\chapter{Hermite functions and exact identities}\label{s:hermite}

In this section, we fix our notation and collect the facts about Hermite functions that will be used in the rest of the paper. The classical facts are recalled in \zcref{s:hermiteclassical}. The identities in \zcref{s:exactidentities} are elementary, but our argument depends on them. They express every global quantity that we need, such as the integral of the Hermite kernel over a half-line, through quantities in the bulk of the spectrum. This allows us to avoid any asymptotic analysis of Hermite functions near the spectral edges.

\section{Conventions}\label{s:conventions}

For real-valued measurable functions $f$ and $g$ on $\R$, we write
\begin{equation}\label{e:ipdef}
\ip{f}{g} = \int_\R f(x) g(x) \dd x
\end{equation}
whenever the integral converges absolutely. All functions in this paper are real-valued, except for the Fourier transforms. For a measurable set $A \subset \R$, we let $\1_A$ denote its indicator function and $|A|$ its Lebesgue measure, and we set $A^c = \R \setminus A$. For a bounded measurable function $Y$ and a function $v$ in a space of functions on which $\ip{Y}{\cdot}$ is defined, we let $v \otimes Y$ denote the rank-one operator
\begin{equation}\label{e:tensordef}
(v \otimes Y) f = \ip{Y}{f} v .
\end{equation}

We set $\sgn(x) = x / |x|$ for $x \neq 0$ and $\sgn(0) = 0$. For $t \in \R$, we define the step function
\begin{equation}\label{e:epstdef}
\eps_t(x) = \frac12 \sgn(x - t) ,
\end{equation}
and for $f \in L^1(\R)$ we define the function $\eps f$ by
\begin{equation}\label{e:epsdef}
(\eps f)(x) = \frac12 \int_\R \sgn(x - z) f(z) \dd z .
\end{equation}
We call $\eps$ the \emph{sign operator}. The following properties are immediate from its definition. First, if $f \in L^1(\R)$ is continuous, then $\eps f$ is continuously differentiable with $(\eps f)' = f$, and
\begin{equation}\label{e:epslimits}
\lim_{x \to \pm \infty} (\eps f)(x) = \pm \frac12 \int_\R f(z) \dd z .
\end{equation}
Further, if $f \in L^1(\R)$ is continuously differentiable with $f' \in L^1(\R)$ and $f(x) \to 0$ as $|x| \to \infty$, then $\eps(f') = f$. Finally, for all $f, g \in L^1(\R)$, we have $\ip{f}{\eps g} = - \ip{\eps f}{g}$.

Throughout the paper, we let $C$ and $c$ denote large and small positive constants, respectively. They may depend on parameters that will be specified in each statement, but not on $N$ or on the positions of the sets considered, and their values may change from line to line.

\section{Classical facts}\label{s:hermiteclassical}

For every integer $n \ge 0$, we define the \emph{Hermite function}
\begin{equation}\label{e:hndef}
h_n(x) = (2^n n! \sqrt{\pi})^{-1/2} (-1)^n e^{x^2/2} \frac{\mathrm d^n}{\mathrm d x^n} e^{-x^2} ,
\end{equation}
and we set $h_{-1} = 0$. The following facts are classical; see, for instance, \cite{Szego1975}. The functions $(h_n)_{n \ge 0}$ form an orthonormal basis of $L^2(\R)$, each $h_n$ is a polynomial of degree $n$ times $e^{-x^2/2}$, and $h_n(-x) = (-1)^n h_n(x)$. For all $n \ge 0$, we have the recurrence relations
\begin{equation}\label{e:recurrence}
\begin{gathered}
x h_n(x) = \sqrt{\frac{n + 1}{2}} \, h_{n + 1}(x) + \sqrt{\frac{n}{2}} \, h_{n - 1}(x) , \\
h_n'(x) = \sqrt{\frac{n}{2}} \, h_{n - 1}(x) - \sqrt{\frac{n + 1}{2}} \, h_{n + 1}(x) .
\end{gathered}
\end{equation}
For $f \in L^1(\R) \cap L^2(\R)$, let
\begin{equation}\label{e:fourierdef}
(\mathcal F f)(\xi) = \frac{1}{\sqrt{2 \pi}} \int_\R f(x) e^{- i x \xi} \dd x
\end{equation}
denote its Fourier transform, extended as usual to a unitary operator on $L^2(\R)$. The Hermite functions are eigenfunctions of this operator, with
\begin{equation}\label{e:hermitefourier}
\mathcal F h_n = (-i)^n h_n .
\end{equation}
We also recall that $h_{2m + 1}(0) = 0$ and
\begin{equation}\label{e:hnzero}
h_{2m}(0) = (-1)^m \pi^{-1/4} \frac{\sqrt{(2m)!}}{2^m m!}
\end{equation}
for all $m \ge 0$. Evaluating \eqref{e:hermitefourier} at $\xi = 0$, we have $\int_\R h_{2m + 1} = 0$ and
\begin{equation}\label{e:hnintegral}
\int_\R h_{2m}(x) \dd x = \sqrt{2 \pi} \, (-1)^m h_{2m}(0) = \sqrt 2 \, \pi^{1/4} \frac{\sqrt{(2m)!}}{2^m m!} .
\end{equation}

We set $V_0 = \{0\}$. For every integer $N \ge 1$, we let $V_N$ denote the span of $h_0, \dots, h_{N - 1}$, and we define
\begin{equation}\label{e:KNdef}
K_N(x, y) = \sum_{n = 0}^{N - 1} h_n(x) h_n(y) .
\end{equation}
We also let $K_N$ denote the integral operator with kernel $K_N(x, y)$. This operator is the orthogonal projection of $L^2(\R)$ onto $V_N$. For every bounded measurable $f$, we define $K_N f$ by the same formula. Then $K_N f \in V_N$. The kernel satisfies the Christoffel--Darboux formula
\begin{equation}\label{e:CD}
K_N(x, y) = \sqrt{\frac{N}{2}} \, \frac{h_N(x) h_{N - 1}(y) - h_{N - 1}(x) h_N(y)}{x - y} , \qquad x \neq y ,
\end{equation}
which follows from the first identity in \eqref{e:recurrence}. Further, for $x \in \R$, we have $K_N(x, x) = \sum_{n < N} h_n(x)^2$, and the Cauchy--Schwarz inequality gives
\begin{equation}\label{e:KCS}
|K_N(x, y)| \le K_N(x, x)^{1/2} K_N(y, y)^{1/2} , \qquad |f(x)|^2 \le K_N(x, x) \| f \|_2^2 \quad \text{for } f \in V_N .
\end{equation}

We will need a crude bound on Hermite functions far outside the spectrum.

\begin{lemma}\label{l:hermitetail}
For all integers $N \ge 1$ and $0 \le n \le N$, and all $x \in \R$ such that $|x| \ge 8 \sqrt N$, we have $h_n(x)^2 \le e^{-x^2/4}$.
\end{lemma}

\begin{proof}
Let $P_n$ denote the Hermite polynomial of degree $n$, normalized so that
\begin{equation}
h_n(x) = \frac{P_n(x) e^{-x^2/2}}{(2^n n! \sqrt \pi)^{1/2}} .
\end{equation}
These polynomials have the generating function (see \cite{Szego1975})
\begin{equation}
\sum_{n \ge 0} P_n(x) z^n / n! = e^{2 x z - z^2} .
\end{equation}
By Cauchy's estimate on the circle $|z| = \sqrt N$, we have
\begin{equation}
|P_n(x)| \le n! \, N^{-n/2} e^{2 |x| \sqrt N + N} .
\end{equation}
It follows that
\begin{equation}
h_n(x)^2 = \frac{P_n(x)^2 e^{-x^2}}{2^n n! \sqrt \pi} \le \frac{n!}{N^n} e^{4 |x| \sqrt N + 2N - x^2} \le e^{4 |x| \sqrt N + 2N - x^2} .
\end{equation}
In the last inequality, we used
\begin{equation}
n! \le n^n \le N^n
\end{equation}
for $1 \le n \le N$. The same bound holds for $n = 0$ by direct inspection. Finally, when $|x| \ge 8 \sqrt N$, we have $4 |x| \sqrt N \le x^2 / 2$ and $2N \le x^2/32$, so the exponent on the right side is bounded above by $- x^2 / 4$.
\end{proof}

\section{The GOE in Hermite coordinates}\label{s:hermitecoordinates}

Recall that $H_N$ denotes an $N \times N$ GOE matrix, normalized as in \eqref{e:goedensity}, with eigenvalues $\lambda_1 < \dots < \lambda_N$. We will work with the rescaled eigenvalues $\sqrt{N/2} \, \lambda_i$. By \eqref{e:goedensity}, the vector $(\sqrt{N/2} \, \lambda_1, \dots, \sqrt{N/2} \, \lambda_N)$ has the density
\begin{equation}\label{e:goehermite}
\frac{1}{\widetilde Z_N} \prod_{1 \le i < j \le N} |x_j - x_i| \prod_{i = 1}^N e^{- x_i^2/2}
\end{equation}
on $\{ x_1 < \dots < x_N \}$, where $\widetilde Z_N$ is a normalizing constant. We call the variable
\begin{equation}\label{e:hermitecoordinates}
x = \sqrt{\frac N2} \, \lambda
\end{equation}
the \emph{Hermite coordinate}, and the variable $\lambda$ the \emph{eigenvalue coordinate}. In Hermite coordinates, the semicircle law is supported on $[-\sqrt{2N}, \sqrt{2N}]$, and the expected number of points per unit length near $x_0$ is approximately $\pi^{-1} \sqrt{2N - x_0^2}$.

\begin{definition}\label{d:holeprob}
For a measurable set $J \subset \R$ of finite measure, we let $K_J$ denote the integral operator on $L^2(J)$ with kernel $K_N(x, y)$, $x, y \in J$, and we define
\begin{equation}\label{e:PDdef}
\Pgoe_N(J) = \Prob\Big( \sqrt{\frac N2} \, \lambda_i \notin J \text{ for all } i \in \{1, \dots, N\} \Big) , \qquad \Dgue_N(J) = \det(\Id - K_J) .
\end{equation}
We call $\Pgoe_N(J)$ the \emph{GOE hole probability} of $J$.
\end{definition}

The operator $K_J$ is positive and trace class, with trace $\int_J K_N(x, x) \dd x < \infty$, since $x \mapsto K_N(x, x)$ is bounded. A nonzero element of $V_N$ cannot vanish on the complement of a set of finite measure, which implies that $\| K_J \| < 1$. Hence $\Dgue_N(J)$ is a well-defined Fredholm determinant in $(0, 1]$. It is the probability that the GUE, with eigenvalue density proportional to $\prod_{i < j} |x_j - x_i|^2 \prod_i e^{-x_i^2}$, has no eigenvalue in $J$ (see, for instance, \cite{AndersonGuionnetZeitouni2010}), but we will not use this interpretation. Expanding $K_N$ in the basis $(h_n)_{n < N}$, we also have
\begin{equation}\label{e:Dfinite}
\Dgue_N(J) = \det\big( \1_{m = n} - \ip{h_m}{\1_J h_n} \big)_{m, n = 0}^{N - 1} .
\end{equation}
Finally, for every measurable set $A \subset \R$ of finite measure, we have $E_{1, N}(A) = \Pgoe_N( \sqrt{N/2} \, A)$, where we recall that $E_{1, N}(A)$ is the hole probability defined in \eqref{e:E1def}.

\section{Top modes and exact identities}\label{s:exactidentities}

We now define the functions that measure how the GOE differs from the GUE. For every integer $N \ge 2$, we define the \emph{top modes}
\begin{equation}\label{e:topmodes}
\varphi_N = \Big( \frac N2 \Big)^{1/4} h_N , \qquad \psi_N = \Big( \frac N2 \Big)^{1/4} h_{N - 1} .
\end{equation}
With this normalization, $\varphi_N(x) \psi_N(y) = \sqrt{N/2} \, h_N(x) h_{N - 1}(y)$, which is the factor appearing in \eqref{e:CD}. We also define the constant $c_N$ and the function $\omega_N$ by
\begin{equation}\label{e:cNomegadef}
c_N = \begin{cases} \big( \int_\R \psi_N \big)^{-1} & \text{if } N \text{ is odd}, \\ 0 & \text{if } N \text{ is even}, \end{cases} \qquad \omega_N = \eps \varphi_N + c_N ,
\end{equation}
and the function
\begin{equation}\label{e:pNdef}
p_N(y) = \int_\R K_N(x, y) \dd x = (K_N \1_\R)(y) .
\end{equation}
For odd $N$, the integer $N - 1$ is even, so $\int \psi_N \neq 0$ by \eqref{e:hnintegral}, and $c_N$ is well defined. Note that $p_N \in V_N$.

\begin{lemma}\label{l:derivative}
For all $N \ge 2$ and $x, y \in \R$, we have
\begin{equation}\label{e:derivativeidentity}
\partial_x K_N(x, y) + \partial_y K_N(x, y) = - \varphi_N(x) \psi_N(y) - \psi_N(x) \varphi_N(y) .
\end{equation}
\end{lemma}

\begin{proof}
By the second identity in \eqref{e:recurrence}, we have
\begin{equation}
\sum_{n = 0}^{N - 1} h_n'(x) h_n(y) = \sum_{n = 1}^{N - 1} \sqrt{\frac n2} \, h_{n - 1}(x) h_n(y) - \sum_{n = 1}^{N} \sqrt{\frac{n}{2}} \, h_{n}(x) h_{n - 1}(y) ,
\end{equation}
where we shifted the index in the second sum. We add the same identity with $x$ and $y$ interchanged. Then all terms with $n \le N - 1$ cancel, and the remaining terms with $n = N$ give
\begin{equation}
- \sqrt{N/2} \, \big( h_N(x) h_{N - 1}(y) + h_{N - 1}(x) h_N(y) \big) ,
\end{equation}
which is the right side of \eqref{e:derivativeidentity}.
\end{proof}

\begin{lemma}\label{l:pN}
For all $N \ge 2$, we have
\begin{equation}\label{e:pNderivative}
p_N' = - \Big( \int_\R \varphi_N \Big) \psi_N - \Big( \int_\R \psi_N \Big) \varphi_N
\end{equation}
and
\begin{equation}\label{e:pNzero}
p_N(0) = \sqrt 2 \sum_{0 \le m \le (N - 1)/2} (-1)^m \binom{2m}{m} 4^{-m} , \qquad |p_N(0) - 1| \le 2 N^{-1/2} .
\end{equation}
Moreover, if $N$ is even, then $\int \psi_N = 0$ and $1 \le \int \varphi_N \le 2$, and if $N$ is odd, then $\int \varphi_N = 0$ and $1 \le \int \psi_N \le 2$. In particular, $c_N \in [1/2, 1]$ when $N$ is odd.
\end{lemma}

\begin{proof}
Integrating \eqref{e:derivativeidentity} in $x$ over $\R$, and using the fact that
\begin{equation}
\int_\R \partial_x K_N(x, y) \dd x = 0 ,
\end{equation}
we obtain \eqref{e:pNderivative}.

Next, since $h_n(0) = 0$ for odd $n$, we have
\begin{equation}
p_N(0) = \sum_{2m < N} h_{2m}(0) \int h_{2m} .
\end{equation}
Inserting \eqref{e:hnzero} and \eqref{e:hnintegral} into this identity, we obtain the formula in \eqref{e:pNzero}.

We now turn to the bound. Set
\begin{equation}
\varpi_m = \binom{2m}{m} 4^{-m} .
\end{equation}
An elementary induction using $\varpi_{m + 1} = \varpi_m (2m + 1)/(2m + 2)$ shows that
\begin{equation}\label{e:betabounds}
\frac{1}{2 \sqrt m} \le \varpi_m \le \frac{1}{\sqrt{2 m + 1}} , \qquad m \ge 1 .
\end{equation}
In particular, $(\varpi_m)_{m \ge 0}$ decreases to $0$, so the series $\sum_{m \ge 0} (-1)^m \varpi_m$ converges. By Abel's theorem applied to the expansion
\begin{equation}
(1 - 4z)^{-1/2} = \sum_{m \ge 0} \varpi_m (4z)^m
\end{equation}
at $z = -1/4$, its value is $2^{-1/2}$. The alternating series bound then gives $|p_N(0) - 1| \le \sqrt 2 \, \varpi_{M + 1}$ with $M = \lfloor (N - 1)/2 \rfloor$, and \eqref{e:betabounds} implies that
\begin{equation}
\sqrt 2 \, \varpi_{M + 1} \le \sqrt 2 (2M + 3)^{-1/2} \le 2 N^{-1/2} .
\end{equation}

Finally, we have
\begin{equation}
\int h_n = 0
\end{equation}
for odd $n$. When $N = 2m$ is even, \eqref{e:hnintegral} gives
\begin{equation}
\int \varphi_N = m^{1/4} \sqrt 2 \, \pi^{1/4} \varpi_m^{1/2} ,
\end{equation}
and \eqref{e:betabounds} shows that this number belongs to $[\pi^{1/4}, 2^{1/4} \pi^{1/4}] \subset [1, 2]$. The same computation with $(N/2)^{1/4} = (m + 1/2)^{1/4}$ shows that $\int \psi_N \in [1, 2]$ when $N = 2m + 1$ is odd.
\end{proof}

The next lemma records which antiderivatives of Hermite functions lie in $V_N$.

\begin{lemma}\label{l:antiderivatives}
Let $n \ge 0$. Then the derivative maps $V_n$ onto $\{ f \in V_{n + 1} : \int_\R f = 0 \}$. Consequently, the following statements hold.
\begin{enumerate}[label=(\alph*)]
\item If $n$ is odd, then $\eps h_n \in V_n$.
\item If $N$ is even, then $K_N(\eps h_N) = 0$.
\item If $N$ is odd, then $\eps h_N \in V_N$ and $K_N(\eps h_{N - 1}) = 0$.
\end{enumerate}
\end{lemma}

\begin{proof}
By \eqref{e:recurrence}, the derivative maps $V_n$ into $V_{n + 1}$. The image consists of functions with zero integral, and the derivative is injective on $V_n$, so the image has dimension $n$. The space $\{ f \in V_{n + 1} : \int f = 0 \}$ also has dimension $n$, since $\int h_0 \neq 0$. Hence the image is all of this space, which is the first claim.

To prove (a), suppose that $n$ is odd. Then $h_n \in V_{n + 1}$ and $\int h_n = 0$, so the first claim gives $h_n = g'$ for some $g \in V_n$. The functions $g$ and $\eps h_n$ vanish at infinity and have derivative $h_n$, so they are equal, and (a) follows.

We now prove (b) and (c), using the identity $\ip{h_m}{\eps f} = - \ip{\eps h_m}{f}$. Let $m < N$. For odd $m$, (a) implies that $\eps h_m \in V_m$, so $\eps h_m$ is orthogonal to $h_N$ and $h_{N - 1}$ when $m \le N - 2$. For even $m$, the function $\eps h_m$ is odd.

Suppose first that $N$ is even. When $m < N$ is odd, we have $m \le N - 1$, so $\ip{\eps h_m}{h_N} = 0$. When $m$ is even, the function $\eps h_m \, h_N$ is odd, so again $\ip{\eps h_m}{h_N} = 0$. We conclude that $K_N (\eps h_N) = 0$, which is (b).

Finally, suppose that $N$ is odd. Then $\eps h_N \in V_N$ by (a). The same parity argument applied to $h_{N - 1}$, which is even, gives $\ip{\eps h_m}{h_{N - 1}} = 0$ for all $m < N$. This proves (c).
\end{proof}

The following lemma contains the cancellation between the two terms in $\omega_N$ for odd $N$, which is the main reason for the definition \eqref{e:cNomegadef}.

\begin{lemma}\label{l:parity}
Let $N \ge 2$.
\begin{enumerate}[label=(\alph*)]
\item If $N$ is even, then $\eps \psi_N = - p_N / \int \varphi_N$ and $\omega_N = \eps \varphi_N$.
\item If $N$ is odd, then $\eps \varphi_N = - c_N p_N$, and therefore $\omega_N = c_N (1 - p_N)$.
\item In both cases, $K_N \omega_N = 0$, $\ip{\psi_N}{\omega_N} = 0$, and $\ip{\psi_N}{\eps \varphi_N} = - c_N \int \psi_N$.
\end{enumerate}
\end{lemma}

\begin{proof}
Suppose first that $N$ is even. By \zcref{l:pN}, we have
\begin{equation}
p_N' = - (\int \varphi_N) \psi_N ,
\end{equation}
so the derivative of $\eps \psi_N + p_N / \int \varphi_N$ vanishes. As $y \to \infty$, we have $p_N(y) \to 0$, since $p_N \in V_N$, and \eqref{e:epslimits} gives
\begin{equation}
(\eps \psi_N)(y) \to (1/2) \int \psi_N = 0 .
\end{equation}
Therefore,
\begin{equation}
\eps \psi_N + p_N / \int \varphi_N = 0 ,
\end{equation}
and (a) follows, since $c_N = 0$.

Next, suppose that $N$ is odd. Then \zcref{l:pN} gives $p_N' = - c_N^{-1} \varphi_N$, so the derivative of $\eps \varphi_N + c_N p_N$ vanishes. As $y \to \infty$, we have $p_N(y) \to 0$ and
\begin{equation}
(\eps \varphi_N)(y) \to (1/2) \int \varphi_N = 0 ,
\end{equation}
which gives (b).

We now prove (c). When $N$ is even, \zcref{l:antiderivatives}(b) implies that
\begin{equation}
K_N \omega_N = K_N \eps \varphi_N = 0 .
\end{equation}
For odd $N$, we have
\begin{equation}
p_N = K_N \1_\R \in V_N ,
\end{equation}
so $K_N p_N = p_N$, and therefore
\begin{equation}
K_N \omega_N = c_N (p_N - K_N p_N) = 0 .
\end{equation}
Since $\psi_N \in V_N$ and $\omega_N$ is bounded, we find
\begin{equation}
\ip{\psi_N}{\omega_N} = \ip{K_N \psi_N}{\omega_N} = \ip{\psi_N}{K_N \omega_N} = 0 .
\end{equation}
Finally, we have
\begin{equation}
\ip{\psi_N}{\eps \varphi_N} = \ip{\psi_N}{\omega_N} - c_N \int \psi_N = - c_N \int \psi_N .
\end{equation}
\end{proof}

Observe that for odd $N$, \zcref{l:parity}(b) and \zcref{l:pN} show that $\eps \varphi_N$ is close to the constant $-c_N$ wherever $p_N$ is close to $1$. We will see in \zcref{s:bulk} that this holds in the bulk of the spectrum, so that $\omega_N$ is small there, although the two terms in its definition are of order one.

The last identity of this section concerns the sign function. It shows that, up to the product $(\eps \varphi_N)(\eps \psi_N)$, the projection $K_N$ applied to a step function only involves the kernel between the jump and the point of evaluation.

\begin{lemma}\label{l:signkernel}
For all $N \ge 2$ and $y \in \R$, we have
\begin{equation}\label{e:signdiagonal}
\frac12 \int_\R \sgn(x - y) K_N(x, y) \dd x = (\eps \varphi_N)(y) \, (\eps \psi_N)(y) .
\end{equation}
Consequently, for all $t, y \in \R$, we have
\begin{equation}\label{e:Kstep}
(K_N \eps_t)(y) = (\eps \varphi_N)(y) \, (\eps \psi_N)(y) + \int_t^y K_N(x, y) \dd x ,
\end{equation}
where the integral is oriented, so that $\int_t^y = - \int_y^t$ when $y < t$.
\end{lemma}

\begin{proof}
Let $G(y)$ denote the left side of \eqref{e:signdiagonal}. Then
\begin{equation}
2 G(y) = \int_y^\infty K_N(x, y) \dd x - \int_{-\infty}^y K_N(x, y) \dd x .
\end{equation}
Differentiating this identity, we have
\begin{equation}
G'(y) = - K_N(y, y) + \frac12 \int_\R \sgn(x - y) \, \partial_y K_N(x, y) \dd x .
\end{equation}
We now insert \eqref{e:derivativeidentity} into the integral. Integrating over $(y, \infty)$ and $(-\infty, y)$ separately, we see that the term with $-\partial_x K_N(x, y)$ gives $K_N(y, y)$. The remaining terms give
\begin{equation}
- \frac12 \int_\R \sgn(x - y) \big( \varphi_N(x) \psi_N(y) + \psi_N(x) \varphi_N(y) \big) \dd x = \psi_N(y) (\eps \varphi_N)(y) + \varphi_N(y) (\eps \psi_N)(y) ,
\end{equation}
where we used
\begin{equation}
\int \sgn(x - y) f(x) \dd x = - 2 (\eps f)(y) .
\end{equation}
Therefore, $G'(y) = \big( (\eps \varphi_N) (\eps \psi_N) \big)'(y)$. As $y \to \infty$, we have $G(y) \to 0$ and
\begin{equation}
(\eps \varphi_N)(y) (\eps \psi_N)(y) \to (1/4) \int \varphi_N \int \psi_N = 0 ,
\end{equation}
since one of the two integrals vanishes by \zcref{l:pN}. This completes the proof of \eqref{e:signdiagonal}.

We now prove \eqref{e:Kstep}, whose two sides are continuously differentiable in $t$, with derivative $-K_N(t, y)$ by the symmetry of $K_N$. For $t = y$, the left side reduces to $G(y)$ and the right side to $(\eps \varphi_N)(y) (\eps \psi_N)(y)$, so \eqref{e:Kstep} follows from \eqref{e:signdiagonal}.
\end{proof}