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\textbf{\textit{Abstract}:} This note shows that the prime values of the Ramanujan tau function $\tau(n)=\pm p$ misses every prime $p\leq 8.0\times 10^{25}$.
\let\thefootnote\relax\footnote{ \today \date{} \\
\textit{AMS MSC}: Primary 11F35; Secondary 11D45 \\
\textit{Keywords}: Tau Function; Modular Function; Lehmer Conjecture.}
\tableofcontents

\newpage

\section{The Result } \label{s9099}
The Fourier coefficient of the discriminant function
\begin{equation} \label{eq9999.003}
\Delta(z)=\sum_{n \geq1} \tau(n)q^n=q-24q^2+252q^3-1472q^4+\cdots,
\end{equation}
where $z\in \mathbb{C}$ is a complex number in the upper half plane, $q=e^{i2\pi }$, is known as the Ramanujan tau function. The classification as a level $N=1$ and weight $k=12$ , and other information appears in \cite{LMFDB}. The prime values of the tau function $\tau: \mathbb{N} \longrightarrow \mathbb{Z}$ is a topic of interest in many papers. The first prime value
\begin{equation} \label{eq9999.005}
\tau\left (251^2\right )=-80561663527802406257321747
\end{equation}
was discovered by Lehmer in \cite{LD65}. About half a century later, many other prime values have been discovered by many authors. The authors in \cite{LR13} extended the numerical works of Lehmer and other authors to the range of integers $n\leq 10^{600}$, and beyond. But it remains unknown if $\tau(n)$ has small prime values for some extremely large integers $n\geq1$. A very recent paper has proved that
\begin{equation} \label{eq9999.023}
\tau\left (n\right )\ne \pm3, \pm5, \pm7,  \pm691,
\end{equation}
see \cite[Theorem 1.1]{BO20}, and a slightly larger range of primes $|q|\leq 100$ in \cite[Theorem 6]{BS21}. This note offers a simpler proof, and extends the range of primes missed by the Ramanujan tau function.

\begin{thm} \label{thm9999.001} Let $q< \tau(251^2)$ be a prime, then $\tau\left (p^{2n}\right )\ne  \pm q$.
\end{thm}

\begin{proof}Let $q\leq 8.0 \times 10^{25}< \tau(251^2)$ be a fixed prime. The verification that
\begin{equation} \label{eq9999.025}
\tau\left (p^{2n}\right )=\pm q
\end{equation}
has no solutions $p^{2n} \in \mathbb{N}$ splits into two cases.\\

\textit{Case {\normalfont I}. Small Solutions $p^{2n}\leq 10^{600}$.} By Theorem \ref{thm7799.002}, the prime values $\tau(p^{2n})=\pm q$ require integers of the form $p^{2n}$, where $p\geq 3$ and $2n+1\geq 3$ are primes. By the numerical works in \cite{LD65} and \cite[Table 2]{LR13} for the range $p\leq10^6$ and $2n+1\leq 100$, it follows that equation \eqref{eq9999.025} has no small integer solutions $p^{2n}\leq 10^{600}$.\\

\textit{Case {\normalfont II}. Large Solutions $p^{2n}\geq 10^{600}$.} Assume that equation \eqref{eq9999.025} has a large solution. By Theorem \ref{thm9999.050},
\begin{eqnarray} \label{eq9999.033}
\left |\tau\left (p^{2n}\right )\right |
&\geq& (\log p^{2n})^{10}\\
&\geq& \left (\log 10^{600}\right )^{10}\nonumber\\
&\geq& 2.5231\times 10^{31}\nonumber.
\end{eqnarray}
In particular, for the small bounded subset of primes $q$, it reduces to
\begin{eqnarray} \label{eq9999.037}
8.0\times 10^{25}&\geq&q\\
&=&\left |\tau\left (p^{2n}\right )\right | \nonumber\\
&\geq& 2.5231\times 10^{31}\nonumber.
\end{eqnarray}

Clearly, this is false. Hence, equation \eqref{eq9999.025} has no large integer solutions $p^{2n}\geq 10^{600}$.
\end{proof}

\section{Structures Of Even And Odd Coefficients}\label{S7799}
The form of the integers $m$ required for odd and prime values of the function $\tau(m)$ was initiated in \cite{LD65}. A complete and simplified version was derived in \cite{LR13}.

\begin{thm} \label{thm7799.002} {\normalfont(\cite[Theorem 1]{LR13})} Let $m\geq 1$ be an integer such that $\tau(m)=p$ is an odd prime. Then, $m=q^{2n}$, where $q\geq 3$ and $2n+1\geq3$ are primes, and $p\nmid \tau(p)$.
\end{thm}

The nonvanishing $\tau(m)\ne0$ of the coefficients was initiated in \cite{LD47}, currently it has been verified for approximately $m\leq 10^{23}$, see \cite{SJ03}, \cite[p.\ 2]{DZ13}.

\section{General Bounds}\label{S9999}
For any prime power $p^n\geq 2$, Deligne theorem claims the upper bound
\begin{equation} \label{eq9999.020}
\left |\tau\left (p^n\right )\right | \leq 2p^{11n/2+\varepsilon},
\end{equation}
and the Atkin-Serre conjecture claims the lower bound
\begin{equation} \label{eq9999.022}
\left |\tau\left (p^n\right )\right | \geq p^{9n/2-\varepsilon},
\end{equation}
where $\varepsilon >0$ is an arbitrary small number, see \cite{RJ05}, \cite[Equation 1.3]{GT20}, et alii. A weaker, almost trivial, but unconditional lower bound is given by the next result.
\begin{thm} \label{thm9999.004} {\normalfont (\cite[Theorem 1]{MS87})} There exists an effectively computable absolute constant $c>0$ such that for all nonnegative integers $m$ for which $\tau(m)$ is odd, then
\begin{equation} \label{eq9999.033}
\left |\tau\left (m\right )\right | \geq (\log m)^c.
\end{equation}
\end{thm}

Set $m=p^n$. Jointly, these results satisfy the inequalities
\begin{equation} \label{eq9999.024}
(\log p^n)^{c}\leq p^{9n/2-\varepsilon}\leq \left |\tau\left (p^n\right )\right | \leq 2p^{11n/2+\varepsilon}.
\end{equation}

\section{Explicit Lower Bound}
A different unconditional and completely explicit lower bound is proved in this Section.

\begin{thm} \label{thm9999.050} Let $n\geq 1$ be an integer, and let $p\geq 2$ be a prime. If $p^{2n}\geq 10^{600}$, then,
\begin{equation} \label{eq9999.050}
\left |\tau\left (p^{2n}\right )\right | \geq (\log p^{2n})^{10}.
\end{equation}
\end{thm}
\begin{proof} On the contrary, suppose that
\begin{equation}\label{eq9999.058}
(\log p^{2n})^{10}> \left |\tau\left (p^{2n}\right )\right |,
\end{equation}
for $p^{2n}\geq 10^{600}$. By Theorem \ref{thm2277.106},
\begin{eqnarray} \label{eq9999.060}
(\log p^{2n})^{10}
&>& \left |\tau\left (p^{2n}\right )\right |\\
&\geq& A^{-1}p^{7n-2-\varepsilon}\nonumber,
\end{eqnarray}
where $A=(2n+1)^32^{2n+1}$ is a constant. Equivalently,
\begin{equation}\label{eq9999.058}
(2n+1)^32^{2n+1}(\log p^{2n})^{10}
>p^{7n-2-\varepsilon}
\end{equation}
Taking logarithm across the board yields
\begin{equation} \label{eq9999.060}
3\log (2n+1)+(2n+1)\log 2+10(\log n+\log \log p+\log 2)>(7n-2-\varepsilon)\log p.
\end{equation}
Division by $n\log p$ yields
\begin{eqnarray} \label{eq9999.060}
\frac{3\log (2n+1)+(2n+1)\log 2+10(\log n+\log \log p+\log 2)}{n\log p}&>&7-\frac{2+\varepsilon}{n}\nonumber\\
&>& 6.
\end{eqnarray}
Numerical calculations shows that this false for $p^{2n}\geq 10^{600}$, where $2n\geq2$, $p\geq2$, and a small number $\varepsilon\leq 1$.
\end{proof}

Inequality \eqref{eq9999.050} was checked against the known numerical data for both primes and probable primes, \cite[Table 3]{LR13}, every entry $\left |\tau\left (p^{2n}\right )\right |=\left |\tau\left (p^{q-1}\right )\right |$ is well above the estimated lower bound.

\section{Fourier Coefficients Estimate I}\label{s9977}
The lower bound considered here is based on the oldest result for the irrationality measure of algebraic real numbers. Nevertheless, it produces an effective result, of nearly the same strength as the Atkin-Serre conjecture, see \eqref{eq9999.022}. The concept of measures of irrationality of real numbers is discussed in \cite[Section 7.4]{NI56}, et alii.

\begin{thm} {\normalfont}\label{thm2277.106} Let $p\geq 2$ be a prime, and let $n\geq 1$ be an integer. Then,
\begin{equation}\label{eq9977.180}
\left |\tau(p^{n})\right |>A^{-1}p^{7n/2-2-\varepsilon},
\end{equation}
where $A=(n+1)^32^{n+1}$, and $\varepsilon>0$ is a small number.
\end{thm}

\begin{proof} Consider the algebraic number $\beta=\alpha_p^{n+1} \overline{\alpha_p}^{-(n+1)}$ of degree $n+1$. Replacing the estimate in Theorem \ref{thm2002.530} in equation \eqref{eq9977.082}, yields
\begin{eqnarray} \label{eq9977.182}
\left |\tau(p^n)\right |
&\geq& p^{11n/2}\left | \alpha_p^{n+1} \overline{\alpha_p}^{-(n+1)}-1 \right |\\
&\geq& p^{11n/2}\cdot \frac{1}{Ap^{2(n+1)}}\nonumber,
\end{eqnarray}
where $A=(n+1)^32^{n+1}$, for all $p^{2n}\geq 10^{600}$.
\end{proof}

Inequality \eqref{eq9977.180} was checked against the known numerical data for both primes and probable primes, \cite[Table 3]{LR13}, every entry $\left |\tau\left (p^{2n}\right )\right |=\left |\tau\left (p^{q-1}\right )\right |$ is well above the estimated lower bound.

\begin{exa}{\normalfont A few cases were computed to illustrate the inequality in previous result.
\begin{enumerate}
\item The largest known prime coefficient $\left |\tau\left (p^{2n}\right )\right |=\left |\tau\left (p^{q-1}\right )\right |\approx 10^{d}$ has the parameters $p=157$, $q=2n+1=2207$, $d=26643$ digits, and $\varepsilon=1$. The inequality is
\begin{equation}
\left |\tau\left (p^{q-1}\right )\right |\approx 10^{26643}\geq \frac{p^{7n-2-\varepsilon}}{(2n+1)^32^{2n+1}}\approx 10^{16275}.
\end{equation}
\item The largest known probable prime coefficient $\left |\tau\left (p^{2n}\right )\right |=\left |\tau\left (p^{q-1}\right )\right |\approx 10^{d}$ has the parameters $p=41$, $q=2n+1=28289$, $d=250924$ digits, and $\varepsilon=1$. The inequality is
\begin{equation}
\left |\tau\left (p^{q-1}\right )\right |\approx 10^{250924}\geq \frac{p^{7n-2-\varepsilon}}{(2n+1)^32^{2n+1}}\approx 10^{151146}.
\end{equation}
\end{enumerate}
}
\end{exa}

\section{Fourier Coefficients Estimate II}\label{s9977}
An explicit lower bound for the Tau function at the prime powers, based on linear forms in logarithm, is derived below. This lower bound is similar to \cite[Lemma 1]{MS87}, but a full proof is given here.

\begin{lem} {\normalfont}\label{lem2277.006} Let $p\geq 2$ be a prime, and let $n\geq 2$ be an integer. If $\tau(p)\ne0$, then
\begin{equation}
\left |\tau(p^n)\right |>p^{11n/2-c_0\log n},
\end{equation}
where $c_0=6.8 \times 10^{10}$.
\end{lem}

\begin{proof} Modify the Binet formula into the following form
\begin{eqnarray} \label{eq9977.080}
\tau(p^n)&=&\frac{\alpha_p^{n+1}-\overline{\alpha_p}^{n+1}}{\alpha_p-\overline{\alpha_p}}\\
&=&\frac{\alpha_p+\overline{\alpha_p}}{\alpha_p+\overline{\alpha_p}} \cdot\frac{\alpha_p^{n+1}-\overline{\alpha_p}^{n+1}}{\alpha_p-\overline{\alpha_p}}\nonumber\\
&=&\left (\alpha_p+\overline{\alpha_p}\right ) \cdot\frac{\alpha_p^{n+1}-\overline{\alpha_p}^{n+1}}{\alpha_p^2-\overline{\alpha_p}^2}\nonumber\\
&=&\left (\alpha_p+\overline{\alpha_p}\right ) \cdot\frac{\overline{\alpha_p}^{n+1}}{\alpha_p^2-\overline{\alpha_p}^2} \left (\alpha_p^{n+1} \overline{\alpha_p}^{-(n+1)}-1\right )\nonumber.
\end{eqnarray}

Replace the algebraic integer $\alpha_p=p^{11/2}e^{i\theta_p}$, where $0< \theta_p< \pi$. Taking absolute value and simplifying yield

\begin{eqnarray} \label{eq9977.082}
\left |\tau(p^n)\right |&=&\left |p^{11/2}\cos(\theta_p)\right |
\left |\frac{p^{11(n+1)/2}e^{i\theta_p}}{2p^{11}\sin(2\theta_p)}\right |
\left | \alpha_p^{n+1} \overline{\alpha_p}^{-(n+1)}-1 \right | \\
&\geq& p^{11n/2}\left | \alpha_p^{n+1} \overline{\alpha_p}^{-(n+1)}-1 \right |\nonumber,
\end{eqnarray}
since $\tau(p)\ne0$ implies $\theta\ne\pi/2$.
Replacing the estimate in Lemma \ref{lem2222.006} for $n\geq 1$, yields
\begin{eqnarray} \label{eq9977.084}
p^{11n/2}\left | \alpha_p^{n+1} \overline{\alpha_p}^{-(n+1)}-1 \right |
&\geq & p^{11n/2}\cdot \left( e^{-c_0 \left (\log n\right ) \left (\log p\right )}\right )
\\
&\geq & p^{11n/2-c_0\log n} \nonumber,
\end{eqnarray}
where $c_0=6.8 \times 10^{10}$.
\end{proof}

\section{Basic Diophantine Approximations Results} \label{S2002}
\begin{thm}{\normalfont (\cite[Theorem 7.8]{NI56})} \label{thm2002.500} A real algebraic number $\alpha>0$ of degree $n\geq 1$ is not approximable to order $n+1$ or higher. In addition, there is a constant $A\geq 1$ such that
\begin{enumerate} [font=\normalfont, label=(\roman*)]
\item $ \displaystyle
\left |\alpha-\frac{p_m}{q_m}\right |>\frac{1}{A q_m^n},
$
\item $ \displaystyle
\left |\alpha-\frac{p_m}{q_m}\right |\leq\frac{1}{A q_m^{n+1}},$
\end{enumerate}
as $q_m \to \infty.$
\end{thm}
\begin{proof} The proof is widely available in the literature, including the cited reference.
\end{proof}

\begin{cor} \label{cor2002.500}Let $\beta$ be an algebraic number such that $\beta\approx 1$, and let $\{p_m/q_m:m\geq1\}$ be the sequence of convergents. Then,
\begin{enumerate} [font=\normalfont, label=(\roman*)]
\item $ \displaystyle
\left |\beta-\frac{p_m}{q_m}\right |>\frac{1}{n^32^n q_m^n},
$
\item $ \displaystyle
\left |\beta-\frac{p_m}{q_m}\right |\leq\frac{1}{n^32^nq_m^{n+1}},$
\end{enumerate}
as $q_m \to \infty.$
\end{cor}
\begin{proof} Let $\beta$ be a root of the polynomial $f(x)=a_0x^{n}+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$ of degree $\deg f=n$. Since $x=p_m/q_m\approx 1$, the rational approximations are in the range $0<1-p_m/q_m\leq |x|\leq 1+p_m/q_m<2$. In addition, the coefficients $a_k$ coincides with the symmetric polynomials $s_k(\beta_1,\ldots, \beta_n)=a_k$ in terms of the roots $\beta=\beta_1,\ldots, \beta_n$ of the polynomial $f(x)$. Thus, $|a_k|\leq 2n$. In summary, the absolute value of the derivative satisfies
\begin{eqnarray}\label{eq2002.522}
|f^{\prime}(x)|&=&\left |na_0x^{n-1}+(n-1)a_1x^{n-2}+\cdots+a_{n-1}\right |\\
&<&n^32^n=A\nonumber,
\end{eqnarray}
see \cite[Equation 7.4]{NI56} for more details.
\end{proof}

\begin{thm} \label{thm2002.530} Let $\beta$ be an algebraic integer of degree $n+1$, such that $\beta\approx 1$, and let $\{p_m/q_m:m\geq1\}$ be the sequence of convergents. Then,
\begin{equation}\label{eq2002.532}
\left |\beta-1\right |>\frac{1}{(n+1)^32^{n+1} p^{2(n+1)} },
\end{equation}
as $q_{m_p} \to \infty.$
\end{thm}

\begin{proof} Let $\beta=\alpha_p^{n+1} \overline{\alpha_p}^{-(n+1)}$. The inverse triangle inequality, and Corollary \ref{cor2002.500} lead to
\begin{eqnarray}\label{eq2002.534}
\left| \beta-1\right | &=&\left| \beta-1+ \frac{p_m}{q_m}-\frac{p_m}{q_m}\right |\\
&\geq&\left| \left |\beta- \frac{p_m}{q_m}\right |- \left| 1-\frac{p_m}{q_m}\right |\right |\nonumber \\
&\geq& \left |\beta- \frac{p_m}{q_m}\right |\nonumber \\
&\geq&\frac{1}{(n+1)^32^{n+1} q_m^{n+1}} \nonumber.
\end{eqnarray}
For large prime $p\geq 2$, let $p_{m_p}/q_{m_p}$ be a subsequence of convergents indexed by $p$ such that $q_{m_p}\leq p^2$. Then,
\begin{equation}\label{eq2002.536}
\left| \beta-1\right | >\frac{1}{(n+1)^32^{n+1} p^{2(n+1)} },
\end{equation}
as $q_{m_p} \to \infty.$
\end{proof}

\section{Result For Linear Forms In Logarithms}\label{s2222}

\begin{thm}\label{thm2222.002} {\normalfont(\cite[Theorem 3.1]{BA75})} Let $\alpha_1, \alpha_2, \ldots, \alpha_d$ be nonnegative algebraic numbers, and let $b_1,b_2, \ldots, b_d$ be rational integers. If
\begin{equation} \label{eq2222.024}
\Lambda= b_1\log \alpha_1+ b_2\log \alpha_2^{b_2}+ \cdots +b_d\log \alpha_d\ne0,
\end{equation}
is not a null period, then
\begin{equation} \label{eq2222.024}
|\Lambda|>e^{-C},
\end{equation}
where
\begin{equation} \label{eq2222.026}
B\geq \max \left \{ | b_1|,| b_2|,\ldots ,| b_d| \right \} ,
\end{equation}
and
\begin{equation} \label{eq2222.028}
A_i\geq \max \left \{ dh(\alpha_i),| h(\alpha_i)|,0.16 \right \}
\end{equation}
for $i=1,2, \ldots d$.
\end{thm}

Explicit versions of Baker theorem are widely used in Diophantine analysis. The height of a polynomial of degree $d=\deg f\geq 1$,

\begin{equation} \label{eq2222.020}
f(x)=a_d\prod_{1\leq i\leq d}\left (x-\alpha_i \right ),
\end{equation}
where $a_d>0$, is defined by
\begin{equation} \label{eq2222.022}
h(\alpha)= \frac{1}{d} \left ( \log a_0+\sum_{1\leq i\leq d}\log \left (\max \{ | \alpha_i|,1 \}  \right )\right ).
\end{equation}

\begin{thm} \label{thm2222.002} {\normalfont(Matveev)} Let $\alpha_1, \alpha_2, \ldots, \alpha_d$ be nonnegative algebraic numbers in a number field $\mathcal{K}$ of degree $k=[\mathcal{K}:\Q]$, and let $b_1,b_2, \ldots, b_d$ be rational integers. If
\begin{equation} \label{eq2222.024}
\Lambda= \alpha_1^{b_1} \alpha_2^{b_2} \cdots \alpha_d^{b_d}-1
\end{equation}
is not zero, then
\begin{equation} \label{eq2222.024}
|\Lambda|>e^{-1.4\times30^{d+3}d^{4.5}k^2\left (1+\log d\right ) \left (1+\log B\right )A_1 A_2 \cdots A_d},
\end{equation}
where
\begin{equation} \label{eq2222.026}
B\geq \max \left \{ | b_1|,| b_2|,\ldots ,| b_d| \right \} ,
\end{equation}
and
\begin{equation} \label{eq2222.028}
A_i\geq \max \left \{ dh(\alpha_i),| h(\alpha_i)|,0.16 \right \}
\end{equation}
for $i=1,2, \ldots d$.
\end{thm}

The version stated above appears in \cite[Theorem 2]{AT19}, and \cite[p.\ 4]{AF12}. Observe that
\begin{equation}\label{eq2222.030}
\left |\Lambda \right |= \left |\alpha_1^{b_1} \alpha_2^{b_2} \cdots \alpha_d^{b_d}-1\right |
\geq \left | \left | \alpha_1\right |^{b_1} \left |\alpha_2\right |^{b_2} \cdots \left |\alpha_d\right |^{b_d}-1\right
|.
\end{equation}
Thus, the result is applicable to any algebraic integers $\alpha_1, \alpha_2, \ldots, \alpha_d$.\\

Under the condition $\alpha \ne 1$, an estimate of the lower bound of the expression $\alpha^n-1$ can be achieved via the basic Diophantine inequality
\begin{eqnarray}\label{eq2222.034}
\left| \alpha^n-1\right | &=&\left| \alpha^n-1+ \frac{p_m}{q_m}-\frac{p_m}{q_m}\right |\\
&\geq&\left| \left |\alpha^n- \frac{p_m}{q_m}\right |- \left| 1-\frac{p_m}{q_m}\right |\right |\nonumber \\
&\geq&\frac{1}{2q_{m+1}} \nonumber,
\end{eqnarray}
where $p_m/q_m$ is the sequence of convergents of the irrational number $\alpha^n$. But, Theorem \ref{thm2222.002} provides an explicit value for any fixed integer $n\geq1$.

\begin{lem}\label{lem2222.006} Let $\alpha_p$ be a root of the polynomial $X^2-\tau(p) X+p^{11}$. Then,
\begin{equation}\label{eq2222.027}
|\Lambda_p(n)|=\left |  \alpha_p^{n+1} \overline{\alpha_p}^{-(n+1)}-1\right |>p^{-c_0\log n}
\end{equation}
where $c_0=6.8 \times 10^{10}$, and any integer $n\geq 1$.
\end{lem}

\begin{proof} Let $\alpha_p=p^{11/2}e^{i\theta_p}$, where $0\leq \theta_p\leq \pi$, be the root of the polynomial $f(X)=a_2X+a_1X+a_0=X^2-\tau(p) X+p^{11}$. Consider the algebraic integer
\begin{equation} \label{eq2222.024}
\Lambda_p(n)= \alpha_1^{b_1} \alpha_2^{b_2} \cdots \alpha_n^{b_d}-1= \alpha_p^{n+1} \overline{\alpha_p}^{-(n+1)}-1.
\end{equation}
Here, the parameters are:
\begin{enumerate}
\item $[\mathcal{K}:\Q]=k=2$
\item $\deg f=d=2$
\item $\displaystyle B\geq \max \left \{ | b_1|,| b_2|,\ldots ,| b_d| \right \}=n+1$
\item $\displaystyle h(\alpha)= \frac{1}{d} \left ( \log a_0+\sum_{1\leq i\leq d}\log \left (\max \{ | \alpha_i|,1 \}  \right )\right )\leq 2\log p^{11}$
\item $\displaystyle A_i\geq \max \left \{ dh(\alpha_i),| h(\alpha_i)|,0.16 \right \}=\log p^{11/2}$.
\end{enumerate}
The lower bound is
\begin{eqnarray}\label{eq2222.032}
|\Lambda_p(n)|&>&e^{-1.4\times30^{d+3}d^{4.5}k^2\left (1+\log d\right ) \left (1+\log B\right )A_1 A_2 \cdots A_d}\\
&\geq&e^{-c_0\left (\log n\right ) \left (\log p\right )}\nonumber\\
&\geq&p^{-c_0\log n}\nonumber,
\end{eqnarray}
where $c_0=6.8 \times 10^{10}$.
\end{proof}

\section{Fourier Coefficients Formulas}\label{s9922}
Some of the theory of modular forms
\begin{equation} \label{eq9999.003}
f(s)=\sum_{n \geq1} \lambda(n)q^n=\prod_{p \mid N} \left (\frac{1}{1-p^{(k-1)s}}\right )\prod_{p \nmid N} \left (\frac{1}{1-\lambda(p)-p^{(k-1)s}}\right ),
\end{equation}
where $s\in \mathbb{H}$ is a complex number in the upper half plane, $q=e^{i2\pi }$, are introduced in \cite{KN84}, et alii.
\begin{lem} {\normalfont(\cite{ML17})}\label{lem2222.002} Let $m,n\geq 1$ be a pair of integers, and let $p\geq 2$ be a prime. If $\lambda(n)$ is the $n$th Fourier coefficient of a modular form $f(z)=\sum_{n\geq 1}\lambda(n)q^n$ of weight $k\geq 4$, then
\begin{enumerate}[font=\normalfont, label=(\roman*)]
\item $\displaystyle \lambda(mn)=\lambda(m)\lambda(n)$ if $\gcd(m,n)=1$.
\item $\displaystyle \lambda(p^n)=\lambda(p)\lambda(p^{n-1})-p^{(k-1)/2}\lambda(p^{n-2})$ if $n\geq 2$.
\end{enumerate}
\end{lem}
Finer details appears in \cite[Proposition 32]{KN84} and similar references. These are compactly expressed by Mordell formula
\begin{equation}\label{eq2222.040}
\lambda(m)\lambda(n)=\sum_{d\mid \gcd(m,n)}d^{k-1} \lambda(mn/d^2).
\end{equation}

\begin{lem} {\normalfont}\label{lem2222.044} For any prime $p\geq 2$, and an integer $n\geq 1$, $\lambda(p^n)\in \Z[p^{(k-1)/2},\lambda(p)]$ has an expansion as a polynomial
\begin{equation}\label{eq2222.040}
\lambda(p^n)=\sum_{0\leq k\leq n/2}(-1)^k \binom{n-k}{n-2k}p^{11k}\lambda(p)^{n-2k}
\end{equation}
of one variable $\lambda(p)$ of degree $n$.
\end{lem}

\begin{exa} \label{exa2222.009}{\normalfont The weight $k=12$ coefficient $\tau(p^n)$. The first few polynomials are:
\begin{enumerate}
\item $\displaystyle \tau(p)$
\item $\displaystyle \tau(p^2)=\tau(p)^2-p^{11}$
\item $\displaystyle \tau(p^3)=\tau(p)\tau(p^2)-p^{11}\tau(p)=\tau(p)\left( \tau(p)^2-p^{11}\right )-p^{11}\tau(p)=\tau(p)^3-2p^{11}\tau(p)$.
\item $\displaystyle \tau(p^4)=\tau(p)\tau(p^3)-p^{11}\tau(p^2)=\tau(p)^4-3p^{11}\tau(p)^2+p^{22}$.
\end{enumerate}

}
\end{exa}

\begin{dfn}\label{dfn2222.000} A weight $k\geq2$ coefficient $\lambda(p^n)$ has a representation as
\begin{equation}
\lambda\left (p^{2n}\right )=2p^{(k-1)n}\cos \theta,
\end{equation}
where $0\leq \theta<\pi$.
\end{dfn}

\section{Smallest Prime Values}
The absolute smallest prime value is
\begin{equation} \label{eq9999.105}
\tau\left (251^2\right )=-80561663527802406257321747,
\end{equation}
the primality test analysis appears in \cite{LD65}. Recent computer studies have unearthed more primes. Some information on for the first few prime values are tabulated below. A larger table appears in \cite[Table 3]{LR13}.\\

 \begin{table}[h!]
\centering
\caption{Smallest Prime Values of $\tau(p^{2n})$} \label{t9001}
\begin{tabular}{c|c|c|c}
$\tau(p^{2n})$&Number of Decimal Digits&$\tau(p^{2n})$&Number of Decimal Digits\\
\hline
 $\tau(251^2)$ & 26&$\tau(677^2)$ & 32\\
 \hline
 $\tau(971^2)$ & 33&$\tau(983^2)$ & 33\\
\hline
 $\tau(47^4)$ & 37&$\tau(197^4)$ & 50\\
\end{tabular}
\end{table}

\section{Problems}\label{p9988}
\begin{exe} {\normalfont Let $k=12, 14, 16, 18, 20, 22, 26$ be the weight, and let $N=1$ be the level of the Fourier coefficient $\lambda(n)$. Prove that $\lambda(n)\ne 0$ for $n \geq 1$.
}
\end{exe}

\begin{exe} {\normalfont Let $k\geq 1$ be the weight, and let $N\geq1$ be the level of the Fourier coefficient $\lambda(n)$. Prove or disprove that the sign of $\lambda(n)$ is a random variable. Is there a deterministic algorithm to compute it?
}
\end{exe}

\begin{exe} {\normalfont Let $k\geq 1$ be the weight, and let $N\geq1$ be the level of the Fourier coefficient $\lambda(n)$. Prove or disprove that the number of prime values
$$\pi_{\lambda}(x,k,N)=\#\{\lambda(n)\leq x\}=c_{\lambda}(k,N)\log \log x+o(\log \log x,$$
where $c_{\lambda}(k,N)\geq 0$ is a constant. The heuristic argument for $\lambda(n)=\tau(n)$ appears in \cite{LR13}.
}
\end{exe}
\begin{exe} {\normalfont Let $k\geq 1$ be the weight, and let $N\geq1$ be the level of the Fourier coefficient $\lambda(n)$. Estimate the maximal prime gap
$$\left |\lambda(p^m)\right |-\left |\lambda(p^n)\right |,$$
where $p\in \tP=\{2,3,5, \ldots\}$, and $m,n \in \N=\{1,2,3, \ldots\}$. Does it depends on the rational prime gap $p_{n+1}-p_n$?
}
\end{exe}

\begin{exe} {\normalfont Let $k\geq 1$ be the weight, and let $N\geq1$ be the level of the Fourier coefficient $\lambda(n)$, and no complex multiplication. Does $\lambda(p)\ne0$ for any prime $p\in \tP=\{2,3,5, \ldots\}$ if and only if $k\geq 4$?
}
\end{exe}
\newpage

config: arxiv


TITLE: Matrix Algebra Questions
QUESTION [0 upvotes]: I would like you to help me with two questions I am stuck in. You can call these homework questions. It would be helpful if you can give me non-trivial hints instead of complete solution.
1) Let $A$ be a matrix such that $Rowspace(A)=Coulmnspace(A)$. Prove or disprove that A is symmetric.
2) Suppose $A$ and $B$ are two matrices such that $Columnspace(A)=Columnspace(B)$ and $BA=AB$. Prove or disprove that $A=B$

REPLY [1 votes]: Counterexample for both:
$$
A = \pmatrix{0&1&0\\0&0&1\\1&0&0}, \quad B = \pmatrix{1&0&0\\0&1&0\\0&0&1}
$$


TITLE: What is the meaning of synchronization in special relativity?
QUESTION [0 upvotes]: Suppose there are two observers $A$ and $B$ (or reference frames if you will), let us label their time coordinates by $t_A$ and $t_B$ respectively.
When  we say we synchronize the clocks of $A$ and $B$, we impose $ t_A= t_B $, wouldn't this mean that if $t_A=t_B$ for the same events, what is simultaneous for $A$ is simultaneous for $B$?

REPLY [1 votes]: No, if A and B are moving relative to each other, you can only make their two times the same at a given point- everywhere else they will be out of synch. You mind find it easier to visualise if you forget about time for a moment and consider a spatial analogy as follows...
Suppose you and I are each make a paper map of map a town using x and y coordinates, with a particular landmark being our common origin. When we have finished, we realised that we have each used a different baseline, so our x and y frames are tilted relative to each other. At the landmark we each agree that our respective x and y coordinates are zero, but everywhere else your x and y coordinates are different to mine.
Now carry that principle over to time. We can have a common origin of time at a particular point, but our two planes of simultaneity will be tilted (just as our maps were in the spatial example), so everywhere away from our common origin we will disagree upon the time.


text:
\begin{document}

\title[Spectral Properties]{Some spectral properties of pseudo-differential operators on the Sierpinski Gasket}

\begin{abstract} We prove versions of the strong Sz\"ego limit theorem for certain classes of pseudodifferential operators defined on the Sierpi\'nski gasket. Our results used in a fundamental way the existence of localized eigenfunctions for the Laplacian on this fractal.

\end{abstract}

\date{\today}
\author{Marius Ionescu}
\address{Department of Mathematics,  United States Naval Academy, Annapolis,
  MD, 21402-5002, USA}

\email{felijohn@gmail.com}
\thanks{This work was partially supported by a grant from the Simons Foundation (\#209277 to Marius Ionescu).}
\thanks{The first author would like to thank Kasso Okoudjou and the Department
  of Mathematics at the University of Maryland, College Park, and the
  Norbert-Wiener Center for Harmonic Analysis and Applications for
  their hospitality.}

 \author{Kasso~A. Okoudjou}
\address{Department of Mathematics, University of Maryland, College
  Park, MD 20742-4015, USA}
\email{kasso@math.umd.edu}

\author{Luke~G. Rogers}
\address{Department of Mathematics, University of Connecticut, Storrs,
  CT 06269-3009, USA}
\email{rogers@math.uconn.edu}
\maketitle

\section{Introduction}
The results of this paper have their origin in a celebrated theorem of
Szeg\"{o}, who proved that if $P_{n}$ is projection onto the span of
$\{e^{im\theta},0\leq m\leq n\}$ in $L^{2}$ of the unit circle and
$[f]$ is multiplication by a  positive $C^{1+\alpha}$  function for
$\alpha>0$ then $(n+1)^{-1}\log\det P_{n}[f]P_{n}$ converges to
$\int_{0}^{2\pi} \log f(\theta)\,d\theta/2\pi$.  Equivalently,
$(n+1)^{-1} \operatorname{Trace} \log P_{n}[f]P_{n}$ has the same
limit.  The literature which has grown from this result is vast,
see~\cite{Simon}; here we are interested in the generalizations that
replace $[f]$ with a pseudodifferential operator defined on fractal sets, in which case the
Szeg\"{o} limit theorem may be viewed as a way to obtain asymptotic
behavior of the operator using only its symbol. In the setting of
Riemannian manifolds there are important results of this latter type  due to Widom
\cite{Wid_JFA79} and Guillemin \cite{Gui_AnnMathSt79}.

In~\cite{OkRoStr_JFAA10} a standard generalization of the classical
Szeg\"{o} theorem was proved in the setting where the underlying space
is the Sierpi\'{n}ski Gasket and $P_{\Lambda}$ is the projection onto the
eigenspace with eigenvalues less than $\Lambda$ of the Laplacian
obtained using analysis on fractals.  In this situation the projection
is not Toeplitz, so most classical techniques fail and the proofs
in~\cite{OkRoStr_JFAA10} rely instead on the fact that most
eigenfunctions are localized.  The present work continues this
development to consider the case where $[f]$ is replaced by a
pseudodifferential operator.  Our results are of three related types.
Our main result, Theorem~\ref{thm:Mainthm}, gives asymptotics when one
considers $\operatorname{Trace} F( P_{n}p(x,-\Delta)P_{n})$ where
$p(x,-\Delta)$ is the pseudodifferential operator and $F$ is
continuous on an interval containing the spectrum of
$P_{n}p(x,-\Delta)P_{n}$.  Theorem~\ref{thm:GeneralSzego}, which is in
some sense a special case of the main result but for which we give a
different proof, considers the classical case of $\log\det
P_{n}p(x,-\Delta)P_{n}$.
Theorem~\ref{thm:specclusters} gives the asymptotics of clusters of
eigenvalues of a pseudodifferential operator in terms of the symbol.
   In all cases the proofs rely on the
predominance of localized eigenfunctions in the Laplacian spectrum on
the Sierpi\'{n}ski Gasket.

\section{\label{sec:Background}Background}

\subsection{Analysis on the Sierpi\'nski gasket}\label{ssec:analysisonSG}
The Sierpi\'{n}ski gasket $X$ is the unique non-empty compact fixed
set of the iterated function system
$\{F_{j}=\frac{1}{2}(x-a_{j})+a_{j}\}$, $j=1,2,3$,
where $\{a_{j}\}$ are not co-linear in $\mathbb{R}^{2}$.  For
$w=w_{1}\dotsm w_{N}$ a word of length $N$ with all
$w_{j}\in\{1,2,3\}$ let $F_{w}=F_{w_{1}}\circ\dotsm\circ F_{w_{N}}$
and call $F_{w}(X)$ an $N$-cell. We will sometimes write the
decomposition into $N$-cells as
$X=\cup_{i=1}^{3^{N}}C_{i}$ where each $C_{i}=F_{w}(X)$ for a word $w$ of
length $N$.

Equip $X$ with the unique probability measure $\mu$ for which an
$N$-cell has measure $3^{-N}$ and the symmetric self-similar
resistance form $\mathcal{E}$ in the sense of Kigami~\cite{Kig_CUP01}.
The latter is a Dirichlet form on $L^{2}(\mu)$ with domain
$\mathcal{F}\subset C(X)$ and by the theory of such forms
(see~\cite{FOT}) there is a negative definite self-adjoint Laplacian
$\Delta$ defined by $\mathcal{E}(u,v)=\int (-\Delta u)v\,d\mu$ for all
$v\in\mathcal{F}$ such that $v(a_{j})=0$ for $j=1,2,3$.  This
Laplacian is often called the Dirichlet Laplacian to distinguish it
from the Neumann Laplacian which is defined in the same way but
omitting the condition $v(a_{j})=0$.

A complete description of the spectrum of $\Delta$ was given in~\cite{FukShi_92}
using the spectral decimation formula of~\cite{rammal1983random}.
The eigenfunctions of $\Delta$ are described in~\cite{DalStrVin_FDESG_JFAA99}
(see \cite{Tep_JFA98,OkRoStr_JFAA10,OkStr_PAMS07,BaKi_JLMS97} or
\cite{Str_Prin06} for proofs).   Without going into details, the features needed for our work are as follows.
The spectrum  is discrete and decomposes naturally into three sets
called the $2$-series, $5$-series and $6$-series eigenvalues, which
further decompose according to the \emph{generation of birth}, which
is a number $j\in\mathbb{N}$.  All $2$-series eigenvalues have
$j=1$ and multiplicity $1$, each $j\in\mathbb{N}$ occurs in the
$5$-series and the corresponding eigenspace has multiplicity
$(3^{j-1}+3)/2$, and each $j\geq2$ occurs in the $6$-series with
multiplicity $(3^{j}-3)/2$.  Moreover the $5$ and $6$-series
eigenvalues are localized.  Suppose $j>N\ge1$ and
$\{C_{i}\}_{i=1}^{3^{N}}$ is the $N$-cell decomposition as above.   If
$\lambda$ is a $5$-series eigenvalue with generation of birth $j$ and
eigenspace $E_{j}$ then there is a basis for $E_{j}$ in which
$(3^{j-1}-3^{N})/2$ eigenfunctions are localized at level $N$, meaning
that they are each supported on a single $N$-cell, and the remaining
$(3^{N}-3)/2$ are not localized at level $N$; the number localized on
a
single $N$-cell is $(3^{j-N-1}-1)/2$.  If $\lambda$ is a $6$-series
eigenvalue with generation of birth $j$ and eigenspace $E_{j}$ then
there is a basis for $E_{j}$ in which $(3^{j}-3^{N+1})/2$ of the
eigenfunctions are localized at level $N$, the
remaining $(3^{N+1}-3)/2$ eigenfunctions are not localized at level
$N$, and the number supported on a single $N$-cell is $(3^{j-N}-3)/2$  (see, for example~\cite{OkRoStr_JFAA10}).

\subsection{Pseudo-differential operators on the Sierpi\'nski gasket}

We recall some of the theory developed in~\cite{IoRoStr_PSDO_2011} and use it to define
pseudo-differential operators on the Sierpi\'nski gasket.  Let $\Delta$ be
the Dirichlet Laplacian. Then the
spectrum $\operatorname{sp}(-\Delta)$ of $-\Delta$ consists of finite-multiplicity
eigenvalues that accumulate only at $\infty$ (\cite{Kig_CUP01,Str_Prin06}).  Arranging them as
$\operatorname{sp}(-\Delta)=\{\lambda_{1}\le\lambda_{2}\le\dots\le\lambda_{n}\le\dots\}$
with $\lim_{n\to\infty}\lambda_{n}=\infty$, let $\{\varphi_{n}\}_{n\in\mathbb{N}}$
be an orthonormal basis of $L^{2}(\mu)$ such that $\varphi_{n}$
is an eigenfunction with eigenvalue $\lambda_{n}$ for all $n\ge1$. The
set $D$ of finite linear combinations of $\varphi_{n}$ is dense in
$L^{2}(\mu)$.

If $p:(0,\infty)\to\mathbb{C}$ is a measurable then
\[
p(-\Delta)u=\sum_{n}p(\lambda_{n})\langle u,\varphi_{n}\rangle\varphi_{n}
\]
for $u\in D$ gives a densely defined operator on $L^{2}(\mu)$
called a \emph{constant coefficient pseudo-differential operator}.  If $p$ is bounded, then $p(-\Delta)$ extends
to a bounded linear operator on $L^{2}(\mu)$ by the spectral theorem.

If $p$ is a $0$-symbol in the sense of
\cite[Definition 3.1]{IoRoStr_PSDO_2011} then $p(-\Delta)$ is a pseudo-differential
operator of order $0$.  Moreover, it is a singular
integral operator on $L^{2}(\mu)$ (\cite[Theorem
3.6]{IoRoStr_PSDO_2011}) and therefore extends to a bounded operator
on $L^q(\mu)$ for all $q\in (1,\infty)$.

If $p:X\times(0,\infty)\to\mathbb{C}$ is measurable
we define a \emph{variable coefficient pseudo-differential operator}
$p(x,-\Delta)$ as in \cite[Definition 9.2]{IoRoStr_PSDO_2011}:
\begin{equation}
p(x,-\Delta)u(x)=\sum_{n}\int_{X}p(x,\lambda_{n})P_{\lambda_{n}}(x,y)u(y)d\mu(y)\label{eq:psdo}
\end{equation}
for $u\in D$, where $\{P_{\lambda_{n}}\}_{n\in\mathbb{N}}$ is
the spectral resolution of $-\Delta$. If $p$ is a 0-symbol in the
sense of \cite[Definition 9.1]{IoRoStr_PSDO_2011}, then $p(x,-\Delta)$
extends to a bounded linear operator on $L^{q}(\mu)$ for all $q\in
(1,\infty)$ (\cite[Theorem 9.3 and Theorem 9.6]{IoRoStr_PSDO_2011}).
If $p:X\to\mathbb{R}_{+}$ is bounded and \emph{independent} of $\lambda$,
then the operator that $p$ determines on $L^{2}(\mu)$ is multiplication
by $p(x)$. In this case, we write $[p]$ for this operator, following the notation in \cite{OkRoStr_JFAA10}.

We describe next the relationship between the spectrum of a constant
coefficient pseudo-differential operator $p(-\Delta)$ and $\operatorname{sp}(-\Delta)$
for a continuous map $p$. Of course, the result is valid for all
the spaces considered in \cite{IoRoStr_PSDO_2011}, not only the Sierpi\'nski
gasket.
\begin{prop}
\label{prop:spectrum_psdo}If  $p:(0,\infty)\to\mathbb{C}$
is continuous then $\operatorname{sp}p(-\Delta)=\overline{p(\operatorname{sp}(-\Delta))}$.\end{prop}
\begin{proof}
Let $\lambda$ be in the resolvent $\rho(p(-\Delta))$ of $p(-\Delta)$.
Then $\lambda I-p(-\Delta)$ has bounded inverse, so there
is $C>0$ such that  $\Vert(\lambda I-p(-\Delta))^{-1}u\Vert\le C\Vert u\Vert$
for all $u\in L^{2}(\mu)$. Let $v\in\operatorname{dom}p(-\Delta)$.
Then $(\lambda I-p(-\Delta))v\in L^{2}(\mu)$ and $\Vert(\lambda I-p(-\Delta))v\Vert\ge\frac{1}{C}\Vert v\Vert$.
 In particular, if $v=\varphi_{n}$ is an element of our $L^{2}(\mu)$ basis of eigenfunctions
of $-\Delta$ we obtain $\Vert(\lambda I-p(-\Delta))\varphi_{n}\Vert\ge1/C$. Since $p(-\Delta)\varphi_{n}=p(\lambda_{n})\varphi_{n}$
and $\Vert\varphi_{n}\Vert=1$ for all $n\in\mathbb{N}$, it follows
that $\lambda\notin\overline{p(\operatorname{sp}(-\Delta))}$.

For the converse, suppose $z\notin\overline{p(\operatorname{sp}(-\Delta))}$.
Then there is $K>0$ such that $\vert p(\lambda)-z\vert\ge K>0$ for
all $\lambda\in\operatorname{sp}(-\Delta)$ and thus $\vert p(\lambda)-z\vert^{-1}\le K^{-1}$
for all $\lambda\in\operatorname{sp}(-\Delta)$. Therefore $(p(-\Delta)-z)^{-1}$
is bounded on $L^{2}(\mu)$ and  $z\in\rho(p(-\Delta))$.
\end{proof}

\begin{cor}
\label{cor:spectrumifpinfty}If $p:(0,\infty)\to\mathbb{R}$ is
continuous and $\lim_{\lambda\to\infty}p(\lambda)=\infty$
then $\operatorname{sp}p(-\Delta)=p(\operatorname{sp}(-\Delta))$.\end{cor}
\begin{proof}
The hypothesis implies that the only accumulation point for $\{p(\lambda_{n})\}_{n\in\mathbb{N}}$
is $\infty$ so we can apply  Proposition \ref{prop:spectrum_psdo}.
\end{proof}

\section{Szeg\"{o} Limit Theorems for pseudo-differential operators on the
Sierpi\'nski gasket\label{sec:Szeg-Limit-Theorems}}

Let $X$, $\mu$ and $\Delta$  be Section \ref{sec:Background}.
We also follow the notation of Section 4 of \cite{OkRoStr_JFAA10}:
for $\Lambda>0$,  let $E_{\Lambda}$ be the span of all eigenfunctions
corresponding to eigenvalues $\lambda$ of $-\Delta$ with $\lambda\le\Lambda$, let $P_{\Lambda}$ be the orthogonal projection onto $E_{\Lambda}$, and
set $d_{\Lambda}$ to be the dimension of $E_{\Lambda}$. For an eigenvalue
$\lambda$ of $-\Delta$, write $E_{\lambda}$ for the eigenspace
of $\lambda$, $P_{\lambda}$ for the orthogonal projection onto $E_{\lambda}$,
and $d_{\lambda}$ for the dimension of $E_{\lambda}$.

\subsection{Trace-type Szeg\"{o} limit theorems}\label{subsec:trace}
\global\long\def\Tr{\operatorname{Tr}}
Fix a measurable map $p:X\times(0,\infty)\to\mathbb{R}$ and
let $p(x,-\Delta)$ be the densely defined operator as in \eqref{eq:psdo}.
We assume, unless otherwise stated, that $p(\cdot,\lambda)$ is continuous
for all eigenvalues $\lambda$ of $-\Delta$ and that there is a continuous
map $q:X\mapsto\mathbb{R}$ such that the following limit exists and
is uniform in $x$:
\begin{equation}
\lim_{\lambda\in\operatorname{sp}(-\Delta),\lambda\to\infty}p(x,\lambda)=q(x)\label{eq:limit_un}.
\end{equation}
In this section we  study the asymptotic behavior of $\Tr F(P_{\Lambda}p(x,-\Delta)P_{\Lambda})$
as $\Lambda\to\infty$ for continuous functions $F$.  Our results are clearly true if $\Vert p\Vert_{\infty}=0$;  henceforth
we assume that $\Vert p\Vert_{\infty}>0$.

\begin{lem}
\label{lem:A_B}The eigenvalues of $P_{\Lambda}p(x,-\Delta)P_{\Lambda}$
are contained in a bounded interval $[A,B]$ for all $\Lambda>0$.
\end{lem}
\begin{proof}
Let $\varepsilon>0$. Then there is some $\overline{\Lambda}$ such that
if $\lambda\in\operatorname{sp}(-\Delta)$ and $\lambda>\overline{\Lambda},$
then $q(x)-\varepsilon<p(x,\lambda)<q(x)+\varepsilon$. Since
\[
\langle P_{\lambda}p(x,-\Delta)P_{\lambda}\varphi,\varphi\rangle=\int p(x,\lambda)\varphi(x)^{2}d\mu(x)
\]
for all $\varphi\in E_{\lambda}$, it follows that, as operators,
$P_{\lambda}[q-\varepsilon]P_{\lambda}\le P_{\lambda}p(x,-\Delta)P_{\lambda}\le P_{\lambda}[q+\varepsilon]P_{\lambda}$.
Now $[q+\varepsilon]$ and $[q-\varepsilon]$ are bounded
on $L^{2}(\mu)$, $P_{\overline{\Lambda}}L^{2}(\mu)$ is finite dimensional,
and
\[
P_{\Lambda}L^{2}(\mu)=\oplus_{\lambda\le\Lambda}P_{\lambda}L^{2}(\mu)=\Bigl( \oplus_{\lambda\le\overline{\Lambda}}P_{\lambda}L^{2}(\mu)\Bigr)\oplus\Bigl(\oplus_{\overline{\Lambda}<\lambda\le\Lambda}P_{\lambda}L^{2}(\mu)\Bigr),
\]
so the assertion of the lemma holds with $A$ the minimum of the smallest eigenvalue of $[q-\varepsilon]$ and the
smallest eigenvalue of
$P_{\overline{\Lambda}}p(x,-\Delta)P_{\overline{\Lambda}}$,
and $B$ the maximum of the $L^2$ norm of $[q+\varepsilon]$  and
the largest eigenvalue of $P_{\overline{\Lambda}}p(x,-\Delta)P_{\overline{\Lambda}}$.
\end{proof}

\begin{lem}
\label{lem:posfunctional}Let $\Lambda>0$. Let $p:X\times(0,\infty)\mathbb{\to}\mathbb{R}$
be a bounded measurable function such that $p(\cdot,\lambda_{n})$
is continuous for all $n\in\mathbb{N}$. Then the map on $C[A,B]$ defined by
\[
F\mapsto\frac{1}{d_{\Lambda}}\Tr F(P_{\Lambda}p(x,-\Delta)P_{\Lambda})
\]
is a continuous non-negative functional, where $A$ and $B$ are as in
Lemma \ref{lem:A_B}.\end{lem}
\begin{proof}
The map is clearly linear and non-negative. Continuity follows
immediately from the fact that $P_{\Lambda}L^{2}(X,\mu)$ is finite
dimensional.
\end{proof}

In preparation for our main result, Theorem \ref{thm:Mainthm}, we prove a version for an increasing sequence
$\{\lambda_{j}\}$ of 6-series eigenvalues where $\lambda_{j}$ has generation of birth $j$.  $E_{j}$ is the eigenspace
corresponding to $\lambda_{j}$ and has  dimension $d_{j}=(3^{j}-3)/2$. For each $1\le N<j$ we have an orthonormal basis
$\{u_{k}\}_{k=1}^{d_{j}}=\{\overline{u}_{k}\}_{k=1}^{d_{j}^{N}}\cup\{v_{k}\}_{k=1}^{\alpha^{N}}$
for $E_{j}$ , where the $\overline{u}_{k}$ are localized at scale $N$
and the $v_{k}$ are not localized at scale $N$. Then $d_{j}^{N}=(3^{j}-3^{N+1})/2$,
$\alpha^{N}=(3^{N+1}-3)/2$ and the number of eigenvalues supported
on a single $N$-cell is $m_{j}^{N}=(3^{j-N}-3)/2$. As remarked in
Section 3 of \cite{OkRoStr_JFAA10}, an analogous construction may
be done for a $5$-series eigenfunction with $d_{j}=(3^{j-1}+3)$,
$d_{j}^{N}=(3^{j-1}-3^{N})/2$, $\alpha^{N}=(3^{N}-3)/2$, and
$m_{j}^{N}=(3^{j-N-1}-1)/2$.  It follows that the results that we prove for $6$-series
eigenvalues in Lemma~\ref{lem:simplefunction} and Theorem~\ref{thm:powers_6series} below are also true for the 5-series eigenvalues
with essentially the same proofs.

Let $P_{j}$ denote the projection
onto $E_{j}$. The matrix $\Gamma_{j}:=P_{j}p(x,-\Delta)P_{j}$ is
a $d_{j}\times d_{j}$ matrix whose entries are given by
\[
\gamma_{j}(m,l):=\int p(x,\lambda_{j})u_{m}(x)u_{l}(x)d\mu(x).
\]
Our first version of a Szeg\"{o} theorem is for the operator of multiplication by a simple function.
\begin{lem}
\label{lem:simplefunction} Let $N\ge1$ be fixed, and suppose $f=\sum_{i=1}^{3^{N}}a_{i}\chi_{C_{i}}$
is a simple function. Then for all $k\geq0$
\begin{equation}
\lim_{j\to\infty}\frac{\Tr(P_{j}[f]P_{j})^{k}}{d_{j}}=\int f(x){}^{k}d\mu(x).\label{eq:simple-function}
\end{equation}
\end{lem}
\begin{proof}
The case $k=0$ is trivial since $\Tr(I_{E_{j}})=d_{j}$. Let $k>0$
and fix $j>N$. The matrix $P_{j}[f]P_{j}$ has the following structure
with respect to the basis $\{u_{m}\}_{m=1}^{d_{j}}$:
\[
\left[\begin{array}{cc}
R_{j} & 0\\

0 & N_{j}
\end{array}\right],
\]
where $R_{j}$ is a $d_{j}^{N}\times d_{j}^{N}$-matrix corresponding
to the localized eigenvectors and $N_{j}$ is an $\alpha^{N}\times\alpha^{N}$-matrix
corresponding to the non-localized eigenvalues (see also equation
(11) of \cite{OkRoStr_JFAA10} -- notice, however, the small typographical error
in \cite{OkRoStr_JFAA10} where $\star$ should be $0$). Moreover,
the matrix $R_{j}$ consists of $3^{N}$ diagonal blocks; each block
is of the form $a_{i}I_{m_{j}^{N}}$, $i=1,\dots,3^{N}$. We have
that
\[
\Tr(P_{j}[f]P_{j}])^{k}=\Tr(R_{j})^{k}+\Tr(N_{j})^{k}.
\]
We can compute  the first term explicitly:
\[
\Tr(R_{j})^{k}=\sum_{i=1}^{3^{N}}m_{j}^{N}a_{i}^{k}=d_{j}^{N}\sum_{i=1}^{3^{N}}\frac{a_{i}^{k}}{3^{N}}=d_{j}^{N}\int f(x)^{k}d\mu(x).
\]
For the second term we use that each element in $N_{j}$
is smaller in absolute value than $\Vert f\Vert_{\infty}$. Hence
$\vert\Tr(N_{j})^{k}\vert\le(\alpha^{N})^{k}\Vert f\Vert_{\infty}^k$.
Using the fact that $d_{j}-d_{j}^{N}=\alpha^{N}$ we then obtain
that
\[
\left\vert \frac{\Tr(P_{j}[f]P_{j})^{k}}{d_{j}}-\int f(x)^{k}d\mu(x)\right\vert \le\frac{\alpha^{N}}{d_{j}}\int\vert f(x)\vert^{k}d\mu(x)+\frac{(\alpha^{N})^{k}}{d_{j}}\Vert f\Vert_{\infty}.
\]
Since $\lim_{j\to\infty}(\alpha^{N})^{k}/d_{j}=0$
for all $k\ge0$ and $f$ is bounded on the compact
set $X$ the result follows.
\end{proof}
We use this lemma to prove the following Szeg\"{o}  theorem for pseudodifferential operators for a
sequence of $6$-series eigenvalues, extending \cite[Theorem 1]{OkRoStr_JFAA10}.
\begin{thm}
\label{thm:powers_6series} Let $p:X\times(0,\infty)\mathbb{\to}\mathbb{R}$
be a bounded measurable function such that $p(\cdot,\lambda_{j})$
is continuous for all $j\in\mathbb{N}$. Assume that $\lim_{j\to\infty}p(x,\lambda_{j})=q(x)$
is uniform in $x$. Then
\begin{equation}
\lim_{j\to\infty}\frac{1}{d_{j}}\Tr(P_{j}p(x,-\Delta)P_{j})^{k}=\int_{X}q(x){}^{k}d\mu(x)\label{eq:powsingeeigenv}
\end{equation}
for all $k\ge0$. Consequently,
\begin{equation}
\lim_{j\to\infty}\frac{1}{d_{j}}\Tr F(P_{j}p(x,-\Delta)P_{j})=\int_{X}F(q(x))d\mu(x)\label{eq:funcsingeeigenv}
\end{equation}
for any continuous $F$ supported on $[A,B]$, where
$A$ and $B$ are as in Lemma~\ref{lem:A_B}.
\end{thm}
\begin{proof}
We prove \eqref{eq:powsingeeigenv} by induction. The case $k=0$ is trivial since
$(P_{j}p(x,-\Delta)P_{j})^{0}=I_{E_{j}}$ and $\Tr(I_{E_{j}})=d_{j}$.
Let $k\ge1$ and assume that \eqref{eq:powsingeeigenv} is true for
all $0\le m<k$. We claim that it suffices to prove the result in
the case when there is $C>0$ such that $p(x,\lambda)\ge C$ for all
$(x,\lambda)\in X\times(0,\infty)$. Indeed, $p(x,\lambda)+2\Vert p\Vert_{\infty}\ge\Vert p\Vert_{\infty}>0$
and, using the induction hypothesis, the result for $p(x,\lambda)$
follows if one proves \eqref{eq:powsingeeigenv} for $p(x,\lambda)+2\Vert p\Vert_{\infty}$.

Let $\varepsilon>0$ and assume that $\varepsilon<C/2$. Since the
function $\lambda\mapsto\lambda^{k}$ is uniformly continuous on $[A,B]$,
there is $0<\delta<\varepsilon$ such that if $\vert\lambda-\lambda^{\prime}\vert<\delta$
then $\vert\lambda^{k}-(\lambda^{\prime})^{k}\vert<\varepsilon$.
Since $q$ is continuous, we can find $N\ge1$ and a simple function
$f_{N}=\sum_{i=1}^{3^{N}}a_{k}\chi_{C_{i}}$, where $\{C_{i}\}_{i=1}^{3^{N}}$
is the decomposition of $X$ into $N$-cells, such that $\Vert q(\cdot)-f_{N}(\cdot)\Vert_{\infty}<\delta/2$.
Therefore, $\vert\int q(x)^{k}d\mu(x)-\int f_{N}(x)^{k}d\mu(x)\vert<\varepsilon$
since $\mu(X)=1$. Moreover, we can find $J\ge1$ such that $\Vert p(\cdot,\lambda_{j})-q(\cdot)\Vert_{\infty}<\delta/2$
for all $j\ge J$. It follows that
\begin{equation}
\frac{C}{2}\le f_{N}(x)-\delta\le p(x,\lambda_{j})\le f_{N}(x)+\delta\label{eq:fnp}
\end{equation}
for all $x\in X$ and $j\ge J$. By increasing $J$, if necessary,
we may assume by Lemma \ref{lem:simplefunction} that
\[
\left\vert \frac{\Tr(P_{j}[f_{N}]P_{j})^{k}}{d_{j}}-\int(f(x_{N}))^{k}d\mu(x)\right\vert <\varepsilon
\]
for all $j\ge J$.

Let $j\ge J$. Equation \eqref{eq:fnp} implies that $0\le P_{j}[f_{N}-\delta]P_{j}\le P_{j}p(x,-\Delta)P_{j}\le P_{j}[f_{N}+\delta]P_{j}$.
We conclude that $\vert\sigma_{m}^{j}-\sigma_{m,N}^{j}\vert<\delta$,
$m=1,\dots,d_{j}$, where $\sigma_{m}^{j}$ are the eigenvalues of
$P_{j}p(x,-\Delta)P_{j}$ and $\sigma_{m,N}^{j}$ are the eigenvalues
of $P_{j}[f_{N}]P_{j}$. Therefore $\vert(\sigma_{m}^{j})^{k}-(\sigma_{m,N}^{j})^{k}\vert<\varepsilon$
for all $m=1,\dots,d_{j}$. Since $\Tr(P_{j}p(x,-\Delta)P_{j})^{k}=\sum_{m}(\sigma_{m}^{j})^{k}$
and $\Tr(P_{j}[f_{N}]P_{j})^{k}=\sum_{m}(\sigma_{m,N}^{j})^{k}$,
we have that
\begin{align*}
\left\vert \frac{\Tr(P_{j}p(x,-\Delta)P_{j})^{k}}{d_{j}}-\int_{X}q(x)^{k}d\mu(x)\right\vert  & \le\left\vert \frac{\Tr(P_{j}p(x,-\Delta)P_{j})^{k}}{d_{j}}-\frac{\Tr(P_{j}[f_{N}]P_{j})^{k}}{d_{j}}\right\vert \\
 & +\left\vert \frac{\Tr(P_{j}[f_{N}]P_{j})^{k}}{d_{j}}-\int_{X}f_{N}(x)^{k}d\mu(x)\right\vert \\
 & +\vert\int_{X}f_{N}(x)^{k}d\mu(x)-\int_{X}q(x)^{k}d\mu(x)\vert\\
 & <\left\vert \frac{\sum_{m=1}^{d_{j}}\bigl((\sigma_{m}^{j})^{k}-(\sigma_{m,N}^{j})^{k}\bigr)}{d_{j}}\right\vert +2\varepsilon\\
 & <3\varepsilon.
\end{align*}
This finishes the proof of \eqref{eq:powsingeeigenv}.

For the last statement of the theorem, a proof similar to that of Lemma \ref{lem:posfunctional} shows
\[
F\mapsto\frac{1}{d_{j}}\Tr F(P_{j}p(x,-\Delta)P_{j})
\]
is a non-negative functional on the set of continuous functions supported
on $[A,B]$ for all $j\ge 1$. The Stone-Weierstrass
theorem implies the result.
\end{proof}

\begin{rem}
The multiplication operator $[f]$ is the special case $p(x,\lambda)=f(x)$ for a continuous function  $f:X\to\mathbb{R}$.  Then for $F$ continuous with support on $[-\Vert[f]\Vert,\Vert[f]\Vert]$
\[
\lim_{j\to\infty}\frac{1}{d_{j}}\Tr F(P_{j}[f]P_{j})=\int_{X}F(f(x))d\mu(x).
\]
Spectral multipliers occur in the special case where $p:(0,\infty)\to\mathbb{R}_{+}$ is a bounded measurable function
such that $\lim_{n\to\infty}p(\lambda_{n})=q$ for some constant $q$.  In this situation if $F$ is continuous with support on $[-\Vert p(-\Delta)\Vert,\Vert p(-\Delta)\Vert]$ then
\[
\lim_{j\to\infty}\frac{1}{d_{j}}\Tr F(P_{j}p(-\Delta)P_{j})=F(q).
\]
\end{rem}

The main goal of this section is to prove the following Szeg\"{o}-type
theorem for pseudo-differential operators on the Sierpi\'nski gasket.
It is an analogue of classical results like \cite[Theorem
29.1.7]{Hor_LPDEIV}, which seem to have originated in \cite{Wid_JFA79}, see
also \cite{Gui_AnnMathSt79}.
\begin{thm}
\label{thm:Mainthm}Let $p:X\times(0,\infty)\mathbb{\to}\mathbb{R}$
be a bounded measurable function such that $p(\cdot,\lambda_{n})$
is continuous for all $n\in\mathbb{N}$. Assume that

$\lim_{n\to\infty}p(x,\lambda_{n})=q(x)$
is uniform in $x$. Then, for any continuous function $F$ supported
on $[A,B]$, we have that
\begin{equation}
\lim_{\Lambda\to\infty}\frac{1}{d_{\Lambda}}\Tr F(P_{\Lambda}p(x,-\Delta)P_{\Lambda})=\int_{X}F(q(x))d\mu(x), \label{eq:GeneralizedSzego}
\end{equation}
where the interval $[A,B]$ is chosen as in Lemma~\ref{lem:A_B}.
\end{thm}

\begin{rem}\ \\
\begin{enumerate}
\item[(a)]
If, in addition to the hypotheses of Theorem \ref{thm:Mainthm}, $p$ is a $0$-symbol in the sense
of  \cite[Definition 9.1]{IoRoStr_PSDO_2011}, then for any continuous $F$ on $[-\Vert p(x,-\Delta)\Vert,\Vert p(x,-\Delta)\Vert]$
we have
\[
\lim_{\Lambda\to\infty}\frac{1}{d_{\Lambda}}\Tr F(P_{\Lambda}p(x,-\Delta)P_{\Lambda})=\int_{X}F(q(x))d\mu(x).
\]

\item[(b)] Spectral mulitpliers are the special case where $p:(0,\infty)\to\mathbb{R}$ is a bounded measurable map
such that $\lim_{j\to\infty}p(\lambda_{j})=q$. Then for any continuous $F$ supported on $[-\Vert p(-\Delta)\Vert,\Vert p(-\Delta)\Vert]$
we have
\[
\lim_{\Lambda\to\infty}\frac{1}{d_{\Lambda}}\Tr F(P_{\Lambda}p(-\Delta)P_{\Lambda})=F(q).
\]

\item[(c)] Multiplication by a continuous $f:X\to\mathbb{R}$ is the special case $p(x,\lambda)=f(x)$. Then for any continuous $F$ supported on $[-\Vert[f]\Vert,\Vert[f]\Vert]$
we have
\[
\lim_{\Lambda\to\infty}\frac{1}{d_{\Lambda}}\Tr F(P_{\Lambda}[f]P_{\Lambda})=\int_{X}F(f(x))d\mu(x).
\]
If $\min f(x)>0$ then the above formula  can be obtained
from \cite[Theorem 3]{OkRoStr_JFAA10}. Specifically, the authors
of \cite{OkRoStr_JFAA10} proved that if $F>0$ is continuous, then
\[
\lim_{j\to\infty}\frac{\sum_{m=1}^{d_{j}}F(\sigma_{m}^{(j)})}{d_{j}}=\int F(f(x))d\mu(x),
\]
where $\sigma_{m}^{(j)}$ are the eigenvalues of $P_{j}[f]P_{j}$.
One can obtain our result by taking $F(x)=x^{k}$ and using the fact
that $\Tr(P_{j}[f]P_{j})^{k}=\sum_{m=1}^{d_{j}}(\sigma_{m}^{(j)})^{k}$.
\end{enumerate}
\end{rem}

\begin{proof}
[Proof of Theorem \ref{thm:Mainthm} ] We prove first that
\begin{equation}
\lim_{\Lambda\to\infty}\frac{1}{d_{\Lambda}}\Tr(P_{\Lambda}p(x,-\Delta)P_{\Lambda})^{k}=\int_{X}q(x)^{k}d\mu(x)\label{eq:Szego_poly}
\end{equation}
for all $k\ge0$. It is easy to prove the formula for  $k=0$.
Let $k\ge1$ and let $\varepsilon>0$. Clearly
\[
\Tr(P_{\Lambda}p(x,-\Delta)P_{\Lambda})^{k}=\sum_{\lambda\le\Lambda}\Tr(P_{\lambda}p(x,-\Delta)P_{\lambda})^{k}
\]
and $d_{\Lambda}=\sum_{\lambda\le\Lambda}d_{\lambda}$. The proof
of Theorem~\ref{thm:powers_6series} and its equivalent for $5$-series
imply that there is $J> 1$ such that if $\lambda$ is a $6$-series
or a $5$-series eigenvalue with generation of birth at least $J$,
then
\begin{equation}\label{eqn:56seriesTrapprox}
\left\vert \frac{\Tr(P_{\lambda}p(x,-\Delta)P_{\lambda})^{k}}{d_{\lambda}}-\int q(x)^{k}d\mu(x)\right\vert <\varepsilon.
\end{equation}
\global\long\def\genbirth{\#}
We will write in this proof $\genbirth\lambda$ for the generation
of birth of the eigenvalue $\lambda$. For each $\Lambda>0$ we let
$\Gamma_J(\Lambda)$ be the set of eigenvalues $\lambda\le \Lambda$
such that $\genbirth\lambda> J$ and $\tilde{\Gamma}_J(\Lambda)$ be
the set of eigenvalues $\lambda\le \Lambda$ such that
$\genbirth\lambda\le J$. Notice that $\Gamma_J(\Lambda)$ consists only
of $5$- and $6$-series eigenvalues.

Now fix $\Lambda_{1}>0$ such that $\Gamma_J(\Lambda_1)\ne \emptyset$
and $\tilde{\Gamma}_J(\Lambda_1)\ne \emptyset$. Then, for
all $\Lambda>\Lambda_{1}$,
\begin{align}
\lefteqn{ \left\vert \frac{\Tr(P_{\Lambda}p(x,-\Delta)P_{\Lambda})^{k}}{d_{\Lambda}}-\int q(x)^{k}d\mu(x)\right\vert}\quad& \notag\\
& \le\frac{\sum_{\lambda\in  \tilde{\Gamma}_J(\Lambda) }\vert\Tr(P_{\lambda}p(x,-\Delta)P_{\lambda})^{k}-d_{\lambda}\int q(x)^{k}d\mu(x)\vert}{d_{\Lambda}} \notag\\
& \quad + \frac{\sum_{\lambda\in  \Gamma_J(\Lambda)}\vert\Tr(P_{\lambda}p(x,-\Delta)P_{\lambda})^{k}-d_{\lambda}\int q(x)^{k}d\mu(x)\vert}{d_{\Lambda}}\notag\\
&=I+II.\label{eq:first_step}
\end{align}

The proof of Theorem 2 of \cite{OkRoStr_JFAA10} implies that
$\lim_{\Lambda\to\infty}\sum_{\lambda\in
  \tilde{\Gamma}_J(\Lambda)}d_{\lambda}/d_{\Lambda}=0$ (see
inequality (22) and the one following it from \cite{OkRoStr_JFAA10}).
Hence, there is $\Lambda_{2}>\Lambda_{1}$ such that, if $\Lambda>\Lambda_{2}$,
we have that
\[
\frac{\sum_{\lambda\in \tilde{\Gamma}_J(\Lambda)}d_{\lambda}}{d_{\Lambda}}<\frac{\varepsilon}{\Vert p\Vert_{\infty}^{k}+\Vert q\Vert_{\infty}^{k}}.
\]
Since $\Tr(P_{\lambda}p(x,-\Delta)P_{\lambda})^{k}\le d_{\lambda}\Vert p\Vert_{\infty}^{k}$
for all $\lambda$ and $\mu(X)=1$ we obtain for $\Lambda>\Lambda_{2}$
\begin{equation*}
I
\le(\Vert p\Vert_{\infty}^{k}+\Vert
q\Vert_{\infty}^{k})\frac{\sum_{\lambda\in \tilde{\Gamma}_J(\Lambda)}d_{\lambda}}{d_{\Lambda}}<\varepsilon.
\end{equation*}
Finally, by~\eqref{eqn:56seriesTrapprox}, for $\Lambda>\Lambda_{2}$:
\begin{equation*}
II <\varepsilon\frac{\sum_{\lambda\in\Gamma_J(\Lambda)}d_{\lambda}}{d_{\Lambda}}  <\varepsilon.
\end{equation*}
Substitution into~\eqref{eq:first_step} gives
\[
\left\vert \frac{\Tr(P_{\Lambda}p(x,-\Delta)P_{\Lambda})^{k}}{d_{\Lambda}}-\int q(x)^{k}d\mu(x)\right\vert <2\varepsilon.
\]
and an application of the Stone-Weierstrass theorem completes the proof.
\end{proof}

We conclude this subsection  by showing that the hypothesis of uniform convergence of $\lim_{j\to\infty}p(x,\lambda_{j})=q(x)$ in Theorem \ref{thm:powers_6series} may be relaxed if we assume some smoothness conditions on $p$.  The result gives convergence of~\eqref{eq:funcsingeeigenv} along a subsequence of $\{\lambda_{j}\}_{j\in\mathbb{N}}$.
\begin{prop}
\label{prop:subsequence_AA}Let $\{\lambda_{j}\}_{j\in\mathbb{N}}$
be an increasing sequence of $6$-series eigenvalues such that $\lambda_{j}$
has generation of birth $j$, for all $j\ge1$. Assume that $p(\cdot,\lambda_{j})\in\operatorname{Dom}(\Delta)$
for all $j\in\mathbb{N}$. Assume that
\[
\lim_{j\to\infty}p(x,\lambda_{j})=q(x)\quad\text{for all }x\in X
\]
and that both $p(\cdot,\lambda_{j})$ and $\Delta_{x}p(\cdot,\lambda_{j})$
are bounded uniformly in $j$. Then there is a subsequence $\{\lambda_{k_{j}}\}$
of $\{\lambda_{j}\}$ such that
\[
\lim_{j\to\infty}\frac{1}{d_{k_{j}}}F(P_{k_{j}}p(x,-\Delta)P_{k_{j}})=\int_{X}F(q(x))d\mu(x)
\]
for all continuous functions $F$ supported on $[A,B]$.
\end{prop}
\begin{proof}
Recall from the proof of \cite[Lemma 3.3]{IR_2010} that there is
a constant $C^{\prime}>0$ such that for any $f\in\operatorname{dom}(\Delta)$
and for any $x,y\in X$,
\[
\vert f(x)-f(y)\vert\le C^{\prime}R(x,y)\left(\sup_{z\in X}\vert\Delta f(z)\vert+\max_{p,q\in\partial X}\vert f(p)-f(q)\vert\right).
\]
Since, by our hypothesis, $p(x,\lambda_{j})$ and $\Delta_{x}p(x,\lambda_{j})$
are bounded uniformly in $j$, using the above estimate, we can find
a constant $C>0$ such that
\[
\vert p(x,\lambda_{j})-p(y,\lambda_{j})\vert\le CR(x,y)
\]
for all $j\in\mathbb{N}$. Hence the sequence$\{p(\cdot,\lambda_{j})\}$
is equicontinuous. Since the sequence is also uniformly bounded, the
Arzel\'a-Ascoli theorem implies that there is a subsequence $\{p(\cdot,\lambda_{k_{j}})\}$
such that
\[
\lim_{j\to\infty}p(x,\lambda_{k_{j}})=q(x)
\]
uniformly in $x$. Therefore we can apply Theorem \ref{thm:powers_6series}
to the subsequence $\{p(x,\lambda_{k_{j}})\}$ to obtain the conclusion.
\end{proof}

\subsection{Determinant  Szeg\"{o}-type limit theorems}
Proposition~\ref{thm:Szego_uniflimit} below is a generalization of \cite[Theorem 1]{OkRoStr_JFAA10}, and can be proved using Theorem~\ref{thm:powers_6series}. Specifically,
the result would follow if we set $F(x)=\log(x)$ and consider a positive
map $p$. We  provide an alternative proof  following
the steps of the proof of \cite[Theorem 1]{OkRoStr_JFAA10}. We
hope that our argument will shed light on some of the  technical
details that are sketched in the above mentioned proof. We still assume
that $\{\lambda_{j}\}_{j\ge1}$ is a fixed increasing sequence of
6-series or $5$-series eigenvalues such that $\lambda_{j}$ has generation
of birth $j$ for all $j\in\mathbb{N}$.
\begin{prop}
\label{thm:Szego_uniflimit}Assume that $p:X\times(0,\infty)\to\mathbb{R}_{+}$
is a measurable function such that $p(\cdot,\lambda_{j})$ is continuous
for all $j\in\mathbb{N}$ and such that there is $C>0$ so that $p(x,\lambda_{j})\ge C$
for all $(x,\lambda_{j})$. Assume that
\begin{equation}
\lim_{j\to\infty}p(x,\lambda_{j})=q(x)\label{eq:limit}
\end{equation}
exists and the limit is uniform in $x$. Then
\begin{equation}
\lim_{j\to\infty}\frac{1}{d_{j}}\log\det P_{j}p(x,-\Delta)P_{j}=\int_{X}\log q(x)d\mu(x).\label{eq:Szego}
\end{equation}
\end{prop}
\begin{proof}
First we notice that since the limit \eqref{eq:limit} is uniform
in $x$, the function $q$ is continuous on $X$ and $q(x)\ge C>0$
for all $x\in X$. Hence $\log q(x)$ is integrable.

Let $\varepsilon>0$ and assume that $\varepsilon<\min(1,C/4)$. There
exists $N\ge1$ and a simple function $f_{N}=\sum_{i=1}^{3^{N}}a_{i}\chi_{C_{i}}$,
where $\{C_{i}\}_{i=1}^{3^{N}}$ is a decomposition of $X$ into cells
of order $N$ as in Section~\ref{ssec:analysisonSG}, such that $\Vert q(\cdot)-f_{N}(\cdot)\Vert_{\infty}<(1/2)\min\bigl(\varepsilon,\varepsilon C/2\bigr)$.
Therefore $f_{N}(x)\ge 3C/4>0$ for all $x\in X$. Since $\log$ is
a continuous function, by increasing $N$, if necessary, we can also
assume that
\[
\vert\int\log q(x)d\mu(x)-\int_{X}\log f_{N}(x)d\mu(x)\vert<\varepsilon.
\]
Since we assume that the limit in \eqref{eq:limit} is uniform, there
is $J\ge1$ such that $\frac{\alpha^{N}}{d_{j}}\log\Vert f_{N}\Vert<\varepsilon$
and $\Vert p(x,\lambda_{j})-q(x)\Vert_{\infty}<(1/2)\min\bigl(\varepsilon,\varepsilon C/2\bigr)$
for all $j\ge J$. Hence $\Vert p(\cdot,\lambda_{j})-f_{N}(\cdot)\Vert_{\infty}<\min(\varepsilon,\varepsilon C/2)$
and
\begin{equation}
1-\varepsilon<\frac{p(x,\lambda_{j})}{f_{N}(x)}<1+\varepsilon\label{eq:poverf}
\end{equation}
 for all $x\in X$ and $j\ge J$.

Let $j\ge J$ be fixed. Recall that the operator $P_{j}p(x,-\Delta)P_{j}$
has a block structure
\[
\Gamma_{j}=\begin{bmatrix}R_{j} & *\\
* & N_{j}
\end{bmatrix}
\]
with respect to the basis $\{u_{k}\}_{k=1}^{d_{j}}$. The entries
of $\Gamma_{j}$ are given by
\[
\gamma_{j}(i,k)=\int_{X}p(x,\lambda_{j})u_{i}(x)u_{k}(x)d\mu(x).
\]
Similarly, the operators $P_{j}[f_{N}(1-\varepsilon)]P_{j}$ and $P_{j}[f_{N}(1+\varepsilon)]P_{j}$
have block structures (see also \cite{OkRoStr_JFAA10})
\[
\Gamma_{j,N}(\pm\varepsilon)=\begin{bmatrix}R_{j,N}(\pm\varepsilon) & 0\\
0 & N_{j,N}(\pm\varepsilon)
\end{bmatrix}
\]
respectively. The blocks $R_{j}$, $R_{j,N}(-\varepsilon)$, and $R_{j,N}(\varepsilon)$
are $d_{j}^{N}\times d_{j}^{N}$ blocks corresponding to the ``localized''
part, while $N_{j}$, $N_{j,N}(-\varepsilon)$, and $N_{j,N}(\varepsilon)$
correspond to the ``nonlocalized'' part. The inequality \eqref{eq:poverf}
implies that
\[
0\le\langle\Gamma_{j,N}(-\varepsilon)g,g\rangle\le\langle\Gamma_{j}g,g\rangle\le\langle\Gamma_{j,N}(\varepsilon)g,g\rangle
\]
 for all $g\in E_{j}$. Thus, as operators, $0\le\Gamma_{j,N}(-\varepsilon)\le\Gamma_{j}\le\Gamma_{j,N}(\varepsilon)$.
 \cite[Corollary 7.7.4]{HoJo_MatAn90} implies that
\[
\det\Gamma_{j,N}(-\varepsilon)\le\det\Gamma_{j}\le\det\Gamma_{j,N}(\varepsilon).
\]
Hence
\[
\log\det\Gamma_{j,N}(-\varepsilon)\le\log\det\Gamma_{j}\le\log\det\Gamma_{j,N}(\varepsilon).
\]
The block $R_{j,N}(\varepsilon)$ consists of $3^{N}$ blocks of the
form $a_{i}(1+\varepsilon)I_{m_{j}^{N}}$, $i=1,\dots,3^{N}$. Hence
\[
\log\det\Gamma_{j,N}(\varepsilon)=d_{j}^{N}\int\log f_{N}(x)d\mu(x)+d_{j}^{N}\log(1+\epsilon)+\log\det N_{j,N}(\varepsilon).
\]
An estimate as in \cite{OkRoStr_JFAA10} shows that $\vert\log\det N_{j,N}(\varepsilon)\vert\le\alpha^{N}\log\Vert f_{N}\Vert_{\infty}+\alpha^{N}\log(1+\varepsilon)$.
Similarly,
\[
\log \det\Gamma_{j,N}(-\varepsilon)=d_{j}^{N}\int\log f_{N}(x)d\mu(x)+d_{j}^{N}\log(1-\epsilon)+\log\det N_{j,N}(-\varepsilon)
\]
and $\vert\log\det N_{j,N}(\varepsilon)\vert\le\alpha^{N}\log\Vert f_{N}\Vert_{\infty}+\alpha^{N}\log(1-\varepsilon)$.
Using the fact that $d_{j}=d_{j}^{N}+\alpha^{N}$, we obtain that
\begin{align*}
\left\vert \frac{\log\det P_{j}p(x,-\Delta)P_{j}}{d_{j}}-\int\log q(x)d\mu(x)\right\vert  & \le\vert\int\log q(x)d\mu(x)-\int\log f_{N}(x)d\mu(x)\vert\\
 & \quad+\frac{\alpha^{N}}{d_{j}}(\Vert\log f_{N}\Vert_{1}+\log\Vert f_{N}\Vert_{\infty})+\varepsilon.
\end{align*}
This implies the conclusion.
\end{proof}

The proof of Proposition \ref{prop:subsequence_AA} can be easily
adapted to the proof of the following corollary.
\begin{cor}
Assume that $p:X\times(0,\infty)\to\mathbb{R}_{+}$ is a measurable
function such that $p(\cdot,\lambda_{j})$ is continuous for all $j\in\mathbb{N}$
and such that there is $C>0$ so that $p(x,\lambda_{j})\ge C$ for
all $(x,\lambda_{j})$. Moreover, assume that $p(\cdot,\lambda)\in\operatorname{Dom}(\Delta)$
for all $\lambda\in \operatorname{sp}(-\Delta)$ and that
\[
\lim_{j\to\infty}p(x,\lambda_{j})=q(x)\quad\text{for all }x\in X
\]
 and that both $p(x,\lambda)$ and $\Delta_{x}p(x,\lambda)$ are bounded
on $X\times \operatorname{sp}(-\Delta)$. Then there is a subsequence $\{\lambda_{k_{j}}\}$
of $\{\lambda_{j}\}$ such that
\[
\lim_{j\to\infty}\frac{1}{d_{k_{j}}}\log\det P_{k_{j}}p(x,-\Delta)P_{k_{j}}=\int_{X}\log q(x)d\mu(x).
\]
\end{cor}
More generally, we have:
\begin{prop}
\label{thm:GeneralSzego} Assume that $p:X\times(0,\infty)\to\mathbb{R}_{+}$
is a measurable function such that $p(\cdot,\lambda)$ is continuous
for all $\lambda\in \operatorname{sp}(-\Delta)$ and such that there is $C>0$ so that
$p(x,\lambda)\ge C$ for all $(x,\lambda)\in X\times\operatorname{sp}(-\Delta)$. Assume
that
\begin{equation}
\lim_{\lambda\to\infty}p(x,\lambda)=q(x)\label{eq:limit-1}
\end{equation}
exists and the limit is uniform in $x$. Then
\begin{equation}
\lim_{\Lambda\to\infty}\frac{1}{d_{\Lambda}}\log\det P_{\Lambda}p(x,-\Delta)P_{\Lambda}=\int_{X}\log q(x)d\mu(x).\label{eq:Szego-1}
\end{equation}
\end{prop}
\begin{proof}
The proof is  similar to the proof of Theorem \ref{thm:Mainthm}
and we omit the details.
\end{proof}

\section{Examples and Applications}\label{sec:ExamplesApp}
In this section we consider some applications of our main results.
Our examples show how one can extract asymptotic behaviour of
operators from elementary integrals, but we also give a result on
spectral clustering for generalized Schr\"{o}dinger operators
following a line of argument due to  Widom \cite{Wid_JFA79}.

\subsection{Examples}
\label{subsec:Example}
 Our results can be used to obtain asymptotics of  Riesz and Bessel operators, as well as
some specific generalized Schr\"{o}dinger operators.
\begin{example}
Let $p:(0,\infty)\to\mathbb{R}$ be $p(\lambda)=1+\lambda^{-\beta}$,
where $\beta$ is a positive real number. Then $p(-\Delta)=I+(-\Delta)^{-\beta}$.
We know from \cite{IR_2010} that $p(-\Delta)$ is bounded on $L^{p}(\mu)$
for all $p>1$. Let $F$ be a continuous function supported on $(-\Vert p(-\Delta)\Vert,\Vert p(-\Delta)\Vert)$.
Since $\lim_{\lambda\to\infty}p(\lambda)=1$, we have that
\[
\lim_{j\to\infty}\frac{\Tr F(P_{j}p(-\Delta)P_{j})}{d_{j}}=F(1)
\]
for any increasing sequence $\{\lambda_{j}\}$ of $6$-series or $5$-series
eigenvalues such that $\lambda_{j}$ has generation of birth $j$.
Moreover
\[
\lim_{\Lambda\to\infty}\frac{\Tr F(P_{\Lambda}p(-\Delta)P_{\Lambda})}{d_{\Lambda}}=F(1).
\]
Since $p(\lambda)\ge 1$ for all $\lambda>0$, we can apply Proposition
\ref{thm:Szego_uniflimit} and Proposition \ref{thm:GeneralSzego} and   obtain that
\[
\lim_{j\to\infty}\frac{1}{d_{j}}\log\det P_{j}p(-\Delta)P_{j}=0
\]
and
\[
\lim_{\Lambda\to\infty}\frac{1}{d_{\Lambda}}\log\det P_{\Lambda}p(-\Delta)P_{\Lambda}=0.
\]

Let $p:(0,\infty)\to\mathbb{R}$ be $p(\lambda)=1+(1+\lambda)^{-\beta}$,
where $\beta$ is a positive real number. Then the corresponding operator is  the Bessel operator given by $p(-\Delta)=I+(I-\Delta)^{-\beta}$. All the conclusions of the previous examples hold for this operator as well.
\end{example}

\begin{example}
Let $p(x,\lambda)=q(\lambda)+\chi(x)$, where $q$ is a bounded measurable
function on $(0,\infty)$ such that $\lim_{\lambda\to\infty}q(\lambda)=l$
exists and $\chi$ is continuous on $X$. Then $p(x,-\Delta)$
is a generalized Schr\"{o}dinger operator.  Since it is the sum of two bounded
operators it is bounded on $L^{2}(\mu)$. Let $F$ be a continuous function supported on $(-\Vert p(x,-\Delta)\Vert,\Vert p(x,-\Delta)\Vert)$.
Then, if $\{\lambda_{j}\}_{j\ge1}$ is an increasing sequence of $ $6-series
or, respectively, $5$-series eigenvalues, we have both of the
following equalities:
\[
\lim_{j\to\infty}\frac{\Tr F(P_{j}p(x,-\Delta)P_{j})}{d_{j}}=\int F(l+\chi(x))d\mu(x)=\lim_{\Lambda\to\infty}\frac{\Tr F(P_{\Lambda}p(x,-\Delta)P_{\Lambda})}{d_{\Lambda}}.
\]

In particular, if $l=0$, we have that
\[
\lim_{j\to\infty}\frac{\Tr F(P_{j}p(x,-\Delta)P_{j})}{d_{j}}=\int F(\chi(x))d\mu(x)=\lim_{\Lambda\to\infty}\frac{\Tr F(P_{\Lambda}p(x,-\Delta)P_{\Lambda})}{d_{\Lambda}}.
\]

If we further assume $p(x,\lambda)\ge C>0$ for all
$(x,\lambda)\in X\times(0,\infty)$ then
\[
\lim_{j\to\infty}\frac{1}{d_{j}}\log\det P_{j}p(x,-\Delta)P_{j}=\int\log(l+\chi(x))d\mu(x)
\]

and the same is true if we replace $P_{j}$ with $P_{\Lambda}$ and
$d_{j}$ with $d_{\Lambda}$, so in particular if $l=0$,
\[
\lim_{j\to\infty}\frac{1}{d_{j}}\log\det P_{j}p(x,-\Delta)P_{j}=\int\log(\chi(x))d\mu(x)=\lim_{\Lambda\to\infty}\frac{1}{d_{\Lambda}}\log\det P_{\Lambda}p(x,-\Delta)P_{\Lambda}.
\]

\end{example}

\subsection{Application: Asymptotics of eigenvalue clusters
for general Schr\"{o}dinger operators}\label{subsec:asymptotics-of-eigenvalue}

Let $p:(0,\infty)\to\mathbb{R}$ be a measurable function and let
$\chi$ be a real-valued bounded measurable function on $X$. We call
the operator $H=p(-\Delta)+[\chi]$ a \emph{generalized Sch}r\emph{\"{o}dinger
operator} with  \emph{potential} $\chi$.  We study
the asymptotic behavior of spectra of generalized Schr\"{o}dinger
operators with continuous potentials and continuous $p$, generalizing
some results of \cite{OkStr_PAMS07}.

We begin with a lemma that is a generalization of the key  \cite[Lemma 1]{OkStr_PAMS07}.
\begin{lem}
\label{lem:minmaxlemma}Let $p:(0,\infty)\to\mathbb{R}$ be a continuous
function such that there is $A\in\mathbb{R}$ with $p(\lambda)\ge A$
for all $\lambda\ge\lambda_{1}$, where $\lambda_{1}$ is the smallest
positive eigenvalue of $-\Delta$. For $i=1,2$, let $\chi_{i}$ be
real-valued bounded measurable functions on $X$. Let $H_{i}=p(-\Delta)+[\chi_{i}]$
denote the corresponding generalized Schr\"{o}dinger operators. For
$n\ge1$, the $n$th eigenvalues $\nu_{n}^{i}$ of $H_{i}$, $i=1,2$,
satisfy the following inequality:
\[
\vert\nu_{n}^{1}-\nu_{n}^{2}\vert\le\Vert\chi_{1}-\chi_{2}\Vert_{L^{\infty}}.
\]
\end{lem}
\begin{proof}
The hypothesis implies that
\[
\langle H_{i}f,f\rangle\ge(A+\min_{X}\chi_{i})\Vert f\Vert_{2}^{2}
\]
for all $f\in D$, $i=1,2$. Hence $H_{i}$ is bounded from below,
$i=1,2$. The remainder of the proof is identical to the proof of \cite[Lemma 1]{OkStr_PAMS07}.

\end{proof}

Assume that $p:(0,\infty)\to\mathbb{R}$
is a continuous function, that there is $\overline{\lambda}>0$
such that $p$ is increasing on $[\overline{\lambda},\infty)$ and
\begin{equation}
\vert p(\lambda)-p(\lambda^{\prime})\vert\ge c\vert\lambda-\lambda^{\prime}\vert^\beta\label{eq:pcondition}
\end{equation}
 for all $\lambda,\lambda^{\prime}\ge\overline{\lambda}$ and some constants $c>0$ and $\beta>0$.
Let $\chi$ be a continuous function on $X$ and $H=p(-\Delta)+[\chi]$
be the corresponding Schr\"{o}dinger operator. Let
$\{\lambda_{j}\}$ be a sequence of $6$-series eigenvalues of $-\Delta$
such that the separation between $\lambda_{j}$
and the next higher and lower eigenvalues of $-\Delta$ grows exponentially
in $j$. For example, if $\lambda_{1}$ is any 6-series eigenvalue with
generation of birth $1$ and if $\lambda_{j}=5^{j-1}\lambda_{1}$,
then the sequence $\{\lambda_{j}\}_{j\ge1}$ satisfies our assumption
(\cite{OkStr_PAMS07}, \cite[Chapter 3]{Str_Prin06}). Let $\tilde{\Lambda}_{j}$ be the portion of the spectrum
of $H$ lying in $[p(\lambda_{j})+\min\chi,p(\lambda_{j})+\max\chi]$.
Lemma \ref{lem:minmaxlemma} implies that, for large $j$, $\tilde{\Lambda}_{j}$
contains exactly $d_{j}$ eigenvalues $\{\nu_{i}^{j}\}_{i=1}^{d_{j}}$.
We call this the $p(\lambda_{j})$ \emph{cluster} of the eigenvalues
of $H$. If we translate by $p(\lambda_j)$ units to the left we obtain
clusters that lie in a fixed interval.  Define the characteristic
measure of the $p(\lambda_{j})$ cluster
of $H$ by
\[
\Psi_{j}(\lambda)=\frac{1}{d_{j}}\sum_{i=1}^{d_{j}}\delta(\lambda-(\nu_{i}^{j}-p(\lambda_{j})).
\]
and observe that integrating the function $x^k$ against this measure yields
\begin{equation}
  \label{eq:psi_j}
  \langle\Psi_j,x^k\rangle=\frac{1}{d_j}\Tr \bigl(\overline{P}_j
  (p(-\Delta)+[\chi]-p(\lambda_j))\overline{P}_j\bigr)^k,
\end{equation}
for all $k\ge 0$, where $\overline{P}_j$ is the spectral projection for
$p(-\Delta)+[\chi]$ associated with the $p(\lambda_j)$
cluster. This allows us to analyze the weak limit of $\Psi_j$ using
Theorem \ref{thm:Mainthm}.

\begin{thm}\label{thm:specclusters}
The sequence $\{\Psi_{j}\}_{j\ge1}$ converges weakly to the pullback
of the measure $\mu$ under $\chi$ defined for all continuous
functions $f$ supported on $[\min \chi, \max \chi]$ by
\[
\langle\Psi_{0},f\rangle=\int_{X}f(\chi(x))d\mu(x).
\]
\end{thm}

\begin{lem}{\label{lem:pow_clust}}
Assume that $N>0$ and that $\chi_N=\sum_{i=1}^N a_i \chi_{C_i}$ is a
simple function,  where $\{C_{i}\}_{i=1}^{3^{N}}$ is a partition
of $X$ into $N$-cells. Let $H_{N}=p(-\Delta)+[\chi_{N}]$
be the corresponding generalized Schr\"{o}dinger operator,
$\tilde{\Lambda}_{j}^{N}$ the $p(\lambda_j)$ cluster of $H_N$, and let
$\overline{P}_{j}^{N}$ be the spectral projection for $H_N$ associated
with the $p(\lambda_j)$ cluster. Then

\[
  \lim_{j\to \infty}\frac{\Tr \bigl(\overline{P}_j^N
    (p(-\Delta)+[\chi_N]-p(\lambda_j))\overline{P}_j^N\bigr)^k}{d_j}=\lim_{j\to
    \infty}\frac{(P_j[\chi_N]P_j)^k}{d_j}=\int_X \chi_N(x)^kd\mu(x),
  \]

  for all $k\ge 0$.
\end{lem}
\begin{proof}
   Consider
  the first equality.
If $\overline{u}$ is an eigenfunction of the basis
of $E_{j}$ that is localized in a single $N$-cell $C_{i}$, then
$H_{N}\overline{u}=(p(\lambda_{j})+a_{i})\overline{u}$. Thus
$\lambda_{i}^{j}:=p(\lambda_{j})+a_{i}$ is an eigenvalue of $H_{N}$
with multiplicity at least $m_{j}^{N}$.
Doing this identifies $d_{j}^{N}$ eigenvalues in $\tilde{\Lambda}_{j}^{N}$
and we let $A_{j}^{N}$ be the set of the remaining $\alpha^{N}$
eigenvalues. Hence
\[
\Tr(\overline{P}_j^N
    (p(-\Delta)+[\chi_N]-p(\lambda_j))\overline{P}_j^N\bigr)^k=m_{j}^{N}\sum_{i=1}^{3^N}a_i^k+\sum_{\nu\in
    A_{j}^{N}} \nu^k.
\]
Recall from the proof of Lemma \ref{lem:simplefunction} that
$\Tr(P_j[\chi_N]P_j)^k=m_{j}^{N}\sum_{i=1}^{3^N}a_i^k +\Tr (N_j)^k$. Since both $\sum_{\nu\in
    A_{j}^{N}} \nu^k$ and $\Tr (N_j)^k$ are bounded by a constant
  times $(\alpha^N)^k$,  the equality of the two limits follows.
The second equality is from Theorem \ref{thm:Mainthm}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:specclusters}]
Since $\chi$ is continuous, it can be approximated uniformly by a
sequence of simple functions $\chi_{N}=\sum_{i=1}^{3^{N}}a_{i}\chi_{C_{i}}$, of the type previously described.

We can choose $\chi_{N}$
such that $\min\chi\le\min\chi_{N}$ and $\max\chi_{N}\le\max\chi$
for all $N$ from which  $\tilde{\Lambda}_{j}^{N}$ is contained
in $[p(\lambda_{j})+\min\chi,p(\lambda_{j})+\max\chi]$. Moreover there
is $J_1$ depending only $\Vert \chi\Vert_\infty$ such that $j\ge J_1$
implies  $\tilde{\Lambda}_{j}^{N}$  contains
$d_{j}$ eigenvalues $\{\tilde{\lambda}_{i}^{j}\}_{i=1}^{d_{j}}$.

From Lemma \ref{lem:minmaxlemma}, for all $i$
and $j$,
$
\vert\nu_{i}^{j}-\tilde{\lambda}_{i}^{j}\vert\le\Vert\chi-\chi_{N}\Vert_{\infty}
$. Therefore, when $j\ge J_1$,
\begin{align*}
&\Bigl|\Tr \bigl(\overline{P}_j
    (p(-\Delta)+[\chi]-p(\lambda_j))\overline{P}_j\bigr)^k-
\Tr \bigl(\overline{P}_j^N
(p(-\Delta)+[\chi_N]-p(\lambda_j))\overline{P}_j^N\bigr)^k\Bigr|\\
&\quad
=\sum_{i=1}^{d_j}\bigl\vert(\nu_i^{j}-p(\lambda_j))^k-(\tilde{\lambda}_{i}^{j}-p(\lambda_j))^k\bigr\vert
\le d_j k\Vert \chi\Vert_{\infty}^{k-1}\Vert \chi-\chi_N\Vert_{\infty}.
  \end{align*}
Then, with $j\ge J_1$,
\begin{align*}
  \biggl\vert \langle \Psi_j,x^k\rangle-\int_{X} \chi^kd\mu\biggr\vert
  &\le k\Vert \chi\Vert_{\infty}^{k-1}\Vert \chi-\chi_N\Vert_{\infty}
  +\int_X\vert \chi_N^{k}-\chi^{k}\vert d\mu\\
 &\quad  +d_{j}^{-1}\biggl\vert \Tr \bigl(\overline{P}_j^N
(p(-\Delta)+[\chi_N]-p(\lambda_j))\overline{P}_j^N\bigr)^k - \int_{X}
\chi_N^{k}d\mu\biggr\vert.
\end{align*}
For sufficiently large $N$ both $\Vert\chi-\chi_N\Vert_\infty$ and
$\Vert \chi-\chi_N\Vert_{1}$ are less than $\varepsilon$ and we can
take $j>J_1$ so large that the last term is smaller than $\varepsilon$
by Lemma  \ref{lem:pow_clust}.
Applying the Stone-Weierstrass theorem completes the proof.
\end{proof}

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