# Sharp Schatten asymptotics and Sonin trace localization for semilocal Fourier cutoffs

**Author:** bramble_trace

## 1. Introduction and conventions

Connes and Consani's semilocal trace-formula program compares the arithmetic Weil form with positive Fourier compressions and Sonin traces [1,2]. The associated ratios of local factors are generally not inner. Their product with the archimedean ratio is quasi-inner: the relevant Hardy off-diagonal block is compact [3]. For one finite prime, [3, Theorem 4.4] gives compactness order $1/2$; for $m$ finite primes, [3, Theorem 4.8] gives order $1/(2m)$ by distributing the gamma factor among the primes. Connes, Consani, and Moscovici subsequently proved that the Euler embedding is onto and described the resulting weighted de Branges spaces [4, Theorem 4.6 and Section 4.8]. Their isomorphism is bounded with bounded inverse, not isometric; the changing norm is essential to the trace asymptotics below.

We give an explicit cosine expansion which yields $\mathcal S_q$ membership for every $q>2$ for each fixed finite prime set. For every nonempty set this threshold is sharp: a lacunary family of cosine blocks has nondecaying trace pairings with the overlap, which rules out Hilbert–Schmidt membership. A sharper fixed-set calculation gives an explicit $(\log N)^m$ asymptotic for the sum of the first $N$ squared singular values, where $m$ is the number of primes. This uses the Baker–Wüstholz theorem to control near collisions among smooth dilation scales. We then keep the test-function smoothing in the estimate. A single-prime squared increment has the scale $(\log p)/p$, but the complete overlap for primes up to $X$ has a squared smoothed norm asymptotic to a positive constant times $\sqrt X$. We also prove that the actual Sonin trace has growth exponent $1/2$, with an explicit logarithmic loss in its lower bound and an $O_f(\sqrt X/(\log X)^2)$ upper bound. The bounds match in order when the test transform is nonzero at the origin. The generic two-projection term therefore carries the full leading positive-compression coefficient. The proof uses weighted polynomial concentration, the classical Vinogradov–Korobov zero-free region, and quantitative analytic restriction. A uniform bound for the weighted archimedean evaluations also gives an exponential-scale density law: the Sonin trace concentrates at Mellin frequency zero, while the projections converge strongly to zero. The trace has limsup at most $4|F(0)|^2$ on the $\sqrt X/(\log X)^2$ scale. No sharp coefficient is asserted. Finally, an explicit unimodular phase adds the Weil pole term to the signed trace and removes the leading growth from the positive overlap. It preserves the fixed-set spectral asymptotic. The trace identities, projection geometry, and gamma-factor normalization are established ingredients; the quantitative estimates and their application to the growing prime set are the contribution studied here.

Subtracting the continuous PNT background from an arithmetic symbol is a familiar operation; for a recent numerical study in the Weil-operator setting, see [5, Sections 3.5–3.7]. Our compensation is specified as an actual unitary factor and is exact on the stated support class. The estimates do not rely on a bounded residual plateau or on numerical zero locations.

On $L^2(0,\infty;dx)$ let

$$
\mathcal Cg(x)=2\int_0^\infty\cos(2\pi xy)g(y)\,dy
$$

be the unitary cosine transform, and let $R=1_{(0,1)}$. For $c>0$, write $C_c$ for the operator on $L^2(0,1)$ with kernel $2\cos(2\pi cxy)$. Let $L_cg(x)=g(cx)$; thus $\|L_c\|=c^{-1/2}$ and $\mathcal CL_c=c^{-1}L_{1/c}\mathcal C$.

For a prime $p$, set

$$
U_p=(I-L_p)(I-p^{-1}L_{1/p})^{-1}.
$$

In logarithmic coordinates its multiplier is

$$
u_p(t)=\frac{1-p^{-1/2+it}}{1-p^{-1/2-it}},
$$

so $U_p$ is unitary. For a finite prime set $\mathcal P$, define

$$
U_{\mathcal P}=\prod_{p\in\mathcal P}U_p,\qquad
K_{\mathcal P}=R\mathcal C U_{\mathcal P}R.
$$

The factors commute. The unitary map $g(x)\mapsto e^{y/2}g(e^y)$ identifies this compression with an operator on $L^2(\mathbb R,dy)$, with both cutoffs at $y<0$. We use the same notation after this identification.

For $f\in C_c^\infty(\mathbb R)$, let $T_f$ be convolution in $y$ and use

$$
F(t)=\int_{\mathbb R}f(y)e^{-ity}\,dy,\qquad
\|F\|_2^2=2\pi\|f\|_2^2.
$$

The norm $\|f\|_{H^1}^2$ means $\|f\|_2^2+\|f'\|_2^2$. All constants and trace normalizations below use these conventions.

## 2. A Schatten exponent for every finite prime set

**Theorem 2.1.** For every finite $\mathcal P$ and every $q>2$,

$$
K_{\mathcal P}\in\mathcal S_q,
$$

and

$$
\boxed{\|K_{\mathcal P}\|_{\mathcal S_q}
\le4^{1/q}\prod_{p\in\mathcal P}
\frac{1+p^{-1/2-1/q}}{1-p^{-1/2+1/q}}.}\tag{2.1}
$$

**Proof.** Let $A=\prod_{p\in\mathcal P}p$ and let $\mathcal N_{\mathcal P}$ be the positive integers whose prime divisors belong to $\mathcal P$. Expanding the finite numerator and the geometric denominators gives

$$
U_{\mathcal P}=
\sum_{a\mid A}\mu(a)\sum_{b\in\mathcal N_{\mathcal P}}b^{-1}L_{a/b}.
$$

This converges absolutely in operator norm, because $\|b^{-1}L_{a/b}\|=(ab)^{-1/2}$ and the prime set is finite. Dilation covariance therefore gives the exact expansion

$$
\boxed{K_{\mathcal P}=
\sum_{a\mid A}\frac{\mu(a)}a
\sum_{b\in\mathcal N_{\mathcal P}}C_{b/a}.}\tag{2.2}
$$

No pointwise convergence of the cosine kernels is required.

Unitarity and direct kernel integration give

$$
\|C_c\|\le c^{-1/2},\qquad
\|C_c\|_{\mathcal S_2}^2
=2+2\frac{\operatorname{Si}(4\pi c)}{4\pi c}\le4.
$$

Writing $\alpha=1/2-1/q>0$, interpolation of the singular values yields

$$
\|C_c\|_{\mathcal S_q}\le4^{1/q}c^{-\alpha}.
$$

The Schatten triangle inequality applied to (2.2) now gives

$$
\|K_{\mathcal P}\|_{\mathcal S_q}
\le4^{1/q}\sum_{a\mid A}a^{\alpha-1}
\sum_{b\in\mathcal N_{\mathcal P}}b^{-\alpha},
$$

which is (2.1). The series converges in $\mathcal S_q$, and its limit agrees with its operator-norm limit. $\square$

For each fixed prime set and every $\epsilon>0$, the theorem implies

$$
s_n(K_{\mathcal P})=O_{\mathcal P,\epsilon}(n^{-1/2+\epsilon}).
$$

For two or more finite primes, this improves the exponent stated in [3, Theorem 4.8]. For one prime, the weak order $1/2$ result of [3] remains more precise at the endpoint. The explicit constant in (2.1) is not uniform as the prime set grows.

**Theorem 2.2.** If $\mathcal P$ is nonempty, then $K_{\mathcal P}\notin\mathcal S_2$. Consequently, for $0<q<\infty$,

$$
K_{\mathcal P}\in\mathcal S_q\quad\Longleftrightarrow\quad q>2.
$$

**Proof.** Collect equal scales in (2.2). For $n\in\mathcal N_{\mathcal P}$ put

$$
a(n)=\prod_{p\mid n}(1-p^{-1})
\prod_{\substack{p\in\mathcal P\\p\nmid n}}(-p^{-1}).
$$

The local coefficient is $-1/p$ when the exponent of $p$ in $n$ is zero, and $1-1/p$ otherwise. Hence

$$
K_{\mathcal P}=\sum_{n\in\mathcal N_{\mathcal P}}a(n)C_{n/A},\tag{2.3}
$$

absolutely in operator norm, since $|a(n)|\le1$ and $\sum_{n\in\mathcal N_{\mathcal P}}n^{-1/2}<\infty$.

Let $n_k=A^k$, $p_* =\min\mathcal P$, and $d_k=\min(p_*^k,n_k/2)$. No other smooth integer lies at distance less than $d_k$ from $n_k$. Indeed, if $m=n_k+d$ with $0<|d|<d_k$, then $v_p(d)<k=v_p(n_k)$ for every $p\in\mathcal P$, so $v_p(m)=v_p(d)$. Smoothness of $m$ would imply $m\mid |d|$, whereas $m\ge n_k-|d|>|d|$. This is a contradiction. Also

$$
a(n_k)=c_{\mathcal P}:=\prod_{p\in\mathcal P}(1-p^{-1})>0.
$$

Write $I(t)=\operatorname{Si}(t)/t$, with $I(0)=1$. Direct integration gives

$$
\langle C_c,C_d\rangle_{\rm HS}
=2I(2\pi(c-d))+2I(2\pi(c+d)).\tag{2.4}
$$

The blocks $B_k=C_{n_k/A}=C_{A^{k-1}}$ form a Bessel sequence in the Hilbert–Schmidt space: their diagonal Gram entries are bounded and their off-diagonal entries are $O_A(A^{-\max(j,k)})$, so the absolute row sums are uniformly bounded. Each $B_k$ is trace class, by the absolutely nuclear-convergent rank-one Taylor expansion of its cosine kernel.

Thus (2.3) can be paired term by term with $B_k$. The term $n=n_k$ tends to $2c_{\mathcal P}$. The total of all other terms tends to zero. For $n\le2n_k$, there are $O_{\mathcal P}((1+\log n_k)^{|\mathcal P|})$ smooth integers; their distance from $n_k$ is at least $d_k$, so (2.4) bounds their total by $O_{\mathcal P}((1+\log n_k)^{|\mathcal P|}/d_k)=o(1)$. For $n>2n_k$, the total is bounded by a constant times

$$
\sum_{\substack{n\in\mathcal N_{\mathcal P}\\n>2n_k}}n^{-1}
\le(2n_k)^{-1/2}\sum_{n\in\mathcal N_{\mathcal P}}n^{-1/2}=o(1).
$$

Therefore $\operatorname{Tr}(K_{\mathcal P}B_k^*)\to2c_{\mathcal P}\ne0$. If $K_{\mathcal P}$ were Hilbert–Schmidt, its pairings with a Bessel sequence would be square summable. This contradiction proves the endpoint assertion. The inclusion $\mathcal S_q\subset\mathcal S_2$ for $q<2$ completes the classification. $\square$

The empty prime set is different: its smooth cosine kernel is trace class. Theorem 2.2 determines the strong Schatten threshold without a Diophantine separation theorem. The next result quantifies that endpoint using a classical bound for linear forms in logarithms.

**Theorem 2.3.** Let $m=|\mathcal P|\ge1$ and $c_{\mathcal P}=\prod_{p\in\mathcal P}(1-p^{-1})$. Then

$$
\boxed{\sum_{j=1}^N s_j(K_{\mathcal P})^2
\sim\frac{2c_{\mathcal P}^2}{m!\prod_{p\in\mathcal P}\log p}
(\log N)^m.}\tag{2.5}
$$

In particular, $K_{\mathcal P}$ does not belong to weak $\mathcal S_2$ when $m\ge2$.

**Proof.** All constants in this proof may depend on the fixed prime set. For distinct $b,c\in\mathcal N_{\mathcal P}$ with $b>c$, the nonzero linear form

$$
\log(b/c)=\sum_{p\in\mathcal P}(v_p(b)-v_p(c))\log p
$$

has coefficients $O_{\mathcal P}(\log b)$. The Baker–Wüstholz theorem [6], in the integer specialization stated in [7, Theorem 3.1], gives constants $\eta,\kappa>0$ such that

$$
|b-c|\ge\eta\frac{b}{(1+\log b)^\kappa}.\tag{2.6}
$$

For $c\le b/2$ this is elementary. Otherwise it follows from the lower bound for $\log(b/c)$ and $b-c=c(e^{\log(b/c)}-1)$. Zero coefficients can be omitted, leaving only finitely many possible subsets of the fixed prime set.

Since $|I(t)|\le\min(1,2/|t|)$, (2.4) and (2.6) imply

$$
\sum_{\substack{b,c\in\mathcal N_{\mathcal P}\\b\ne c}}
|\langle C_{b/A},C_{c/A}\rangle_{\rm HS}|<\infty.\tag{2.7}
$$

Indeed, for $2^{\ell-1}<\max(b,c)\le2^\ell$ each entry is $O(2^{-\ell}(1+\ell)^\kappa)$ and there are $O((1+\ell)^{2m})$ pairs. The diagonal entries equal $2+O(A/b)$, with summable errors.

Let $H_R=\sum_{b\le R}a(b)C_{b/A}$, using (2.3). Counting exponent lattice points in the simplex $\sum e_p\log p\le\log R$ gives

$$
\#\{b\in\mathcal N_{\mathcal P}:b\le R\}
=\frac{(\log R)^m}{m!\prod_{p\in\mathcal P}\log p}
+O((1+\log R)^{m-1}).
$$

For example, unit cubes around the exponent vectors compare this count with simplex volumes at $\log R$ and $\log R+\log A$. The terms missing at least one prime lie on exponent faces and number $O((1+\log R)^{m-1})$; all other terms have $a(b)=c_{\mathcal P}$. Thus, writing $\mathfrak c=2c_{\mathcal P}^2/(m!\prod\log p)$,

$$
\|H_R\|_{\rm HS}^2=\mathfrak c(\log R)^m
+O((1+\log R)^{m-1}).\tag{2.8}
$$

Here (2.7) bounds the total cross contribution uniformly.

There are $O((1+\log R)^{m-1})$ smooth integers in $[R,2R]$: fix all but one exponent, then the last has a bounded number of choices. Dyadic summation gives the operator-norm tail

$$
\|K_{\mathcal P}-H_R\|\le C R^{-1/2}(1+\log R)^{m-1}.\tag{2.9}
$$

For $S_N(K)=\sum_{j\le N}s_j(K)^2$, the square root is a unitarily invariant norm. Choose $R=N(\log N)^{2m}$. Then

$$
\sqrt{S_N(K_{\mathcal P})}\le\|H_R\|_{\rm HS}
+\sqrt N\|K_{\mathcal P}-H_R\|
=\sqrt{\mathfrak c}(\log N)^{m/2}(1+o(1)),\tag{2.10}
$$

since the tail term is $O(1/\log N)$.

For the matching lower bound, let $E_R$ project onto the span of $x^{2j}$, $0\le j\le J=\lceil10\pi R\rceil$. This has rank $O(R)$. The rank-one Taylor expansion of $C_{b/A}$ has nuclear tail beyond $J$ bounded by

$$
2\sum_{j>J}\frac{(2\pi R)^{2j}}{(2j)!(4j+1)}=O(e^{-R})
\qquad(b\le R).
$$

The truncated polynomial operator is fixed by compression with $E_R$, so

$$
\|H_R-E_RH_RE_R\|_{\mathcal S_1}=o(1).\tag{2.11}
$$

Moreover, operator-norm convergence in (2.3), trace-class membership of $H_R$, and (2.7) give

$$
\operatorname{Tr}(K_{\mathcal P}H_R^*)
=\|H_R\|_{\rm HS}^2+o(1).\tag{2.12}
$$

The omitted pairs have $b>R\ge c$, and form a tail of the absolutely summable off-diagonal array. Replacing $H_R$ by $E_RH_RE_R$ in (2.12) changes it by $o(1)$, by (2.11). Cauchy–Schwarz and contraction of the Hilbert–Schmidt norm under compression imply

$$
\|E_RK_{\mathcal P}E_R\|_{\rm HS}^2
\ge\|H_R\|_{\rm HS}^2+o(1).\tag{2.13}
$$

The left side is at most $S_{\operatorname{rank}E_R}(K_{\mathcal P})$. Since $\operatorname{rank}E_R\le C_0R$, choosing $R=N/(2C_0)$ supplies the required lower bound for every large $N$. Equations (2.8), (2.10), and (2.13) prove (2.5).

Finally, weak $\mathcal S_2$ membership would give $s_j\le Mj^{-1/2}$, hence $S_N\le M^2(1+\log N)$, contradicting (2.5) when $m\ge2$. $\square$

For one prime, [3] supplies weak $\mathcal S_2$ membership. Thus the prime count appears as a logarithmic power in (2.5), while the strong Schatten threshold stays at $2$. We do not infer an individual singular-value asymptotic from this partial-sum law. The Diophantine constants in its proof are not uniform in a growing prime set.

## 3. Mellin smoothing of a single dilation block

**Theorem 3.1.** For $c\ge1$,

$$
\boxed{\left|\|T_f C_c\|_{\rm HS}^2
-\frac{\log c}{c}\|f\|_2^2\right|
\le\frac{24}{c}\|f\|_{H^1}^2.}\tag{3.1}
$$

For $0<c\le1$, $\|T_fC_c\|_{\rm HS}^2\le16\|f\|_2^2$.

Here and below $C_c$ is logarithmically conjugated and extended by zero outside the two negative half-lines before applying $T_f$.

**Proof.** Its logarithmic kernel is

$$
1_{u<0}1_{v<0}\,2e^{(u+v)/2}\cos(2\pi c e^{u+v}).
$$

Put

$$
A_t(v)=\int_0^v w^{-1/2-it}\cos(2\pi w)\,dw,\qquad
H_c(t)=\int_0^c|A_t(v)|^2\frac{dv}{v}.
$$

Fourier transformation in the output variable, followed by Parseval and Tonelli, gives

$$
\boxed{\|T_fC_c\|_{\rm HS}^2
=\frac2{\pi c}\int_{\mathbb R}|F(t)|^2H_c(t)\,dt.}\tag{3.2}
$$

Indeed, for fixed input $v<0$, the output transform before convolution is

$$
2e^{itv}c^{-1/2+it}A_t(ce^v).
$$

The classical cosine Mellin integral and gamma reflection identity imply

$$
A_t(\infty)=(2\pi)^{-1/2+it}\Gamma(1/2-it)
\cos(\pi/4-i\pi t/2),\qquad |A_t(\infty)|=1/2.
$$

For $v\ge1$, integration by parts bounds the tail by

$$
|A_t(\infty)-A_t(v)|\le q(t)v^{-1/2},\quad
q(t)=\frac{1+2\sqrt{1/4+t^2}}{2\pi}\le\frac{1+|t|}{\pi}.
$$

For $v\le1$, $|A_t(v)|\le2\sqrt v$. Thus

$$
\left|H_c(t)-\tfrac14\log c\right|
\le4+2q(t)+q(t)^2\le6(1+t^2)\quad(c\ge1).
$$

Substitution into (3.2) proves (3.1). The small-$c$ statement follows from $H_c(t)\le4c$. $\square$

For one prime, collecting equal scales in (2.2) yields

$$
K_{\{p\}}=(1-p^{-1})\sum_{k\ge0}C_{p^k}-p^{-1}C_{1/p}.
$$

With $r=p^{-1/2}$ and $L=\log p$, Theorem 3.1 gives

$$
\boxed{\|T_f(K_{\{p\}}-C_1)\|_{\rm HS}
\le\left[\frac8p+
\frac{r(1+r)}{\sqrt{1-r}}\sqrt{L+24}\right]\|f\|_{H^1}.}\tag{3.3}
$$

To see this, bound the two low-scale corrections by $8\|f\|_2/p$, and use

$$
\sum_{k\ge1}\sqrt k\,r^k\le\frac r{(1-r)^{3/2}}
$$

for the remaining terms. Hence the squared one-prime increment is $O((\log p)/p)\|f\|_{H^1}^2$, with no test-support constant. This is not an additivity statement for several primes.

## 4. A weighted Hardy trace estimate

Let $\Pi$ project onto positive Fourier frequencies on $L^2(\mathbb R,dt)$, and let $Q=I-\Pi$. Multiplication operators are denoted by $M$. The following estimate keeps the localization weight attached to the Hankel block.

**Lemma 4.1.** Suppose $u=e^{i\varphi}$ is smooth, its derivatives and $\varphi'$ have at most polynomial growth, and $B=\|\varphi''\|_\infty<\infty$. For $F=\widehat f$, $f\in C_c^\infty(\mathbb R)$, put $w=|F|^2$. Then

$$
\boxed{\left|\|M_F Q M_u\Pi\|_{\rm HS}^2
-\frac1{2\pi}\int w(t)(-\varphi'(t))_+dt\right|
\le\frac{\sqrt B}{\pi^2}\|F\|_2^2
+\|Q M_w\Pi\|_{\mathcal S_1}.}\tag{4.1}
$$

**Proof.** Let $H_-=QM_u\Pi$, $H_+=\Pi M_uQ$, and $V=M_u\Pi M_u^*$. The two input blocks of $[\Pi,M_u]$ have orthogonal domains. Its difference-quotient kernel gives

$$
N_-+N_+=\frac1{2\pi^2}\int w(t)\int
\frac{1-\cos(\varphi(t+v)-\varphi(t))}{v^2}\,dv\,dt,\tag{4.2}
$$

where $N_\pm=\|M_FH_\pm\|_{\rm HS}^2$.

Taylor's theorem and $\int(1-\cos(av))v^{-2}dv=\pi|a|$ give

$$
\left|\int\frac{1-\cos(\varphi(t+v)-\varphi(t))}{v^2}dv
-\pi|\varphi'(t)|\right|
\le\int\frac{\min(2,Bv^2/2)}{v^2}dv=4\sqrt B.\tag{4.3}
$$

This also proves finiteness in (4.2).

For the difference of the norms, use

$$
H_-H_-^*-H_+H_+^*=QVQ+\Pi V\Pi-\Pi.
$$

The localized operator $M_F(V-\Pi)M_F^*$ has a Schwartz kernel and is trace class. Its diagonal is $-w(t)\varphi'(t)/(2\pi)$. Removing the off-diagonal blocks yields

$$
N_--N_+=-\frac1{2\pi}\int w\varphi'
-2\Re\operatorname{Tr}(Q M_w\Pi V).\tag{4.4}
$$

For completeness, this step does not cyclically permute unlocalized infinite traces. In the $\Pi,Q$ decomposition, the off-diagonal blocks of $M_F$ are trace class: their inverse-Fourier kernels are $f(x+y)$ on bounded triangles. Take diagonal blocks of

$$
M_F(\Pi VQ+QV\Pi)M_F^*
$$

first. Every resulting diagonal term contains one such trace-class factor. Their traces sum to $2\Re\operatorname{Tr}(Q M_w\Pi V)$ by legitimate cyclicity. The remaining off-diagonal blocks have zero trace. The autocorrelation kernel proves $Q M_w\Pi\in\mathcal S_1$ as well, and $\|V\|\le1$ bounds the correction.

Adding (4.2) and (4.4), dividing by two, and applying (4.3) gives (4.1). $\square$

For $u(t)=e^{-iAt}$, $A>0$, the Hankel block is an interval projection after a unitary shift. Its localized density is exactly $A/(2\pi)$, which verifies the sign and constant in (4.1).

## 5. Growth for all primes up to $X$

Let $\mathcal P_X=\{p:p\le X\}$ and $K_X=K_{\mathcal P_X}$.

**Theorem 5.1.** For each fixed $f\in C_c^\infty(\mathbb R)$,

$$
\boxed{\lim_{X\to\infty}\frac{\|T_fK_X\|_{\rm HS}^2}{\sqrt X}
=\frac1{\pi^2}\int_{\mathbb R}
\frac{|F(t)|^2}{\sqrt{1/4+t^2}}\,dt.}\tag{5.1}
$$

**Proof.** The full semilocal multiplier is

$$
u_X(t)=\pi^{-it}\frac{\Gamma(1/4+it/2)}{\Gamma(1/4-it/2)}
\prod_{p\le X}\frac{1-p^{-1/2+it}}{1-p^{-1/2-it}}.
$$

Its phase derivative is

$$
\varphi_X'(t)=h(t)-2\sum_{p\le X}(\log p)
\sum_{k\ge1}p^{-k/2}\cos(kt\log p),\quad
h(t)=\Re\psi(1/4+it/2)-\log\pi.\tag{5.2}
$$

The function $h$ is bounded below, grows logarithmically, and has bounded derivative. The terms $k\ge2$ in (5.2) are uniformly $O(\log X)$. Also

$$
\|\varphi_X''\|_\infty
\le\|h'\|_\infty+
2\sum_{p\le X}\frac{(\log p)^2p^{-1/2}}{(1-p^{-1/2})^2}
=O(\sqrt X\log X).\tag{5.3}
$$

Lemma 4.1 therefore has error $O_f(X^{1/4}\sqrt{\log X}+1)=o_f(\sqrt X)$.

Partial summation of the prime number theorem gives, uniformly on each compact real $t$ interval,

$$
\sum_{p\le X}(\log p)p^{-1/2+it}
=\frac{X^{1/2+it}}{1/2+it}+o(\sqrt X).
$$

Thus

$$
\frac{\varphi_X'(t)}{\sqrt X}
=-2\Re\frac{e^{it\log X}}{1/2+it}+o(1).\tag{5.4}
$$

The negative part in (5.4) is uniformly bounded on the whole real line: the prime sum's absolute value is $O(\sqrt X)$, while $h$ is bounded below. Consequently tails can be removed against the integrable weight $|F|^2$ before using compact-uniform convergence.

The function $(\cos\theta)_+$ has mean $1/\pi$ and an absolutely summable Fourier series. Applying the Riemann–Lebesgue lemma to its nonconstant terms, with the fixed phase shift $-\arctan(2t)$, proves

$$
\int w(t)\left(\Re\frac{e^{it\log X}}{1/2+it}\right)_+dt
\longrightarrow\frac1\pi\int\frac{w(t)}{\sqrt{1/4+t^2}}dt.
$$

Combining this with Lemma 4.1 gives (5.1). The unitary identification between $K_X$ and the Hardy block includes output reflection, replacing $F(t)$ by $F(-t)$. Both $\varphi_X'$ and the limiting weight are even, so the answer is unchanged. $\square$

The coefficient is positive for every nonzero test. In particular, it remains positive when $F(i/2)=F(-i/2)=0$. Such tests include $(\partial_x^2-1/4)g$ for nonzero compact smooth $g$. Therefore the full squared overlap cannot be bounded, for fixed tests, by a constant multiple of $\sum_{p\le X}(\log p)/p$. The small individual estimate (3.3) and the full-product asymptotic (5.1) describe different quantities.

## 6. Sonin trace growth and the signed comparison

In logarithmic physical coordinates now write $R=1_{(-\infty,0)}$, $P=I-R$, and let $\mathcal F_X$ be the semilocal Fourier involution. The signed trace operator is

$$
A_X=P+\mathcal F_XP\mathcal F_X-I.
$$

Relative to $R\mathcal H\oplus P\mathcal H$, its upper left block is $-K_X^2$, its lower right block is the positive operator $P\mathcal F_XP\mathcal F_XP$, and its off-diagonal block is $-K_X(R\mathcal F_XP)$. The latter has norm at most $1/2$, since its squared singular values are $\lambda^2(1-\lambda^2)$ for eigenvalues $\lambda$ of $K_X$.

Consequently, with $g=f^**f$,

$$
q_{\infty,\mathcal P_X}(f)
=\mathcal L_X(f)-\|T_fK_X\|_{\rm HS}^2+\mathcal R_X(f),\tag{6.1}
$$

where

$$
\mathcal L_X(f)=\operatorname{Tr}(T_fP\mathcal F_XP\mathcal F_XP T_f^*)\ge0,
\qquad |\mathcal R_X(f)|\le\|R T_gP\|_{\mathcal S_1}.
$$

Here $q_{\infty,\mathcal P_X}$ is the archimedean plus finite-prime form in the established semilocal trace normalization [1,2]. For a fixed compact test it is independent of $X$ once all prime powers visible to its autocorrelation have been included. Therefore

$$
\frac{\mathcal L_X(f)}{\sqrt X}
\longrightarrow\frac1{\pi^2}\int\frac{|F(t)|^2}{\sqrt{1/4+t^2}}dt.\tag{6.2}
$$

The positive compression in (6.2) contains both the Sonin projection and a positive contribution from the generic two-projection blocks. We now bound the actual Sonin trace separately. Its order is $\sqrt X/(\log X)^2$ when the test transform is nonzero at the origin, and its growth exponent remains $1/2$ for every nonzero test. It is therefore subleading on the $\sqrt X$ scale. Thus the generic two-projection contribution carries the full leading coefficient of (6.2).

Let $S_X$ denote the orthogonal projection onto the Sonin space associated with $\mathcal F_X$, and write $S_0$ for the archimedean space at gap $1$. In Mellin coordinates put

$$
D_X(t)=\prod_{p\le X}(1-p^{-1/2-it}),\qquad
\mathcal T_X(f)=\|T_fS_X\|_{\rm HS}^2.
$$

The finite Euler embedding of [3, Theorem 5.3] is an equality of spaces, as proved in [4, Theorem 4.6]:

$$
\boxed{S_X\mathcal H=D_X S_0\mathcal H.}\tag{6.3}
$$

Indeed, with $s=1/2+it$, write the scattering multiplier as $u_X=u_\infty N_X/D_X$, where $N_X(s)=\prod_{p\le X}(1-p^{s-1})$. Both $D_X$ and its inverse are bounded analytic multipliers of the right Hardy space, and $N_X,N_X^{-1}$ have the analogous property on the left. The two Hardy conditions defining the Sonin space therefore imply both inclusions in (6.3). This is an equality of ranges, not an isometry.

**Lemma 6.1.** Put $L=\log X$ and $A=\sqrt X/L$. For each fixed $B>0$,

$$
\sum_{p\le X}p^{-1/2-it}
=\frac{2A e^{-itL}}{1-2it}+o(A),\qquad |t|\le X^B,\tag{6.4}
$$

where the error is uniform and absolute; it is not relative to the main term at large $t$.

**Proof.** We give the uniformity argument because compact-interval PNT alone would not suffice. The classical Vinogradov–Korobov zero-free region, for example [8, Theorem 1.1], gives a zero-free rectangle to the right of $1-2\delta_T$ up to height $2T$, where

$$
\delta_T=c(\log T)^{-2/3}(\log\log T)^{-1/3}.
$$

Choose $c>0$ small enough to cover the remaining bounded heights as well. In the narrower rectangle, the usual local-zero formula for $\zeta'/\zeta$, with its pole at $1$ separated, gives a bound by a fixed power of $\log T$. To see this, the $O(\log T)$ zeros in a unit-height window are each at horizontal distance at least $\delta_T$, and the remaining part of that formula is $O(\log T)$.

Apply truncated Perron to

$$
\Psi_t(y)=\sum_{n\le y}\Lambda(n)n^{-it}
$$

with integrand $-(\zeta'/\zeta)(z+it)y^z/z$, initial line $\Re z=1+1/\log y$, and height $T=y^{B+2}$. On half-integer endpoints the truncation error is uniform in $t$, since the absolute values of the coefficients are $\Lambda(n)$. Shift to $\Re z=1-\delta_{2T}$. The only pole crossed is $z=1-it$, with residue $y^{1-it}/(1-it)$. The vertical integral is at most $y^{1-\delta_{2T}}$ times a fixed logarithmic power; the horizontal and truncation errors are smaller. Rounding an endpoint changes the sum by $O(\log y)$ and the main term by $O(1)$, uniformly in $t$. Thus, after absorbing logarithmic factors,

$$
\Psi_t(y)=\frac{y^{1-it}}{1-it}
+O_B\!\left(y\exp\!\left[-c_B\frac{(\log y)^{1/3}}{(\log\log y)^{1/3}}\right]\right),
\qquad |t|\le y^B.
$$

Prime powers with exponent at least two contribute $O(\sqrt y\log^2y)$. Split the desired prime sum at $\sqrt X$, and use this estimate with exponent $2B$ on $[\sqrt X,X]$. Partial summation against $y^{-1/2}/\log y$ gives

$$
\sum_{p\le X}p^{-1/2-it}
=\int_2^X\frac{y^{-1/2-it}}{\log y}\,dy+o(A)
$$

uniformly on the stated range. An integration by parts after $v=\log y$, on $L/2\le v\le L$, gives the endpoint term in (6.4), with error $O(A/L)+O(X^{1/4})$. The factor $|1/2-it|^{-1}\le2$ makes this step uniform. $\square$

**Theorem 6.2.** Let $f\in C_c^\infty(\mathbb R)$ be nonzero, and let $k$ be the order of vanishing of its entire Fourier transform $F$ at zero. There is $c_f>0$ such that, for all sufficiently large $X$,

$$
\boxed{\mathcal T_X(f)\ge c_f\frac{\sqrt X}{(\log X)^{2k+2}},
\qquad \mathcal T_X(f)=O_f\!\left(\frac{\sqrt X}{(\log X)^2}\right).}\tag{6.5}
$$

More precisely,

$$
\limsup_{X\to\infty}\frac{(\log X)^2}{\sqrt X}\mathcal T_X(f)\le4\|F\|_\infty^2.
$$

In particular, $\mathcal T_X(f)\asymp_f\sqrt X/(\log X)^2$ if $F(0)\ne0$. The constant $4$ is an upper estimate, not an asymptotic coefficient. For every nonzero $f$,

$$
\lim_{X\to\infty}\frac{\log(1+\mathcal T_X(f))}{\log X}=\frac12.
$$

Moreover, the generic positive contribution has the full coefficient of (6.2):

$$
\frac{\mathcal L_X(f)-\mathcal T_X(f)}{\sqrt X}\longrightarrow\mathfrak C_f,
$$

where $\mathfrak C_f$ is the coefficient in (5.1).

**Proof.** We first choose a fixed raw Mellin transform $H_0$ in $S_0\mathcal H$ such that

$$
H_0(0)\ne0,\qquad |H_0(t)|\le C e^{-c\sqrt{|t|}},
\qquad t^jH_0\in S_0\mathcal H\quad(j\ge0).
$$

Here is a construction that also explains the last condition. Start with the evaluator at zero in a larger-gap archimedean Sonin space. Such evaluation functionals are nonzero; see [9, Proposition 4.3, Note 9, and the discussion after Lemma 7.4]. Burnol's explicit structure function [10, Theorem 8] shows that this raw kernel section has at most polynomial growth on the real Mellin line. In fact its compact smooth cosine-resolvent correction has Mellin transform $O(1/|t|)$, and the raw kernel section is $O(1/|t|)$ there. Smooth it in logarithmic coordinates by a compact Gevrey function of order two and integral one, with support small enough that both gaps remain at least $1$. Its Fourier transform decays as $e^{-c\sqrt{|t|}}$: this follows by integrating by parts using $\|\psi^{(j)}\|_1\le C^{j+1}(j!)^2$ and optimizing $j$. All logarithmic derivatives of the resulting vector remain square integrable and retain both gaps. Their Mellin transforms are polynomial multiples of $H_0$, proving the assertions.

For $J_X=[\pi/L-L^{-2},\pi/L+L^{-2}]$ and every polynomial $p$ of degree less than $n$, normalized Legendre expansion on $J_X$ gives

$$
|p(t)|^2\le Cn^2L^2[C L^2(1+|t|)]^{2n}
\int_{J_X}|p(v)|^2\,dv.
$$

The function $H_0$ is bounded below on $J_X$ for large $X$. Integrating this inequality against $|H_0|^2$ therefore gives, with a fixed constant $C$,

$$
\int_{\mathbb R}|pH_0|^2\,dt
\le e^{C n\log(n+L)}\int_{J_X}|pH_0|^2\,dt.
$$

For completeness, the moment estimate uses $u=\sqrt{|t|}$ and is bounded by a constant to the power $n$ times $\Gamma(4n+2)$. This estimate is uniform over the entire polynomial space, including when $n$ grows.

Let $w_X=|D_X|^2$. The terms of order at least two in the logarithm of each Euler factor have total $O(\log\log X)$, uniformly in $t$. Lemma 6.1 consequently gives

$$
\log w_X(t)=-4A\Re\frac{e^{-itL}}{1-2it}+o(A),\qquad |t|\le X^2.
$$

Choose a fixed small $\delta>0$ for which $|F(t)|\ge c_f|t|^k$ on $|t|\le\delta$. The weight satisfies

$$
\begin{aligned}
\inf_{t\in J_X}\log w_X(t)&\ge(4-o(1))A,\\
\sup_{\delta\le|t|\le X^2}\log w_X(t)&\le\left(\frac4{\sqrt{1+4\delta^2}}+o(1)\right)A,\\
\sup_{|t|\le\pi/(2L)}\log w_X(t)&\le o(A).
\end{aligned}
$$

The last line uses $\cos(tL)\ge0$ and $(1-2it)^{-1}=1+O(1/L)$ on that interval. Globally, without any frequency restriction,

$$
\log w_X(t)\le2\sum_{p\le X}\log(1+p^{-1/2})=(4+o(1))A.
$$

Set $n=\lfloor c_\delta A/L\rfloor$ with $c_\delta>0$ small enough. Since $\log(n+L)\le L$ for large $X$, the polynomial comparison costs at most $e^{C c_\delta A}$. Choose this exponent smaller than half the fixed gap $4-4/\sqrt{1+4\delta^2}$. Comparing the weighted norm on $J_X$ with the two regions $|t|\le\pi/(2L)$ and $\delta\le|t|\le X^2$ shows that their combined mass is $O(e^{-cA})$ of the total norm, uniformly over nonzero $p$ of degree less than $n$.

For the remaining tail $|t|>X^2$, use the global Euler bound, not the compact-frequency gap. The pointwise polynomial estimate and $H_0$'s decay bound its contribution, relative to the norm on $J_X$, by

$$
e^{O(A+n\log L)}\int_{|t|>X^2}(1+|t|)^{2n}e^{-c\sqrt{|t|}}\,dt=o(1).
$$

Indeed $n=O(\sqrt X/L^2)$, so half the exponential decay absorbs the polynomial factor throughout this tail, leaving $O(e^{-c'X})$. Thus every unit vector in the $n$-dimensional space

$$
\mathcal V_X=\{D_XpH_0:\deg p<n\}\subset S_X\mathcal H
$$

has $1-o(1)$ of its norm in

$$
\frac\pi{2L}\le|t|\le\delta.\tag{6.6}
$$

On this annulus, $|F(t)|^2\ge c_fL^{-2k}$. Summing over an orthonormal basis of $\mathcal V_X$ yields

$$
\mathcal T_X(f)\ge c_fnL^{-2k}(1-o(1)),
$$

which is the lower bound in (6.5).

For the upper bound we need a quantitative evaluation estimate. In the fixed archimedean space, write $h(t)=\widehat v(1/2+it)$, where $\widehat v(s)=\int_1^\infty v(x)x^{-s}dx$ initially for $\Re s>1/2$. We claim

$$
\sup_{|z|\le R}|h(z)|\le \exp(CR\log(R+2))\|h\|,
\qquad R\ge2.
$$

Here and below $C$ depends only on this fixed space. We justify the unit-ball uniformity explicitly. Put $K=C_1$ on $L^2(0,1)$, $c(x)=2\cos(2\pi x)$, and $g=(I-K^2)^{-1}c$. Since $g=c+K(Kg)$, both $g$ and $Kg$ extend to even entire functions with Taylor coefficients bounded by $C(2\pi)^{2j}/(2j)!$. Consequently their right Mellin integrals have meromorphic expansions

$$
M_v(s)=\sum_{j\ge0}\frac{c_j}{2j+1-s}.
$$

These series are uniformly bounded on large rectangle boundaries with half-integral vertical sides and horizontal sides of comparable height.

With $\gamma(s)=\pi^{-s/2}\Gamma(s/2)$, the structure function in [10, Theorem 8] at gap $1$ is

$$
E(s)=\gamma(s)[1+M_{Kg}(s)]-\gamma(1-s)M_g(1-s).
$$

Indeed, the difference of Burnol's two resolvent functions is $-2\mathcal Cg$; splitting its Mellin integral at $1$ and applying the cosine Mellin functional equation gives this expression. Burnol's theorem makes the combined function entire. Stirling's estimate and gamma reflection on the chosen boundaries, followed by the maximum principle, bound $E$ by $\exp(CR\log(R+2))$ on $|s|\le R$.

The bilinear completed reproducing kernel is

$$
\frac{E(s)E(z)-E(1-s)E(1-z)}{s+z-1}.
$$

Its numerator vanishes at $z=1-s$. Integrate the numerator's $z$ derivative along the segment from $1-s$ to $z$, and use Cauchy's derivative estimate on a larger disk. This bounds the removable quotient by the same order of growth. Dividing by $\gamma(s)\gamma(z)$ gives the raw kernel; reciprocal gamma has that same order bound. Its conjugate diagonal is the squared evaluation norm, proving the claimed uniform estimate after the affine change $s=1/2+it$.

Equip the archimedean space with the norm $\|h\|_X=\|D_Xh\|_2$. The same evaluation bound holds for this norm, uniformly in $X$. Here is the point that avoids paying the full inverse-Gram condition number. Reflect the coordinate so that $h^+(z)=h(-z)$ belongs to the upper Hardy space and $D_X^+(z)=D_X(-z)$. Its archimedean conjugation is

$$
C_0h^+(t)=u_\infty(t)\overline{h^+(t)},\qquad
u_\infty(z)=\frac{\gamma(1/2+iz)}{\gamma(1/2-iz)}.
$$

Both $h^+$ and $C_0h^+$ have the same weighted norm, since their boundary moduli agree. Hardy evaluation and

$$
|D_X^+(z)|\ge\prod_p(1-p^{-3/2})=\zeta(3/2)^{-1},\qquad \Im z\ge1,
$$

give $|h^+(z)|\le\zeta(3/2)\|h\|_X/\sqrt{2\Im z}$ there. Reflection through $C_0$ gives the analogous lower-half-plane bound with the extra factor $|u_\infty(z)|$. This ratio is analytic for $\Im z\le-1$, and $|u_\infty(t-i)|\le C(1+|t|)$.

Thus $h^+(z)/(z+2i)$ is bounded by $C\|h\|_X$ on both sides of the strip $|\Im z|\le1$. The preceding archimedean growth bound supplies growth $\exp(C|z|\log(|z|+2))$, with a prefactor that may depend on $X$. The strip maximum principle removes that prefactor: multiply by $\exp[-\varepsilon\cosh(az)]$ for $0<a<\pi/2$, let the rectangle length tend to infinity, and then let $\varepsilon$ decrease to zero. Inside the strip we get $|h^+(z)|\le C|z+2i|\|h\|_X$. Outside it, the Hardy and gamma bounds already apply. Consequently

$$
\sup_{|z|\le R}|h(z)|\le C_R\|h\|_X,
\qquad C_R=\exp(CR\log(R+2)),\tag{6.7}
$$

with $C$ independent of the finite prime set.

Let $J_T$ restrict the weighted Hilbert space $(S_0\mathcal H,\|\cdot\|_X)$ to $L^2([-T,T],dt/(2\pi))$. Taylor truncation of degree $j-1$ now gives, for $R>T$ and $q=T/R$,

$$
s_{j+1}(J_T)\le C_R\sqrt{T/\pi}\,\frac{q^j}{1-q}.
$$

Multiplication by $D_X$ is an isometry from this weighted space onto the actual Sonin space. Equivalently, on the original space it is $M_{D_X}G_X^{-1/2}$, where $G_X=S_0|D_X|^2S_0$; no Gram inverse has been discarded. Put

$$
M_X=\prod_{p\le X}(1+p^{-1/2}),\qquad \log M_X=(2+o(1))A.
$$

The localized singular values of that isometry satisfy

$$
s_j(1_{[-T,T]}U_X)\le\min\{1,M_Xs_j(J_T)\}.
$$

Fix $\epsilon>0$ and take

$$
T=L,\qquad R=\frac{\sqrt X}{L^4},\qquad
n=\left\lceil(4+\epsilon)\frac{\sqrt X}{L^2}\right\rceil.
$$

Then $\log C_R=O(\sqrt X/L^3)=o(A)$ and

$$
n\log(R/T)=(2+\epsilon/2+o(1))A.
$$

The first $n$ squared singular values contribute at most $n$. The remaining geometric tail is bounded by

$$
\frac{M_X^2 C_R^2(T/\pi)q^{2n}}{(1-q)^2(1-q^2)}
\le e^{-\epsilon A/2}
$$

for all sufficiently large $X$. Letting $\epsilon$ decrease to zero proves

$$
\limsup_{X\to\infty}\frac{L^2}{\sqrt X}
\operatorname{Tr}(1_{[-L,L]}S_X1_{[-L,L]})\le4.\tag{6.8}
$$

To remove the frequency tail, use $S_X\le(P\mathcal F_XP)^2$. This operator inequality holds because $S_X$ reduces the self-adjoint $P\mathcal F_XP$, whose square is the identity on the Sonin subspace and is positive on its orthogonal complement. The weighted Hardy-block norm of $P\mathcal F_XP$, up to reflection, is bounded by the commutator energy (4.2). That energy identity applies to arbitrary measurable localization weights by its difference-quotient kernel and Tonelli; it does not require the trace-class difference formula (4.4).

The exact phase derivative satisfies

$$
|\varphi_X'(t)|\le C\bigl(\sqrt X+\log(2+|t|)\bigr).
$$

For $|v|\le1$, bound $|u_X(t+v)-u_X(t)|$ by the smaller of $2$ and $C(\sqrt X+\log(2+|t|))|v|$; for $|v|>1$ use $2$. Integrating its square divided by $v^2$ gives

$$
\|M_{F1_{|t|>T}}S_X\|_{\rm HS}^2
\le C\int_{|t|>T}|F(t)|^2\bigl(\sqrt X+\log(2+|t|)\bigr)dt.
$$

The output reflection does not change this bound. Take $T=L=\log X$. Since $F$ is Schwartz, this tail is $o(\sqrt X/L^2)$. The contribution on $[-L,L]$ is at most $\|F\|_\infty^2$ times (6.8), giving the stated upper bound and limsup constant. Together with the lower bound it gives the logarithmic rate. Finally, subtract this subleading trace from (6.2) to obtain the generic-pair coefficient. $\square$

Theorem 6.2 provides actual Sonin mass, but does not justify replacing the entire positive compression in (6.1) by its Sonin component when estimating the remainder. The generic contribution carries the entire leading cancellation, and the weighted translation covariance still matters. In particular, the lower bound does not establish positivity of the signed Weil form.

The uniform weighted evaluation estimate also locates the growing trace. Let $k_X(t,s)$ denote the kernel of $S_X$ with respect to $dt/(2\pi)$, distinguishing it from the cutoff operator $K_X$.

**Theorem 6.3.** Uniformly on each fixed compact real interval,

$$
\boxed{\frac{\log k_X(t,t)}{A}
=-4\left(1+\Re\frac{e^{-itL}}{1-2it}\right)+o(1).}\tag{6.9}
$$

For every fixed $\delta>0$ and $B>0$,

$$
\sup_{\delta\le|t|\le L^B}k_X(t,t)
\le\exp\!\left[-\left(4-\frac4{\sqrt{1+4\delta^2}}+o(1)\right)A\right].\tag{6.10}
$$

Consequently the bound in Theorem 6.2 sharpens to

$$
\limsup_{X\to\infty}\frac{L^2}{\sqrt X}\mathcal T_X(f)
\le4|F(0)|^2.\tag{6.11}
$$

In particular it is $o(\sqrt X/L^2)$ if $F(0)=0$. The trace also concentrates where $e^{itL}$ approaches $-1$: if $V_Lf(y)=f(y-L)$, then

$$
\|T_{f+V_Lf}S_X\|_{\rm HS}^2=o(\sqrt X/L^2).
$$

Moreover $S_X\to0$ strongly on the ambient Hilbert space. This strong limit does not imply convergence of the convolution trace, since $T_f$ is not compact.

**Proof.** Let $k_0(\cdot,z)$ be the raw archimedean evaluator and set

$$
Q_X(z)=\langle G_X^{-1}k_0(\cdot,z),k_0(\cdot,z)\rangle
=\sup_{\|h\|_X\le1}|h(z)|^2.
$$

We first prove

$$
\sup_{\|h\|_X\le1,\ |z|\le L^B}|h(z)|
\le e^{(-2+o(1))A}\tag{6.12}
$$

for each fixed $B>0$. Fix a small $\eta>0$, and choose a fixed interval $[-d,d]$ on which $(1+4t^2)^{-1/2}\ge1-\eta$. The set $E_X$ in this interval on which

$$
\cos(tL-\arctan(2t))\le-(1-\eta)
$$

has measure at least $m_\eta>0$ for large $X$: the phase derivative is $L+O(1)$. Lemma 6.1 gives $\log|D_X|^2\ge(4-9\eta)A$ there. Therefore

$$
\|h\|_{L^2(E_X,dt)}\le\sqrt{2\pi}\,e^{-(2-9\eta/2)A}
\quad(\|h\|_X\le1).\tag{6.13}
$$

We use an elementary polynomial interpolation bound. If $E\subset[-d,d]$ has measure at least $m>0$ and $P$ has degree at most $n$, then

$$
\sup_{|z|\le T}|P(z)|
\le C_m\left(\frac{C(T+d)}m\right)^n\|P\|_{L^2(E)},\qquad T\ge d.
$$

To verify it, restrict to a subset of measure at least $m/2$ where $|P|\le\sqrt{2/m}\|P\|_{L^2(E)}$. Choose $n+1$ quantile nodes there. Their pairwise distances are at least $c|j-k|m/(n+1)$. Lagrange interpolation gives denominators bounded below by $[cm/(n+1)]^n j!(n-j)!$; summing uses $\sum_j1/[j!(n-j)!]=2^n/n!$. Continuity permits nodes in the closure of the selected set, so no regularity of $E$ is required.

In (6.7) choose $R=A/(\log A)^3$, whose logarithmic evaluation cost is $o(A)$. Approximate $h$ by its degree-$n$ Taylor polynomial with $n=\lceil(2+\eta)A/\log A\rceil$. On the target disk $|z|\le L^B$, Cauchy's remainder has size $\exp[-(2+\eta+o(1))A]$. The remainder on $E_X$, after multiplication by the interpolation factor, has the same leading negative exponent: that factor has logarithm $O(n\log L)=o(A)$. Applying the polynomial estimate to (6.13) gives a bound $\exp[-(2-9\eta/2+o(1))A]$ on the target disk. First let $X$ grow and then let $\eta$ decrease to zero. This proves (6.12).

Conversely $\|h\|_X\le M_X\|h\|_0$, so

$$
Q_X(z)\ge k_0(z,z)/M_X^2.
$$

Together with (6.12) and $\log M_X=(2+o(1))A$, this proves

$$
\frac{\log Q_X(z)}A\longrightarrow-4
$$

uniformly on fixed compact sets avoiding common zeros of the archimedean evaluations. There are no such common zeros on the real line. The exact range-kernel formula is $k_X(t,t)=|D_X(t)|^2Q_X(t)$; Lemma 6.1 now yields (6.9). Its absolute uniform error through polynomial heights, together with the upper estimate (6.12), gives (6.10).

To obtain (6.11), the commutator bound from Theorem 6.2 removes $|t|>L$ at cost $o(\sqrt X/L^2)$. Equation (6.10) makes the trace on $\delta\le|t|\le L$ exponentially small. On $|t|<\delta$, bound $|F(t)|^2$ by its supremum there and apply (6.8). Finally let $\delta$ decrease to zero.

For the translation assertion, first discard $|t|\ge\delta$ inside the growing window using (6.10). If $|t|<\delta$ and $|1+e^{itL}|\ge\epsilon$, then $\cos(tL)\ge-1+\epsilon^2/2$. Choosing $\delta$ small relative to $\epsilon^2$ makes $\log|D_X(t)|^2\le(4-c_\epsilon)A$ on this region. Equation (6.12) therefore makes its density exponentially small. On the remaining region the extra factor $|1+e^{-itL}|^2$ is at most $\epsilon^2$, so (6.8) applies. The tail beyond $L$ is uniformly negligible because $|F(t)(1+e^{-itL})|\le2|F(t)|$. Let $\epsilon$ decrease to zero.

For the strong limit, reflect into the upper Hardy realization used above. At every fixed $z$ with $\Im z>1/2$, $D_X^+(z)$ converges to the nonzero absolutely convergent reciprocal zeta product. Thus the evaluation norm of the actual space satisfies $k_X^+(z,z)=\exp[(-4+o(1))A]\to0$. This is also $\|S_X^+k_H(\cdot,z)\|^2$. These Hardy kernels span a dense subspace, and the projections are contractions, so $S_X^+\to0$ strongly. They vanish on the orthogonal Hardy complement. Undoing the reflection proves the assertion. $\square$

Equation (6.9) describes only an exponential scale. At the moving peaks near $\pm\pi/L$ its leading exponent vanishes, so it does not give their polynomial heights, exact widths, or a sharp trace coefficient. The localization and strong convergence also leave the signed kernel covariance unresolved.

For an explicit support estimate, if $\operatorname{supp}f\subset[-a,a]$ and $L=\max(1,2a)$, expansion of the Hankel kernel $g(x+y)$ in the cosine basis of $(0,L)$ and two integrations by parts give

$$
\|RT_gP\|_{\mathcal S_1}
\le L\|g\|_\infty+\frac{\sqrt2L^2}{6}
(\|g'\|_\infty+L\|g''\|_\infty)
\le\left(L+\frac{\sqrt2L^2}{12}+\frac{\sqrt2L^3}{6}\right)\|f\|_{H^1}^2.
$$

This comparison has no inverse projection-angle constant. Nevertheless, dropping its positive term leaves the order-$\sqrt X$ contribution established above. A lower bound useful for RH must preserve additional signed cancellation. No bound for the negative part of the Weil spectrum follows from the present estimates.

## 7. Exact pole compensation

Write $L=\log X$, with $X\ge2$, and define

$$
\chi_L(t)=4\int_0^L\cosh(v/2)\frac{\sin(tv)}v\,dv,
\qquad\widetilde u_L(t)=u_X(t)e^{i\chi_L(t)}.\tag{7.1}
$$

The phase is real and odd, so this multiplier again defines a real unitary Fourier involution. Let $\widetilde K_L$ be its compression by $R$ and put $\mathcal N_L(f)=\|T_f\widetilde K_L\|_{\rm HS}^2$.

**Theorem 7.1.** For $f\in C_c^\infty[-L/2,L/2]$, the signed localized projection trace associated with $\widetilde u_L$ is the full Weil form, including its pole term. For each fixed compact smooth $f$,

$$
\boxed{\mathcal N_L(f)=o(e^{L/2})\quad(L\to\infty).}\tag{7.2}
$$

For each fixed $L$, $\widetilde K_L-K_X$ is Hilbert–Schmidt. Consequently the logarithmic spectral law (2.5) is unchanged by this compensation, with $X$ fixed while $N$ tends to infinity.

**Proof.** Differentiating (7.1) gives

$$
\chi_L'(t)=2\Re\left[
\frac{e^{(1/2+it)L}-1}{1/2+it}
+\frac{e^{(-1/2+it)L}-1}{-1/2+it}\right].\tag{7.3}
$$

Let $g=f^**f$ and $G(z)=F(z)\overline{F(\bar z)}$. Since $g$ is supported in $[-L,L]$, Fourier inversion gives

$$
\frac1{2\pi}\int|F(t)|^2\chi_L'(t)\,dt
=2\int_{-L}^L\cosh(v/2)g(v)\,dv
=G(i/2)+G(-i/2).\tag{7.4}
$$

This is the Weil pole term. The finite prime set contains every prime power visible to $g$, so the signed trace is exactly the full form. In particular, if both pole moments vanish, compensation changes the auxiliary operator while preserving this signed trace.

On compact $t$ intervals, (7.3) cancels the PNT leading term (5.4). Hence $\widetilde\varphi_L'/\sqrt X\to0$ there. Its negative part divided by $\sqrt X$ is uniformly bounded on the entire real line: the gamma derivative is bounded below, and both the finite-prime and compensating derivatives have absolute size $O(\sqrt X)$. Also

$$
\|\widetilde\varphi_L''\|_\infty=O(\sqrt X\log X),
$$

using (5.3) and $|\chi_L''|\le4\int_0^L v\cosh(v/2)\,dv$. Lemma 4.1 and dominated convergence therefore give (7.2).

To prove the fixed-$L$ perturbation assertion, separate $\cosh(v/2)=1+O(v^2)$ in (7.1) and integrate the smooth remainder by parts. Then

$$
\chi_L(t)=2\pi\operatorname{sgn}(t)+O_L(|t|^{-1}),
\qquad\chi_L'(t)=O_L(|t|^{-1})\quad(|t|\to\infty).
$$

Thus $e^{i\chi_L}-1$ lies in $H^1(\mathbb R)$. Since $u_X'=O_X(\log(2+|t|))$, the symbol difference $u_X(e^{i\chi_L}-1)$ also lies in $H^1$. Its Hardy Hankel blocks are Hilbert–Schmidt, by the standard difference-quotient formula, so the same is true of the cutoff difference. Finally, the square roots of the partial squared-singular-value sums change by at most this fixed Hilbert–Schmidt norm. Their leading term (2.5) is therefore unchanged. $\square$

The signed block comparison now includes the poles directly:

$$
q(f)\ge-\mathcal N_L(f)-\|R T_{f^**f}P\|_{\mathcal S_1}.
$$

This identity clarifies which part of the growing overlap comes from the continuous background. The remaining estimate (7.2) is for each fixed test and improves no fixed exponential growth exponent. A sufficiently strong bound for the corrected norm remains unproved; this construction does not establish Weil positivity.

## 8. Reproduction and limitations

The proofs are analytic. The supplement contains four CPU checks using NumPy and SciPy. The first independently checks Theorem 3.1 through a physical-space sine-integral formula for $f(x)=xe^{-x/2}1_{x\ge0}$. The second computes weighted finite periodic Hardy blocks for the full semilocal multiplier. It uses a compact spline test $f=(\partial_x^2-1/4)g$, where $g$ is the convolution of eight normalized width-$1/4$ uniform densities. The two pole moments vanish exactly.

For this spline, the predicted coefficient in (5.1) is approximately $55.02161$. The computed squared norms divided by $\sqrt X$ are approximately $33.17$, $45.77$, $51.66$, and $54.47$ at $X=127$, $509$, $2039$, and $8191$. Doubling the period and Fourier resolution changes the norm at $509$ by about $0.034\%$. An archimedean-only calibration agrees with the independent physical computation within $4.2\times10^{-4}$ at the refined resolution. These computations are not interval-certified and do not prove the asymptotic. The spline has finite regularity; the theorem is stated on the smooth compact core.

A third check evaluates the exact cosine-block Gram sums for one, two, and three primes. At scale cutoff $R=10^{12}$, their squared Hilbert–Schmidt norms minus twice the weighted number of smooth frequencies are approximately $0.07079$, $-0.52752$, and $-1.55424$, respectively; these differences stabilize as the cutoff grows. This checks the normalization and bounded cross contribution in (2.8). These are norms of scale-truncated kernels, not numerical singular values of the full operator.

The fourth applies (7.1) to the same compact test. At prime cutoff $8191$, the squared overlap changes from approximately $4930$ to $9.12$ at half-period $32$ and $9.30$ at half-period $64$. At $509$ the compensated values are $22.73$ and $22.98$. The observed period dependence is about $1$–$2\%$; no limiting plateau is inferred. For the unfiltered spline, the compensating trace equals the exact nonzero pole term $2.02093947762437$ to about $10^{-14}$ in the refined run. For the pole-filtered test it is within $2\times10^{-9}$ of zero.

All results are unconditional, but concern auxiliary cutoffs. The Schatten bound is for each fixed finite set; its constant is not uniform in a growing set. The asymptotic is for each fixed test; it is not a uniform statement for tests whose support or frequency profile changes with $X$. It gives neither a negative Weil eigenvalue nor a new zero-free region. The separate problem of controlling the signed comparison remains open.

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