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\title{New criteria to identify spectrum}
\author{A Jensen \\ Department of Mathematical Sciences and 
MaPhySto\thanks{MaPhySto---A Network in Mathematical 
Physics and Stochastics, funded by the Danish National Research Foundation}\\
Aalborg University, 
Fr. Bajers Vej 7G \\DK-9220 Aalborg \O{}, Denmark \\ and \\
M Krishna\\ Institute of Mathematical Sciences \\ Taramani, Chennai 600 113, India}
%\date{27 April 2003}
\begin{document}
\maketitle
\begin{abstract}In this paper we give some  new criteria 
for identifying the components
of a probability measure, in its Lebesgue decomposition.  This enables us to
give new criteria to identify spectral 
 types of self adjoint operators on Hilbert
spaces, especially those of interest.
\end{abstract}

\section{Introduction}
In the spectral theory of self adjoint operators it is of interest to identify
the type of the spectrum. This problem is equivalent to identifying the
components of the spectral measures. 
The components of a probability measure can be identified via a transform of
the measure. Two of these are well known,  viz.
the Fourier transform and the Borel transform. In this paper we address the
question of identifying the components using a more general transform.
We give results using a general approximate identity, and an associated 
continuous wavelet transform.

Concerning the literature, the connection between an approximate identity 
and the continuous wavelet transform was discussed in the 
book by Holschneider~\cite{hol95}, while
wavelet coefficients of fractal measures were studied
by Strichartz in \cite{MR93c:42036}.
In the theory of selfadjoint operators finer decomposition of spectra
with respect to Hausdorff measures was first used by Last \cite{yl} and
general criteria for recovering a measure from its Borel transform was done
by Simon \cite{si1}.



\section{The criteria}
We need to introduce
conditions on  our function $\psi$. Several of these can be relaxed in some of
the results. We use the standard notation $\jap{x}=(1+x^2)^{1/2}$.

\begin{assumption}\label{a1.1}
Assume that $\psi\in C^1(\rr)$, $\psi(0)=1$, $\psi$ is even, and there exist
$C>0$ and $\delta>1$, such that
\begin{equation}\label{eqassume}
|\psi(x)|+|x\psi'(x)|\leq C \jap{x}^{-\delta},\quad x\in\rr.
\end{equation}
We set $A_\psi = \int_\rr \psi(x) ~ dx$ and assume that $A_\psi \neq 0$.
\end{assumption}
In the sequel we always impose this assumption on $\psi$.
We introduce the notation
\begin{equation}
\psi_a(x)=\psi(x/a)\quad\text{and}\quad \tilde{\psi}_a(x)=\frac{1}{a}\psi_a(x),
\quad a>0.
\end{equation}
In particular, the family $\{A_{\psi}^{-1}\tilde{\psi}_a\}$ is an approximate identity.
Let $\mu$ be a probability measure on $\rr$ in what follows, 
with Lebesgue decomposition $\mu=\mu_{\rm s}+\mu_{\rm ac}$. Let $f$ be a 
function. We recall that
the convolution $(f\ast \mu)(x)=\int f(x-y)d\mu(y)$ is defined, when the
integral converges. Since $\psi$ is bounded, the convolution $\psi_a\ast\mu$ is
defined for all $a>0$. 

For $0\leq\alpha\leq1$ we define
\begin{equation}
(d_{\alpha}\mu)(x)=\lim_{\eps\downarrow0}
\frac{\mu((x-\eps,x+\eps))}{(2\eps)^{\alpha}},
\end{equation}
whenever the limit on the right hand side exists.


We can now state the results. We first give results based on $\psi_a$
and $\tilde{\psi}_a$, and then on an associated continuous wavelet transform.

\begin{thm}
\label{thm1}
Let $\mu$ be a probability measure.  Then 
\begin{itemize}
\item[\rm 1.] Let $\psi$ satisfy Assumption~{\rm\ref{a1.1}}.  Then
for every continuous function $f$ of compact support, the
following is valid.
$$
\lim_{a \rightarrow 0} \int (\tilde{\psi}_a * \mu)(x) f(x) dx = 
A_\psi \int f(x) d\mu(x).
$$
\item[\rm 2.] $\displaystyle{\lim_{a \rightarrow 0}} 
(\psi_{a} * \mu)(x) = \mu(\{x\})$.
\item[\rm 3.] Assume $0<\alpha\leq1$ and 
$(d_\alpha \mu)(x)$ finite.  Then we have
\begin{equation}
\label{eqn1.1}
\lim_{a \rightarrow 0} a^{-\alpha} (\psi_{a}*\mu)(x) = c_\alpha (d_\alpha\mu)(x),
\end{equation}
where 
$c_\alpha = 
\int_0^{\infty} \alpha 2^\alpha y^{\alpha-1}\psi(y)dy$.
\end{itemize}
\end{thm}

\begin{remark}
(1) Equation \eqref{eqn1.1} implies that if $\mu$ is
purely singular, then the limit of $\tilde{\psi}_a * \mu(x)$ is zero almost everywhere
with respect to the Lebesgue measure, since the derivative $(d_1\mu)(x) = 0$ 
almost everywhere for purely singular $\mu$.

\noindent
(2) If $x$ is not in the topological support of $\mu$, then
for each $0\leq \alpha \leq 1$,
$$
\lim_{a \rightarrow 0} a^{-\alpha} \psi_{a}*\mu(x) = 0.
$$
\end{remark}


Our next theorem is a bit more and the first part is analogous to 
Wiener's theorem and its extension by Simon \cite{si1}.

\begin{thm}
\label{thm2}
Let $\mu$ be a probability measure.
Then for any bounded
interval $(c,d)$ the following are valid. 
\begin{itemize}
\item[\rm 1.] Let 
$$
C = 
\displaystyle\int_\rr |\psi(x)|^2dx,  
$$
then
\begin{multline}
\lim_{a \rightarrow 0} \frac{1}{a} 
\int_c^d |(\psi_a * \mu)(x)|^2dx\\
= C \Bigl(\sum_{x \in (c,d)} \mu(\{x\})^2 +
\tfrac{1}{2} \left[\mu(\{c\})^2 + \mu(\{d\})^2 \right]\Bigr).
\end{multline}

\item[\rm 2.] For $0 < p < 1$, we have
\begin{equation}
\lim_{a \rightarrow 0} \int_c^d |(\tilde{\psi}_a * \mu) (x)|^p  dx 
= |A_{\psi}|^p\int_c^d \bigl|\frac{d\mu_{\rm ac}}{dx}(x)\bigr|^p dx.
\end{equation}

\end{itemize}
\end{thm}

This theorem has the following corollary.

\begin{cor} 
\label{cor1}
Let $\mu$ be a probability measure. Then we have the following results
\begin{itemize}
\item[\rm 1.] $\mu$ has no point part in $[c,d]$, if and only if
\begin{equation}
\liminf_{a\to0}\frac{1}{a}\int_c^d|(\psi_a\ast\mu)(x)|^2dx=0.
\end{equation}
\item[\rm 2.] If $\mu$ has no absolutely continuous
 part in $(c,d)$, if and only if for some
 $p$, $0<p<1$, 
\begin{equation}
\liminf_{a\to0}\int_c^d|(\tilde{\psi}_a\ast\mu)(x)|^pdx=0.
\end{equation}
\end{itemize}
\end{cor}

Now to state the results in terms of the continuous wavelet transform, 
we introduce
\begin{equation}\label{cwt}
h(x)=\psi(x)+x\psi'(x).
\end{equation}
Under Assumption~\ref{a1.1} we clearly have
\begin{equation}\label{hdecay}
|h(x)|\leq C\jap{x}^{-\delta},
\end{equation}
with the $\delta$ from the assumption. Integration by parts and \eqref{cwt}
imply that
$h$ satisfies the admissibility condition 
for a continuous wavelet, i.e. $\int_{-\infty}^{\infty}h(x)dx=0$.

Thus we can define the continuous wavelet transform of a 
probability measure $\mu$
as
\begin{equation}\label{cwttrans}
W_h(\mu)(b,a)=\frac{1}{a}\int_{-\infty}^{\infty}h((b-y)/a)d\mu(y).
\end{equation}
The connection between the approximate identity and this transform is
\begin{equation}\label{relation}
-a\frac{\partial}{\partial a}(\tilde{\psi}_a*\mu)(b)=W_h(\mu)(b,a).
\end{equation}
This result follows from
$$
-a\frac{\partial}{\partial a}(\frac{1}{a}\psi(\frac{x}{a}))=
\frac{1}{a}\Bigl(\psi(\frac{x}{a})+\frac{x}{a}\psi'(\frac{x}{a})\Bigr),
$$
and the definitions.

We have the following analogue of Theorem~\ref{thm1}:

\begin{thm}\label{thm3}
Let $\mu$ be a probability measure. 
Then we have the following results:
\begin{enumerate}
\item We have
\begin{equation}
\lim_{\varepsilon\downarrow0}\varepsilon
\int_{\varepsilon}^{\infty}W_h(\mu)(b,a)\frac{da}{a}
=\mu(\{b\}).
\end{equation}
\item
Let $0<\alpha\leq1$. Assume that 
$(d_{\alpha}\mu)(b)$ exists.
Then
\begin{equation}
\lim_{\varepsilon\downarrow0}\varepsilon^{1-\alpha}
\int_{\varepsilon}^{\infty}W_h(\mu)(b,a)\frac{da}{a}
=c_{\alpha}(d_{\alpha}\mu)(b),
\end{equation}
where $c_{\alpha}$ was defined in Theorem~{\rm\ref{thm1}}.
\end{enumerate}
\end{thm}
\begin{remark}
We note that for $0<\alpha<1$ we can replace $\int_{\eps}^{\infty}$
by $\int_{\eps}^M$ for any $M>0$. See the proof of the Theorem.
\end{remark}

We also have the following analogue of Theorem~\ref{thm2}(1).

\begin{thm}\label{thm4}
Let $\mu$ be a probability measure. Then for any bounded interval $(c,d)$ we
have the following result. Let
$$
C_h=
\displaystyle\int_{\rr}|h(x)|^2dx,
$$
Then we have
\begin{multline}
\lim_{a\downarrow0}\int_c^d|W_h(\mu)(b,a)|^2db\\ =
C_h\Bigl(\sum_{x\in(c,d)}\mu(\{x\})^2 +\tfrac{1}{2}\bigl(\mu(\{c\})^2 + 
\mu(\{d\})^2\bigr)\Bigr).
\end{multline}
\end{thm}

Even when the quantity $(d_\alpha \mu)(x)$ does not exist, it is 
possible to say something on the wavelet transforms, to cover
the cases of measures which are not supported on the sets
where such limits exist.  Set
$$
C_{\mu, \psi}^\alpha(x) = 
\displaystyle\limsup_{a \rightarrow 0}  
\frac{\psi_a*\mu}{a^\alpha}(x) \quad \text{and} \quad 
D_\mu^\alpha(x) = \displaystyle\limsup_{\varepsilon \rightarrow 0}  
\frac{\mu((x-\varepsilon, x+\varepsilon))}{(2\varepsilon)^\alpha}. 
$$
Then we have
\begin{thm}
\label{thm5}
Let $\mu$ be a probability measure, and let $\psi$ satisfy  
Assumption {\rm\ref{a1.1}}.  Then $C_{\mu, \psi}^\alpha(x)$ is finite for 
any $x$, whenever $D_\mu^\alpha(x)$ is
finite for the same $x$, and, if $\psi$ is non-negative, they are both 
finite or both infinite.
\end{thm}
\begin{remark}
\label{rem-new-1}
The above theorem implies that if 
$\limsup_{a \to0} |(\tilde{\psi}_a*\mu)(x)| < \infty $ for all
$x \in (c, d)$, then there is no singular part of $\mu$ supported in $(c, d)$.
\end{remark}

Finally as an application of the above theorems we consider $\hh$ to be 
a separable Hilbert space and $A$ a selfadjoint operator.  Then

\begin{thm}
\label{thm6}
Suppose $A$ is a selfadjoint operator on $\hh$.  Consider
a function $\psi$ satisfying Assumption~{\rm\ref{a1.1}}. 
Then
\begin{enumerate}
\item[\rm 1.] $\lambda$ is in the point spectrum of $A$, if for some
$f \in \hh$,  $\|f\| = 1$, 
$$
\lim_{a\rightarrow 0} \jap{f, \psi_a(A-\lambda)f} = 0.
$$
\item[\rm 2.] Let $B \subset \rr$ be a Borel set of positive Lebesgue measure.
Then $B \cap \sigma_{ac}(A) \neq \emptyset$, if for some $f \in \hh$, 
$ \|f\|=1$,
$$
\lim_{a \rightarrow 0} \jap{f, \tilde{\psi}_a(A-\lambda)f} \neq 0, 
\quad\text{for a.e. $\lambda\in B$}.
$$
\item[\rm 3.] The point spectrum of $A$ in $(c, d)$ is empty, if and only 
if for some
orthonormal basis $\{f_n\}$, of $\hh$, one has for every $n$, 
$$
 \liminf_{a \rightarrow 0} \frac{1}{a}
\int_c^d |\jap{f_n, \psi_a(A-\lambda)f_n}|^2  d\lambda = 0.
$$
\item[\rm 4.] The absolutely continuous spectrum  of $A$ in $(c, d)$ is 
empty, if and only if for some orthonormal basis $\{f_n\}$ of $\hh$, 
one has for every $n$ and some $0 < p < 1$, 
$$
 \liminf_{a \rightarrow 0} 
\int_c^d \bigl|\frac{1}{a}\jap{f_n, \psi_a(A-\lambda)f_n}\bigr|^p  d\lambda = 0.
$$
\end{enumerate}
\end{thm}


\section{Proofs}
Throughout the computations below the letter $C$ denotes a constant, whose
value may vary from line to line.

\paragraph{Proof of Theorem \ref{thm1}:} 
Part (1): Since $f$ is a continuous function of compact support and $\psi_a$ is bounded for
each $a >0$, $f(x) \psi_a(x-y)$ is absolutely integrable
and the integral is uniformly bounded in $y \in \rr$.
Therefore, by an application of
Fubini, a change of variable $x \rightarrow ax+y$ and
dominated convergence theorem, in that order, it follows that
\begin{align*}
\lim_{a\rightarrow 0 } \int\!\! dx \, f(x) \bigl(\tilde{\psi}_a*\mu\bigr)(x)
&=
\lim_{a\rightarrow 0 } \int\!\! dx \, f(x) \int\!\! \tilde{\psi}_a(x-y) d\mu(y)\\
&= \lim_{a\rightarrow 0} \int\!\! d\mu(y) \int\!\! f(x) \tilde{\psi}_a(x-y) dx \\
&= \lim_{a\rightarrow 0} \int\!\! d\mu(y) \int\!\! f(ax+y) \psi(x) dx \\
&= \int\!\! d\mu(y) \int\! (\lim_{a\rightarrow 0} f(ax+y)) \psi(x) dx \\
&=\int\!\! f(y)d\mu(y) \cdot \int\!\! \psi(x) dx.
\end{align*}
Part (2): This is a direct consequence of the 
definition of the integral noting that pointwise we have
\begin{equation*}
\lim_{a \rightarrow 0} \psi_{a}(x) =  
\begin{cases}
0, &\text{if $x \neq 0$}, \\
1, &\text{if $x = 0$}.
\end{cases} 
\end{equation*}
We also need to use the dominated convergence theorem
to interchange the limit and the
integral.

\noindent
Part (3): Let $\Phi_\mu$ denote the distribution function of $\mu$.
Then we have 
\begin{align}
\frac{1}{a^\alpha}\int_\rr \psi_a(x&-y) d\mu(y)\notag\\ 
&= - \frac{1}{a^\alpha}\int_\rr \frac{d}{dy}\psi((x-y)/a)
\Phi_\mu(y)  dy 
\notag\\
&=  \frac{1}{a^\alpha}\int_\rr \psi^\prime (y) \Phi_\mu(x-ay)  dy 
\notag\\
&= - \int_0^\infty \psi^\prime (y)
(2y)^\alpha \frac{\Phi_\mu(x+ay) - \Phi_\mu(x - ay)}{(2ay)^\alpha} dy,
\label{eqnnt}
\end{align}
where in the first step we used integration by parts, the next step changed
variables and in the last step used the oddness of $\psi^\prime$ to
split the integral into the positive and negative half lines and multiplied
by appropriate powers.  


We observe that
$$
(d_\alpha \mu)(x) = 
{\lim_{a \rightarrow 0}} 
\frac{\Phi_\mu(x + ay) - \Phi_\mu(x- ay)}{(2ay)^\alpha}
$$
for each $y \in \rr$, and is finite by assumption. Furthermore, the function
$(\Phi_\mu(x + ay) - \Phi_\mu(x- ay))(2ay)^{-\alpha} $ is a bounded measurable
function, such that we due to \eqref{eqassume} we can take the limits inside
the integral sign in \eqref{eqnnt}  and use the dominated convergence theorem.

Now doing
an integration by parts gives the value of the integral as stated
in the theorem. 


\paragraph{Proof of Theorem \ref{thm2}:} 
Part (1): We have 
\begin{equation*}
\frac{1}{a}\int_c^d |\psi_a * \mu(x)|^2  dx 
= \iint d\mu(y_1)\, d\mu(y_2) \int_c^d dx 
\frac{1}{a}\overline{\psi_a(x - y_1)} \psi_a(x-y_2). 
\end{equation*}
Since the function $\psi_a$ is bounded, the interval $(c,d)$ is bounded,
and $\mu$ is a probability measure, the right hand side integral converges
absolutely, 
so we used Fubini to interchange integrals to get the equality above.  Let
$$
h_a(y_1, y_2) = \int_c^d dx ~ \frac{1}{a}\overline{\psi_a(x - y_1)} \psi_a(x-y_2).
$$ 
Suppose $y_1 \neq y_2$, then using the bound 
$|\psi(x)| \leq C\jap{x}^{-\delta}$, we
see that the bound
\begin{align*}
|h_a(y_1, y_2)| &\leq \frac{C}{a}\int_{-\infty}^{\infty}
\jap{(x+y_2-y_1)/a}^{-\delta}
\jap{x/a}^{-\delta}dx\\
&= \frac{C}{a}
\big(\int_{|x|\leq|y_1-y_2|/2}+\int_{|x|\geq|y_1-y_2|/2}\bigr)(\cdots)dx\\
&\leq
\frac{Ca^{\delta}}{|y_1-y_2|^{\delta}}
\int_{-\infty}^{\infty}\jap{x/a}^{-\delta}
d(x/a)\\
&\leq \frac{Ca^{\delta}}{|y_1-y_2|^{\delta}}
\end{align*}
is valid.  
It follows that $\lim_{a\to0}h_a(y_1,y_2)=0$ for $y_1\neq y_2$.
It remains to consider $y_1=y_2$.
This is done by noting
that  
$$
h_a(y_1, y_1) = \int_c^d   \frac{1}{a} |\psi_a(x - y_1)|^2 dx
 = \int_{(c-y_1)/a}^{(d-y_1)/a}   |\psi(x)|^2 dx, 
$$
from which taking limits, we obtain the stated value for the coefficient,
either $C$ or $C/2$, based on whether $c < y_1 < d$ or $y_1 = c, d$,
using the evenness of $\psi$.  
Now to complete the proof, we note the estimate
$$
|h_a(y_1, y_2)|\leq C\int_{\rr}\jap{x/a}^{-\delta}d(x/a)\leq C_0,
$$
where the constant $C_0$ is independent of $a$, $y_1$, and $y_2$. Thus the
proof is completed used the dominated convergence theorem.

\noindent
Part (2): We adapt the arguments in \cite{si1} to the case at hand.
We split the measure in three components: $\mu=\mu_1+\mu_2+\mu_3$. Here
$d\mu_1=(1-\chi_{[c-1,d+1]})d\mu$, $d\mu_2=gdx$ with $g\in L^1([c-1,d+1])$, and
$\mu_3$ is purely singular, and supported on $[c-1,d+1]$.
We have for $x\in[c,d]$ the estimate
$$
|(\tilde{\psi}_a\ast\mu_1)(x)|\leq 
C\int_{\rr\setminus[c-1,d+1]}a^{-1}\jap{(x-y)/a}^{-\delta}d\mu_1(y)
\leq C a^{\delta-1}.
$$
We now look at the $\mu_2$ part.
We have, for $0 < p <1$, by the reverse H\"older inequality
$$
\int_c^d \bigl|(\tilde{\psi}_a *g)(x) - A_\psi g(x)\bigr|^p dx \leq 
\Bigl(\int_c^d \bigl|
(\tilde{\psi}_a *g)(x) - A_\psi g(x)\bigr| dx\Bigr)^{p} (d-c)^{1-p}, 
$$  
which implies that $\tilde{\psi}_a*g \rightarrow A_{\psi} g$ in 
$L^p((c, d))$, $0 < p \leq 1$. 

Now we will show that the singular part $\mu_3$ does not contribute
to the limit.  So assume that $\mu_3$ is purely singular and that 
its support $S$ is contained in $[c-1, d+1]$.  
Since $\mu_3$ is singular, by the definition of support, 
$S$ satisfies $\mu_3(\RR \setminus S) = 0$ 
and $|S| = 0$, with $|\cdot|$ denoting the Lebesgue measure.  By
the regularity of the Lebesgue measure, given an $\varepsilon >0$, 
there is an open set $O\subset(c-2,d+2)$, such that 
$S \subset O$, with $|O\setminus S| < \varepsilon$.
We also have
$|O| \leq |O\setminus S| + |S| < \varepsilon$.  For the same
$\varepsilon$, since the measure $\mu_3$ is regular, we also have
a compact $K \subset S$, such that $\mu_3(S\setminus K) < \varepsilon.$
In addition, since $K \subset S$, and $S$ has Lebesgue measure zero, $K$ also
has Lebesgue measure zero.


The above reverse H\"older inequality gives 
\begin{align*}
\int_c^d \bigl|(\tilde{\psi}_a*\mu_3)(x)\bigr|^p dx 
&= \int_O \bigl|(\tilde{\psi}_a*\mu_3)(x)\bigr|^p dx   
+\int_{(c,d)\setminus O} \bigl|(\tilde{\psi}_a*\mu_3)(x)\bigr|^p dx\\
&\leq |O|^{1-p} \mu_3((c,d))^p \|\psi\|_1^p \\
{}&\quad
+|d-c|^{1-p} \Bigl(\int_{(c,d)\setminus O}
\bigl|(\tilde{\psi}_a*\mu_3)(x)\bigr| dx\Bigr)^p \\ 
&\leq C \varepsilon^{1-p} +   
|d-c|^{1-p} \Bigl(\int_{(c,d)\setminus O} \bigl|(\tilde{\psi}_a*\mu_3)(x)\bigr| dx\Bigr)^p. 
\end{align*}
Now consider a bounded continuous function $h$
which is $1$ on $(c,d) \setminus O$, and  $0$ on $K$.



Then using  Assumption \ref{a1.1}, that 
$|\psi(x)| \leq C \jap{x}^{-\delta}$, and setting $\phi(x) = \jap{x}^{-\delta}$, 
\begin{align*}
\int_{(c,d)\setminus O} \bigl|(\tilde{\psi}_a*\mu_3)(x)\bigr| dx 
&\leq \int_{(c,d) \setminus O} \frac{1}{a} \int_\RR |\psi_a(x-y)| d\mu_3(y) dx\\
&\leq C\int_{(c,d) \setminus O} \frac{1}{a} 
\int_\RR \langle{(x-y)/a}\rangle^{-\delta} d\mu_3(y)  dx\\
&\leq C\int_{(c,d) \setminus O}  h(x)(\tilde{\phi}_a*\mu_3)(x)  dx.
\end{align*}
The  function $\phi$ satisfies Assumption \ref{a1.1}, so
the Theorem \ref{thm1}(1) is applicable with $\psi$ replaced
by $\phi$ there.  Therefore the last term, which has positive 
integrand, converges to $\int_{(c,d)\setminus O} h(x) d\mu(x)$
as $a$ goes to zero, which is bounded by
$\int_{(c,d)\setminus K} d\mu(x)$,
$$
\int_{(c,d)\setminus O} h(x) d\mu(x) \leq 
\mu((c,d)\setminus K) \leq 
\mu((c,d)\setminus S) + \mu(S\setminus K) < \varepsilon,
$$
using the facts that $\mu((c,d)\setminus S) = 0$ 
and $\mu(S\setminus K) < \varepsilon$.

Using the inequality $(a+b+c)^p\leq a^p+b^p+c^p$ for $0<p<1$ and non-negative
numbers $a$, $b$, $c$, we have
\begin{align*}
 \int_c^d |(\tilde{\psi}_a * \mu) (x)-A_{\psi}g(x)|^p  dx 
&\leq
\int_c^d |(\tilde{\psi}_a * \mu_1) (x)|^p  dx\\
{}&\quad+\int_c^d |(\tilde{\psi}_a * \mu_2) (x)-A_{\psi}g(x)|^p  dx\\
{}&\quad+\int_c^d |(\tilde{\psi}_a * \mu_3) (x)|^p  dx
\end{align*}
Putting the above  estimates together and using that $\varepsilon$ is
arbitrary, one gets 
$$
\lim_{a\to0} \int_c^d |(\tilde{\psi}_a * \mu) (x)-A_{\psi}g(x)|^p  dx=0.
$$
Now the spaces $L^p((c,d))$, $0<p<1$, are metric spaces with the metric
$d(f,g)=\|f-g\|_p^p$. It then follows from the triangle inequality for this
metric that 
$$
\lim_{a \rightarrow 0} \int_c^d \bigl|(\tilde{\psi}_a * \mu)(x)\bigr|^p  dx 
= \left|A_\psi\right|^p\int_c^d |g(x)|^p dx.
$$
Since $g=\frac{d\mu_{\rm ac}}{dx}$, the result follows.


\paragraph{Proof of Theorem~\ref{thm3}:} Let $0<\eps<M<\infty$.
It follows from \eqref{relation} that we have
$$
\int_{\eps}^MW_h(\mu)(b,a)\frac{da}{a}=
(\tilde{\psi}_{\eps}\ast\mu)(b)-(\tilde{\psi}_M\ast\mu)(b).
$$
The results now follow from Theorem~\ref{thm1}. \qed

\paragraph{Proof of Theorem~\ref{thm4}:} The proof is entirely analogous 
to the proof of Theorem~\ref{thm2}, replacing $\psi$ by $h$ and 
adjusting the powers of $a$. \qed


\paragraph{Proof of Theorem~\ref{thm5}:} Consider the case when 
$D_\mu^\alpha(x)$ is finite
for some $x$ and for some fixed $\alpha$.  Then
for any $0 < y < 1$, $\mu(x-y, x+y) \leq C |y|^\alpha$ for 
some finite constant $C$.  So, using the last line in 
equation \ref{eqnnt} and estimating the right hand side there,
one has, by assumption \ref{a1.1},
\begin{equation*}
\bigl|\frac{1}{a^\alpha}(\psi_a*\mu)(x)\bigr| 
\leq C \int_0^{\infty} |\psi^\prime (y)|(2y)^\alpha dy 
\leq C \int_0^{\infty} \jap{y}^{-\delta}|y|^{-1+\alpha} dy <\infty.
\end{equation*}
Now taking the $\limsup$ of the left hand side the finiteness
of $C_{\mu, \psi}^\alpha$ follows.  

On the other hand, since $\psi$ is positive continuous with 
$\psi(0) = 1$, there is a $\beta >0 $ such that $\psi(y) > 1/2, ~ 
-\beta < y < \beta$.  Using this and the evenness~of~$\psi$,  
\begin{align*}
\frac{1}{a^\alpha}(\psi_a*\mu)(x) 
& = \frac{1}{a^\alpha}\int\! \psi_a(x-y) d\mu(y) = \int\! \psi(y/a) d\mu(y+x)\\
& \geq \frac{1}{a^\alpha}\int_{-\beta a}^{\beta a} \tfrac{1}{2} d\mu(y+x) \\
& \geq  \frac{1}{2a^\alpha} [\mu(x + a\beta ) - \mu(x-a\beta)],
\end{align*}
where $\psi\geq0$ is used to get the first 
inequality above. 
The above inequalities immediately imply, since $\beta$ is fixed, that 
$D_\mu^\alpha(x) =\infty$ implies the same for $C_{\mu, \psi}^\alpha(x)$. \qed


\paragraph{Proof of Theorem~\ref{thm6}:} Parts (1) and (2) are a direct 
application of Theorem \ref{thm1}(2) and (3) respectively.
Parts (3) and (4) are a direct application of Corollary \ref{cor1} (1) and (3) 
respectively. 
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