TITLE: Fourier transform of 3D Sinc function
QUESTION [3 upvotes]: What is the Fourier trasnform of the function
$$\frac{\sin(P|\mathbf{x-y}|)}{|\mathbf{x-y}|}$$
where $P$ is a real parameter and $\mathbf{y}$ is a fixed point in three-dimensional space?

REPLY [0 votes]: I found it is
$i\sqrt{\frac{\pi}{2}}\frac{e^{-i\mathbf{k}\cdot\mathbf{y}}}{|\mathbf{k}|} \delta(P-|\mathbf{k}|)$


text:
\begin{document}

\begin{abstract}
We prove a general ampliation homogeneity result for the quasicentral modulus of an $n$-tuple of operators with respect to the $(p,1)$ Lorentz normed ideal. We use this to prove a formula involving Hausdorff measure for the quasicentral modulus of $n$-tuples of commuting Hermitian operators the spectrum of which is contained in certain Cantor-like self-similar fractals.
\end{abstract}

\maketitle

\section{Introduction}
\label{sec1}

The quasicentral modulus $k_{\sJ}(\tau)$ is a number associated with an $n$-tuple $\tau$ of Hermitian operators relative to a normed ideal $(\sJ,|\ |_{\sJ})$ of compact operators. It underlies many questions on normed ideal perturbations of $n$-tuples of operators (see the recent survey \cite{11}), and it also had applications in non-commutative geometry in work on the spectral characterization of manifolds \cite{1}.

We proved in \cite{7} that in the case of $\tau$ an $n$-tuple of commuting Hermitian operators and if the normed ideal is the $(n,1)$-Lorentz ideal, which we denote by $\sC_n^-$, the corresponding quasicentral modulus $k^-_n(\tau)$ has the property that $(k_n^-(\tau))^{1/n}$ is proportional to the integral w.r.t.\ $n$-dimensional Lebesgue measure of the multiplicity function of $\tau$.

Here we prove a similar result in fractional dimension. More precisely instead of a cube in $\bR^n$ which contains the spectrum of $\tau$ we assume there is a fixed self-similar fractal in $\bR^n$ of Hausdorff dimension $p > 1$ containing the spectrum $\sigma(\tau)$. The analogous formula we prove has the exponent $n$ replaced by $p$ and the integral of the multiplicity function is with respect to $p$-Hausdorff measure. For technical reasons the class of fractals is rather restricted, only certain totally disconnected sets, that is Cantor-like fractals are considered. One should certainly expect this can be extended to a larger class of fractals, the present paper being only a first step.

\bigskip
\noindent
\underline{\qquad\qquad}

2020 {\em Mathematics Subject Classification}. Primary: 47A55; Secondary: 28A78, 28A80, 47B10.

{\em Key words and phrases}. Lorentz (p, 1) normed ideal, p-Hausdorff measure, quasicentral modulus.

\vfill
\noindent
${}^*$Research supported in part by NSF-Grant DMS-1665534.

\newpage
What made the extension of the formula to fractional dimension possible is a completely general ampliation homogeneity result $k^-_p(\tau \otimes I_m) = m^{1/p}k^-_p(\tau)$. Such a result was previously known only for $p = \infty$ (\cite{9}) and for $p = 1$ (\cite{7}). Note that in \cite{7} we obtained easily an ampliation homogeneity result for $k_p(\tau)$, that is with $\sC^-_p$ replaced by the Schatten--v.~Neumann class $\sC_p$. For $p = 1$ we have $\sC_1 = \sC^-_1$, but for $p > 1$ the result for $k_p(\tau)$ turned out to be trivial after we showed in \cite{8} that in this case $k_p(\tau) \in \{0,\infty\}$. The interesting quantity which replaces $k_p(\tau)$ is $k_p^-(\tau)$.

The paper has six sections including this introduction. Section~2 contains preliminaries about the quasicentral modulus. Section~3 is devoted to the ampliation homogeneity theorem. In section~4 we collected preliminaries concerning the class of Cantor-type fractals we consider. The formula for $k_p^-(\tau)$ in the fractal setting is obtained in section~5. Section~6 deals with concluding remarks.

\section{Operator preliminaries}
\label{sec2}

By $\sH$ we denote a separable complex Hilbert space of infinite dimension and by $\sB(\sH)$, $\sK(\sH)$, $\sR(\sH)$ the bounded operators, the compact operators and the finite rank operators. When no confusion will arise we will simply write $\sK,\sR$ and we will denote by $\sR_1^+(\sH)$ or $\sR_1^+$ the finite rank positive contractions $0 \le A \le I$ on $\sH$. The $(p,1)$ Lorentz normed ideal of compact operators will be denoted by $(\sC^-_p,|\ |_p^-)$. We recall that the norm is
\[
|T|^-_p = \sum_{j \in {\mathbb N}} s_j\ j^{-1+1/p}
\]
where $s_1 \ge s_2 \ge \dots$ are the eigenvalues of $|T| = (T^*T)^{1/2}$ in decreasing order. If $(\sC_p,|\ |_p)$ is the Schatten--von~Neumann $p$-class, then $\sC_1 = \sC_1^-$. More on normed ideals can be found in \cite{5} and \cite{6}.

We shall also use the following notation for operations on $n$-tuples of operators, in line with \cite{7}. If $\tau = (T_i)_{1 \le i \le n} \in (\sB(\sH))^n$ and $X,Y \in \sB(\sH)$ then we use
\[
\begin{aligned}
X\tau Y &= (XT_iY)_{1 \le i \le n} \\
[X,\tau] &= ([X,T_i])_{1 \le i \le n} \\
\tau^* &= (T_i^*)_{1 \le i \le n}.
\end{aligned}
\]
If also $\sigma = (S_i)_{1 \le i \le n} \in (\sB(\sH))^n$ then we write
\[
\begin{aligned}
\sigma+\tau &= (S_i+T_i)_{1 \le i \le n} \\
\sigma\oplus\tau &= (S_i \oplus T_i)_{1\le i \le n} \\
\tau\otimes I_m &= (T_i \otimes I_m)_{1 \le i \le n}
\end{aligned}
\]
where $I_m$ is the identity operator on ${\mathbb C}^m$. When we identify $\sH \otimes {\mathbb C}^m$ and $\sH \oplus \dots \oplus \sH$ we also have
\[
\tau \otimes I_m \cong \underset{\text{$m$-times}}{\underbrace{\tau \oplus \dots \oplus \tau}}\,.
\]
Further, we consider norms
\[
\begin{aligned}
\|\tau\| &= \max_{1 \le i \le n} \|T_i\| \\
|\tau|_{\sJ} &= \max_{1 \le i \le n} |T_i|_{\sJ}.
\end{aligned}
\]

The quasicentral modulus of an $n$-tuple $\tau = (T_i)_{1 \le i \le n}$ with respect to a normed ideal $(\sJ,|\ |_{\sJ})$ (see \cite{7}, \cite{9}) is the number
\[
k_{\sJ}(\tau) = \liminf_{A \in \sR_1^+} |[\tau,A]|_{\sJ}
\]
where the $\liminf$ is w.r.t.\ the natural order on $\sR_1^+$. This definition is also equivalent to $k_{\sJ}(\tau)$ being the
\[
\begin{aligned}
\inf\{\alpha \in [0,\infty]| \alpha &= \lim_{k \to \infty} |[A_k,\tau]|_{\sJ},\ A_k \uparrow I \\
&A_k \in \sR_1^+\}
\end{aligned}
\]
or the same with $w - \lim_{k \to \infty} A_k = I$ instead of $A_k \uparrow I$. If $\sJ = \sC_p^-$ we denote $k_{\sJ}(\tau)$ by $k_p^-(\tau)$.

We should also record as the next proposition the results in \cite{7} Prop.~$1.4$ and Prop.~$1.6$.

\medskip
\noindent
{\bf Proposition 2.1.}
{\em If $\tau^{(j)} \in \sB(\sH)^n$, $j \in {\mathbb N}$ then we have
\[
\max_{j=1,2} k_{\sJ}(\tau^{(j)}) \le k_{\sJ}(\tau^{(1)} \oplus \tau^{(2)}) \le k_{\sJ}(\tau^{(1)}) + k_{\sJ}(\tau^{(2)})
\]
and
\[
k_{\sJ}\Big( \underset{j \in {\mathbb N}}{\bigoplus} \tau^{(j)}\Big) = \lim_{m_i \to \infty} k_{\sJ}\Big( \underset{1 \le j \le m}{\bigoplus} \tau^{(j)}\Big).
\]
If $\lambda^{(j)} \in {\mathbb C}^n$ and $\lambda^{(j)} \otimes I_{\sH} \in \sB(\sH)^n$ then we have
\[
\begin{aligned}
k_{\sJ}(\tau^{(1)} \oplus \dots \oplus \tau^{(m)}) &= \\
&= k_{\sJ}\left((\tau^{(1)}-\lambda^{(1)} \otimes I_{\sH}) \oplus \dots \oplus (\tau^{(m)} - \lambda^{(m)} \otimes I_{\sH})\right).
\end{aligned}
\]
}

\medskip
Finally, if $\tau$ is a $n$-tuple of commuting Hermitian operators we denote by $\sigma(\tau) \subset {\mathbb R}^n$ the joint spectrum and by $E(\tau;\omega)$ the spectral projection of $\tau$ for the Borel set $\omega \subset {\mathbb R}^n$.

\section{Ampliation homogeneity}
\label{sec3}

\medskip
\noindent
{\bf Theorem 3.1.}
{\em If $\tau$ is a $n$-tuple of bounded operators and $1 \le p \le \infty$ then
\[
k_p^-(\tau \otimes I_m) = m^{1/p} k_p^-(\tau).
\]
}

\medskip
The cases $p = 1$ and $p = \infty$ have already been proved (\cite{7} Prop.~1.5 and \cite{9} Prop.~3.9).

We begin the proof with a couple of lemmas.

\medskip
\noindent
{\bf Lemma 3.1.}
{\em Let $X_j \in {\mathcal C}^-_p$, $j \in {\mathbb N}$, $p \in [1,\infty]$ be so that $|X_j|^-_p \le C$. If $\lim_{j \to \infty} \|X_j\| = 0$ then we have
\[
\lim_{j \to \infty} (|X_j \otimes I_m|^-_p - m^{1/p}|X_j|_p^-) = 0.
\]
}

\begin{proof}
Let $s_1^{(j)} \ge s_2^{(j)} \ge \dots$ be the eigenvalues of $(X_j^*X_j)^{1/2}$. Then
\[
\begin{aligned}
|X_j \otimes I_m|^-_p &= \sum_{k \in {\mathbb N}} s_k^{(j)}((m(k-1)+1)^{-1+1/p} + \dots + (mk)^{-1+1/p}) \\
&\ge m^{1/p} \sum_{k \in {\mathbb N}} s_k^{(j)}k^{-1+1/p} = m^{1/p}|X_j|^-_p.
\end{aligned}
\]
On the other hand, given $\epsilon > 0$ there is $N$ so that
\[
k \ge N \Rightarrow (m(k-1)+1)^{-1+1/p} + \dots + (mk)^{-1+1/p} \le (1+\epsilon)m^{1/p}k^{-1+1/p}.
\]
This gives
\[
\begin{aligned}
|X_j \otimes I_m|_p^- &\le (1+\epsilon)m^{1/p} \sum_{k \ge N} s_k^{(j)}k^{-1+1/p} + N\|X_j\| \\
&\le (1+\epsilon)m^{1/p}|X_j|^-_p + N\|X_j\|.
\end{aligned}
\]
Thus we have
\[
0 \le |X_j \otimes I_m|^-_p - m^{1/p}|X_j|^-_p \le \epsilon m^{1/p}|X_j|_p^- + N\|X_j\|.
\]
Since $\epsilon > 0$ is arbitrary and $\|X_j\| \to 0$ we get the desired result when $j \to \infty$.
\end{proof}

\medskip
\noindent
{\bf Corollary 3.1.}
{\em Let $X_j = (X_{ji})_{1 \le i \le n}$ be $n$-tuples of operators so that $|X_j|_p^- \le C$ where $p \in [1,\infty]$ and $\lim_{j \to \infty} \|X_j\| = 0$. Then we have
\[
\lim_{j \in \infty}(m^{1/p} |X_j|_p^- - |X_j \otimes I_m|^-_p) = 0.
\]
}

\medskip
\noindent
{\bf Lemma 3.2.}
{\em If $\tau \in (\sB(\sH))^n$ and $(\sJ,|\ |_{\sJ})$ is a normed ideal so that $k_{\sJ}(\tau) < \infty$, then there are $B_j \in \sR^+_1$ so that $B_j \uparrow I$ and
\[
\begin{aligned}
\lim_{j \to \infty} |[\tau,B_j]|_{\sJ} = k_{\sJ}(\tau) \\
\lim_{j \to \infty} \|[\tau,B_j]\| = 0.
\end{aligned}
\]
}

\begin{proof}
It suffices to show that given $\epsilon > 0$ and $P$ a finite rank Hermitian projector we can find $B \in \sR_1^+$ so that $B \ge P$ and
\[
\begin{aligned}
|[B,\tau]|_{\sJ} &\le k_{\sJ}(\tau) + \epsilon \\
\|[B,\tau]\| &\le \epsilon.
\end{aligned}
\]
Such that $B$ can be constructed as follows. We find recursively $P = P_1 \le P_2 \le P_3 \le \dots$ finite rank Hermitian projectors and $A_j \in \sR_1^+$ so that
\[
\begin{aligned}
&A_j \ge P_j,\ |[A_j,\tau]|_{\sJ} \le k_{\sJ}(\tau) + \epsilon \\
&\tau P_j = P_{j+1}\tau P_j,\ \tau^*P_j = P_{j+1}\tau^*P_j \\
&P_{j+1} \ge A_j.
\end{aligned}
\]
If we put $Q_j = P_{j+1} - P_j$ if $j \ge 1$ and $Q_0 = P_1 = P$ we have
\[
Q_r\tau Q_s \ne 0 \Rightarrow |r-s| \le 1
\]
and
\[
A_j = (Q_0 + \dots + Q_{j-1}) + Q_jA_jQ_j.
\]
This gives
\[
Q_r[\tau,A_j]Q_s \ne 0
\]
$\Rightarrow |r-s| \le 1$ and $j-1 \le r,s \le j+1$. It follows that if $|k-j| \ge 4$ $([\tau,A_j])^*[\tau,A_k] = 0$ so that
\[
\|[\tau,A_4 + A_8 + \dots + A_{4N}]\| \le 2\|\tau\|.
\]
Thus if $B_N = N^{-1}(A_4 + A_8 + \dots + A_{4N})$ we have $B_N \ge P$, $B_N \in \sR_1^+$, $|[B_N,\tau]|_{\sJ} < k_{\sJ}(\tau) + \epsilon$ and
\[
\|[B_N,\tau]\| \le 2N^{-1}\|\tau\|.
\]
Thus if $2N^{-1}\|\tau\| < \epsilon$ we may take $B = B_N$.
\end{proof}

\medskip
\noindent
{\bf Lemma 3.3.}
{\em If $m \in {\mathbb N}$ and $(\sJ,|\ |_{\sJ})$ is a normed ideal and $\tau \in (B(\sH))^n$ is so that $k_{\sJ}(\tau) < \infty$, then there are $A_j \in \sR^+_1$, $A_j \uparrow I$ so that
\[
k_{\sJ}(\tau \otimes I_m) = \lim_{j \to \infty} |[\tau \otimes I_m,A_j \otimes I_m]|_{\sJ}
\]
and $\lim_{j \to \infty} \|[\tau,A_j]\| = 0$.
}

\begin{proof}
Let $G$ be the group $\sS_m \rtimes {\mathbb Z}_2^m$ of permutation matrices with $\pm 1$ entries and $g \to U_g$ its representation on $\sH \oplus \dots \oplus \sH$ which is $\otimes I_{\sH}$ the representation on ${\mathbb C}^m$. Then the commutant $\{U_g: g \in G\}'$ is $B(\sH) \otimes I_m$ and the map $\Phi: B(\sH^m) \to \sB(\sH) \otimes I_m$ given by $\Phi(X) = |G|^{-1} \sum_{g \in G} U_gXU_g^*$ is the projection of norm one which preserves the trace. If $B \in R_1^+(\sH^m)$ we have
\[
|[B,\tau \otimes I_m]|_{\sJ} = |U_g[B,\tau\otimes I_m]U_g^*|_{\sJ} = |[U_gBU_g^*,\tau \otimes I_m]|_{\sJ},
\]
which gives by taking the mean over $G$
\[
|[\Phi(B),\tau \otimes I_m]|_{\sJ} \le |[B,\tau \otimes I_m]|_{\sJ}
\]
and clearly $B_j \uparrow I \otimes I_m$ implies $\Phi(B_j) \uparrow I \otimes I_m$. Thus if $A_j \otimes I_m = \Phi(B_j)$ and $B_j \uparrow I \otimes I_m$ are so that
\[
\lim_{j \to \infty} |[B_j,\tau \otimes I_m]|_{\sJ} = k_{\sJ}(\tau \otimes I_m)
\]
and
\[
\lim_{j \to \infty} |[B_j,\tau \otimes I_m]\| = 0
\]
then we have
\[
\lim_{j \to \infty} \sup|[A_j \otimes I_m,\tau \otimes I_m]|_{\sJ} \le k_{\sJ}(\tau \otimes I_m)
\]
and
\[
\lim_{j \to \infty} \|[A_j \otimes I_,\tau \otimes I_m]\| = 0.
\]
Since on the other hand
\[
\liminf_{j \to \infty} |[A_j \otimes I_m,\tau \otimes I_m]|_{\sJ} \ge k_{\sJ}(\tau \otimes I_m)
\]
we conclude that
\[
\lim_{j \to \infty} |[A_j \otimes I_m,\tau \otimes I_m]|_{\sJ} = k_{\sJ}(\tau).
\]
\end{proof}

\medskip
\noindent
{\bf Proof of Theorem~3.1.} Using Lemma~$3.3$ we can find $A_j \in \sR_1^+$, $A_j \uparrow I$ when $j \to \infty$ so that
\[
\lim_{j \to \infty} |[A_j,\tau] \otimes I_m]|^-_p = k^-_p(\tau \otimes I_m)
\]
and
\[
\lim_{j \to \infty} \|[A_j,\tau]\| = 0.
\]
Using Corollary~$3.1$ we infer that
\[
\lim_{j \to \infty} m^{1/p}|[A_j;\tau]|^-_p = k_p^-(\tau \otimes I_m)
\]
which implies that
\[
m^{1/p}k^-_p(\tau) \le k^-_p(\tau \otimes I_m).
\]
On the other hand, Lemma~$3.2$ shows that there are $A_j \uparrow I$, $A_j \in \sR_1^+$ so that
\[
\lim_{j \to \infty} |[\tau,A_j]|^-_p = k^-_p(\tau)
\]
and
\[
\lim_{j \to \infty} \|[\tau,A_j]\| = 0.
\]
Then by Corollary~$3.1$ we get that
\[
\begin{aligned}
m^{1/p}k^-_p(\tau) &= \lim_{j \to \infty} |[\tau,A_j] \otimes I_m|^-_p \\
&= \lim_{j \to \infty} |[\tau \otimes I_m,A_j \otimes I_m]|^-_p \\
&\ge k^-_p(\tau \otimes I_m)
\end{aligned}
\]
which concludes the proof.\qed

\section{Fractal preliminaries}
\label{sec4}

To keep things simple the fractal context will be certain totally disconnected Cantor-like self-similar sets and on which a certain Hausdorff measure is a Radon measure on Borel sets.

We consider a non-empty compact set $K \subset \bR^n$ and a $N$-tuple of maps
\[
F_i(x) = \lambda (x - b(i)) + b(i),\ 1 \le i \le N
\]
where $0 < \lambda < 1$ and $b(i) \in \bR^n$, so that
\[
K = \bigcup_{1 \le i \le N} F_iK
\]
and we assume that
\[
i_1 \ne i_2 \Rightarrow F_{i_1}K \cap F_{i_2}K = \emptyset.
\]
Note that the open set condition for the contractions $F_i$ (see \cite{4}, page~121) in this case can be satisfied with an open neighborhood of $K$. The Hausdorff and box-dimension of $K$ are equal to
\[
p = \frac {\log N}{\log(1/\lambda)}
\]
by \cite{4} Thm.~$8.6$, and the $p$-Hausdorff measure of $K$ is finite and non-zero. This Hausdorff measure is often referred to as the Hutchinson measure of $K$. Note also the uniqueness of $K$ given the maps $F_i$, $1 \le i \le N$ (\cite{4} Thm.~$8.3$).

If $w \in \{1,\dots,N\}^m$ we put $|w| = m$ and define $F_w = F_{w_1} \circ \dots \circ F_{w_m}$ and $K_w = F_wK$. In particular if $|w| = |w'|$ then $K_w$ and $K_{w_1}$ are congruent and in particular have the same $p$-Hausdorff measure and diameter. Moreover
\[
K = \bigcup_{|w|=L} K_w
\]
and for any $L \in {\mathbb N}$ the union is disjoint. On $K$ there is a unique Radon measure $\mu$ so that
\[
\mu(K_w) = N^{-|w|} = \lambda^{|w|p}
\]
and on Borel sets $\mu = cH_p$ where $H_p$ denotes the $p$-Hausdorff measure and $c$ is a constant.  We use the $l^{\infty}$-norm $|(x_1,\dots,x_n)| = \max_{1 \le i \le n} |x_i|$ on $\bR^n$. Note that
\[
\mbox{diam}(K_w) = c \cdot \lambda^{|w|}
\]
for some constant $c$.

We shall also assume that the Hausdorff dimension $p \ge 1$.

If $\tau$ is a $n$-tuple of commuting Hermitian operators on $\sH$ with spectrum $\sigma(\tau) \subset K$, like with Lebesgue measure here on $K$ with $\mu$ that is with $H_p$ the Hilbert space splits
\[
\sH = \sH_{psing} \oplus \sH_{pac}
\]
where $\sH_{psing}$, $\sH_{pac}$ are reducing subspaces for $\tau$ and consist of vectors $\xi$ so that
\[
\langle E(\tau;\cdot)\xi,\xi\rangle
\]
is singular w.r.t.\ $\mu$ and respectively absolutely continuous w.r.t.\ $\mu$, that is w.r.t.\ $H_p$. (Use for instance $1.6.3$ in \cite{3}.) For more on Hausdorff-measure and on fractals see \cite{3} and \cite{4}.

\section{$k^-_p$ in the fractal setting}
\label{sec5}

In this section we study $k^-_p(\tau)$ where $\tau$ is a $n$-tuple of commuting Hermitian operators with $\sigma(\tau) \subset K$, in the context of Section~4. We assume $p \ge 1$ (and for certain results we will require $p > 1$).

\medskip
\noindent
{\bf Lemma 5.1.}
{\em Assume $\tau$ is a $n$-tuple of commuting Hermitian operators with $\sigma(\tau) \subset K$ and with a cyclic vector $\xi$. Then for some constant $C$ depending only on $K$, we have
\[
k^-_p(\tau) \le C(H_p(\sigma(\tau)))^{1/p}.
\]
}

\begin{proof}
Let $\Omega(L) = \{w \mid |w| = L,\ K_w \cap \sigma(\tau) \ne \emptyset\}$ and $G(L) = \bigcup_{w \in \Omega(L)} K_w$. Then $G(L)$ is open in $K$ and there is $L_0$ so that for a given $\epsilon > 0$ we have $L \ge L_0 \Rightarrow H_p(G(L)) \le H_p(\sigma(\tau)) + \epsilon$. Let further $E_w = E(\tau;K_w)$ and observe that since $G(L) \supset \sigma(\tau)$ we have
\[
\sum_{w \in \Omega(L)} E_w = I.
\]
We also have
\[
H_p(G_L) = |\Omega(L)|H_p(K)\lambda^{L_p}.
\]
Let $P_L$ be the orthogonal projection onto $\sum_{w \in \Omega(L)} {\mathbb C}E_w\xi$. and $P_w$ the orthogonal projection onto ${\mathbb C}E_w\xi$. Remark that the non-zero $E_w\xi$ are an orthogonal basis on $P_L\sH$ and rank $P_L \le |\Omega_L|$. We have $[P_L,E_w] = 0$ if $|w| = L$ and $L_1 \ge L_2 \Rightarrow P_{L_1} \ge P_{L_2}$ since each  $E_w$ with $|w| = L_2$ is the sum of $E_{w'}$ with $|w'| = L_1$. Remark also that
\[
\begin{aligned}
\|[P_L,\tau]\| &= \max_{|w|=L} \|[P_L,\tau]E_w\| \\
&= \max_{|w| = L}\|[P_w,E_w\tau]\| \\
&\le \max_{|w|=L} 2\mbox{diam}(\sigma(E_w\tau \mid E_w\sH)) \\
&\le \max_{|w|=L} 2\mbox{diam}(K_2) \le 2\lambda^L\mbox{diam}(K).
\end{aligned}
\]
If $P_L \uparrow P$ then $P\xi = \xi$ and $[P,\tau] = 0$ since $\|[P_L,\tau]\| \to 0$. The vector $\xi$ being cyclic this gives $P = I$, so $P_L \uparrow I$ as $L \to \infty$. Hence we get
\[
\begin{aligned}
|[P_L,\tau]|^-_p &\le p(\mbox{rank}[P_L,\tau])^{1/p} \cdot \|[P_L,\tau]\| \\
&\le p(2|\Omega(L)|)^{1/p} \cdot 2\lambda^L\mbox{diam}(K) \\
&= p \cdot 2^{1+1/p} \cdot \left( \frac {H_p(G_L)}{H_p(K)} \lambda^{-Lp}\right)^{1/p} \cdot \lambda^L\mbox{diam } K \\
&\le p \cdot 2^{1+1/p}(H_p(K))^{-1/p} \cdot \mbox{diam } K(H_p(\sigma(\tau))+\epsilon)^{1/p}.
\end{aligned}
\]
Since $\epsilon > 0$ we get
\[
|[P_L,\tau]|^-_p \le c(H_p(\sigma(\tau)))^{1/p}
\]
where $c = p \cdot 2^{1+1/p}\cdot\mbox{diam } K(H_p(K))^{-1/p}$. Letting $L \to \infty$ we have
\[
k_p^-(\sigma(\tau)) \le c(H_p(\sigma(\tau)))^{1/p}.
\]
\end{proof}

\medskip
\noindent
{\bf Lemma 5.2.}
{\em Assume $\sigma(\tau) \subset K$ and assume that the spectral measure $E(\tau;\cdot)$ is singular w.r.t.\ $H_p$. Then we have
\[
k^-_p(\tau) = 0.
\]
}

\begin{proof}
Since $\tau$ is the orthogonal sum of $n$-tuples with cyclic vector, it suffices to prove the lemma when $\tau$ has a cyclic unit vector $\xi$. The absolute continuity class on $K$ of $E(\tau;\cdot)$ is then the same as the absolute continuity class of the scalar measure $\nu =\langle E(\tau;\cdot)\xi,\xi\rangle$.

Given $\epsilon > 0$ we can find a compact set $C_m$ which is a finite union of $K_w$ so that $H_p(C_m) \le \epsilon$ and $\nu(K\backslash C_m) < 2^{-m}$. Since $\sigma(\tau \mid E(\tau;C_m)\sH) \subset C_m$ the preceding lemma gives that
\[
\begin{aligned}
&k^-_p(\tau \mid E(\tau;C_m)\sH) \le \\
&\le c \cdot H_p(C_m) \le c \cdot \epsilon.
\end{aligned}
\]
On the other hand $\nu(K\backslash C_m) \le 2^{-m}$ gives
\[
\|\xi - E(\tau;C_m)\xi\|^2 \le 2^{-m}.
\]
Since $(\tau)''$ is a maximal abelian von~Neumann algebra in ${\mathcal B}({\mathcal H})$, $\xi$ being cyclic is also separating and we infer $E(\tau;C_m)\ {\overset{w}{\rightarrow}}\ I$ and hence
$E(\tau;C_m)\ {\overset{s}{\rightarrow}}\ I$ as $m \to \infty$. This gives that we can find $A_m \in \sR_1^+$, $A_m \le E(\tau;C_m)$, $|[A_m,\tau]|^-_p \le k_p^-(\tau \mid E(\tau;C_m))$ and $A_m \uparrow I$. It follows that
\[
k^-_p(\tau) \le c \cdot \epsilon
\]
and $\epsilon$ being arbitrary $k^-_p(\tau) = 0$.
\end{proof}

Note that on $K$ the $p$-Hausdorff measure satisfies an Ahlfors regularity condition
\[
C^{-1}r^p \le H_p(B(x,r)) \le Cr^p
\]
if $r \le 1$ for some $C > 0$. The right half of this $H_p(B(x,r)) \le Cr^p$ if $r \le 1$ is the sub-regularity condition where $p > 1$ required in \cite{2} Cor.~$4.7$ to show that $k^-_p(\tau_K) > 0$ where $\tau_K$ is the $n$-tuple of multiplication operators by the coordinate functions in $L^2(K,H_p \mid K)$. Thus we have

\medskip
\noindent
{\bf Lemma 5.3.} ([2]) Assume $p > 1$ then $k^-_p(\tau_K) > 0$.

\medskip
More generally if $\omega \subset K$ is a Borel set let $\tau_{\omega}$ be the $n$-tuple of multiplication operators by the coordinate functions in $L^2(\omega,H_p \mid \omega)$ (this is the same as $\tau_K \mid L^2(\omega,H_p \mid \omega)$ since $L^2(\omega,H_p \mid \omega) \subset L^2(K,H_p \mid K))$. A key part of the proof of the main theorem will be to evaluate $k^-_p(\tau_{\omega})$ for increasingly general $\omega$, along lines similar of Lebesgue measure on ${\mathbb R}^n$ considered in \cite{7}.

We also define a constant $\gamma_K = \frac {(k^-_p(\tau_K))^p}{H_p(K)}$ where $p$ is the Hausdorff dimension of $K$. Lemma~$5.1$ and Lemma~$5.3$ imply that $0 < k^-_p(\tau_K) < \infty$ so that $0 < \gamma_K < \infty$.

\medskip
\noindent
{\bf Theorem 5.1.}
{\em Let $\tau$ be a $n$-tuple of commuting Hermitian operators with $\sigma(\tau) \subset K$ and assume $p > 1$. Then we have
\[
(k^-_p(\tau))^p = \gamma_K \int_K m(x)dH_p(x)
\]
where $m$ is the multiplicity function of $\tau$.
}

\begin{proof}
Using Lemma~$5.2$ and the decomposition $\sH = \sH_{psing} \oplus \sH_{pac}$ the proof reduces to the case when the spectral measure of $\tau$ is absolutely continuous w.r.t.\ $H_p$, that is when $\sH = \sH_{pac}$. In view of Proposition~$2.1$ a further reduction is possible to the case when $\tau$ has finite cyclicity, that is when the multiplicity function is bounded. Since when $\tau$ has a cyclic vector and $H_p$-absolutely continuous spectral measure it is unitarily equivalent to a $\tau_{\omega}$, it means that the proof reduces to the case when $\tau = \tau_{\omega_1} \oplus \dots \oplus \tau_{\omega_m}$ for some Borel sets $\omega_j \subset K$, $1 \le j \le m$. In view of the last assertion in Proposition~$2.1$ the theorem holds for $\tau_{\omega_1} \oplus \dots \oplus \tau_{\omega_m}$ iff it holds for
\[
\tau_{F_{w_1}(\omega_1)} \oplus \dots \oplus \tau_{F_{w_m}(\omega_m)}
\]
where $|w_1| = \dots = |w_m|$ because
\[
\tau_{F_{w_j}(\omega_j)} \simeq F_{w_j}(\tau_{\omega_j}).
\]
We may then choose $|w_j|$ sufficiently large and so that the $F_{w_j}(\omega_j)$, $1 \le j \le m$ are disjoint, which implies that
\[
\tau_{F_{w_1}(\omega_1)} \oplus \dots \oplus \tau_{F_{w_m}(\omega_m)} \simeq \tau_{\omega}
\]
where
\[
\omega = F_{w_1}(\omega_1) \cup \dots \cup F_{w_m(\omega_m)}.
\]
Thus the proof has been reduced to showing that
\[
(k^-_p(\tau_{\omega}))^p = \gamma_KH_p(\omega).
\]

First, assume $\omega$ is a finite union of $K_w$. Since $K_w$ is a disjoint union of $K_{w'}$, with $|w'| \ge |w|$ we may assume
\[
\omega = K_{w_1} \cup \dots \cup K_{w_m}
\]
where $|w_1| = \dots = |w_m|$ and $w_1,\dots,w_m$ are distinct. These $K_{w_j}$ are congruent and using again the last assertion in Prop.~$2.1$ the proof of this case reduces to proving the theorem for $\tau = \tau_{K_w} \otimes I_m$. The multiplicity function is $m$ times the indicator function of $K_w$ so that the right-hand side in the formula we want to prove is
\[
\begin{aligned}
&\gamma_KmH_p(K_w) \\
&= \frac {(k^-_p(\tau_K))^p}{H_p(K)} \cdot m \cdot \lambda^{p|w|} \cdot H_p(K) \\
&= m \cdot (\lambda^{|w|}k_p^-(\tau_K))^p = m(k^-_p(\tau_{K_w}))^p.
\end{aligned}
\]
On the left-hand side we have
\[
(k^-_p(\tau_{K_w} \otimes I_m))^p = (m^{1/p}k^-_p(\tau_{K_w}))^p
\]
by Thm.~$3.1$ which equals the right-hand side.

Next we prove the theorem for $\tau_{\omega}$ when $\omega \subset K$ is a general open subset. Let $\omega^{(L)}$ be the union of the $K_w \subset \omega$ with $|w| \le L$. The $\omega^{(L)}$ are clopen subsets of $K$ and are finite unions of $K_w$ so that the theorem holds for $\tau_{\omega^{(L)}}$ and the theorem for $\tau_{\omega}$ is obtained using Prop.~$2.1$, which gives $k^-_p(\tau_{\omega^{(L)}}) \uparrow k^-_p(\tau_{\omega})$.

Finally let $\omega \subset K$ be a Borel set and let $C$ be compact and $G$ in $K$ be open so that $C \subset \omega \subset G$ and $H_p(G\backslash C) < \epsilon$ for a given $\epsilon > 0$. We have $|k^-_p(\tau_{\omega}) - k^-_p(\tau_G)| \le |k^-_p(\tau_{G\backslash C})| = (\gamma_k\epsilon)^{1/p}$ using the fact that $G\backslash C$ is open in $K$ and Prop.~$2.1$. Thus
\[
\begin{aligned}
&|k^-_p(\tau_{\omega}) - (\gamma_KH_p(\omega))^{1/p}| \\
&\le |k^-_p(\tau_{\omega}) - k^-_p(\tau_G)| + |k^-_p(\tau_G) - (\gamma_KH_p(\omega))^{1/\omega}| \\
&\le (\gamma_K\epsilon)^{1/p} + |(\gamma_KH_p(G))^{1/p} - (\gamma_KH_p(\omega))^{1/p}| \\
&\le (\gamma_K\epsilon)^{1/p} + |(\gamma_K(H_p(\omega)+\epsilon))^{1/p} - (\gamma_KH_p(\omega))^{1/p}|.
\end{aligned}
\]
Since $\epsilon > 0$ was arbitrary, we get $k^-_p(\tau_{\omega}) = (\gamma_KH_p(\omega))^{1/p}$.
\end{proof}

\medskip
\noindent
{\bf Corollary 5.1.}
{\em Assume $\sigma(\tau) \subset K$ and $p > 1$. Then $k^-_p(\tau) = 0$ iff the spectral measure of $\tau$ is singular w.r.t.\ $H_p$.
}

\medskip
\noindent
{\bf Remark 5.1.}
In \cite{8} we showed for a $n$-tuple $\tau$ and a normed ideal $\sJ$ that there is a largest reducing subspace for $\tau$ on which $k_{\sJ}$ vanishes. In the case of commuting $n$-tuples of Hermitian operators and $\sJ = \sC^-_n$ this subspace is the subspace where the spectral measure is singular w.r.t.\ Lebesgue measure. The theorem we proved in this section shows that if $\sigma(\tau) \subset K$ and $p > 1$, then the largest reducing subspace on which $k^-_p$ vanishes for the restriction of $\tau$ is precisely $\sH_{psing}$.

\section{Concluding remarks}
\label{sec6}

\medskip
\noindent
{\bf Remark 6.1.}
It is natural to wonder whether in general
\[
(k^-_p(\tau_1 \oplus \tau_2))^p = (k^-_p(\tau_1))^p + (k_p^-(\tau_2))^p
\]
which would be much more than the ampliation homogeneity we proved. If $p = 1$ this is known to be true \cite{7}. For $1 < p \le \infty$ this is an open problem. While a negative answer would not be surprising, it is certainly desirable to clarify this issue.

\medskip
\noindent
{\bf Remark 6.2.}
To get results for more general self-similar fractals than the Cantor-like $K$ we considered it may be useful to replace $k^-_p$ by ${\tilde k}_p^-$ the variant of $k_p^-$ considered in \cite{7} pages 13--16. This amounts to extending the norms of normed ideals to $n$-tuples, not by the max of norm on the components but by the norm of $(T^*_1T_1 + \dots + T_n^*T_n)^{1/2}$ that is the modulus in the polar decomposition of the column $\left( \begin{matrix}
T_1 \\
\vdots \\
T_m
\end{matrix}
\right)$. This $|\tau|_{\sJ}^{\sim}$ has the advantage over $|\tau|_{\sJ}$ of being invariant under rotations, that is if $(u_{ij})_{1 \le i,j \le n}$ is a unitary matrix en the $n$-tuple $\left(\sum_j u_{ij}T_j\right)_{1 \le i \le n}$ has the same $\sim\sJ$-norm as $\tau = (T_i)_{1 \le i \le n}$. In particular ${\tilde k}^-_p$ may be better suited to handle self-similar sets $K$ when we use more general $F_i(x) = \lambda U_i(X - b(i)) + b(i)$ where $U_i \in O(n)$. In particular it is quite straightforward to use $\sim$-norms in \S 3 and to see that ampliation homogeneity still holds for ${\tilde k}_p^-$ which we record as the next theorem.

\medskip
\noindent
{\bf Theorem 6.3.}
{\em If $\tau$ is a $n$-tuple of bounded operators and $1 \le p \le \infty$ then
\[
{\tilde k}^-_p(\tau \otimes I_m) = m^{1/p}{\tilde k}_p^-(\tau).
\]
}

\medskip
\noindent
{\bf Remark 6.4.}
In \cite{10} we give an extension in another direction to the formula for $k^-_n(\tau)$ in \cite{7} to hybrid perturbations.

config: arxiv


TITLE: If $a_1,a_2,....,a_n$ are in A.P, then prove that:
QUESTION [1 upvotes]: We have to prove that $\frac{1}{a_1a_n} +\frac{1}{a_2a_{n-1}}+....+\frac{1}{a_{n-1}a_2} +\frac{1}{a_aa_1} = \frac{2}{a_1+a_n}(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n})$
The middle term will depend on the fact that $n$ is even or odd. So, I can't understand how to proceed. If I add the common terms in $L.H.S$, then my final term will be the central term.

REPLY [0 votes]: Case 1: $n=2m-1$. Note that:
$$\frac{1}{a_1a_{2m-1}}=\frac{1}{2a_m}\left(\frac{1}{a_{m}-(m-1)d}+\frac{1}{a_{m}+(m-1)d}\right).$$
Hence:
$$\sum_{k=1}^n\frac 1{a_ka_{n-k+1}}=\sum_{k=1}^{2m-1}\frac 1{a_ka_{2m-k}}=\\
2\cdot \frac{1}{2a_m}\sum_{k=1}^{m} \left(\frac{1}{a_{m}-(k-1)d}+\frac{1}{a_{m}+(k-1)d}\right)=\\
\frac{2}{a_1+a_{2m-1}}\sum_{k=1}^{2m-1} \frac{1}{a_{k}}=\frac{2}{a_1+a_n}\sum_{k=1}^n \frac{1}{a_k}.$$
Case 2: $n=2m$. You can do it similarly.


text:
\begin{document}

\title {The enclosure method for the heat equation using time-reversal invariance for a wave equation}
\author{Masaru IKEHATA\footnote{
Laboratory of Mathematics,
Graduate School of Engineering,
Hiroshima University,
Higashihiroshima 739-8527, JAPAN}}

\maketitle

\begin{abstract}
The heat equation does not have {\it time-reversal invariance}.  However, using a solution
of an associated wave equation which has time-reversal invariance, one can establish an explicit
extraction formula of the {\it minimum sphere} that is centered at an arbitrary given point and encloses an unknown
cavity inside a heat conductive body.
The data employed in the formula consist of a special heat flux depending on
a large parameter prescribed on the surface of the body over an
arbitrary fixed finite time interval and the corresponding temperature field.
The heat flux never blows up as the parameter tends to infinity. This is different from a previous formula for the heat equation
which also yields the minimum sphere.  In this sense, the prescribed heat flux is {\it moderate}.

\noindent
AMS: 35R30, 35L05, 35K05

\noindent KEY WORDS: enclosure method, inverse obstacle problem, cavity, heat equation, wave equation, Kirchhoff's formula,
time-reversal invariance, time-reversal operation,
non-destructive testing.
\end{abstract}

\section{Introduction}
This paper is concerned with the {\it methodology} for so-called inverse obstacle problems
which aims at reconstructing or extracting an unknown discontinuity such as an
obstacle, inclusion, cavity, etc. embedded or occurred in a reference medium for various observation data.
As one of direct analytical methods which have an exact mathematical base the author
has introduced the {\it enclosure method} in \cite{E00} and \cite{E001} using infinitely many data and a single set of data, respectively.
The enclosure method enables us to find a domain that encloses the unknown discontinuity.

Originally the method treats a signal which does not depend on a time variable.
Clearly, this restricts the possible application of the enclosure method since
there are a lot of inverse problems which employs various data depending on a time variable.
However, in \cite{I4} the author found an idea how to treat a signal which depends on a time variable
in a {\it bounded interval}.
Now we have various realizations of this {\it time domain enclosure method}
in three-space dimensions for the signals governed by
the heat and wave equations \cite{IW00, IFR, IEO2, IEO3, Iwall, IK1, IK2, IK3},
a system of equations in a linearized viscoelastic body \cite{II},
the Maxwell system \cite{IMax, IMax2} and references therein.

Quite recently, in \cite{EV} the author considered an inverse obstacle problem for the wave
governed by the wave equation in a bounded domain and introduced an idea which combines
the {\it time-reversal invariance} of the wave equation with the newest version of enclosure method \cite{EIV}
which employs a {\it single point} on the graph of the {\it response operator} associated with the inverse obstacle problem.
Note that the idea itself in \cite{EIV} has been applied also to a system of equations in the linear theory of thermoelasticity in \cite{Ithermo}.

The key point of the idea introduced in \cite{EV} is:  making a {\it time-reversal operation} on the solution
of the free space wave equation supported on an arbitrary given ball at the initial time.
This gives us an explicit input Neumann data depending on the ball such that
the corresponding Dirichlet data as the output over a finite time interval yields the minimum sphere having the same center point as the ball
and enclosing the obstacle.

It is curious to know whether there is an analogous approach to inverse obstacle problems
for the signal governed by the heat equation.  However, one can easily understand that the heat equation does not have time-reversal invariance
and thus everyone may think that there is no hope to establish any corresponding formula.

The aim of this paper is to show that the hope can be realized {\it partially} with the help
of time-reversal invariance of an {\it associated wave equation with a growing speed}.
One can establish an explicit
extraction formula of the minimum sphere that is centered at an arbitrary given point and encloses an unknown
cavity inside a heat conductive body.

The data employed in the formula consists of a special heat flux depending on
a large parameter prescribed on the surface of the body over an
arbitrary fixed finite time interval and the corresponding temperature field.
The heat flux is given by taking the time-reversal operation of the normal derivative of a solution of the associated wave equation.
This idea, that is, a combination of the enclosure method and prescribing a heat flux coming from a wave equation
was initiated in \cite{PW}.

Remarkably enough, the flux never blows up as the parameter tends to infinity.
This is different from a previous formula established
in \cite{IK2} for the heat equation with a discontinuity coefficient, which also yields the minimum sphere.
The flux used there depends on a large parameter, this is a common point, however, blows up on the whole surface of the body
as the parameter tends to infinity.
In this sense, the prescribed heat flux in this paper is {\it moderate}.
In fairness, however, it should be pointed out that the formula using the special heat flux in this paper is not a complete answer to the original question itself since the flux depends on a parameter.  This is the meaning of the partial realization of the hope mentioned above.

Note that the time-reversal operation appears in the BC-method \cite{BEL, B2}, related works \cite{BKLS, DKL, O}
and a numerical approach \cite{dBK}.  However, these are concerned with inverse problems for the wave equation.
To the author's best knowledge, no one considers any application of time-reversal invariance to
inverse obstacle problem for the heat equation.

Now let us describe the inverse obstacle problem to be considered.
We consider the problem for the signal governed by the heat equation formulated as
below. Let $\Omega$ be a bounded domain of $\Bbb R^3$ with $C^2$-boundary.
Let $D$ be a nonempty bounded open set of $\Bbb R^3$ with $C^2$-boundary such that
$\overline D\subset\Omega$ and $\Omega\setminus\overline D$ is connected.
The set $D$ is our target which is a mathematical model of an unknown cavity in a heat conductive body $\Omega$.

Given an arbitrary positive number $T$ and $f=f(x,t)$, $(x,t)\in\partial\Omega\times\,[0,\,T]$,
let $u=u_{f}(x,t)$, $(x,t)\in(\Omega\setminus\overline D)\times\,[0,\,T]$ denote the solution of the following initial
value problem for the heat equation
$$\left\{
\begin{array}{ll}
\displaystyle
(\partial_t-\Delta)u=0 & \text{in $(\Omega\setminus\overline D)\times\,]0,\,T[$,}
\\
\\
\displaystyle
u(x,0)=0 & \text{in $\Omega\setminus\overline D$,}\\
\\
\displaystyle
\frac{\partial u}{\partial\nu}=0 & \text{on $\partial D\times\,]0,\,T[$,}\\
\\
\displaystyle
\frac{\partial u}{\partial\nu}=f(x,t) & \text{on $\partial\Omega\times\,]0,\,T[$.}
\end{array}
\right.
\tag {1.1}
$$
The $f$, $u$ and $\partial_t u$ belong to $L^2(0,T;H^{-1/2}(\partial\Omega))$, $L^2(0,\,T;H^1(\Omega\setminus\overline D))$
and $L^2(0,\,T;H^1(\Omega\setminus\overline D)')$, respectively.
The $u$ satisfies, for a positive constant $C_T$ beging independent of $f$
$$\displaystyle
\Vert u\Vert_{L^2(0,T;H^1(\Omega\setminus\overline D))}
+\Vert\partial_tu\Vert_{L^2(0,T;H^1(\Omega\setminus\overline D)')}
+\Vert u(\,\cdot\,,T)\Vert_{L^2(\Omega\setminus\overline D)}
\le C_T\Vert f\Vert_{L^2(0,T;H^{-1/2}(\partial\Omega))}.
\tag {1.2}
$$
See \cite{DL} for more information.

The inverse obstacle problem to be considered here is: find an extraction formula
of information about the geometry of $D$ from $u_f(x,t)$ for $(x,t)\in\partial\Omega\times\,]0,\,T[$
corresponding to finitely or infinitely many $f$.

Let $B$ be an open ball centered at an arbitrary point $p\in\Bbb R^3$ with radius $\eta$.
The aim of this paper is to show that: there exists a suitable Neumann data $f$
depending on $B$ and a large parameter $\tau$ such that

$\bullet$  $f=O(1)$ in $L^2(0,T; H^{-1/2}(\partial\Omega))$ as $\tau\rightarrow\infty$;

$\bullet$  the $u_f$ on $\partial\Omega\times\,]0,\,T[$ with an aribitrary fixed $T$
as $\tau\rightarrow\infty$ yields $R_D(p)$, where
$$\displaystyle
R_D(p)=\sup_{x\in D}\,\vert x-p\vert.
$$
Note that the quantity $R_D(p)$ gives the radius of the minimum sphere centered at $p$
that encloses the unknown cavity $D$.

From (1.2) we have, for such $f$ as $\tau\rightarrow\infty$
$$\displaystyle
\Vert u_f\Vert_{L^2(0,T;H^1(\Omega\setminus\overline D))}
+\Vert\partial_tu_f\Vert_{L^2(0,T;H^1(\Omega\setminus\overline D)')}
+\Vert u_f(\,\cdot\,,T)\Vert_{L^2(\Omega\setminus\overline D)}
=O(1).
\tag {1.3}
$$
In this sense, the temperature field $u_f$ never blows up inside the body
as $\tau\rightarrow\infty$.

Note that from a combination of \cite{IK1} and \cite{IK2}, one can easily show that:
if the heat flux
$f$ takes the form
$$\begin{array}{lll}
\displaystyle
f(x,t)=\varphi(t)\frac{\partial}{\partial\nu}v_{\tau}(x;p), & x\in\partial\Omega, & 0<t<T,
\end{array}
$$
where
$$\displaystyle
v_{\tau}(x;p)=\left\{
\begin{array}{ll}
\displaystyle
\frac{\sinh\sqrt{\tau}\,\vert x-p\vert}{\vert x-p\vert} & \text{ if $x\in\Bbb R^3\setminus\{p\}$,}\\
\\
\displaystyle
\sqrt{\tau} & \text{if $x=p$,}
\end{array}
\right.
$$
and $\varphi\in L^2(0,\,T)$ satisfies, say $\varphi(t)\sim t^m$ with an integer $m\ge 0$ as $t\downarrow 0$, then
one can extract $R_D(p)$ from $u_f$ on $\partial\Omega$ over time interval $]0,\,T[$ as $\tau\rightarrow\infty$.
Note that the function $v_{\tau}(x;p)$ of $x\in\Bbb R^3$ is an entire solution of the modified Helmholtz equation
$(\Delta-\tau)v=0$.  However, the $f$ given above blows up as $\tau\rightarrow\infty$ in the order of exponential
everywhere on $\partial\Omega$ if $p\in\Omega$, which is the most interesting case.  Thus, this choice does not satisfy
the requirement of moderateness of the state as $\tau\rightarrow\infty$.

\section{Statement of the result and its proof}

The construction of $f$ is the following.

First solve the initial value problem for the wave equation:
$$\displaystyle
\left\{
\begin{array}{ll}
(\partial_s^2-\Delta)v=0 & \text{in $\Bbb R^3\times\,]0,\,\infty[$,}
\\
\\
\displaystyle
v(x,0)=0 & \text{in $\Bbb R^3$,}
\\
\\
\displaystyle
\partial_s v(x,0)=\Psi_B(x) & \text{in $\Bbb R^3$,}
\end{array}
\right.
\tag {2.1}
$$
where
$$\begin{array}{ll}
\displaystyle
\Psi_B(x)=(\eta-\vert x-p\vert)\chi_B(x), & x\in\Bbb R^3
\end{array}
$$
and the function $\chi_B(x)$ denotes the characteristic function of $B$.
Note that one can find the $v$ in the class
$$\displaystyle
C^2([0, \infty[, L^2(\Bbb R^3))\cap C^1([0,\,\infty[, H^1(\Bbb R^3))\cap C([0,\,\infty[, H^2(\Bbb R^3)).
$$
This is an application of the theory of $C_0$-semigroups \cite{Y}.
Note that the $v$ has an explicit form which is a consequence of {\it Kirchhoff's formula} \cite{ES}, that is
$$\displaystyle
v(x,s)=\frac{1}{4\pi s}
\,\int_{\partial B_s(x)}\,\Psi_B(y)dS_y,
$$
where $B_s(x)=\{y\in\Bbb R^3\,\vert\,\vert y-x\vert<s\}$.

Let $\tau>0$ and define
$$\begin{array}{lll}
\displaystyle
f_{B,\,T,\,\tau}(x,t)=\frac{1}{\sqrt{\tau}}\cdot\frac{\partial}{\partial\nu}v(x,\sqrt{\tau}(T-t)),
& x\in\partial\Omega, & 0<t<T.
\end{array}
$$
Note that the function $v_{\tau}(x,t), x\in\Bbb R^3, t\in\,[0,\,T]$ defined by
$$\displaystyle
v_{\tau}(x,t)=\frac{1}{\sqrt{\tau}}v(x,\sqrt{\tau}t),
$$
satisfies
$$\displaystyle
\left\{
\begin{array}{ll}
(\partial_t^2-\tau\Delta)v_{\tau}=0 & \text{in $\Bbb R^3\times\,]0,\,T[$,}
\\
\\
\displaystyle
v_{\tau}(x,0)=0 & \text{in $\Bbb R^3$,}
\\
\\
\displaystyle
\partial_t v_{\tau}(x,0)=\Psi_B(x) & \text{in $\Bbb R^3$.}
\end{array}
\right.
\tag {2.2}
$$
Thus one has the expression
$$\begin{array}{lll}
\displaystyle
f_{B,\,T,\,\tau}(x,t)=\frac{\partial}{\partial\nu}v_{\tau}(x,T-t),
& x\in\partial\Omega, & 0<t<T.
\end{array}
$$
This means that $f_{B,\,T,\,\tau}$ is the time-reversal mirror \cite{FWCM} on $\partial\Omega$
of the wave $v_{\tau}(x,t)$.

As can be seen in \cite{PW} it is not difficult to show that
$$\displaystyle
\Vert f_{B,\,T,\,\tau}(\,\cdot\,,t)\Vert_{H^{1/2}(\partial\Omega)}
\le C_{\Omega,B}\sqrt{(T-t)^2+\frac{3}{\tau}}
$$
and hence $f_{B,\,T,\,\tau}=O(1)$ in $L^2(0,T; H^{-1/2}(\partial\Omega))$ as $\tau\rightarrow\infty$.
Therefore from (1.3) we have, for $f=f_{B,\,T,\,\tau}$
$$\displaystyle
\Vert u_f(\,\cdot\,,T)\Vert_{L^2(\Omega\setminus\overline D)}=O(1).
\tag {2.3}
$$

Using $u_f$ with $f=f_{B,\,T,\,\tau}$, we define the indicator function
by the formula
$$\displaystyle
I_{\partial\Omega}(\tau;B,T)
=\int_{\partial\Omega}(w_f-w_f^*)\frac{\partial w_f^*}{\partial\nu}\,dS,
$$
where
$$\displaystyle
\left\{
\begin{array}{ll}
\displaystyle
w_f(x,\tau)=\int_0^T e^{-\tau t} u_f(x,t)dt, & x\in\Omega\setminus\overline D,
\\
\\
\displaystyle
w_f^*(x,\tau)=\int_0^T e^{-\tau t} v_{\tau}(x,T-t)\,dt, & x\in\Bbb R^3.
\end{array}
\right.
$$
Now we state the result of this paper.

\proclaim{\noindent Theorem 2.1.}
Let $\eta$ satisfy
$$\displaystyle
\eta+2R_D(p)>R_{\Omega}(p).
\tag {2.4}
$$

(i)  There exists a positive number $\tau_0$ such that $I_{\partial\Omega}(\tau;B,T)>0$
for all $\tau\ge\tau_0$ and
we have
$$\displaystyle
\lim_{\tau\rightarrow\infty}
\frac{1}{\sqrt{\tau}}\left(\log I_{\partial\Omega}(\tau;B,T)+2\tau T\right)
=2(\eta+R_D(p)).
\tag {2.5}
$$

(ii) We have
$$\displaystyle
\lim_{\tau\rightarrow\infty}
e^{-\sqrt{\tau}\,T}e^{2\tau T}I_{\partial\Omega}(\tau;B,T)
=
\left\{
\begin{array}{ll}
0 & \text{if $T>2(\eta+R_D(p))$,}
\\
\\
\displaystyle
\infty & \text{if $T<2(\eta+R_D(p))$.}
\end{array}
\right.
$$

\endproclaim

{\bf\noindent  Remark 2.2.}
The $\eta$ which is the radius of ball $B$ can not be arbitrary small since we have the constraint (2.4)
and it is really constraint if $2R_D(p)<R_{\Omega}(p)$.
If $2R_D(p)\ge R_{\Omega}(p)$, then $\eta$ can be arbitrary small.
To cover the both cases, one can choose an arbitrary $\eta$ such that $\eta\ge R_{\Omega}(p)$.
This is an advantage of using the initial data $\Psi_B$ in (2.1) with finitely extended support $\overline B$ not like $\{p\}$.

{\it\noindent Proof.}
Note that the assertion (ii) is a direct consequence from (i) since there is no restriction on $T$ in (i).
Let $w^*=w_f^*$.
One can write
$$\displaystyle
w^*(x,\tau)=e^{-\tau T}w_0(x,\tau),
$$
where
$$\displaystyle
w_0(x,\tau)=\int_0^T e^{\tau s}v_{\tau}(x,s)ds.
$$
From (2.2) we have
$$\displaystyle
(\Delta-\tau)w_0
+\frac{1}{\tau}\Psi_B(x)
=\frac{e^{\tau T}}{\tau}
(\partial_tv_{\tau}(x,T)-\tau v_{\tau}(x,T)).
$$
Thus, $w^*$ satisfies
$$\begin{array}{ll}
\displaystyle
(\Delta-\tau)w^*+\frac{1}{\tau}
(\sqrt{\tau} v(x,\sqrt{\tau}\,T)-\partial_sv(x,\sqrt{\tau}\,T))
=e^{-\tau T}F_0(x), & x\in\Bbb R^3
\end{array}
\tag {2.6}
$$
where
$$\displaystyle
F_0(x)=-\frac{\Psi_B(x)}{\tau}.
\tag {2.7}
$$
One the other hand, from (1.1) we see that $w=w_f$ satisfies
$$\displaystyle
\left\{
\begin{array}{ll}
(\Delta-\tau)w=e^{-\tau T}F(x) & \text{in $\Omega\setminus\overline D$,}
\\
\\
\displaystyle
\frac{\partial w}{\partial\nu}=\frac{\partial w^*}{\partial\nu} & \text{on $\partial\Omega$,}
\\
\\
\displaystyle
\frac{\partial w}{\partial\nu}=0 & \text{on $\partial D$,}
\end{array}
\right.
\tag {2.8}
$$
where
$$\begin{array}{ll}
\displaystyle
F(x)=u_f(x,T), & x\in\Omega\setminus\overline D.
\end{array}
\tag {2.9}
$$
Hereafter let $\tau$ satisfy
$$\displaystyle
\sqrt{\tau}\,T-\eta\ge R_{\Omega}(p),
$$
that is
$$\displaystyle
\tau\ge \left(\frac{\eta+R_{\Omega}(p)}{T}\right)^2.
$$
Then, we have $\Omega\subset B_{\sqrt{\tau}\,T-\eta}(p)$.
Since we have, for all $s>\eta$
$$\displaystyle
\text{supp}\,v(\,\cdot\,,s)\cup\text{supp}\,\partial_s v(\,\cdot\,,s)
\subset \Bbb R^3\setminus B_{s-\eta}(p),
\tag {2.10}
$$
from (2.6) we have
$$\begin{array}{ll}
\displaystyle
(\Delta-\tau)w^*=e^{-\tau T}F_0(x),
& \text{$x\in\Omega$}.
\end{array}
\tag {2.11}
$$
Note that (2.10) is a consequence of Kirchhoff's formula.

Set $R=w-w^*$.  Then from (2.8) and (2.11) we have
$$\displaystyle
\left\{
\begin{array}{ll}
(\Delta-\tau)R=e^{-\tau T}(F(x)-F_0(x)) & \text{in $\Omega\setminus\overline D$,}
\\
\\
\displaystyle
\frac{\partial R}{\partial\nu}=0 & \text{on $\partial\Omega$,}
\\
\\
\displaystyle
\frac{\partial R}{\partial\nu}=-\frac{\partial w^*}{\partial\nu} & \text{on $\partial D$.}
\end{array}
\right.
$$
Then, we obtain the following decomposition formula
which corresponds to (2.6) in \cite{PW}:
$$\displaystyle
I_{\partial\Omega}(\tau;B)
=J_h(\tau)+E_h(\tau)+{\cal R}_h(\tau),
\tag {2.12}
$$
where
$$
\displaystyle
\left\{
\begin{array}{l}
\displaystyle
J_h(\tau)=\int_D(\vert\nabla w^*\vert^2+\tau\vert w^*\vert^2)\,dx,\\
\\
\displaystyle
E_h(\tau)
=\int_{\Omega\setminus\overline D}(\vert\nabla R\vert^2+\tau\vert R\vert^2)\,dx
\end{array}
\right.
$$
and
$$\displaystyle
{\cal R}_h(\tau)
=e^{-\tau T}
\left\{\int_DF_0 w^*dx+\int_{\Omega\setminus\overline D}FRdx+
\int_{\Omega\setminus\overline D}
(F_0-F)w^*dx\right\}.
$$
And also similarly to \underline{Lemma 2.2} in \cite{PW} one can derive the estimate
$$\displaystyle
E_h(\tau)=O(\tau J_h(\tau)+e^{-2\tau T}).
\tag {2.13}
$$
From (2.6) and (2.7) we have the expression
$$\displaystyle
w^*=\frac{1}{\tau}\left(w_1^*+e^{-\tau T}w_R^*\right),
\tag {2.14}
$$
where
$$\begin{array}{ll}
\displaystyle
w_1^*(x,\tau)=\frac{1}{4\pi}\int_{\Bbb R^3}\frac{e^{-\sqrt{\tau}\vert x-y\vert}}{\vert x-y\vert}\,
(\sqrt{\tau}v(y,\sqrt{\tau}\,T)-\partial_s v(y,\sqrt{\tau}\,T))dy,
& x\in\Bbb R^3
\end{array}
\tag {2.15}
$$
and $w_R^*=w_R^*(x,\tau)$ satisfies
$$\begin{array}{ll}
\displaystyle
(\Delta-\tau)w_R^*+\Psi_B=0, & x\in\Bbb R^3.
\end{array}
$$
By integration by parts we have immediately, as $\tau\rightarrow\infty$,
$$
\displaystyle
\sqrt{\tau}\Vert w_R^*\Vert_{L^2(\Bbb R^3)}+\Vert\nabla w_R^*\Vert_{L^2(\Bbb R^3)}
=O(1).
\tag {2.16}
$$

Here we note that $v$ satisfies
$$\displaystyle
\text{supp}\,v(\,\cdot\,,s)\cup\text{supp}\,\partial_s v(\,\cdot\,,s)
\subset B_{s+\eta}(p)
\tag {2.17}
$$
for all $s\ge 0$, which is also a consequence of Kirchhoff's formula.
From (2.10) and (2.17) together with \underline{Propositions 3.3 and 3.4} in \cite{EV} we have
the explicit expression of the right-hand side on (2.15):
$$\displaystyle
w_1^*(x,\tau)
=\frac{1}{\tau}
({\cal H}_+(\sqrt{\tau};\sqrt{\tau}\,T,\eta)
+{\cal H}_{-}(\sqrt{\tau};\sqrt{\tau}\,T,\eta))
\,\frac{\sinh\sqrt{\tau}\vert x-p\vert}{\vert x-p\vert},
\tag {2.18}
$$
for all $x\in B_{\sqrt{\tau}\,T-\eta}(p)\setminus\{p\}$, where ${\cal H}_+(\sqrt{\tau};\sqrt{\tau}\,T,\eta)$
and ${\cal H}_{-}(\sqrt{\tau};\sqrt{\tau}\,T,\eta)$ are given by the following ${\cal H}_{+}(\tau;T,\eta)$ and ${\cal H}_{-}(\tau;T,\eta)$, respectively
in which $\tau$ and $T$ are replaced with
$\sqrt{\tau}$ and $\sqrt{\tau}\,T$, respectively:
$$\left\{
\begin{array}{l}
\displaystyle
{\cal H}_{+}(\tau;T,\eta)
=f_{\tau}(T)e^{-\tau T}-f_{\tau}(T+\eta)e^{-\tau (T+\eta)},
\\
\\
\displaystyle
{\cal H}_{-}(\tau;T,\eta)
=g_{\tau}(T-\eta)e^{-\tau (T-\eta)}-g_{\tau}(T)e^{-\tau T}.
\end{array}
\right.
$$
\underline{Here $f_{\tau}(\,\cdot\,)$ and $g_{\tau}(\,\cdot\,)$ are
cubic polynomials given by}
$$\begin{array}{ll}
\displaystyle
f_{\tau}(\xi)
&
\displaystyle
=\frac{\tau}{6}\xi^3+\left\{1-\frac{\tau}{4}(\eta+2T)\right\}\xi^2+\left\{\frac{1}{2}\tau T(\eta+T)-(\eta+2T)+\frac{2}{\tau}\right\}\xi
\\
\\
\displaystyle
&
\displaystyle
\,\,\,
+\left\{\frac{1}{12}\tau(\eta-2T)(\eta+T)^2
+T(\eta+T)-\frac{\eta+2T}{\tau}+\frac{2}{\tau^2}\right\}
\end{array}
$$
and
$$\begin{array}{ll}
\displaystyle
g_{\tau}(\xi)
&
\displaystyle
=-\frac{\tau}{6}\xi^3-\left\{1+\frac{\tau}{4}(\eta-2T)\right\}\xi^2+\left\{\frac{1}{2}\tau T(\eta-T)-(\eta-2T)-\frac{2}{\tau}\right\}\xi\\
\\
\displaystyle
&
\displaystyle
\,\,\,
+\left\{\frac{1}{12}\tau(\eta+2T)(\eta-T)^2
+T(\eta-T)-\frac{\eta-2T}{\tau}-\frac{2}{\tau^2}\right\}.
\end{array}
$$
Note that, in the derivation of (2.18) \underline{we} fully made use of Kirchhoff's formula which gives the solution form
of (2.1) in the $(x,t)$-space.
Furthermore we have
$$\displaystyle
g_{\tau}(T-\eta)=\frac{1}{\tau}\left(\eta-\frac{2}{\tau}\right).
\tag {2.19}
$$
See (3.8) in \cite{EV}.
Using these, we have
$$\displaystyle
e^{\sqrt{\tau}(\sqrt{\tau}\,T-\eta)}(
{\cal H}_+(\sqrt{\tau};\sqrt{\tau}\,T,\eta)
+{\cal H}_{-}(\sqrt{\tau};\sqrt{\tau}\,T,\eta))
=\frac{1}{\sqrt{\tau}}
\left(\eta+O(\tau^{-1/2})\right).
$$
Then, applying \underline{Lemma 2.4} in \cite{EV} to the right-hand side on (2.18), we obtain, as
$\tau\rightarrow\infty$
$$\displaystyle
e^{2\tau T-2\sqrt{\tau}\,\eta}\left(\Vert w_1^*\Vert_{L^2(U)}^2+\Vert\nabla w_1^*\Vert_{L^2(U)}^2\right)=O(\tau^{\mu_1}e^{2\sqrt{\tau}\,R_U(p)})
\tag {2.20}
$$
and
$$\displaystyle
e^{2\tau T-2\sqrt{\tau}\,\eta}
\tau^{\mu_2}e^{-2\sqrt{\tau}\,R_U(p)}\,\Vert w_1^*\Vert_{L^2(U)}^2\ge C,
\tag {2.21}
$$
where $U$ is an arbitrary bounded open subset of $\Bbb R^3$; $p$ an arbitrary point in $\Bbb R^3$;
$\mu_1$ and $\mu_2$ are real numbers and $C$ a positive constant; it is assumed that
$\partial U$ is Lipschitz for (2.21).

From (2.14), (2.16) and (2.20) we obtain, as $\tau\rightarrow\infty$
$$\begin{array}{l}
\displaystyle
\,\,\,\,\,\,
e^{2\tau T-2\sqrt{\tau}\,\eta}(\Vert w^*\Vert_{L^2(U)}^2+\Vert\nabla w^*\Vert_{L^2(U)}^2)\\
\\
\displaystyle
=O(\tau^{-2}\tau^{\mu_1}e^{2\sqrt{\tau}\,R_U(p)})+O(\tau^{-2}e^{2\tau T-2\sqrt{\tau}\,\eta}e^{-2\tau T})\\
\\
\displaystyle
=O(\tau^{\mu_1-2}e^{2\sqrt{\tau}\,R_U(p)})(1+\tau^{-\mu_1}e^{-2\sqrt{\tau}\,\eta}e^{-2\sqrt{\tau}\,R_U(p)}))\\
\\
\displaystyle
=O(\tau^{\mu_1-2}e^{2\sqrt{\tau}\,R_U(p)}).
\end{array}
\tag{2.22}
$$
From (2.14), (2.16) and (2.21) we obtain
$$\begin{array}{l}
\displaystyle
\,\,\,\,\,\,
e^{2\tau T-2\sqrt{\tau}\,\eta}e^{-2\sqrt{\tau}\,R_U(p)}\Vert w^*\Vert_{L^2(U)}^2
\\
\\
\displaystyle
\ge \tau^{-2}\left(\frac{1}{2}e^{2\tau T-2\sqrt{\tau}\,\eta}e^{-2\sqrt{\tau}\,R_U(p)}\Vert w_1^*\Vert_{L^2(U)}^2
-e^{2\tau T-2\sqrt{\tau}\,\eta}e^{-2\sqrt{\tau}\,R_U(p)}e^{-2\tau T}\Vert w_R^*\Vert_{L^2(U)}^2
\right)\\\\
\displaystyle
\ge
\tau^{-2}\left(\frac{C}{2}\,\tau^{-\mu_2}
-C_1e^{-2\sqrt{\tau}\,\eta}e^{-2\sqrt{\tau}\,R_U(p)}\tau^{-1}\right)
\\
\\
\displaystyle
\ge C_2\tau^{-(2+\mu_2)},
\end{array}
\tag {2.23}
$$
where $C_1$, $C_2$ are positive constants being independent of all $\tau\ge\tau_0$ and $\tau_0$ is a large positive constant.

From (2.22) with $U=D$ we have
$$\displaystyle
e^{2\tau T-2\sqrt{\tau}\,\eta}e^{-2\sqrt{\tau}\,R_D(p)}J_h(\tau)
=O(\tau^{\mu_1}).
\tag {2.24}
$$
Then from (2.13) we have
$$\begin{array}{ll}
\displaystyle
e^{2\tau T-2\sqrt{\tau}\,\eta}e^{-2\sqrt{\tau}\,R_D(p)}E_h(\tau)
&
\displaystyle
=O(\tau^{\mu_1+1})
+O(e^{-2\sqrt{\tau}\,\eta}e^{-2\sqrt{\tau}\,R_D(p)})
\\
\\
\displaystyle
&
\displaystyle
=O(\tau^{\mu_1+1}).
\end{array}
\tag {2.25}
$$
Moreover, applying (2.3) to (2.9) and using (2.7), we have
$$\begin{array}{l}
\displaystyle
\,\,\,\,\,\,
\left\vert{\cal R}_h(\tau)\right\vert
\\
\\
\displaystyle
\le
e^{-\tau T}
\left(\Vert F_0\Vert_{L^2(D)}\Vert w^*\Vert_{L^2(D)}+\Vert F\Vert_{L^2(\Omega\setminus\overline D)}
\Vert R\Vert_{L^2(\Omega\setminus\overline D)}
+\Vert F_0-F\Vert_{L^2(\Omega\setminus\overline D)}\Vert w^*\Vert_{L^2(\Omega\setminus\overline D)}\right)
\\
\\
\displaystyle
\le
Ce^{-\tau T}
\left(
\tau^{-1}\Vert w^*\Vert_{L^2(D)}+
\Vert R\Vert_{L^2(\Omega\setminus\overline D)}
+\Vert w^*\Vert_{L^2(\Omega\setminus\overline D)}\right).
\end{array}
$$
Then, from (2.24), (2.25) and (2.22) with $U=\Omega\setminus\overline D$
we obtain
$$
\begin{array}{l}
\displaystyle
\,\,\,\,\,\,
e^{\tau T-\sqrt{\tau}\,\eta}\left\vert{\cal R}_h(\tau)\right\vert
\\
\\
\displaystyle
=
Ce^{-\tau T}
\left(
O(\tau^{\frac{\mu_1}{2}-2}e^{\sqrt{\tau}\,R_D(p)})
+
O(\tau^{\frac{\mu_1}{2}-1})e^{\sqrt{\tau}\,R_{D}(p)}
+O(\tau^{\frac{\mu_1}{2}-1}e^{\sqrt{\tau}\,R_{\Omega\setminus\overline D}(p)})
\right)
\\
\\
\displaystyle
=
Ce^{-\tau T}
\left(
O(\tau^{\frac{\mu_1}{2}-1})e^{\sqrt{\tau}\,R_{D}(p)}
+O(\tau^{\frac{\mu_1}{2}-1}e^{\sqrt{\tau}\,R_{\Omega}(p)})
\right).
\end{array}
$$
This yields
$$
\begin{array}{l}
\displaystyle
\,\,\,\,\,\,
e^{2\tau T-2\sqrt{\tau}\,\eta}e^{-2\sqrt{\tau}\, R_D(p)}\left\vert{\cal R}_h(\tau)\right\vert
\\
\\
\displaystyle
=
O(\tau^{\frac{\mu_1}{2}-1}e^{-\sqrt{\tau}\,\eta}e^{-\sqrt{\tau}\,R_{D}(p)})
+O(\tau^{\frac{\mu_1}{2}-1}e^{-\sqrt{\tau}\,(\eta+2R_D(p)-R_{\Omega}(p)}).
\end{array}
\tag {2.26}
$$
Now we make use of the assumption (2.4).
Then, from (2.26) one concludes
$$\displaystyle
e^{2\tau T-2\sqrt{\tau}\,\eta}e^{-2\sqrt{\tau}\, R_D(p)}{\cal R}_h(\tau)
=O(\tau^{-\infty}).
\tag {2.27}
$$
Applying this together with (2.24) and (2.25) to (2.12), we obtain, as $\tau\rightarrow\infty$
$$\displaystyle
e^{2\tau T-2\sqrt{\tau}\,\eta}e^{-2\sqrt{\tau}\, R_D(p)}I_{\partial\Omega}(\tau;B,T)
=O(\tau^{\mu_1+1}).
\tag {2.28}
$$

From (2.23) with $U=D$ we have
$$\begin{array}{l}
\displaystyle
\,\,\,\,\,\,
e^{2\tau T-2\sqrt{\tau}\,\eta}e^{-2\sqrt{\tau}\, R_D(p)}\tau^{\mu_2}J_h(\tau)
\\
\\
\displaystyle
\ge e^{2\tau T-2\sqrt{\tau}\,\eta}e^{-2\sqrt{\tau}\, R_D(p)}\tau^{2+\mu_2}\Vert w^*\Vert_{L^2(D)}^2
\\
\\
\displaystyle
\ge C_2.
\end{array}
$$
Thus this together with (2.12) and (2.27) gives, for all $\tau\ge\tau_0$
for a sufficiently large $\tau_0>0$
$$\displaystyle
e^{2\tau T-2\sqrt{\tau}\,\eta}e^{-2\sqrt{\tau}\, R_D(p)}\tau^{\mu_2}I_{\partial\Omega}(\tau;B,T)
\ge \frac{C_2}{2}.
\tag {2.29}
$$

Now the assertion (i) follows from (2.28) and (2.29).

\noindent
$\Box$

{\bf\noindent Remark 2.3.}
Let $T=2(\eta+R_D(p))$.  This is the excluded case in Theorem 2.1 (ii).
It follows from (2.28) that, as $\tau\rightarrow\infty$
$$\displaystyle
e^{-\sqrt{\tau}\,T}e^{2\tau T}I_{\partial\Omega}(\tau;B,T)=O(\tau^{\mu_1+1}).
$$
The point is: this right-hand side is at most {\it algebraic}.

{\bf\noindent Remark 2.4.}
It follows from (2.28) and (2.29) that, as $\tau\rightarrow\infty$
$$\displaystyle
\frac{1}{\tau}\,\log I_{\partial\Omega}(\tau;B,T)=-2T+\frac{2}{\sqrt{\tau}}\,(\eta+R_D(p))+O(\frac{\log \tau}{\tau}).
\tag {2.30}
$$
This yields the convergence rate for (2.5):
$$\displaystyle
\frac{1}{\sqrt{\tau}}
\,(\log I_{\partial\Omega}(\tau;B,T)+2\tau T)
=2(\eta+R_D(p))+O(\frac{\log \tau}{\sqrt{\tau}}).
$$

{\bf\noindent Remark 2.5.}
In \cite{PW} instead of $f_{B,T,\tau}$ we have made use of the heat flux $f=f_{B,\,\tau}$ given by
$$\begin{array}{lll}
\displaystyle
f_{B,\,\tau}(x,t)=\frac{\partial}{\partial\nu}v_{\tau}(x,t), & x\in\partial\Omega, & 0<t<T,
\end{array}
$$
where $B$ is the same open ball as above, however, satisfies the constraint $\overline B\cap\overline\Omega=\emptyset$.
Note that there is no additional constraint on $B$ like (2.4).
Therein it is shown that, replacing $w_f^*$ in $I_{\partial\Omega}(\tau;B,T)$ with $w_f^0$
given by
$$\begin{array}{ll}
\displaystyle
w_f^0(x,\tau)=\int_0^T e^{-\tau t}v_{\tau}(x,t)dt, & x\in\Bbb R^3,
\end{array}
$$
one can extract $\text{dist}\,(D,B)$
by the formula
$$\displaystyle
\lim_{\tau\rightarrow\infty}\frac{1}{\sqrt{\tau}}\,\log I_{\partial\Omega}(\tau;B,T)
=-2\,\text{dist}\,(D,B).
$$
Comparing this with (2.30), one may think that the information $R_D(p)$ is {\it hidden} deeply more than $\text{dist}\,(D,B)$.

\section{An alternative proof of (2.18)}

In \underline{Remark 3.5} of \cite{EV} we have already pointed out that: together with (2.19) the formula
$$
\left\{
\begin{array}{l}
\displaystyle
f_{\tau}(T)=\frac{\tau}{12}\eta^3-\frac{\eta}{\tau}+\frac{2}{\tau^2},\\
\\
\displaystyle
f_{\tau}(T+\eta)=\frac{1}{\tau}\left(\eta+\frac{2}{\tau}\right),\\
\\
\displaystyle
g_{\tau}(T)=\frac{\tau}{12}\eta^3-\frac{\eta}{\tau}-\frac{2}{\tau^2}
\end{array}
\right.
$$
is valid.  Therefore we have
$$\begin{array}{l}
\displaystyle
\,\,\,\,\,\,
{\cal H}_+(\tau,T,\eta)
+{\cal H}_{-}(\tau;T,\eta)
\\
\\
\displaystyle
=\frac{4}{\tau^2}e^{-\tau T}
-\frac{1}{\tau}\left(\eta+\frac{2}{\tau}\right)e^{-\tau(T+\eta)}
+\frac{1}{\tau}\left(\eta-\frac{2}{\tau}\right)e^{-\tau(T-\eta)}.
\end{array}
\tag {3.1}
$$
We do not need this explicit formula (3.1) for the present purpose.
However, for future purpose which aims at extending the method presented in this paper to
other partial differential equations, including systems,
we present here an another proof which is based on the solution formula on (2.1) in $(\xi,t)$-space not $(x,t)$-space.
It is a combination of Parseval's identity and the residue theorem and completely different from the original proof
given in \cite{EV} which is a combination of Kirchhoff's formula and explicit computation formula of some volume potentials.

Let us formulate what we give a proof.

\proclaim{\noindent Proposition 3.1.}
Let $v$ be the solution of (2.1).
Let $T>\eta$.
We have
$$\begin{array}{l}
\,\,\,\,\,\,\displaystyle
\frac{\tau^2}{4\pi}
\int_{\Bbb R^3}
\frac{e^{-\tau\vert x-y\vert}}{\vert x-y\vert}
(\tau v(y,T)-\partial_tv(y,T))\,dy\\
\\
\displaystyle
=
\left({\cal H}_+(\tau,T,\eta)
+{\cal H}_{-}(\tau;T,\eta)\right)
\,\frac{\sinh\tau\vert x-p\vert}{\vert x-p\vert}
\end{array}
$$
for all $x\in B_{T-\eta}(p)\setminus\{p\}$,  where ${\cal H}_+(\tau,T,\eta)+{\cal H}_{-}(\tau;T,\eta)$
is given by the right-hand side on (3.1).
\endproclaim

{\it\noindent Proof.}
It is known that
$$\displaystyle
\hat{v}(\xi,T)
=\frac{\sin\vert\xi\vert T}{\vert\xi\vert}\,\hat{\Psi_B}(\xi)
$$
and
$$\displaystyle
\hat{(\partial_tv)}(\xi,T)
=\cos\vert\xi\vert T \,\hat{\Psi_B}(\xi),
$$
where we denote by $\hat{h}$ the Fourier transfom of a fuinction $h(x)$, that is
$$\displaystyle
\hat{h}(\xi)=\int_{\Bbb R^3}e^{-ix\cdot\xi}h(x)dx.
$$
We have also
$$
\displaystyle
\frac{1}{4\pi}\int_{\Bbb R^3}\frac{e^{-\tau\vert x\vert}}{\vert x\vert}e^{-ix\cdot\xi}dx
=\frac{1}{\vert\xi\vert^2+\tau^2}
$$
and thus
$$
\displaystyle
\frac{1}{4\pi}\int_{\Bbb R^3}\frac{e^{-\tau\vert x-y\vert}}{\vert x-y\vert}e^{-iy\cdot\xi}dy
=\frac{1}{\vert\xi\vert^2+\tau^2}e^{-ix\cdot\xi}.
$$
By Parseval's identity, we obtain
$$\begin{array}{l}
\,\,\,\,\,\,
\displaystyle
I_1(\tau)\equiv
\frac{\tau^2}{4\pi}
\int_{\Bbb R^3}
\frac{e^{-\tau\vert x-y\vert}}{\vert x-y\vert}
\tau v(y,T)\,dy\\
\\
\displaystyle
=\frac{\tau^2}{(2\pi)^3}\int_{\Bbb R^3}\tau\frac{\sin\vert\xi\vert T}{\vert\xi\vert}
\,\hat{\Psi_B}(\xi)\frac{1}{\vert\xi\vert^2+\tau^2}e^{ix\cdot\xi}d\xi
\end{array}
$$
and
$$\begin{array}{l}
\,\,\,\,\,\,
\displaystyle
I_2(\tau)\equiv
\frac{\tau^2}{4\pi}
\int_{\Bbb R^3}
\frac{e^{-\tau\vert x-y\vert}}{\vert x-y\vert}
\partial_tv(y,T)\,dy\\
\\
\displaystyle
=\frac{\tau^2}{(2\pi)^3}\int_{\Bbb R^3}\cos\vert\xi\vert T
\,\hat{\Psi_B}(\xi)\frac{1}{\vert\xi\vert^2+\tau^2}e^{ix\cdot\xi}d\xi.
\end{array}
$$
A change of variables yields
$$\begin{array}{ll}
\displaystyle
\hat{\Psi_B}(\xi)
&
\displaystyle
=\frac{4\pi\eta^3e^{-i\xi\cdot p}}{\vert\xi\vert}
\int_0^1r(1-r)\sin\eta\vert\xi\vert r\,dr.
\end{array}
$$
Here we have
$$
\displaystyle
\int_0^1r(1-r)\sin\eta\vert\xi\vert r\,dr
=\frac{1}{(\eta\vert\xi\vert)^2}
\left\{\frac{2}{\eta\vert\xi\vert}\,(1-\cos\eta\vert\xi\vert)-\sin\eta\vert\xi\vert\right\}.
$$
Thus we obtain
$$\begin{array}{ll}
\displaystyle
\hat{\Psi_B}(\xi)
&
\displaystyle
=\frac{4\pi\eta^3e^{-i\xi\cdot p}}{\vert\xi\vert}\cdot\frac{1}{(\eta\vert\xi\vert)^2}
\left\{\frac{2}{\eta\vert\xi\vert}\,(1-\cos\eta\vert\xi\vert)-\sin\eta\vert\xi\vert\right\}\\
\\
\displaystyle
&
\displaystyle
=4\pi\eta e^{-i\xi\cdot p}g(\vert\xi\vert;\eta),
\end{array}
$$
where
$$\displaystyle
g(z;\eta)
=\,\frac{1}{z^3}
\left\{\frac{2}{\eta z}\,(1-\cos\eta z)-\sin\eta z\right\}.
$$
Note that $g(z;\eta)$ is an even function and that $z=0$ is a removable singularity of $g(z;\eta)$.

Therefore we obtain the expression
$$\begin{array}{ll}
\displaystyle
I_1(\tau)
&
\displaystyle
=\frac{\tau^2}{(2\pi)^3}\int_{\Bbb R^3}\tau\frac{\sin\vert\xi\vert T}{\vert\xi\vert}
\cdot 4\pi\eta e^{-i\xi\cdot p}g(\vert\xi\vert;\eta)\frac{1}{\vert\xi\vert^2+\tau^2}e^{ix\cdot\xi}d\xi
\\
\\
\displaystyle
&
\displaystyle
=\frac{\tau^3\eta}{2\pi^2}\int_{\Bbb R^3}\frac{\sin\vert\xi\vert T}{\vert\xi\vert}
\cdot
\frac{g(\vert\xi\vert;\eta)}{\vert\xi\vert^2+\tau^2}e^{i(x-p)\cdot\xi}d\xi\\
\\
\displaystyle
&
\displaystyle
=\frac{\tau^2\eta}{2\pi^2}\int_{\Bbb R^3}
\frac{\sin\vert\xi\vert T}{\vert\xi\vert}
\frac{g(\vert\xi\vert;\eta)}{\vert\xi\vert^2+\tau^2}e^{i\vert x-p\vert\xi_3}d\xi
\\
\\
\displaystyle
&
\displaystyle
=\frac{\tau^2\eta}{2\pi^2}
\int_0^{\infty}r^2 dr
\int_0^{2\pi}d\theta\int_0^{\pi}\sin\varphi d\varphi
\frac{\sin r T}{r}
\frac{g(r;\eta)}{r^2+\tau^2}e^{i\vert x-p\vert r\cos\varphi}\\
\\
\displaystyle
&
\displaystyle
=\frac{\tau^2\eta}{\pi}
\int_0^{\infty}\frac{r^2}{r^2+\tau^2}\frac{\sin r T}{r}g(r;\eta)
dr\int_0^{\pi}e^{i\vert x-p\vert r\cos\varphi}\sin\varphi d\varphi
\\
\\
\displaystyle
&
\displaystyle
=\frac{i\tau^2\eta}{\pi\vert x-p\vert}\int_0^{\infty}\frac{r}{r^2+\tau^2}
\frac{\sin r T}{r}g(r;\eta)
dr\int_0^{\pi}
(e^{i\vert x-p\vert r\cos\varphi})'d\varphi
\\
\\
\displaystyle
&
\displaystyle
=\frac{2\tau^2\eta}{\pi\vert x-p\vert}\int_0^{\infty}\frac{r}{r^2+\tau^2}
\frac{\sin r T}{r}g(r;\eta)
\sin\vert x-p\vert r \,dr
\\
\\
\displaystyle
&
\displaystyle
=\frac{\tau^2\eta}{\pi\vert x-p\vert}\int_{-\infty}^{\infty}\frac{\sin rT}{r^2+\tau^2}g(r;\eta)
\sin\vert x-p\vert r \,dr.
\end{array}
$$
Similarly, we have
$$
\displaystyle
I_2(\tau)
=\frac{\tau^2\eta}{\pi\vert x-p\vert}\int_{-\infty}^{\infty}\frac{r\cos rT}{r^2+\tau^2}
g(r;\eta)
\sin\vert x-p\vert r \,dr.
$$

Since $g(-r;\eta)=g(r;\eta)$, one can rewrite
$$
\left\{
\begin{array}{l}
\displaystyle
\int_{-\infty}^{\infty}\frac{\sin rT}{r^2+\tau^2}
g(r;\eta)
\sin\vert x-p\vert r\,dr
=\frac{1}{i}\int_{-\infty}^{\infty}\frac{e^{irT}}{r^2+\tau^2}
g(r;\eta)
\sin\vert x-p\vert r\,dr,\\
\\
\displaystyle
\int_{-\infty}^{\infty}\frac{r\cos rT}{r^2+\tau^2}
g(r;\eta)
\sin\vert x-p\vert r\,dr
=\int_{-\infty}^{\infty}\frac{re^{irT}}{r^2+\tau^2}
g(r;\eta)\sin\vert x-p\vert r\,dr.
\end{array}
\right.
$$
Therefore we have
$$\left\{
\begin{array}{l}
\displaystyle
I_1(\tau)
=-i\frac{\tau^3\eta}{\pi\vert x-p\vert}\int_{-\infty}^{\infty}\frac{e^{irT}}{r^2+\tau^2}
g(r;\eta)
\sin\vert x-p\vert r\,dr,
\\
\\
\displaystyle
I_2(\tau)
=\frac{\tau^2\eta}{\pi\vert x-p\vert}
\int_{-\infty}^{\infty}\frac{re^{irT}}{r^2+\tau^2}
g(r;\eta)\sin\vert x-p\vert r\,dr.
\end{array}
\right.
$$

Let $z=Re^{i\theta}$ with $0<\theta<\pi$, $R>0$.
We have, as $R\rightarrow\infty$
$$
\left\{
\begin{array}{l}
\displaystyle
\vert g(z;\eta)\vert=O(R^{-3}e^{\eta R\sin\theta}),\\
\\
\displaystyle
\vert\sin\vert x-p\vert z\vert=O(e^{\vert x-p\vert R\sin\theta}),\\
\\
\displaystyle
\vert e^{izT}\vert=O(e^{-TR\sin\theta}).
\end{array}
\right.
$$
Since we have $\displaystyle T>\eta+\vert x-p\vert$,
one can apply the residue theorem by taking a standard closed contour $-R\longrightarrow R\longrightarrow Re^{i\theta}\,(0\le\theta\le\pi)\longrightarrow -R$
with $R>\tau$ and letting $R\rightarrow\infty$, one gets
$$\begin{array}{ll}
\displaystyle
\int_{-\infty}^{\infty}\frac{e^{irT}}{r^2+\tau^2}
g(r;\eta)
\sin\vert x-p\vert r\,dr
&
\displaystyle
=2\pi i\,\text{Res}_{z=i\tau}\left(\frac{e^{izT}}{z^2+\tau^2}
g(z;\eta)
\sin\vert x-p\vert z\right)
\\
\\
\displaystyle
&
\displaystyle
=2\pi i
\left(\frac{e^{-\tau T}}{2i\tau}
g(i\tau;\eta)
\sin\vert x-p\vert i\tau\right)
\\
\\
\displaystyle
&
\displaystyle
=-\frac{\pi}{i\tau}e^{-\tau T}g(i\tau;\eta)\sinh \tau\vert x-p\vert.
\end{array}
$$
Similarly, we have
$$\begin{array}{ll}
\displaystyle
\int_{-\infty}^{\infty}\frac{re^{irT}}{r^2+\tau^2}
g(r;\eta)\sin\vert x-p\vert r\,dr
&
\displaystyle
=2\pi i\frac{i\tau e^{-\tau T}}{2i\tau}
g(i\tau;\eta)\sin\vert x-p\vert i\tau
\\
\\
\displaystyle
&
\displaystyle
=-\pi e^{-\tau T}
g(i\tau;\eta)\sinh\tau\vert x-p\vert.
\end{array}
$$
From these one gets
$$
\displaystyle
I_1(\tau)
=\tau^2\eta
e^{-\tau T}
g(i\tau;\eta)\frac{\sinh\tau\vert x-p\vert}{\vert x-p\vert}
$$
and $I_2(\tau)=-I_1(\tau)$.

Therefore we obtain
$$\displaystyle
I(\tau)=2I_1(\tau)=
2\tau^2\eta e^{-\tau T}g(i\tau;\eta)\frac{\sinh\tau\vert x-p\vert}{\vert x-p\vert}.
$$
Here we have
$$\begin{array}{ll}
\displaystyle
g(i\tau;\eta)
&
\displaystyle
=\frac{1}{(i\tau)^3}
\left\{\frac{2}{\eta i\tau}\,(1-\cos i\eta\tau)-\sin i\eta\tau\right\}
\\
\\
\displaystyle
&
\displaystyle
=\frac{2}{\eta\tau^4}\left(1-\frac{e^{\tau\eta}+e^{-\tau\eta}}{2}\right)-\frac{1}{(i\tau)^3}\frac{e^{-\tau\eta}-e^{\tau\eta}}{2i}\\
\\
\displaystyle
&
\displaystyle
=\frac{2}{\eta\tau^4}\left(1-\frac{e^{\tau\eta}+e^{-\tau\eta}}{2}\right)-\frac{1}{\tau^3}\frac{e^{-\tau\eta}-e^{\tau\eta}}{2}\\
\\
\displaystyle
&
\displaystyle
=\frac{2}{\eta\tau^4}+\left(\frac{1}{2\tau^3}-\frac{1}{\eta\tau^4}\right)e^{\tau\eta}
-\left(\frac{1}{2\tau^3}+\frac{1}{\eta\tau^4}\right)e^{-\tau\eta}.
\end{array}
$$
This completes the proof.

\noindent
$\Box$

$$\quad$$

\centerline{{\bf Acknowledgment}}

The author was partially supported by Grant-in-Aid for
Scientific Research (C)(No. 17K05331) of Japan  Society for
the Promotion of Science.

$$\quad$$


TITLE: How do I prove this sum is not an integer
QUESTION [16 upvotes]: Assume that $k,n\in\mathbb{Z}^+$. Prove that the sum
\begin{equation*}
\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+\ldots +\dfrac{1}{k+n-1}+\dfrac{1}{k+n}
\end{equation*}
is not an integer.
The case $k=0$ is proved in this question: "Is there an elementary proof that $\sum_{k=1}^{n}\frac{1}{k}$ is never an integer?"

REPLY [15 votes]: The proof that I gave in that thread works just as well here. It depends only on the fact that in any contiguous sequence of integers (here denominators) the maximal power $\rm 2^k$ that divides any element occurs in precisely one element. Indeed, after the first (necessarily odd) multiple of $\rm 2^k$, the next multiple would, by contiguity, be an even multiple, so a multiple of $\rm 2^{k+1}$, contra maximality. Here is said proof:
HINT $\;$ Since there is a unique denominator $\rm 2^k$ having maximal power of $2$, upon multiplying all terms through by $\rm 2^{k-1}$ one deduces the contradiction that $\rm\; a/2 = b/c \;$ with $\rm \; a,\ c \:$ odd. As an example:
$\quad\quad\quad\quad\quad\quad m = \frac{1}{2} + \frac{1}{3} +\; \frac{1}{4} \;+\; \frac{1}{5} + \frac{1}{6} + \frac{1}{7} $
$\quad\quad\Rightarrow\quad\;\; 2m = \; 1 + \frac{2}{3} +\; \frac{1}{2} \;+\; \frac{2}{5} + \frac{1}{3} + \frac{2}{7} $
$\quad\quad\Rightarrow\quad -\frac{1}{2}  = \; 1 + \frac{2}{3} - 2m  + \frac{2}{5} + \frac{1}{3} + \frac{2}{7}$
The prior sum has all odd denominators so reduces to a fraction with odd denominator $\rm d\:|3\cdot 5\cdot 7$.
Note $\:$ I purposely avoided any use of valuation theory because Anton requested an "elementary" solution. The above proof can easily be made comprehensible to a high-school student. The proof is trivial to anyone who knows valuation theory: namely the sum has a lone dominant term with 2-adic value $\rm v_2<0\:$ so, by the domination principle, the sum has 2-adic value $\rm v_2<0\:,\:$ i.e. the sum has even denominator in lowest terms, so it is nonintegral.

REPLY [4 votes]: {Differences of harmonic numbers are not integers}
Consider $\sum_{k=m}^n {1\over k}$. Find the
largest power of $2^r$ that divides one of the denominators between $m$ and $n$.
There can be no even multiples of $2^r$ between $m$ and $n$, hence there
 is only one odd multiple.
Therefore $2^{r-1}(1/m+\cdots+1/n)$ is equal to $1/2o_1$ plus a bunch of
fractions with odd denominator. Adding them gives one fraction with
an odd denominator, write it as $x/o_2$.
Here $o_1$ and $o_2$ are odd integers.
We have $2^{r-1}(1/m+\cdots+1/n)=1/2o_1+x/o_2$ so that
$${1\over m}+\cdots+{1\over n}={2xo_1+o_2\over 2^r o_1 o_2}.$$
Therefore $2^r$ divides the denominator.
If $n\not=m$, then $r\geq1$ and the sum is not an integer.
Cf. Exercise 6.21 (page 311) of Graham, Knuth, and Patashnik.