsilon |n-m|$.

On the other hand, the eigenvalues of $T^0(-\kappa)+B$ are easily calculated and one finds that the smallest (in absolute value) eigenvalue is bounded away from zero by a constant times $\kappa^{-1}$ independently of the $b_n$. (To be a bit more explicit, the positive eigenvalue of  $(T^0(-\kappa)+B)_{nn}$ is larger than $\kappa/2$ and the negative eigenvalue is smaller than $-1/2\kappa$.) Hence both $T^0(-\kappa^2) +B$ and $T(-\kappa^2)+B = T^0(-\kappa^2) + B + T^R(-\kappa^2)$ are invertible, and the norms of their inverses are bounded from above by a constant times $\kappa$. We conclude that
\begin{align*}
& \left\| (T(-\kappa^2)+B)^{-1} - (T^0(-\kappa^2)+B)^{-1} \right\| \\
& \qquad = \left\| (T(-\kappa^2)+B)^{-1} \ T^R(-\kappa^2) \ (T^0(-\kappa^2)+B)^{-1} \right\| \\
& \qquad \leq \const \kappa^3 \left( e^{- 2 \kappa t_1} + e^{- \kappa \varepsilon} \right) \,,
\end{align*}
and so by \eqref{eq:mmatrix}
\begin{align*}
m(\kappa,0) & = -\kappa + \sum_{n} e^{-2\kappa t_n }
\begin{pmatrix} 1\\ -\kappa \end{pmatrix} ^*
(T^0(-\kappa^2)+B)^{-1}_{n,n}
\begin{pmatrix} 1\\ -\kappa \end{pmatrix} \\
& \quad \qquad + \mathcal O \left( \kappa^5 e^{-2\kappa t_1} \left( e^{- 2 \kappa t_1} + e^{- \kappa \varepsilon} \right) \right) \\
& = -\kappa - \frac{b_1 -1}{b_1+1}\ 2 \kappa e^{-2\kappa t_1}
+  \mathcal O \left( \kappa^5 e^{-2\kappa t_1} \left( e^{- 2 \kappa t_1} + e^{- \kappa \varepsilon} \right) \right) \,,
\end{align*}
as claimed.
\end{proof}

\begin{remark*}
Looking at the previous proof, we see that the assumption $\inf_{n\neq m} |t_n-t_m|>0$ can be significantly relaxed to
$\sum_{n:\,n\neq m} e^{-\kappa |t_n-t_m|} = o(\kappa^{-4})$ uniformly in $m$.
\end{remark*}

\section{Proof of Theorems \ref{delta-operator}, \ref{no-ev} and \ref{wonderland1d}}

\subsection{A Remling-type theorem}

Roughly speaking, Remling's Theorem states that for a given
one-dimensional Schr\"odinger operator, $A$, (with a natural
boundedness assumption on the potential) any right-limit of $A$ is
reflectionless on the absolutely continuous spectrum of $A$. Clearly, the
two central notions behind this theorem are that of right-limit and
that of a reflectionless operator. We proceed to define these notions in our
setting and formulate the version of Remling's Theorem that will be useful for us.

\begin{definition} \label{right-limit}
Let $\mu, \hat{\mu} \in \mathcal{M}_{\rm a}^{0}$. The measure $\hat\mu$ is said to be a \emph{right-limit} of $\mu$ if there exists a strictly increasing sequence, $s_j \rightarrow \infty$, such that for every continuous,
compactly supported function $f$ on $\R$,
\beq \no
\lim_{j \rightarrow \infty}\int_{\bbR}f(t-s_j)d\mu(t)=\int_{\bbR}f(t)d\hat{\mu}(t).
\eeq
\end{definition}

Let $\mu\in \mathcal{M}_{\rm a}^{0,+}$ with $\mu(\{0\})=0$ and let $A$ and $A^+$ be the corresponding whole-line and half-line operators. Recall that we have introduced the m-functions $m_\pm(z;t)$ before Corollary \ref{mmatrix}. Of course, $m_+(z;0)$ depends only on the restriction of $\mu$ to $[0,\infty)$. Since $m_+(z;0)$ is a Herglotz
function of $z\in\C^+$, its boundary values on the real line exist a.e. We denote by $\Sigma_{\rm ac}(A^+)$ the set
\beq \no
\Sigma_{\rm ac}(A^+)=\{E \in \bbR \mid 0<\Im m_+(E+i0;0)<\infty\}.
\eeq
$\Sigma_{\rm ac}(A^+)$ is an essential support of the
absolutely continuous spectrum of $A^+$. In particular,
$A^+$ has absolutely continuous spectrum iff $\Sigma_{\rm
ac}(A^+)$ has positive Lebesgue measure (see e.g.\ \cite{rankone}).
\iffalse
$m_+(z;0)$ is a Herglotz
function of $z$, and so is associated naturally with a measure
$\nu_+$. We call $\nu_+$ the spectral measure of $A^+$. We
denote by $\Sigma_{\rm ac}(A^+)$ the set \beq \no \Sigma_{\rm
ac}(A^+)=\{E \in \bbR \mid 0<\Im m^+(E+i0)<\infty\}. \eeq
$\Sigma_{\rm ac}(A^+)$ is an essential support of the
absolutely continuous spectrum of $A^+$. In particular,
$A^+$ has absolutely continuous spectrum iff $\Sigma_{\rm
ac}(A^+)$ has positive Lebesgue measure.
\fi

\begin{definition} \label{reflectionless}
Fix $\Lambda \subseteq \bbR$ and let $\mu \in \mathcal{M}_{\rm a}^{0}$.
The operator $A_{\mu}$ is called \emph{reflectionless} on $\Lambda$ if for all $t\in \R\setminus\supp\mu$ and for almost
every $E \in \Lambda$, $m_+(E+i0;t)=-\bar{m}_-(E+i0;t)$.
\end{definition}

We are now in a position to formulate the version of Remling's Theorem
appropriate for our setting.
\begin{theorem}\label{BPR}
Let $A^+=A^+_\mu$ for some $\mu\in \mathcal{M}_{\rm a}^{0,+}$ and let $\hat{\mu}$ be a right-limit of $\mu$. Then
$A_{\hat{\mu}}$ is reflectionless on $\Sigma_{\rm ac}(A^+)$.
\end{theorem}

\begin{proof}[Proof of Theorem \ref{BPR}]
The proof of this theorem is essentially the same as the proof of
Theorem 4.1 in \cite{remling-cont}, but let us make a few remarks.
As stated in the introduction, Remling's Theorem follows from a
result of Breimesser-Pearson (Theorem 1 in \cite{bp1}) which states
that, on $\Sigma_{\rm ac}(A^+_\mu)$, the value distribution of
$-\frac{v'(E+i0;t)}{v(E+i0;t)}$ and $-\bar{m}_+(E+i0;t)$ are
asymptotically equal (as $t \rightarrow \infty$). Here $v(z;t)$ is
the Dirichlet solution to the formal equation $A^+_\mu v=zv$
(generally, $v \notin L^2$), and $m_+(z;t)$ is the m-function for
$A^+_\mu$. (For the concept of value distribution see
\cite{bp1,bp2}.)

Remling's Theorem follows from this if we can prove that if $\hat{\mu}$ is a right-limit of $\mu$ then
$\lim_{j \rightarrow \infty} m_+(z;s_j)= \hat{m}_+(z;0)$ and
$\lim_{j \rightarrow \infty} -\frac{v'(z;s_j)}{v(z;s_j)}=\hat{m}_-(z;0)$ uniformly on compact subsets of
$\bbC^+$ where $\{s_j\}$ is the sequence from Definition \ref{right-limit} and $\hat{m}_\pm$ are the m-functions
for $A_{\hat{\mu}}$.

To show this, consider $\hat{m}_+(z;0)$ and $A^+_{\hat{\mu}}$,
the right half-line restriction of $A_{\hat{\mu}}$. First, note
that Green's formula (see \cite{cod-lev}): \beq \no \int_{t_1}^{t_2}
(-f''(t))\bar{g}(t)dt-\int_{t_1}^{t_2}f(t)(-\bar{g}''(t))dt
=W\left(f,\bar{g} \right)(t_2)-W\left(f,\bar{g} \right)(t_1) \eeq
holds in our case, by integration by parts along intervals which
contain no atoms of $\hat{\mu}$ and then by summing up the resulting
telescoping sum. Therefore, the Weyl nested disk construction (see,
e.g., \cite{cod-lev}) works in our setting to show that, for any
$\delta>0$ there exists $N>0$ such that, if $\ti{\mu}$ agrees with
$\hat{\mu}$ on $(0,N)$, then $\ti{m}_+(z;0)$, the m-function for
$A^+_{\ti{\mu}}$, lies in a disk of radius no bigger than
$\delta$ which also contains $\hat{m}_+(z;0)$. Explicitly, this disk
is the image of $\bbC^+\cup \bbR$ under the M\"obius transformation
given by the matrix $T_z(N)^{-1}$, where $T_z(N)$ is defined by
$T_z(N) \left( \begin{array}{c} f(0) \\ f'(0)
\end{array} \right)= \left( \begin{array}{c} f(N) \\ f'(N)
\end{array} \right)$ for any formal solution, $f$, of
$A^+_{\hat{\mu}}f=zf$. Using \eqref{boundaries} for the atoms
and the solutions of the free equation along the intervals between
them, it is possible to write this matrix as a product of simple
matrices, and so note that its entries are continuous functions of
the parameters defining the restriction of $\hat{\mu}$ to $(0,N)$.
Thus, the center and radius of the disk are continuous functions of
these parameters. Recalling the definition of right-limit, this
implies the convergence of $m_+(z;s_j)$ to $\hat{m}_+(z;0)$. The
convergence of $-\frac{v'(z;s_j)}{v(z;s_j)}$ to $\hat{m}_-(z;0)$ is
established in a similar way, by considering $A^-_{\hat{\mu}}$, the
restriction of $A_{\hat{\mu}}$ to $(-\infty,0)$ with a Dirichlet boundary condition.

As for the applicability of Theorem 1 of Breimesser-Pearson
\cite{bp1}, the only thing that actually depends on the particular
properties of the model is Lemma 3 there and its corollary. Once
again, since Green's formula holds in our case it is easy to see
that the proof goes through with no change.
\end{proof}

\subsection{Proof of Theorem \ref{delta-operator}}
Assume, by contradiction, that $\limsup_{n \rightarrow \infty} (t_{n+1}-t_n)=\infty$ and that $\Sigma_{\rm ac}(A^+)$ has positive Lebesgue measure. Note that $A^+$ is non-negative, so $\Sigma_{\rm ac}(A^+) \subset [0,\infty)$.
Let $\mu=\sum_{n=1}^\infty \beta_n \delta_{t_n}$.
By the hypothesis on $t_n$ and a compactness argument, $\mu$ has a right-limit $\hat{\mu}\in\mathcal M_{\mathrm a}^{0}$ such that
$\hat{\mu}(-\infty,0]=0$ while $\hat{\mu}(0,\infty) \neq 0$.

Now, Theorem \ref{BPR} implies that $A_{\hat{\mu}}$ is reflectionless on $\Sigma_{\rm ac}(A^+)$.
Hence $\hat m_+(E+i0;t)=-\bar{\hat m}_-(E+i0;t)=i\sqrt{E}$ for all $E\in \Sigma_{\rm ac}(A^+)$ and all
$t\notin\supp\hat\mu$.
Since a Herglotz function is uniquely determined by its values on a set of positive measure (see, e.g.\
\cite[Appendix B]{teschl}),
one has $\hat m_+(k^2;t)=ik$ for all $k$ with $\im k>0$ and all $t\notin\supp\hat\mu$, and hence by
Proposition \ref{unique}, $\hat\mu(0,\infty)=0$, a contradiction.
\hfill $\Box$

\subsection{Proof of Theorem \ref{no-ev}}
First note that in view of \eqref{eq:apos}, $0$ is not an eigenvalue of $A^+$. Hence we need to show that any function $f$ satisfying $f(0)=0$ and \eqref{boundaries} and solving
\beq \label{evequation}
-f''(x)=E f(x) \qquad \text{in} \ \R_+\setminus\{t_n\}_{n=1}^\infty
\eeq
for some $E\in(0,\infty)$ is not in $L^2$.

Any solution of \eqref{evequation} satisfies
\beq \no
\left( \begin{array}{c} f(x) \\ f'(x) \end{array} \right)=T(x,y,E)
\left( \begin{array}{c} f(y) \\ f'(y) \end{array} \right)
\eeq
for a certain $2 \times 2$ matrix $T(x,y,E)$ of determinant $1$. It follows from \cite[Thm.~2.1]{simon-stolz} that if
\beq \label{transfer-bound}
\int_0^\infty \frac{dx}{ \parallel T(x,O,E) \parallel^2}=\infty
\eeq
then \eqref{evequation} has no solution in $L^2(0,\infty)$.

If $x, y \in (t_n,t_{n+1})$ for some fixed $n$ then
\beq \no
\left( \begin{array}{c} f(x) \\ f'(x) \end{array} \right)=
\left(\begin{array}{cc} \cos k (x-y) & k^{-1} \sin k(x-y) \\
-k \sin k(x-y) & \cos k (x-y) \end{array} \right)
\left( \begin{array}{c} f(y) \\ f'(y) \end{array} \right)
\eeq
where $k=\sqrt{E}>0$. The norm of this matrix is bounded by
$\max(k,k^{-1})$. Furthermore, the jump condition is taken into account by the
jump matrix
\beq \no
S_n=\left( \begin{array}{cc} \sqrt{b_n} & 0 \\
0 & \frac{1}{\sqrt{b_n}} \end{array}\right).
\eeq
Thus, if $x \in (t_n,t_{n+1})$,
\beq \no
\parallel T(x,O,E) \parallel \leq \left(\max(k,k^{-1}) \right)^{n+1} \prod_{m=1}^n\sqrt{b_m}
\leq n^n,
\eeq
for sufficiently large $n$ (and any fixed $E \in (0, \infty)$).
This implies
\beq \no
\int_{t_n}^{t_{n+1}} \frac{dx}{\parallel T(x,O,E) \parallel^2}\geq \frac{t_{n+1}-t_n}{n^{2n}},
\eeq
which, by \eqref{sparseness}, implies the result.
\hfill $\Box$

\subsection{Proof of Theorem \ref{wonderland1d}}
It is easy to see (cf. also \cite{remling-cont}) that $\mathcal{M}^{\varepsilon,C,+}_\mathrm a$ is a complete
(indeed, compact) metric space where the topology coincides with that of weak convergence of measures.
According to Proposition \ref{strongresconv} weak convergence of measures implies convergence in the strong resolvent
sense for the corresponding operators.
Thus, it follows from \cite[Thm. 1.1]{simon-wonderland} that the
set of operators in $\mathcal{M}^{\varepsilon,C,+}_\mathrm a$ with no eigenvalues in $[0,\infty)$ is a $G_\delta$ set.
Moreover, \cite[Thm. 1.2]{simon-wonderland} implies that the set of operators in $\mathcal{M}^{\varepsilon,C,+}_\mathrm a$
with no absolutely continuous spectrum in $(0,\infty)$ (and so also in $[0,\infty)$) is a $G_\delta$ set.

To complete the proof it suffices to note that the set of measures in
$\mathcal{M}^{\varepsilon,C,+}_\mathrm a$ with $\{t_n\}$ satisfying \eqref{sparseness} is a dense set
(since, for any given measure one may take a measure coinciding with it on a fixed bounded set,
but satisfying condition \eqref{sparseness}), and for such measures the spectrum of the corresponding operator is purely
singular continuous by Theorem \ref{no-ev}.
\hfill $\Box$

\bibliographystyle{amsalpha}

config: arxiv


TITLE: Calculate the sum of reciprocals of elements in a set
QUESTION [1 upvotes]: Given set $A=\{ \frac{i^2+1}{2}\ |\ i\in\mathbb{N},\ i\geq 65,\ 2\nmid i \}$ calculate sum of reciprocals of values of its elements.
So first I simplified the set:
$$
A=\{ \frac{(2j+1)^2+1}{2}\ |\ j\in\mathbb{N},\ j\geq 32 \}=\{ 2j^2+2j+1\ |\ j\in\mathbb{N},\ j\geq 32 \}
$$
And now the sum we are looking can be calculated with the expression:
$$
\sum_{j=32}^{\infty}\frac{1}{2j^2+2j+1}=\sum_{j=0}^{\infty}\frac{1}{2j^2+2j+1}-\sum_{j=0}^{31}\frac{1}{2j^2+2j+1}
$$
However I'm not sure how to carry on. Can you help me solve it?

REPLY [1 votes]: Lets start with:
$$\eqalign{
\sum_{j\in\mathbb{Z}}\frac{1}{2j^2+2j+1}&=\color{red}{\sum_{- \infty \le j \le -1}\frac{1}{2j^2+2j+1}}+\color{blue}{\sum_{0 \le j  \le  \infty  }\frac{1}{2j^2+2j+1}}\cr
&=\color{red}{\sum_{\infty  \ge  -j  \ge 1}\frac{1}{2j^2+2j+1}} +\color{blue}{\sum_{0 \le j  \le  \infty  }\frac{1}{2j^2+2j+1}}\cr
&=\color{red}{\sum_{\infty  \ge  \underbrace{-j-1}_{-j-1:=k}  \ge 0}\frac{1}{2j^2+2j+1}} +\color{blue}{\sum_{0 \le j  \le  \infty  }\frac{1}{2j^2+2j+1}}\cr
&=\color{red}{\sum_{0 \le k \le  \infty }\frac{1}{2(-k-1)^2+2(-k-1)+1}} +\color{blue}{\sum_{0 \le j  \le  \infty  }\frac{1}{2j^2+2j+1}}\cr
&=\color{red}{\sum_{0 \le k \le  \infty }\frac{1}{2k^2+2k+1}} +\color{blue}{\sum_{0 \le j  \le  \infty  }\frac{1}{2j^2+2j+1}}\cr
&=2\color{blue}{\sum_{0 \le j  \le  \infty  }\frac{1}{2j^2+2j+1}}\cr
}$$
therefore:
$$\color{blue}{\sum_{0 \le j  \le  \infty  }\frac{1}{2j^2+2j+1}}=\frac{1}{2}\sum_{j\in\mathbb{Z}}\frac{1}{2j^2+2j+1}$$
due to Infinite Series and the Residue Theorem black sum it is easy to count:
$$\eqalign{\color{blue}{\sum_{0 \le j  \le  \infty  }\frac{1}{2j^2+2j+1}}&= -\frac{1}{2}\left( \text{Res}\left\{\frac{ \pi \cot \left(  \pi z\right) }{2z^2+2z+1} , \frac{-1-i}{2}  \right\} +\text{Res}\left\{\frac{ \pi \cot \left(  \pi z\right) }{2z^2+2z+1} ,\frac{-1+i}{2}  \right\} \right)\cr &=\frac{ \pi }{2}\text{tgh}\left(  \frac{ \pi }{2} \right)}$$
summarizing with the litle help of Wolfram for finite sum
$$\sum_{32 \le j  \le  \infty  }\frac{1}{2j^2+2j+1}=\\ \frac{ \pi }{2}\text{tgh}\left(\frac{ \pi }{2} \right)-\frac{11331201497882268207659413442681772956711855668184103152864}{7951513543813219897041288425928485300865169239957975379625}$$


TITLE: Probability of $2$ runners both finishing in any of the top $3$ positions
QUESTION [0 upvotes]: Given a race of $n$ runners, what is the probability of $2$ chosen runners both finishing in any of the top $3$ positions of a race.
I think the probability is $6(n-2)/n!$, but it seems intuitively wrong.

REPLY [1 votes]: If $n= 2$ the answer is trivially $1$. If $n>2$, the two runners need to be set in one of these three permutations: The two runners are first and second, first and third, or second and third. That's $3$ permutations, and we multiply by $2$ because we can switch the two runners for each permutation. So we get $6$ permutation for the two runners.
There are left $(n-2)$ runners which can be set in $(n-2)!$ different permutations. So overall we have $6(n-2)!$ valid permutations. Without any limitations, all the permutations are $n!$. Thus, the answer is $\frac{6(n-2)!}{n!}$.
I hope it is clear enough.


TITLE: Finding $P (X > kY)$: Require explanation for how solution is finished
QUESTION [0 upvotes]: I have the following problem:

Let $X$ and $Y$ be independent random variables taking values in the positive integers and having the same mass function $P(X = k) = P(Y = k) = 2^{−k}$ for $k = 1, 2, \dots$
Find $P (X > kY)$, for a given positive integer $k$.

My solution is as follows:
$$\begin{align} P(X > ky) &= \sum_{i = 1}^\infty P(X > kY, Y = i) \ \ \text{(Joint PMF of $X$ and $Y$.)} \\ &= \sum_{i = 1}^\infty P(X > kY | Y = i) P(Y = i) \ \ \text{(By the law of total probability.)} \\ &= \sum_{i = 1}^\infty P(X > ki) P(Y = i) \ \ \text{(Since $X$ and $Y$ are independent.)} \\ &= \sum_{i = 1}^\infty \sum_{j = ki + 1}^\infty 2^{-i} 2^{-j} \ \ \text{($j = ki + 1$ since we have $X > ki$.)} \\ &= \sum_{i = 1}^\infty \sum_{j' = 1}^\infty 2^{-i} 2^{-(j' + ki)} \ \ \text{(Set $j' = j - ki = 1$ to setup as geometric series.)} \\ &= \sum_{i = 1}^\infty \sum_{j' = 1}^\infty 2^{-i} 2^{-j' - ki} \\ &= \sum_{i = 1}^\infty \sum_{j' = 1}^\infty 2^{-j'} 2^{-i(k + 1)} \\ &= \sum_{i = 1}^\infty 2^{-i(k + 1)} \ \ \left(\text{Since the geometric series becomes $1 \dfrac{2^{^{-1}}}{1 - 2^{-1}} = 1$}\right) \end{align}$$
The solution proceeds from where I ended as follows:

$$\begin{align} \sum_{i = 1}^\infty 2^{-i(k + 1)} &= 2^{-(k + 1)} \sum_{i = 1}^\infty 2^{-(i - 1)(k + 1)} \\ &= 2^{-(k + 1)} \dfrac{1}{1 - 2^{-(k + 1)}} \\ &= \dfrac{1}{2^{k + 1} - 1} \end{align}$$

It is not clear to me how the author made this leap in their reasoning. I would greatly appreciate it if people could please take the time to clarify this.

REPLY [0 votes]: \begin{align} \sum_{i = 1}^\infty 2^{-i(k + 1)} &= \sum_{i = 1}^\infty 2^{-(i-1+1)(k + 1)} \\&= 2^{-(k + 1)} \sum_{i = 1}^\infty 2^{-(i - 1)(k + 1)} \\ &= 2^{-(k + 1)} \dfrac{1}{1 - 2^{-(k + 1)}} \\ &= \dfrac{1}{2^{k + 1} - 1} \end{align}
The line where they get rid of the summation sign is by identitying that it is a geometric sum of common ration $2^{-(k+1)}$ with the first term being $1$.
$$\sum_{i=0}^\infty r^i=\frac1{1-r}$$
if $|r|<1$.


text:
\section{Conclusions and future work}\label{sec:con}

We investigated the temporal dynamics of interference in Poisson networks with a focus on its auto-correlation function and coherence time.
We see from the auto-correlation function that the correlation of interference for small time lags typically decreases. This decrease strongly depends on the sources of correlation.
When evaluated over a large span of time lags, the correlation approaches zero, follows a damped oscillation around zero or shows a completely different behavior, all depending on the chosen scenario. It is evident that networking protocols and techniques have to be designed and parametrized differently for such different scenarios in order to operate best. Hence, a good knowledge of the auto-correlation function of interference contributes to improved network performance and robustness.
For example, if interference traces are interpreted as time series, their underlying model order is determined by the auto-correlation function. Hence, our work contributes to model design by providing essential input parameters. It is thus a key enabling result for interference prediction based on time series.

The coherence time of interference can be calculated by means of expressions provided in the article at hand. This is a valuable tool for designing network protocols properly. Moreover, the expressions are simple enough to be implemented on networked nodes, allowing them to adapt to a changing network environment. This is a requirement in scenarios with, e.g., high mobility or high fluctuations in the network nodes.

Ongoing work analyzes how, by estimating a small number of network parameters, the node gathers a good estimate of the interference coherence time. Based on this information, we plan to devise a distributed algorithm to optimize the node's successful communication attempts and thus the overall network performance.


TITLE: Lattice points on $x+y=n$ always visible from the origin.
QUESTION [1 upvotes]: Let $m \in \mathbb{Z_{\gt 0}}$ and $\{n \in \mathbb{Z_{\gt 0}}:n\mod 10\}$  = $\{1, 3, 7, 9\}$.
Define
$\Large a_{m} = \frac {5^m-3}2$

Claim
Any lattice points $(x,y)$ of the form
$\Large(\left \lfloor{\frac{n}2}\right \rfloor + a_{m} + 2,\left \lfloor{\frac{n}2}\right \rfloor - a_{m} - 1)$
where $x+y=n$ are always visible from the origin.
Proof
Let
\begin{align*}
x=\frac {n}2 + a_{m} + 2\\
y=\frac {n}2 - a_{m} - 1
\end{align*}
Point $(x,y)$ is coprime when
\begin{align*}
gcd(x,y)=1 \iff gcd(x+y,x-y)=1
\end{align*}
Calculating
\begin{align*}
gcd(x+y,x-y) &= gcd(n,\frac {n}2 + a_{m} + 2 - (\frac {n}2 - a_{m} - 1))\\
&=gcd(n,2a_{m} + 3)\\
&=gcd(n,5^m)\\
&=1\ or \ 5^m\\
\end{align*}
However, the only prime divisor of $5^m$ is $5$ and that is not a factor of $n$, so
\begin{align*}
gcd(x+y,x-y) &= 1\\
\end{align*}
Calculating
\begin{align*}
gcd(x, y) &= gcd(x−y, y),\ if\ (x > y)\\
&= gcd(\frac {n}2 + a_{m} + 2 - (\frac {n}2 - a_{m} - 1),\frac {n}2 - a_{m} - 1)\\
&=gcd(2a_{m} + 3,\frac {n}2 - a_{m} - 1)\\
&=gcd(5^m,\frac {n}2 - a_{m} - 1)\\
&=5^m\ or \ 1\\
\end{align*}
However, for any integers
\begin{align*}
\frac {gcd(x+y, x-y)}{gcd(x,y)}
\end{align*}
because anything that divides both $x$ and $y$ divides their sum and difference.
Substituting yields
\begin{align*}
\frac {1}{gcd(x,y)}
\end{align*}
Therefore
\begin{align*}
gcd(x,y) = 1
\end{align*}
$\blacksquare$

Examples
$\xrightarrow{\text{(n = 31, m = 1)}} (18, 13)$
$\xrightarrow{\text{(n = 31, m = 2)}} (28, 3)$
$\xrightarrow{\text{(n = 77777, m = 1)}} (38891, 38886)$
$\xrightarrow{\text{(n = 77777, m = 2)}} (38901, 38876)$
$\xrightarrow{\text{(n = 123456789, m = 10)}} (66611207, 56845582)$
$\xrightarrow{\text{(n = 123456789, m = 11)}} (86142457,37314332)$
$\xrightarrow{\text{(n = 9876543213, m = 12)}} (5060341919, 4816201294)$
$\xrightarrow{\text{(n = 9876543213, m = 13)}} (5548623169, 4327920044)$

Note: I only know basic high-school level maths.

REPLY [2 votes]: You are overthinking this.
First, for any integers
$$
\gcd(x,y) \text{ divides } \gcd(x+y, x-y)
$$
because anything that divides both $x$ and $y$ divides their sum and their difference.
In this question, $x -y$ = $5^m$. It's a little harder to see that $x+y = n$. (Since $n$ is odd, rounding down $n/2$ and doubling yields $n-1$).
The only prime divisor of $5^m$ is $5$ and $5$ is not a factor of $n$ so $\gcd(n, 5^m) = 1$.
Thus
$$
\gcd(x,y) \text{ divides } \gcd(n, 5^m) = 1
$$
so $x$ and $y$ are relatively prime.
(Note: be a little careful with odds and evens if you want to use this trick again. $\gcd(5,3) = 1$ but $\gcd(5-3,5+3) = 2$.)


TITLE: An Idea to solve a limit involve integral
QUESTION [1 upvotes]: Suppose $f(x)$ is a continuous function over $[0,1]$,and $0<a<b$ find the below limit $$\lim_{t\to 0^+}{\int_{at}^{bt}\frac{f(x)}{x}}=?$$
My work is below
$\int_{at}^{bt}\frac{f(x)}{x}=\int_{at}^{bt}f(x)d(\ln(x))$
Using the mean value theorem, and $f\in c^1[0,1]$ we have $$\exists c:a<c<b \ |f(ct)=\frac{\int_{at}^{bt}f(x)d(\ln(x))}{\ln(bt)-\ln(at)}$$ so $$\lim_{t\to 0^+}{\int_{at}^{bt}\frac{f(x)}{x}}=\lim_{t\to 0^+}{\int_{at}^{bt}f(x)d(\ln(x))}=\\
\lim_{t\to 0^+}f(ct)ln(\frac{bt}{at})=f(0)\ln(\frac ba)
$$
My questions:
1:Is my working true?
2:Is there another Idea or method to solve the problem?
Thanks in advance

REPLY [1 votes]: A couple things. First the assumption only says $f$ is continuous, not that $f\in C^1[0,1]$, so mean value theorem doesn't work.
Second, I believe the bounds in the integrals should be $\ln(at)$ and $\ln(bt)$ when you've switched to $d(\ln(x))$. Though this could just be some convention I am unaware of.
A similar idea is to use inequalities to bound it i.e.
$$\int_{at}^{bt} \frac{f(x)}{x} dx \leq \sup_{at \leq x \leq bt} f(x) \int_{at}^{bt}\frac{1}{x}dx = \ln(b/a)\sup_{at\leq x\leq bt}f(x) $$
and
$$\int_{at}^{bt} \frac{f(x)}{x} dx \geq \inf_{at \leq x \leq bt} f(x) \int_{at}^{bt}\frac{1}{x}dx = \ln(b/a)\inf_{at\leq x\leq bt}f(x) $$
Since we are taking $t\to 0$ and $f$ is continuous, both the supremum and infimum will become $f(0)$ in the limit.


TITLE: How to solve $z^2 - 2(4+i)z+10- 20i = 0$
QUESTION [0 upvotes]: I need to solve this question:
$$ z^2 - 2(4+i)z+10- 20i = 0 $$
Can you help me please , thank you

REPLY [2 votes]: Hint:
The solutions to the equation $az^2 + bz + c = 0$ can be given by $$z = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a}$$
This is the quadratic formula.
In this case, $a = 1$, $b = -2(4+ i)$ and $c=10-20i$, so substitute these values and do the algebra.
The tricky part is evaluating $b^2 - 4ac$, but just remember how to expand brackets and make sure you use $i^2 = -1$.

Edit:
Note you probably do not have to simplify the radial and it will likely be acceptable for you to leave your solutions in the form $z = p \pm \sqrt{q + ri}$

Alt:
If you don't like the quadratic formula, you can try to complete the square. Here would be the first step: $$[x-(4 + i)]^2 - (4 + i)^2 + 10 - 20i = 0$$ Again, expand the brackets and then re-arrange.
The lesson is: solve this quadratic with complex coefficients in the way you would usually solve a quadratic equation using whatever your favourite approach is.


TITLE: How to put a matrix in Jordan canonical form, when it has a multiple eigenvalue?
QUESTION [4 upvotes]: Put the matrix $$\begin{bmatrix} 3 & -4\\ 1 & -1\end{bmatrix}$$ in Jordan Canonical Form.  Moreover, find the appropriate transition matrix to the basis in which the original matrix assumes its Jordan form.

I'm having a lot of trouble with this.  I know that the eigenvalue has multiplicity two and is $\lambda = 1$.  I can find the first eigenvector, which is:
\begin{bmatrix}
        2 \\
        1  \\
        \end{bmatrix}
I'm having trouble finding the second since both eigenvalues tell us the same thing.  But I'm not nearly as concerned about the eigenvectors as I am about what to do after.
If anyone could explain thoroughly the next steps involved (not necessarily the answer but how to obtain it), I would be forever grateful.  This homework is in 2 days and it may determine my grade letter.

REPLY [2 votes]: To put a matrix in Jordan normal form requires to know three matrices such that $A=PJP^{-1}$ i.e: a matrix $P^{-1}$ that transform to the canonical basis $(\mathbf{i},\mathbf{j},\mathbf{k})$ to a new basis in which  the matrix $J$ represents the transformation of a vector $\mathbf{v}$ such that the transformed vector $\mathbf{v'}$ is the same as we find when we transform $\mathbf{v}$ with $A$ in the canonical basis. Last the matrix $P$ returns this result to the canonical basis.
As noted in OP the matrix $A$ has eigenvalues $\lambda_1=\lambda_2=1$ and a single eigenvector
$$
\mathbf{u_1}=\left[
\begin{array}{cccc}
2\\
1
\end {array}
\right]
$$
So the main problem is to find another vector that completes the new basis.
To find such a vector notes that all vectors $\mathbf{x}$ such that $(A-\lambda I)\mathbf{x}=0$ are transformed in the eigenspace generated by the eigenvector $\mathbf{u_1}$, so we want a vector $\mathbf{u_2}$ such that $(A-\lambda I)\mathbf{u_2} \ne 0$, and the way to do this is to find a vector such that : $(A-\lambda I)\mathbf{u_2} = \mathbf{u_1}$. (Note that this equation is the same as $(A-\lambda I)^2\mathbf{u_2}= 0$).
Solving in our case we find:
$$
\left[
\begin{array}{cccc}
2&-4\\
1&-2
\end {array}
\right]
\left[
\begin{array}{cccc}
x\\
y
\end {array}
\right]=
\left[
\begin{array}{cccc}
2\\
1
\end {array}
\right]
$$
so that the components $x$ and $y$ of the searched vector must satisfies $x-2y=1$, and we can find the vector
$$
\mathbf{u_2}=
\left[
\begin{array}{cccc}
1\\
0
\end {array}
\right]
$$
So the matrix $P$ we are searching is
$$
P=[\mathbf{u_1},\mathbf{u_2}]=
\left[
\begin{array}{cccc}
2&1\\
1&0
\end {array}
\right]
$$
And the inverse is:
$$
P^{-1}=
\left[
\begin{array}{cccc}
0&1\\
1&-2
\end {array}
\right]
$$
The matrix $J$ is a typical Jordan block, with the eigenvalues as diagonal elements and an entry $1$ up-right them:
$$
J=
\left[
\begin{array}{cccc}
1&1\\
0&1
\end {array}
\right]
$$
and we can easily verify that $A=PJP^{-1}$.


TITLE: Uncertainty principle and the speed of light
QUESTION [4 upvotes]: If an EM wave only gives us a probability of where a photon may be at a given moment, and the HUP tells us that we can't know the exact location of the photon. Then would it be correct to say that a photon does not travel in a straight line?
If this is true, wouldn't the photon's crooked path mean that it must travel faster than $c$ for us to measure its straight line speed at $c$?
Here's a comparison to explain my question further: If two sprinters run a 100 yard dash in 10 seconds, but one sprinter is required to run in a zigzag manner wouldn't that runner need to run faster than the straight line sprinter for both runners to run the race in 10 seconds.
Does the crooked traveling photon need to exceed $c$ for us to measure its straight line speed at $c$?

REPLY [2 votes]: From the point of view of worry that the HUP might contradict the existence of the speed limit $c$, one can assuage your worries by pointing out that:

$c$ does not primarily mean the speed of light, rather it is a geometric constant that defines our spacetime and experimentally light is observed to move at that speed. See more about this in my answer here and
The ban on speeds amounts to a limit to the speed at which information, or a cause-effect relationship can propagate. So an inferred speed exceeding $c$ is not needfully a problem: it only violates known physics if the speed can be interpreted as a propagation speed for a cause-effect relationship. Thus, for example, phase speeds of waves can happily exceed $c$, e.g. at anomalous dispersion wavelengths for light propagation in dielectric materials. Moreover, gainsaying many texts, even group velocities can exceed $c$ in some wavelength bands and for slightly weird materials: that's not even a problem because, although the group velocity approximates the signal speed propagation, it's not exact and the step response of the medium still propagates at less than $c$.

Any speed inferred from the kind of reasoning you are making falls into the category in my point (2) above. There are two reasons for this:

Lone photons without quantum measurement cannot be used to transmit information. In particle physics, the propagation of the electromagnetic field is calculated from a superposition of paths of photons - and so called "off-shell" photons (those with speeds other than $c$ - above or below)-  enter this calculation. This is not a problem, because information-bearing aspects of the field are still calculated to propagate at $c$;
I don't think you have a sound conception of a photon. Please don't take that as an attack as there are a great many education resources out there that give people these kinds of ideas (I was 35 before I shifted on from these ideas). The modern idea is subtler, but much simpler and much sounder logically than the 1920s Bohrian ideas that are still taught. I recommend you begin with Daniel Sank's glorious answer here. Inspired by his answer, I followed up with this one (which may also be useful to you). Indeed, for massless particles like the photon, one cannot even define a sound notion of "position" as one can for nonrelativistic electrons following the nonrelativistic Schrödinger equation.


TITLE: Give an Example of diffeomorphism from unit circle to plane without circle.
QUESTION [1 upvotes]: Give an Example of diffeomorphism from  circle to plane without
  circle.

Any Ideas?

REPLY [2 votes]: You could try $(x,y) \to (1/(x+101),1/(y+101))
.$ Then each coordinate of image is positive and at most $1/100$ so image strictly inside unit circle.
Added: A much simpler map is $(x,y) \to (rx,ry)$ where $r$ is positive and not $1.$ That maps the unit circle $x^2+y^2=1$ to the concentric (distinct) circle $x^2+y^2=r^2.$


TITLE: Relation between global Ext for sheaves on $\mathbb{P}^n$ and Ext for graded modules
QUESTION [2 upvotes]: Let $S = \mathbb{C}[z_0,\dots,z_n]$, let $F$ be a finitely generated graded $S$-module, and let $\mathcal{F}$ be its associated coherent $\mathcal{O}_{\mathbb{P}^n}$-module.
Question:
What is the relation between $\text{Ext}_S^k(F,S)$ and global Ext of the sheaves $\mathcal{F}$ and $\mathcal{O}_{\mathbb{P}^n}$: $\text{Ext}^k(\mathcal{F},\mathcal{O}_{\mathbb{P}^n})$?
I am confused by the following. By local duality, and Serre duality, it seems that
$$
\text{Ext}_S^k(F,S)_d \cong (H_\mathfrak{m}^{n+1-k}(F)_d)^* \cong (H^{n-k}(\mathbb{P}^n,\mathcal{F}(d)))^* \cong \text{Ext}^k(\mathcal{F}(d),\mathcal{O}_{\mathbb{P}^n})
$$
for $0 \leq k < n$.
This seems strange since, if I understand correctly, $\text{Ext}^k(\mathcal{F}(d),\mathcal{O}_{\mathbb{P}^n})$ cannot be computed starting with a locally free resolution of $\mathcal{F}(d)$. However, $\text{Ext}_S^k(F,S)$ can be computed starting from a graded free resolution of $F$:
$$
0 \to F_n \to \dots \to F_0 \to F.
$$
Since the maps are matrices of polynomials this corresponds to a locally free resolution of $\mathcal{F}(d)$:
$$
0 \to \widetilde{F}_n(d) \to \dots \to \widetilde{F}_0(d) \to \mathcal{F}(d).
$$

REPLY [0 votes]: Whenever $\mathcal F = \tilde M$ is a coherent sheaf given by a graded module $M$ over the ring $S = \mathbb C[x_0,\dots,x_n]$, it is possible to compute the cohomology of $\mathcal F$ from a graded free resolution of $M$. Only, that this is more complicated. The recipe can probably be extracted from Hartshorne's "Residues and Duality". There is a more down-to-earth approach sketched e.g. in the last section of this paper which makes this a little more explicit in terms of Cech cohomology.


TITLE: What is the relationship between the Archimedean Property and Calculus?
QUESTION [4 upvotes]: My textbook begin with a chapter dedicaded to the Real Numbers. It introduces firstly integers and rational numbers, then it introduces the irrational numbers, which appeared in the first place as a necessity of expressing the length of incommesurable line segments. To my understanding, all of this helps to introduce the notion of continuity of a variable, which is put into work when treating the continuity of a variable dependent of another variable (if we restrict ourselves to single-valued functions).
The Archimedean Property is stated as follows (or as it is written in my textbook):

Given any number $c > 0$, there exists  a natural number $n$, such that $n > c$.
Given any positive number $\epsilon$, there always exists a natural
  number $n$ such that the inequality  $1/n < \epsilon$ is fulfilled.

The last statement is the first one with $c = 1/\epsilon$. Although I understand what is meant by it (I think it can readily be proved by contradiction), what I'm not getting is how the Archimedean Property fits in the preparatory material preceding that dealing with supremum, infimum and then with continuity and limits.
Does it have something to do with the concept of limit, perhaps?

REPLY [1 votes]: The classical calculus as developed by the pioneers of mathematical analysis like Leibniz and Euler exploited numbers called infinitesimals (the term was possibly introduced by Leibniz himself). These are numbers that violate the Archimedean property when compared to $1$.  Thus, adding an infinitesimal to itself finitely many times produces a number that is again an infinitesimal.  Thus the number systems used by the pioneers of analysis were non-Archimedean.
At the end of the 19th century a different approach to analysis was developed by Cantor, Dedekind, Weierstrass, and others.  In this approach one only uses the Archimedean complete ordered field. Some of the arguments and procedures of the pioneers of the calculus have to be reformulated in this approach using long-winded paraphrases that make the subject harder to learn.
An approach to calculus using infinitesimals is developed in Keisler's textbook.


TITLE: What is the expectation of a random variable raised to the $n$th power?
QUESTION [5 upvotes]: If $Y=X^n$, with $n$ and the expectation and variance of $X$ known, what is the expectation and variance of $Y$?

REPLY [3 votes]: To give this an answer, as Stefan Hansen comments:

Knowing the expectation and variance of X is not enough information to know the moments of all orders.

As an explicit example, let $X$ be normally distributed with mean zero and variance 1, and let $W$ be +1 or -1 with probability 1/2 (the Rademacher distribution).  Then $EX = EW = 0$ and $E X^2 = E W^2 = 1$, but $E X^4 = 3$ while $E W^4 = 1$.
In the special case of the normal distribution, the distribution of a normal random variable are completely determined by its mean and variance, hence so are its higher moments.  You can find various formulas for those moments on Wikipedia.


TITLE: How to evaluate $\int_{0}^{\infty}\frac{1}{t}\arctan\left(\frac{t}{1+2t^2}\right)\,\mathrm dt$?
QUESTION [3 upvotes]: I entered this integral into Wolframalpha, and got $$\int_{0}^{\infty}\frac{1}{t}\arctan\left(\frac{t}{1+2t^2}\right)\,\mathrm dt=\frac{1}{2}\pi\log{2}.$$ But it doesn't provide step by step solution for this integral.
This integral is a bonus challenge in my Calculus class, and the professor that the key is $\arctan$. But I don't know is there any special about $$\arctan\left(\frac{t}{1+2t^2}\right),$$ so I tried some common integration method, and it doesn't work.

REPLY [4 votes]: By integration by parts we have
$$I=\int_0^\infty\ln x\cdot\frac{2x^2-1}{4x^4+5x^2+1}dx$$
$$=\int_0^\infty\frac{\ln x}{1+x^2}dx-\color{red}{\int_0^\infty\frac{2\ln x}{1+4x^2}dx}$$
$$\overset{\color{red}{2x\mapsto x}}{=}\int_0^\infty\frac{\ln2}{1+x^2}dx=\frac{\pi}{2}\ln2$$


TITLE: Bounded martingale convergence
QUESTION [1 upvotes]: True or False?
There exists a martigale $\left(M_{n}\right)_{n}$ such that $\mathbb{P}\left(\left|M_{n}-M_{n-1}\right| \leq 10\right)=1 \quad \forall n$ and $\mathbb{P}\left(\lim _{n \rightarrow \infty}\left|M_{n}\right|=+\infty\right)=1$
My attempt
I thought of using dominating convergence theorem to prove this statement is false.
$\mathbb{P}\left(\left|M_{n}-M_{n-1}\right| \leq 10\right)$ implies that $E\left[\left|M_{n}-M_{n-1}\right|\right] \leq 10$
Hence, by Jensen's inequality we have: $E\left[M_{n}\right] \leqslant E\left[\left|M_{n}\right|\right] \leqslant 10+E\left[|M_{0}|\right]$
Thus by dominating convergence theorem: $$
\lim _{n \rightarrow \infty} E\left[M_{n}\right]=E\left[M_{\infty}\right]<10+E\left[|M_{0}|\right]$$
Hence $M_n$ converges to $M_{\infty}$ a.s and this statement is false.
I am not sure about using dominating convergence theorem as the theorem requires $M_n$ to converge in order to use. Thank you for your help!

REPLY [0 votes]: A martingale with bounded jumps, like $(M_n)$, has the following property (see Durrett's Probability: Theory and Examples, https://services.math.duke.edu/~rtd/PTE/PTE5_011119.pdf, Theorem 4.3.1): If $C:=\{\omega: \lim_nM_n(\omega)$ exists in $\Bbb R\}$ and $D:=\{\omega:\limsup_n M_n(\omega)=+\infty,\liminf_n M_n(\omega)=-\infty\}$, then $\Bbb P[C\cup D]=1$.
Now ask yourself how $\{\omega:\lim_n|M_n(\omega)|=+\infty\}$ is related to $C$ and $D$.


TITLE: Does the convex hull of a sphere contain every point on its surface?
QUESTION [0 upvotes]: Sorry if this is an obvious question but I have not been able to find a straightforward answer for it.
My intuition tells me this is correct, and have confirmed that the scipy convex hull algorithm will return all samples used to generate a Fibonacci sphere as hull vertices.
I am not a math person, but a coder trying to test the performance scaling of a script, so if someone could confirm this (with a proof if you're feeling generous) I would much appreciate it. Thank you.

REPLY [1 votes]: Recall these two pertinent facts about the norm/length function:
\begin{align*}
\|x + y\| &\le \|x\| + \|y\| \\
\|\lambda x\| &= |\lambda| \|x\|,
\end{align*}
where $x$ and $y$ are vectors and $\lambda$ is a scalar.
The unit ball is defined by $\{x : \|x\| \le 1\}$. Suppose $x$ and $y$ lie in this ball. We wish to show that the line segment between $x$ and $y$, i.e. all points of the form $\lambda x + (1 - \lambda) y$ where $\lambda \in [0, 1]$, also lies in the ball. We have,
\begin{align*}
\|\lambda x + (1 - \lambda)y\| &\le \|\lambda x\| + \|(1 - \lambda)y\| \\
&=|\lambda| \|x\| + |1 - \lambda|\|y\| \\
&\le |\lambda| + |1 - \lambda|
\end{align*}
When $\lambda \in [0, 1]$, then $\lambda, 1 - \lambda \ge 0$, so
$$\|\lambda x + (1 - \lambda)y\| \le \lambda + 1 - \lambda = 1,$$
which is what we wanted to show.


TITLE: Fermi surface and external fields
QUESTION [0 upvotes]: From Wikipedia: "In condensed matter physics, the Fermi surface is the surface in reciprocal space which separates occupied from unoccupied electron states at zero temperature".
What does happen when an external electric or magnetic field is applied? The allowed electron energy levels could change accordingly. Does the Fermi surface modify its shape?

REPLY [1 votes]: To understand some of the weirder properties of metals it is probably necessary to think about Fermi surfaces pretty deeply.
When you think about applying an electric field you are disturbing the way the electrons are distributed at or near the Fermi surface.  For simple cases that seems pretty straightforward and and you can think about fermi gas and relaxation times etc. If you think of the simple case of a Fermi sphere you can think of the Fermi sphere shifting and little bit when the field is applied and the amount of that shift is related to the drift velocity.
When you have a crystal lattice, rather than having a sphere you can have more complicated shapes for the Fermi surface. This depends on crystal lattice and is related to the band structure.
So now think about applying a magnetic field. An electron in a magnetic field moving at a velocity in a vacuum would move in a circle and not do any work. In the crystal with the magnetic field the same kind of thing happens but the electron stays on the Fermi surface and depending on the surface can have an electron like orbit, or could have a hole like orbit or move from Brillioun zone to Brillioun zone in an open orbit.
So for copper which looks kind of like a sphere with bumps you have electron like orbits and the Hall effect behaves like you would expect with the electrons being deflected by the magnetic field (Lorentz force) but in aluminum you can find that the Hall effect has the wrong sign (anomalous Hall effect) because the Fermi surface has the hole like orbits.
But in both cases as you apply the fields the interesting stuff is at the Fermi surface where and electron can move to a nearby unoccupied state.


TITLE: Are the electrons and protons electric potential in bonded state are the same as if they are free particles?
QUESTION [1 upvotes]: During the approach of an electron and a proton into a hydrogen atom, are their electric fields unchanged? Or do they cancel each over out (more or less, only a weak electric dipole alive).

REPLY [1 votes]: The hydrogen atom is analytically solvable in the Schrödinger equation

U(r) is the potential and the solution of the equation exists for all r. If the energy is positive it is a free electron and a free proton, if it is negative it has to be in a quantized state.


text:
\begin{document}

\begin{frontmatter}

\title{A Note on the Isotopism of Commutative Semifields}

\author{Yue Zhou}
\ead{yue.zhou@st.ovgu.de}

\address{Department of Mathematics, Otto-von-Guericke-University
Magdeburg, \\39106 Magdeburg, Germany}

\begin{abstract}
We present an example of two isotopic but not strongly isotopic commutative semifields. This example shows that a recent result of Coulter and Henderson on semifield of order $p^n$, $n$ odd, can not be generalized to the case $n$ even.
\end{abstract}

\begin{keyword}

commutative semifield \sep isotopism \sep planar function \sep projective plane
\end{keyword}

\end{frontmatter}

\section{Introduction}\label{sec_intro}

A \emph{semifield} $F$ is an algebraic structure satisfying all the axioms for a skewfield except (possibly) associativity.
A finite field is a trivial example of a semifield. Furthermore, if $F$ does not necessarily have a multiplicative identity, then it is called a \emph{presemifield}. A semifield is not necessarily commutative or associative. However, by Wedderburn's Theorem \cite{Wedderburn1905}, in the finite case, associativity implies commutativity. Therefore, a non-associative finite commutative semifield is the closest algebraic structure to a finite field.

In the earlier literature, semifields were also called \emph{division rings} or \emph{distributive quasifields}. The study of semifields was initiated by Dickson \cite{Dickson1906}, shortly after the classification of the finite fields. Until now, semifields have become an attracting topic in many different areas of mathematics, such as difference sets, coding theory and finite geometry.

The first non-trivial semifields were constructed by Dickson \cite{Dickson1906}. In \cite{Knuth1963}, Knuth showed that the additive group of a semifield $F$ is an elementary abelian group, and the additive order of the elements in $F$ is called the characteristic of $F$. Hence, any finite semifield can be represented by $(\mathbb{F}_{p^n}, +, *)$. Here $(\mathbb{F}_{p^n}, +)$ is the additive group of the finite field $\mathbb{F}_{p^n}$ and $x*y=\varphi(x,y)$, where $\varphi$ is a mapping from $\mathbb{F}_{p^n}\times \mathbb{F}_{p^n}$ to $\mathbb{F}_{p^n}$.

On the other hand, there is a well-known correspondence, via coordinatisation, between commutative semifields and translation planes of Lenz-Barlotti type V.1, see \cite{HughesPiper}. In \cite{Albert1960}, Albert showed that two semifields coordinatise isomorphic planes if and only if they are isotopic:
\begin{definition}
    Let $F_1=(\mathbb{F}_{p^n}, +, *)$ and $F_2=(\mathbb{F}_{p^n}, +, \star)$ be two presemifields. If there exist three linearized permutation polynomials $L, M, N\in\mathbb{F}_{p^n}[x]$ such that
    $$M(x)\star N(y)=L(x*y)$$
    for any $x,y\in\mathbb{F}_{p^n}$, then $F_1$ and $F_2$ are called \emph{isotopic}, and the triple $(M,N,L)$ is an \emph{isotopism } between $F_1$ and $F_2$. Furthermore, if there exists an isotopism of the form $(N, N, L)$ between $F_1$ and $F_2$, then $F_1$ and $F_2$ are \emph{strongly isotopic}.
\end{definition}

We refer the reader to \cite{LidlNiederreiter} for more background on finite fields, in particular about linearized polynomials. Let $F=(\mathbb{F}_{p^n}, +, *)$ be a presemifield, and $a\in F$. If we define a new multiplication $\star$ by the rule
    $$(x*a)\star(a*y)=x*y,$$
we obtain a semifield $(\mathbb{F}_{p^n}, +, \star)$ with unit $a*a$. There are many semifields associated with a presemifield, but they are all isotopic.

Let $F=(\mathbb{F}_{p^n},+,*)$ be a semifield. The subsets
\begin{equation*}
\begin{aligned}
  N_l(F)=\{a\in F: (a*x)*y=a*(x*y) \text{ for all }x,y\in F\},\\
  N_m(F)=\{a\in F: (x*a)*y=x*(a*y) \text{ for all }x,y\in F\},\\
  N_r(F)=\{a\in F: (x*y)*a=x*(y*a) \text{ for all }x,y\in F\},
\end{aligned}
\end{equation*}

are called the \emph{left, middle} and \emph{right nucleus} of $F$, respectively. It is easy to check that these sets are finite fields. The subset $N(F)=N_l(F)\cap N_m(F) \cap N_r(F)$ is called the \emph{nucleus} of $F$. It is easy to see, if $F$ is commutative, then $N_l(F)=N_r(F)=N(F)$. In \cite{HughesPiper}, the geometry interpretations of these nuclei are presented.

Next, we give the definition of planar functions, which was introduced by Dembowski and Ostrom in \cite{DO1968} to describe affine planes possessing a collineation group with specific properties.
\begin{definition}
    Let $p$ be an odd prime. A function $f: \mathbb{F}_{p^n}\rightarrow\mathbb{F}_{p^n}$ is called a \emph{planar function}, or \emph{perfect nonlinear (PN)}, if for each $a\in \mathbb{F}_{p^n}^*$, $f(x+a)-f(x)$ is a bijection on $\mathbb{F}_{p^n}$.
\end{definition}

For $p=2$, if $x_0$ is a root of $f(x+a)-f(x)=b$, then $x_0+a$ is another one, hence there is no planar functions over $\mathbb{F}_{2^n}$.
A \emph{Dembowski-Ostrom} (DO) polynomial $D\in \mathbb{F}_{p^n}[x] $ is a polynomial
$$D(x)=\sum_{i,j}a_{ij}x^{p^i+p^j}\enspace .$$
Obviously, $D(x+a)-D(x)-D(a)$ is a linearized polynomial for any nonzero $a$. It can be proved that a planar DO polynomial is equivalent to a commutative presemifield with odd characteristic, see \cite{Coulter2008}. In fact, if $*$ is the presemifield product, then the corresponding planar function is $f(x)=x*x$; when the planar DO polynomial $f$ is given, then the corresponding presemifield product can be defined as
\begin{equation}\label{eq_PN->production}
    x*y=\frac{1}{2}(f(x+y)-f(x)-f(y))\enspace .
\end{equation}

A function from a finite field $\mathbb{F}_{p^n}$ to itself is \emph{affine}, if it is defined by the sum of a constant and a linearized polynomial over $\mathbb{F}_{p^n}$. There are several equivalence relations of functions for which the \emph{planar} property is invariant:
\begin{definition}
    Two functions $f$ and $g: \mathbb{F}_{p^n}\rightarrow \mathbb{F}_{p^n}$ are called
    \begin{itemize}
      \item \emph{extended affine equivalent} (EA-equivalent), if $g = l_1 \circ f \circ l_2+l_3$, where $l_1, l_2$ and $l_3$ are affine functions, and where $l_1,l_2$ are permutations of $\mathbb{F}_{p^n}$. Furthermore, if $l_3$ is the zero mapping, then $f$ and $g$ are called \emph{affine equivalent};
      \item \emph{Carlet-Charpin-Zinoviev equivalent} (CCZ-equivalent or graph equivalent), if there is some affine permutation $L$ of $\mathbb{F}_{p}^{2n}$, such that $L(G_f)=G_g$, where $G_f=\{(x,f(x):x\in\mathbb{F}_{p^n})\}$ and $G_g=\{(x,g(x):x\in\mathbb{F}_{p^n})\}$.
    \end{itemize}
\end{definition}

Generally speaking, EA-equivalence implies CCZ-equivalence, but not vice versa, see \cite{Budaghyan06}. However, if planar functions $f$ and $g$ are CCZ-equivalent, then they are also EA-equivalent \cite{B-H2008,EA=CCZ_PN2008}.  Because of the correspondence between commutative presemifields with odd characteristic and planar functions as we mentioned above, the strong isotopism of two commutative presemifields is equivalent to the affine equivalence of the corresponding  planar DO functions, which we call directly the \emph{equivalence} of planar DO functions.

\section{Isotopism $\neq$ Strong Isotopism}
In \cite{Coulter2008}, Coulter and Henderson proved the following theorem.
\begin{theorem}\label{th_CoulterHenderson}
    Let $F_1 =(\mathbb{F}_q, +, \star)$ and $F_2 =(\mathbb{F}_q, +, *)$ be isotopic commutative semifields. Then there exists an isotopism $(M, N, L)$ between $F_1$ and $F_2$ such that either
    \begin{enumerate}
      \item $M=N$, or
      \item $M(x)\equiv \alpha\star N(x) \mod (x^q-x)$, where $\alpha \in N_m(F_1)\setminus N(F_1)$ cannot be written in the form $\alpha=\gamma\star\beta^2$ where $\gamma\in N(F_1)$ and $\beta\in N_m(F_1)$.
    \end{enumerate}
\end{theorem}

It implies that any commutative semifield can generate at most two non-strongly isotopic commutative semifields. Some important corollaries are also presented in \cite{Coulter2008}, for example,
\begin{corollary}
    Any two commutative semifields of order $p^e$ with $e$ odd are isotopic if and only if they are strongly isotopic.
\end{corollary}

Pieper-Seier and Spille  \cite{pieper-seier_remarks_1999} showed that the Cohen-Ganley commutative semifield \cite{Cohen-Ganley1982} has exactly two classes of strong isotopy. In this paper, we present another example\footnote{In the previous version of this paper, we have claimed that our example is the first one. However, later Coulter and Knarr informed us about the result from \cite{pieper-seier_remarks_1999}}.

First, we introduce a family of planar functions:
$$\frac{1}{2}(\mathrm{Tr}(x^2)+G(x^{q^2+1}))$$
over $\mathbb{F}_{q^{2m}}$, where $q$ is a power of an odd prime $p$, $m=2k+1$, $\mathrm{Tr}(\cdot)$ is the trace function from $\mathbb{F}_{q^{2m}}$ to $\mathbb{F}_{q^m}$, and $G(x)=h(x-x^{q^m})$, where $h\in \mathbb{F}_{q^{2m}}[x]$ is defined as
$$h(x) = \sum_{i=0}^{k}(-1)^i x^{q^{2i}}+\sum_{j=0}^{k-1}(-1)^{k+j}x^{q^{2j+1}}\enspace.$$
This planar function family corresponds to Bierbrauer's generalization of the semifield discovered by Lunardon, Marino, Polverino and Trombetti over $q^{6}$, see \cite{Bierbrauer2009-3,LMPT2009}. Hence the corresponding semifield  should be called Lunardon-Marino-Polverino-Trombetti-Bierbrauer (LMPTB) semifields \cite{Bierbrauer2009-3}.

Let $q=3$, $m=3$ and $\mathbb{F}_{3^6}=\mathbb{F}_{3}(\xi)$, where $\xi$ is a root of $x^6-x^4+x^2-x-1\in \mathbb{F}_{3}[x]$. Let $F_1=(\mathbb{F}_{3^6}, +, \star)$ be the LMPTB semifield. By MAGMA\cite{Magma}, we calculate that $| N_m(F_1)| =3^2$ and $| N(F_1)| =3$, and there are four $\alpha \in N_m(F_1)\setminus N(F_1)$, which cannot be written in the form $\alpha=\gamma\star\beta^2$, where $\gamma\in N(F_1)$ and $\beta\in N_m(F_1)$. They are $\lambda, \lambda^3, \lambda^5$ and $\lambda^7$, where $\lambda=\xi^{91}$.

Now, we can define another semifield $F_2$ with the multiplication given by
$$x\odot y=(\lambda\star x)\star y$$
Obviously, $F_1$ and $F_2$ are isotopic. As we mentioned above, to tell whether $F_1$ and $F_2$ are strongly isotopic, we just need to calculate whether $f_1(x)=x\star x$ is equivalent to $f_2(x)=x\odot x$. By Lagrange interpolation, we have
\begin{equation*}
    f_1(x)=x^{270} - x^{246} + x^{90} - x^{82} - x^{54} + x^{30} - x^{10} -
    x^{2},

\end{equation*}
\begin{equation*}
    f_2(x)= \lambda^3(x^{270} - x^{246} - \lambda^2x^{90} + \lambda^2x^{82} -
       x^{54} + x^{30} + \lambda^2x^{10} + \lambda^2x^2)\enspace .

\end{equation*}

Let $f:\mathbb{F}_{p^n}\rightarrow \mathbb{F}_{p^n}$ be any function. Since the additive group of $\mathbb{F}_{p^n}$ is the linear space $\mathbb{F}_p^n$, $f$ can also be considered as a mapping from $\mathbb{F}_p^n$ to itself. Define a matrix $M_f\in \mathbb{F}_p^{(2n+1, p^n)}$ as follows:
\begin{equation}\label{eq_Mf}
  M_f=\left(
               \begin{array}{ccc}
                 \cdots & 1 & \cdots \\
                 \cdots & x & \cdots \\
                 \cdots & f(x) & \cdots \\
               \end{array}
             \right)_{x\in\mathbb{F}_p^n}
\end{equation}
Then we can construct a linear code $C_f$ over $\mathbb{F}_p$ by the generator matrix $M_f$. Furthermore, it can be proved that
\begin{proposition}
     Let $p$ be a prime, and $n$ be an integer. Two functions $f,g:\mathbb{F}_{p^n}\rightarrow \mathbb{F}_{p^n}$ are CCZ-equivalent, if and only if the corresponding codes $C_f$ and $C_g$ are permutation equivalent.
\end{proposition}
\begin{proof}
    Assume that $C_f$ and $C_g$ are permutation equivalent, then we have a permutation matrix $P$ and a $(2n+1)\times(2n+1)$ matrix $L$ with full rank, such that
    $$L\cdot M_f\cdot P=M_g\enspace.$$
    That means there are $u,v\in\mathbb{F}_p^n$ and a matrix $\tilde{L}$ with full rank, such that
    $$\tilde{L}\cdot \left(
               \begin{array}{ccc}
                 \cdots & x & \cdots \\
                 \cdots & f(x) & \cdots \\
               \end{array}
             \right)\cdot P=\left(
               \begin{array}{ccc}
                 \cdots & x & \cdots \\
                 \cdots & g(x) & \cdots \\
               \end{array}
             \right)+\left(
               \begin{array}{c}
                 u \\
                 v \\
               \end{array}
             \right) \enspace.
    $$
    Therefore, by the definition of CCZ-equivalence, $f$ and $g$ are CCZ-equivalent. The proof of the converse is the same. \qed

\end{proof}

For the equivalence of codes, including permutation equivalence
and monomial equivalence, see \cite{HuffmanPless2003}.

It is well-known that function $f$ mapping $\mathbb{F}_{p^n}$ to itself
is planar if and only if for every nonzero $a\in\mathbb{F}_{p^n}$,
the function $\mathrm{Tr}(af(x))$ is generalized bent, see \cite{CarletDubuc1999}. For planar DO-polynomials, it is equivalent to the nonsingularity of $\mathrm{Tr}(af(x))$ as a p-ary quadratic form, for every nonzero $a\in\mathbb{F}_{p^n}$. Therefore, the weight distribution of $C_f$ can be deduced, see \cite{LiQuLing2009}, and there are only $p-1$ code words $(i,i,\cdots,i)$ with weight $p^n$ ($0<i<p$). Thus, for the codes $C_f$ and $C_g$ from the planar functions $f$ and $g$, monomial and permutation equivalence are identical. By MAGMA, we calculated that $C_{f_1}$ is not monomially equivalent to $C_{f_2}$ (MAGMA only offers the command to tell the monomial equivalence of two linear codes, that is why we emphasize the identity of monomial and permutation equivalence between $C_f$ and $C_g$). Therefore, $F_1$ is not strongly isotopic to $F_2$, which means that it is possible to construct inequivalent planar functions from known ones by the isotopism of corresponding presemifield.

\begin{remark}
    For Dickson \cite{Dickson1906}, Albert \cite{Albert1960}, Ganley \cite{Ganley1981} and Cohen-Ganley \cite{Cohen-Ganley1982} commutative semifields, we did not find such $\lambda$ to construct affine-inequivalent functions $f_1$ and $f_2$ as defined above on $\mathbb{F}_{3^{2m}}$, where $m=2, 3$. For the Budaghyan-Helleseth-Bierbrauer (BHB) semifields \cite{Bierbrauer2009-3,B-H2008,BudaghyanHelleseth2010} of order $3^6$, such $\lambda$ can also be found. For any other larger $m$, it is beyond our computation capacity.
\end{remark}

\begin{remark}
    We find that $f_2$ is equivalent to the planar function from BHB semifield of order $3^6$, which means BHB semifield and LMPTB semifield of order $3^6$ are isotopic but not strongly isotopic.
\end{remark}

\section*{Acknowledgement}
    We are grateful to R. Coulter and R. Knarr for pointing out the result in \cite{pieper-seier_remarks_1999}.

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