\chapter{Estimates in the bulk}\label{s:bulk}

In this section, we prove the asymptotic estimates for Hermite functions that we need. They all concern points in the bulk of the spectrum, and they follow from the Plancherel--Rotach asymptotics together with the identities of \zcref{s:exactidentities}. The main results are \zcref{l:localsine}, which compares the Hermite kernel with the sine kernel on growing windows, and \zcref{l:kernelintegral,c:gt}, which control integrals of the kernel against step functions.

For $\delta \in (0, 1/2]$, we define the \emph{bulk}
\begin{equation}\label{e:bulkdef}
\bulk_N(\delta) = \big[ - (1 - \delta) \sqrt{2N}, (1 - \delta) \sqrt{2N} \big] ,
\end{equation}
and for $x_0 \in \bulk_N(\delta)$ we define the \emph{local scale}
\begin{equation}\label{e:qdef}
\varrho_N(x_0) = \sqrt{2N - x_0^2} .
\end{equation}
Note that $\sqrt{2 \delta N} \le \varrho_N(x_0) \le \sqrt{2N}$ for $x_0 \in \bulk_N(\delta)$. We also define the \emph{sine kernel}
\begin{equation}\label{e:sinedef}
S(t) = \frac{\sin t}{\pi t} , \quad t \neq 0 , \qquad S(0) = \frac1\pi .
\end{equation}
Throughout this section, we fix $\delta \in (0, 1/2]$. The constants $C$, $c$, and $N_0$ depend only on $\delta$, and all statements hold for $N \ge N_0$.

\section{Plancherel--Rotach asymptotics}

For $n \ge 1$ and $|x| < \sqrt{2n + 1}$, we define $\theta_n(x) \in (0, \pi)$ by $x = \sqrt{2n + 1} \cos \theta_n(x)$, and we set
\begin{equation}\label{e:PRdef}
\begin{gathered}
\alpha_n(x) = \frac{2^{1/4}}{\sqrt \pi \, n^{1/4}} \big( \sin \theta_n(x) \big)^{-1/2} , \\
\Theta_n(x) = \Big( \frac n2 + \frac14 \Big) \big( \sin 2 \theta_n(x) - 2 \theta_n(x) \big) + \frac{3\pi}{4} .
\end{gathered}
\end{equation}

\begin{lemma}\label{l:PR}
There exists $C > 0$ such that for all integers $n$ with $|n - N| \le 2$ and all $x \in \bulk_N(\delta)$, we have
\begin{equation}\label{e:PR}
h_n(x) = \alpha_n(x) \big( \sin \Theta_n(x) + \rho_n(x) \big) , \qquad |\rho_n(x)| \le \frac{C}{N} .
\end{equation}
Moreover, the following estimates hold for such $n$ and $x$.
\begin{enumerate}[label=(\roman*)]
\item We have $\Theta_n'(x) = \sqrt{2n + 1 - x^2}$, $|\Theta_n''(x)| \le C$, and $|\theta_n'(x)| \le C N^{-1/2}$.
\item We have $C^{-1} N^{-1/4} \le \alpha_n(x) \le C N^{-1/4}$ and $|\alpha_n'(x)| \le C N^{-3/4}$.
\item The functions defined by $\mathfrak a = \alpha_{N - 1} / \alpha_N$ and $\vartheta = \Theta_{N - 1} - \Theta_N$ satisfy $|\mathfrak a(x) - 1| \le C / N$, $|\mathfrak a'(x)| \le C N^{-3/2}$, $|\vartheta(x) - \theta_N(x)| \le C / N$, and $|\vartheta'(x)| \le C N^{-1/2}$.
\end{enumerate}
\end{lemma}

\begin{proof}
When $N \ge N_0$, we have $|x| \le (1 - \delta/2) \sqrt{2n + 1}$ for $|n - N| \le 2$ and $x \in \bulk_N(\delta)$, so $\theta_n(x)$ belongs to a compact subset of $(0, \pi)$ depending only on $\delta$. The expansion \eqref{e:PR} is then the Plancherel--Rotach asymptotic formula \cite{PlancherelRotach1929} in the form of \cite[Theorem~8.22.9]{Szego1975}, multiplied by the normalizing factor $(2^n n! \sqrt \pi)^{-1/2}$ in \eqref{e:hndef}.

For (i), we have
\begin{equation}
\theta_n'(x) = - (2n + 1 - x^2)^{-1/2}
\end{equation}
and
\begin{equation}
\partial_\theta \big( ((2n + 1)/4)(\sin 2 \theta - 2 \theta) \big) = - (2n + 1) \sin^2 \theta .
\end{equation}
Therefore,
\begin{equation}
\Theta_n'(x) = \sqrt{2n + 1} \sin \theta_n(x) = \sqrt{2n + 1 - x^2} ,
\end{equation}
and
\begin{equation}
\Theta_n''(x) = - x (2n + 1 - x^2)^{-1/2} .
\end{equation}
These are bounded as claimed, since $2n + 1 - x^2 \ge c N$.

The bounds on $\alpha_n$ in (ii) follow from $\sin \theta_n(x) \in [c, 1]$. Further, we have
\begin{equation}
(\log \alpha_n)'(x) = x / (2 (2n + 1 - x^2)) ,
\end{equation}
which is bounded by $C N^{-1/2}$ in absolute value.

For (iii), the identity for $(\log \alpha_n)'$ gives
\begin{equation}
(\log \mathfrak a)' = x / (2 (2N - 1 - x^2)) - x / (2 (2N + 1 - x^2)) ,
\end{equation}
which is bounded by $C N^{-3/2}$ in absolute value. The bound on $\mathfrak a - 1$ follows from $(N/(N - 1))^{1/4} = 1 + O(N^{-1})$ and $\sin \theta_{N - 1}(x) / \sin \theta_N(x) = 1 + O(N^{-1})$. Next, we have
\begin{equation}
\vartheta' = \sqrt{2N - 1 - x^2} - \sqrt{2N + 1 - x^2} ,
\end{equation}
which is bounded by $C N^{-1/2}$ in absolute value.

Finally, we regard $n$ as a continuous variable. Differentiating \eqref{e:PRdef} in $n$ at fixed $x$, and using
\begin{equation}
\partial_n \theta_n(x) = \cos \theta_n(x) / ((2n + 1) \sin \theta_n(x)) ,
\end{equation}
we have
\begin{equation}
\partial_n \Theta_n(x) = \frac12 \big( \sin 2 \theta_n - 2 \theta_n \big) - (2n + 1) \sin^2 \theta_n \cdot \frac{\cos \theta_n}{(2n + 1) \sin \theta_n} = - \theta_n(x) .
\end{equation}
Integrating this identity, and using the bound $|\partial_n \theta_n(x)| \le C / N$ for $n \in [N - 1, N]$, we conclude that
\begin{equation}
\vartheta(x) = \int_{N - 1}^N \theta_t(x) \dd t = \theta_N(x) + O(N^{-1}) .
\end{equation}
\end{proof}

\begin{lemma}\label{l:pointwise}
For all $x, y \in \bulk_N(\delta)$, the following estimates hold.
\begin{enumerate}[label=(\alph*)]
\item We have $|h_n(x)| \le C N^{-1/4}$ for $|n - N| \le 2$. In particular, $|\varphi_N(x)| + |\psi_N(x)| \le C$.
\item We have $K_N(x, x) \le C \sqrt N$ and $K_{N + 1}(x, x) \le C \sqrt N$.
\item We have $|K_N(x, y)| \le C \sqrt N \, (1 + \sqrt N |x - y|)^{-1}$.
\item We have $|\partial_x K_N(x, y)| \le C N$.
\end{enumerate}
\end{lemma}

\begin{proof}
The bounds in (a) follow from \zcref{l:PR}. For (b), letting $y \to x$ in \eqref{e:CD} and using the second identity in \eqref{e:recurrence} for $h_N'$ and $h_{N - 1}'$, we find
\begin{equation}
\begin{split}
K_N(x, x) & = \frac N2 \big( h_{N - 1}(x)^2 + h_N(x)^2 \big) - \frac{\sqrt{N (N + 1)}}{2} h_{N + 1}(x) h_{N - 1}(x) \\
& \quad - \frac{\sqrt{N (N - 1)}}{2} h_{N - 2}(x) h_N(x) .
\end{split}
\end{equation}
By (a), each term on the right side is bounded by $C N \cdot N^{-1/2}$. The same argument with $N + 1$ in place of $N$ bounds $K_{N + 1}(x, x)$, since it only involves $h_n$ with $|n - N| \le 2$, which are covered by (a).

To prove (c), we first suppose that $|x - y| \ge N^{-1/2}$. Then \eqref{e:CD} and (a) give
\begin{equation}
|K_N(x, y)| \le C N^{1/2} \cdot N^{-1/2} \cdot |x - y|^{-1} ,
\end{equation}
which implies the claim. In the remaining case $|x - y| < N^{-1/2}$, the claim follows from \eqref{e:KCS} and (b).

For (d), we write
\begin{equation}
\partial_x K_N(x, y) = \sum_{n < N} h_n'(x) h_n(y) ,
\end{equation}
and the Cauchy--Schwarz inequality gives
\begin{equation}
|\partial_x K_N(x, y)| \le \Big( \sum_{n = 0}^{N - 1} h_n'(x)^2 \Big)^{1/2} K_N(y, y)^{1/2} .
\end{equation}
To bound the sum on the right side, we use \eqref{e:recurrence}, which gives
\begin{equation}
h_n'(x)^2 \le n \, h_{n - 1}(x)^2 + (n + 1) h_{n + 1}(x)^2 ,
\end{equation}
so the sum is bounded by
\begin{equation}
2N K_{N + 1}(x, x) \le C N^{3/2} ,
\end{equation}
where we used (b) in the last step. Using (b) again, we have
\begin{equation}
|\partial_x K_N(x, y)| \le C N^{3/4} N^{1/4} .
\end{equation}
\end{proof}

\section{The local sine kernel}

\begin{lemma}\label{l:localsine}
For all $x_0 \in \bulk_N(\delta)$ and all $u, v \in \R$ such that $|u|, |v| \le 2 N^{1/4}$, we have, setting $\varrho = \varrho_N(x_0)$,
\begin{equation}\label{e:localsine}
\Big| \frac{1}{\varrho} K_N\Big( x_0 + \frac{u}{\varrho} , x_0 + \frac{v}{\varrho} \Big) - S(u - v) \Big| \le C N^{-1/2} .
\end{equation}
\end{lemma}

\begin{proof}
Set $x = x_0 + u/\varrho$ and $y = x_0 + v/\varrho$, and define
\begin{equation}
\mathcal K(u, v) = \varrho^{-1} K_N(x, y) .
\end{equation}
Observe that
\begin{equation}
|x - x_0| + |y - x_0| \le C N^{-1/4} .
\end{equation}
Hence $x, y \in \bulk_N(\delta/2)$ when $N \ge N_0$, and we may apply \zcref{l:PR,l:pointwise} with $\delta / 2$ in place of $\delta$.

We first suppose that $|u - v| \ge N^{-1/2}$. Using \eqref{e:CD} and $x - y = (u - v)/\varrho$, we find
\begin{equation}\label{e:Kcalnum}
\mathcal K(u, v) = \sqrt{\frac N2} \, \frac{h_N(x) h_{N - 1}(y) - h_{N - 1}(x) h_N(y)}{u - v} .
\end{equation}
We insert \eqref{e:PR}. The terms involving $\rho_N$ or $\rho_{N - 1}$ contribute to $\mathcal K(u, v)$ at most
\begin{equation}
C N^{1/2} \cdot N^{-1/2} N^{-1} |u - v|^{-1} \le C N^{-1/2} .
\end{equation}
We set $z_1 = \Theta_N(x)$ and $z_2 = \Theta_N(y)$. In the notation of \zcref{l:PR}(iii), the remaining terms in the numerator equal
\begin{equation}
\alpha_N(x) \alpha_N(y) \Big( \mathfrak a(y) \sin z_1 \sin\big(z_2 + \vartheta(y)\big) - \mathfrak a(x) \sin \big( z_1 + \vartheta(x) \big) \sin z_2 \Big) .
\end{equation}

\zcref{l:PR}(iii) implies that
\begin{equation}
|\mathfrak a(y) - \mathfrak a(x)| \le C N^{-3/2} |x - y|
\end{equation}
and
\begin{equation}
|\vartheta(y) - \vartheta(x)| \le C N^{-1/2} |x - y| .
\end{equation}
Replacing $\mathfrak a(y)$ by $\mathfrak a(x)$ and $\vartheta(x)$ by $\vartheta(y)$ therefore changes the numerator by at most $C N^{-1} |x - y|$, and changes $\mathcal K(u, v)$ by at most
\begin{equation}
C N^{1/2} N^{-1} |x - y| / |u - v| = C N^{-1/2} \varrho^{-1} \le C N^{-1} .
\end{equation}
After these replacements, we use the identity
\begin{equation}
\sin z_1 \sin (z_2 + w) - \sin (z_1 + w) \sin z_2 = \sin w \sin (z_1 - z_2) , \qquad z_1, z_2, w \in \R ,
\end{equation}
with $w = \vartheta(y)$, and we find that
\begin{equation}\label{e:Kcalmain}
\mathcal K(u, v) = \sqrt{\frac N2} \, \alpha_N(x) \alpha_N(y) \, \mathfrak a(x) \sin \vartheta(y) \, \frac{\sin(z_1 - z_2)}{u - v} + O(N^{-1/2}) .
\end{equation}

We now evaluate the prefactor. It follows from \zcref{l:PR}(iii) and (i) that
\begin{equation}
\sin \vartheta(y) = \sin \theta_N(y) + O(N^{-1})
\end{equation}
and
\begin{equation}
\sin \theta_N(x) / \sin \theta_N(y) = 1 + O(N^{-3/4}) ,
\end{equation}
since $|x - y| \le C N^{-1/4}$. Using \eqref{e:PRdef}, we have
\begin{equation}
\begin{split}
\sqrt{\frac N2} \, \alpha_N(x) \alpha_N(y) \, \mathfrak a(x) \sin \vartheta(y) & = \frac1\pi \Big( \frac{\sin \theta_N(y)}{\sin \theta_N(x)} \Big)^{1/2} \big( 1 + O(N^{-1}) \big) \\
& = \frac1\pi \big( 1 + O(N^{-3/4}) \big) .
\end{split}
\end{equation}

Next, \zcref{l:PR}(i) and Taylor's theorem give
\begin{equation}
\begin{split}
z_1 - z_2 & = \Theta_N'(x_0)(x - y) + O\big( |x - y| ( |x - x_0| + |y - x_0| ) \big) \\
& = (u - v) \big( 1 + O(N^{-3/4}) \big) ,
\end{split}
\end{equation}
where we used the bound $|x - x_0| + |y - x_0| \le C N^{1/4} / \varrho$ and
\begin{equation}
\Theta_N'(x_0) = \sqrt{\varrho^2 + 1} = \varrho (1 + O(N^{-1})) .
\end{equation}
Using the inequality
\begin{equation}
|\sin(w (1 + w')) - \sin w| \le |w w'| ,
\end{equation}
which holds for all $w, w' \in \R$, we deduce that
\begin{equation}
\sin(z_1 - z_2) / (u - v) = \sin(u - v)/(u - v) + O(N^{-3/4}) .
\end{equation}
Inserting this estimate and the evaluation of the prefactor into \eqref{e:Kcalmain}, we obtain \eqref{e:localsine} when $|u - v| \ge N^{-1/2}$.

Now suppose that $|u - v| < N^{-1/2}$. Let $u' = v - \sgn(v) N^{-1/2}$ if $v \neq 0$, and $u' = N^{-1/2}$ if $v = 0$. Then $|u' - v| = N^{-1/2}$, $|u'| \le 2 N^{1/4}$, and $|u - u'| \le 2 N^{-1/2}$. By \zcref{l:pointwise}(d), we have
\begin{equation}
|\partial_u \mathcal K(u, v)| = \varrho^{-2} |\partial_x K_N(x, y)| \le C ,
\end{equation}
and $|S'| \le 1$. Therefore, by the case already treated,
\begin{equation}
| \mathcal K(u, v) - S(u - v) | \le | \mathcal K(u', v) - S(u' - v) | + (C + 1) |u - u'| \le C N^{-1/2} .
\end{equation}
This completes the proof.
\end{proof}

\section{Oscillatory integrals and global functions}

\begin{lemma}\label{l:oscillatory}
For $n \in \{N - 1, N\}$, all $a < b$ in $\bulk_N(\delta)$, and every continuously differentiable function $f$ on $[a, b]$, we have
\begin{equation}\label{e:oscillatory}
\Big| \int_a^b h_n(x) f(x) \dd x \Big| \le C N^{-3/4} \Big( \sup_{[a, b]} |f| + \int_a^b |f'(x)| \dd x \Big) + C N^{-5/4} \int_a^b |f(x)| \dd x .
\end{equation}
In particular, $\big| \int_a^b \varphi_N \big| + \big| \int_a^b \psi_N \big| \le C N^{-1/2}$.
\end{lemma}

\begin{proof}
By \eqref{e:PR}, the contribution of $\rho_n$ to the integral is at most
\begin{equation}
C N^{-1/4} N^{-1} \int_a^b |f| .
\end{equation}
To bound the main term, we integrate by parts using $\sin \Theta_n = - (\cos \Theta_n)' / \Theta_n'$, and we find
\begin{equation}
\int_a^b \alpha_n f \sin \Theta_n \dd x = \Big[ - \frac{\alpha_n f}{\Theta_n'} \cos \Theta_n \Big]_a^b + \int_a^b \cos \Theta_n \Big( \frac{\alpha_n}{\Theta_n'} \Big)' f \dd x + \int_a^b \cos \Theta_n \frac{\alpha_n}{\Theta_n'} f' \dd x .
\end{equation}
By \zcref{l:PR}, we have $|\alpha_n / \Theta_n'| \le C N^{-3/4}$ and $|(\alpha_n/\Theta_n')'| \le C N^{-5/4}$ on $\bulk_N(\delta)$. Together with the last display, these bounds imply \eqref{e:oscillatory}. The last claim follows by taking $f = 1$ and using $b - a \le C \sqrt N$ and $(N/2)^{1/4} \le N^{1/4}$.
\end{proof}

We now show that the global functions introduced in \zcref{s:exactidentities} are close to their limiting values in the bulk. Recall that $p_N$, $\omega_N$, and $c_N$ were defined in \eqref{e:cNomegadef} and \eqref{e:pNdef}.

\begin{lemma}\label{l:globalfunctions}
For all $y \in \bulk_N(\delta)$, the following estimates hold.
\begin{enumerate}[label=(\alph*)]
\item We have $|p_N(y) - 1| \le C N^{-1/2}$.
\item We have $|(\eps \varphi_N)(y)| + |(\eps \psi_N)(y)| \le C$ and $|(\eps \varphi_N)(y) \, (\eps \psi_N)(y)| \le C N^{-1/2}$. If $N$ is odd, then $|(\eps \psi_N)(y)| \le C N^{-1/2}$.
\item We have $|\omega_N(y)| \le C N^{-1/2}$.
\end{enumerate}
\end{lemma}

\begin{proof}
To prove (a), we use \eqref{e:pNderivative}, which gives
\begin{equation}
p_N(y) - p_N(0) = - \Big( \int_\R \varphi_N \Big) \int_0^y \psi_N(x) \dd x - \Big( \int_\R \psi_N \Big) \int_0^y \varphi_N(x) \dd x .
\end{equation}
\zcref{l:pN,l:oscillatory} show that the right side is bounded by $C N^{-1/2}$ in absolute value, since $[0, y] \subset \bulk_N(\delta)$. Recalling that $|p_N(0) - 1| \le 2 N^{-1/2}$, we obtain (a).

For (b), observe that $(\eps f)(0) = 0$ for every even integrable function $f$, so that
\begin{equation}
(\eps f)(y) = \int_0^y f .
\end{equation}
Suppose first that $N$ is even. Then $\varphi_N$ is even, and \zcref{l:oscillatory} implies that $|(\eps \varphi_N)(y)| \le C N^{-1/2}$. Further, \zcref{l:parity}(a), part (a), and $\int \varphi_N \ge 1$ give
\begin{equation}
|(\eps \psi_N)(y)| = p_N(y) / \int \varphi_N \le 2 .
\end{equation}
Now suppose that $N$ is odd. Then $\psi_N$ is even, so $|(\eps \psi_N)(y)| \le C N^{-1/2}$, while \zcref{l:parity}(b) gives
\begin{equation}
|(\eps \varphi_N)(y)| = c_N p_N(y) \le 2 .
\end{equation}
This proves (b).

It remains to prove (c). If $N$ is even, then $\omega_N = \eps \varphi_N$, which was bounded in the proof of (b). If $N$ is odd, then \zcref{l:parity}(b) gives $\omega_N = c_N(1 - p_N)$, and the claim follows from (a) and $c_N \le 1$.
\end{proof}

\section{Integrals of the kernel}

\begin{lemma}\label{l:kernelintegral}
For all $a, y \in \bulk_N(\delta)$ with $a < y$, the following estimates hold.
\begin{enumerate}[label=(\alph*)]
\item We have
\begin{equation}\label{e:kernelhalf}
\Big| \int_a^y K_N(x, y) \dd x - \frac12 \Big| \le \frac{C}{1 + \sqrt N (y - a)} + C N^{-1/4} .
\end{equation}
\item Set $\varrho = \varrho_N(y)$ and $T = \varrho (y - a)$. If $T \le 2 N^{1/4}$, then
\begin{equation}\label{e:kernellocal}
\Big| \int_a^y K_N(x, y) \dd x - \int_0^T S(t) \dd t \Big| \le C (1 + T) N^{-1/2} .
\end{equation}
\end{enumerate}
The same estimates hold for $\int_y^a K_N(x, y) \dd x$ when $a > y$, with $y - a$ replaced by $a - y$.
\end{lemma}

\begin{proof}
The last sentence of the lemma follows from the first two claims and the symmetry $K_N(-x, -y) = K_N(x, y)$, so we may suppose that $a < y$. For (b), the change of variables $x = y - t/\varrho$ gives
\begin{equation}
\int_a^y K_N(x, y) \dd x = \int_0^T \varrho^{-1} K_N(y - t/\varrho, y) \dd t ,
\end{equation}
and \zcref{l:localsine} with $x_0 = y$, $u = -t$, and $v = 0$ shows that the integrand is $S(t) + O(N^{-1/2})$, since $S$ is even. Integrating, we obtain (b).

For (a), suppose first that $T \le N^{1/4}$. We have
\begin{equation}\label{e:sinetail}
\Big| \frac12 - \int_0^T S(t) \dd t \Big| = \Big| \int_T^\infty \frac{\sin t}{\pi t} \dd t \Big| \le \min \Big( \frac12 , \frac{2}{\pi T} \Big) ,
\end{equation}
where we integrated by parts once for the second bound. Combining \eqref{e:sinetail} with (b), and using $T \ge c \sqrt N (y - a)$ and $(1 + T) N^{-1/2} \le 2 N^{-1/4}$, we obtain \eqref{e:kernelhalf}.

Now suppose that $T > N^{1/4}$, and set
\begin{equation}
a' = y - N^{1/4}/\varrho \in (a, y) .
\end{equation}
The case already treated gives
\begin{equation}
\int_{a'}^y K_N(x, y) \dd x = 1/2 + O(N^{-1/4}) ,
\end{equation}
and \eqref{e:CD} yields
\begin{equation}
\int_a^{a'} K_N(x, y) \dd x = \sqrt{\frac N2} \Big( h_{N - 1}(y) \int_a^{a'} \frac{h_N(x)}{x - y} \dd x - h_N(y) \int_a^{a'} \frac{h_{N - 1}(x)}{x - y} \dd x \Big) .
\end{equation}

We apply \zcref{l:oscillatory} with $f(x) = (x - y)^{-1}$ on $[a, a']$. Here $\sup |f| = (y - a')^{-1}$,
\begin{equation}
\int |f'| \le (y - a')^{-1} ,
\end{equation}
and
\begin{equation}
\int |f| \le \log ((y - a)/(y - a')) \le C \log N .
\end{equation}
We also have
\begin{equation}
(y - a')^{-1} = \varrho N^{-1/4} \le C N^{1/4} ,
\end{equation}
so these bounds give
\begin{equation}
|\int_a^{a'} h_n(x) (x - y)^{-1} \dd x| \le C N^{-1/2}
\end{equation}
for $n \in \{N - 1, N\}$. Together with \zcref{l:pointwise}(a), this implies that
\begin{equation}
|\int_a^{a'} K_N(x, y) \dd x| \le C N^{1/2} N^{-1/4} N^{-1/2} = C N^{-1/4} .
\end{equation}
In this case, we have $\sqrt N (y - a) \ge c N^{1/4}$, and \eqref{e:kernelhalf} follows.
\end{proof}

We now apply these estimates to the projection of a step function. For $t \in \R$, we define
\begin{equation}\label{e:gtdef}
g_t = K_N \eps_t - \eps_t ,
\end{equation}
where $\eps_t$ was defined in \eqref{e:epstdef}, and we define the bounded function
\begin{equation}\label{e:gsinedef}
g_\sine(u) = - \sgn(u) \int_{|u|}^\infty S(v) \dd v , \qquad u \in \R .
\end{equation}
The function $g_t$ measures how far the projection $K_N$ is from preserving the step function $\eps_t$. The next corollary shows that it is concentrated near $t$, where it is approximated by $g_\sine$ on the local scale.

\begin{corollary}\label{c:gt}
For all $t, y \in \bulk_N(\delta)$ with $y \neq t$, the following statements hold.
\begin{enumerate}[label=(\alph*)]
\item We have
\begin{equation}\label{e:gtformula}
g_t(y) = (\eps \varphi_N)(y) \, (\eps \psi_N)(y) + \sgn(y - t) \Big( \int_{\min(t, y)}^{\max(t, y)} K_N(x, y) \dd x - \frac12 \Big) .
\end{equation}
\item We have $|g_t(y)| \le C \big( 1 + \sqrt N |y - t| \big)^{-1} + C N^{-1/4}$.
\item Set $\varrho = \varrho_N(t)$, and write $y = t + u/\varrho$. If $0 < |u| \le N^{1/4}$, then $|g_t(y) - g_\sine(u)| \le C (1 + |u|) N^{-1/2}$.
\end{enumerate}
\end{corollary}

\begin{proof}
The identity (a) is a restatement of \eqref{e:Kstep}, since $\eps_t(y) = \sgn(y - t)/2$ and the oriented integral $\int_t^y$ equals $\sgn(y - t) \int_{\min(t, y)}^{\max(t, y)}$. The bound (b) follows from (a), \zcref{l:globalfunctions}(b), and \zcref{l:kernelintegral}(a).

To prove (c), we apply \zcref{l:kernelintegral}(b) with the local scale $\varrho_N(y)$ in place of $\varrho$. Set
\begin{equation}
T = \varrho_N(y) |y - t| = |u| \, \varrho_N(y) / \varrho .
\end{equation}
Since $\varrho \ge c \sqrt N$ and
\begin{equation}
|\varrho_N(y) - \varrho| \le C |y - t| \le C |u| / \varrho ,
\end{equation}
we have
\begin{equation}
|T - |u|| \le C |u|^2 / N \le C N^{-1/2} ,
\end{equation}
and in particular $T \le 2 N^{1/4}$. By \zcref{l:kernelintegral}(b) and $|S| \le 1/\pi$, we find
\begin{equation}
\int_{\min(t, y)}^{\max(t, y)} K_N(x, y) \dd x = \int_0^{|u|} S(v) \dd v + O\big( (1 + |u|) N^{-1/2} \big) .
\end{equation}
Inserting this into \eqref{e:gtformula}, and using \zcref{l:globalfunctions}(b) and the identity
\begin{equation}
\int_0^{|u|} S - 1/2 = - \int_{|u|}^\infty S ,
\end{equation}
we obtain (c).
\end{proof}