bibkey: frankliebseiringer2006hardy authors: Rupert L. Frank, Elliott H. Lieb, and Robert Seiringer year: 2006 title: Hardy-Lieb-Thirring inequalities for fractional Schrodinger operators doi: null url: https://arxiv.org/abs/math/0610593v2 claim: Nonlocal IMS retains a joint pole-prime correction. Below the first prime shift, refinement leaves the prime atoms unchanged; actual local contributions minus the actual Gamma defect approach the negative shifted-digamma baseline on slow dilations. Positivity requires a positive arithmetic reserve, which is not supplied by the mesh or the known optimal local floor. The classical Stechkin reserve alone fails the universal reduced certificate; the published xi null vector also obstructs dropping only the exact Stechkin residual, with the auxiliary line and other terms retained. strata_touched: [] license: citation-only triage: anchor

Nonlocal localization and the actual Weil correction

The primary source is Frank–Lieb–Seiringer, arXiv:math/0610593v2, 27 October 2006, §3.3, Lemma 3.5 and equation (3.13), printed p.12. For $u\in C_c^{\infty }({\mathbb{R}}^d)$, $0<s<\min (1,d/2)$ and a finite real Lipschitz square partition ${\sum }_j{\chi }_j(x{)}^2=1$ for all $x\in {\mathbb{R}}^d$, it writes the fractional Hardy form as the sum of the localized forms minus an integral-operator correction. That correction has kernel

$$a_{s,d}|x-y{|}^{-d-2s}\sum _j({\chi }_j(x)-{\chi }_j(y){)}^2.$$

The paper’s fractional kernel is not the Weil Gamma kernel and has no prime atoms or zeta pole term. The transferable ingredient is the localization calculation; its arithmetic application below is a separate paper derivation, not an additional theorem attributed to that source or a newly compiled general Weil theorem. A nonnegative pointwise correction kernel need not define a positive semidefinite operator.

Common test and normalization

Use the project’s actual energy decomposition. Let $f\in C_c^{\infty }(\mathbb{R};\mathbb{C})$ be even, with support in $[-L,L]$, $L>0$. Let ${\chi }_1,\dots ,{\chi }_m$ be real, even, smooth functions satisfying ${\sum }_i{\chi }_i(x{)}^2=1$ on $[-L,L]$, and put $f_i={\chi }_{i}f$. Each $f_i$ is an admissible even test. Use the same global $L$ for every term. Write

$$E_t(f)={\int }_{\mathbb{R}}|f(x)-f(x-t){|}^{2}dx,k_{\Gamma }(t)=\frac{e^{-t/2}}{1-e^{-2t}},P(f)=2{\left|{\int }_{\mathbb{R}}{e}^{x/2}f(x)dx\right|}^2.$$

With $w_n=\Lambda (n)/\sqrt{n}$, the source form is

$$Q(f)=P(f)+{\int }_0^{\infty }{k}_{\Gamma }(t)E_t(f)dt+\sum _{2\le n\le e^{2L}}{w}_n{E}_{\log n}(f)-c_L\Vert f{\Vert }_2^2,c_L=2\sum _{2\le n\le e^{2L}}{w}_n-c_{\Gamma }.$$

Here $c_{\Gamma }=\Re \mathrm{digamma}(1/4)-\log \pi$, and $\Lambda$ includes prime powers. A linear partition $\sum {\rho }_i=1$ is not automatically a square partition. Taking $\sqrt{{\rho }_i}$ also needs a smoothness argument at its zeros. The ordinary smooth partition construction in ExternalSupportInvisibility does not by itself settle these obligations.

Define the real defect correlation

$$D_{\chi }(x,t)=\sum _i({\chi }_i(x)-{\chi }_i(x-t){)}^2,K_f(t)={\int }_{\mathbb{R}}{D}_{\chi }(x,t)\Re (f(x)\overline{f(x-t)})dx.$$

Square expansion gives the one-shift IMS identity

$$\sum _i{E}_t(f_i)-E_t(f)=K_f(t).$$

There is no extra $1/2$ in this one-shift convention. The mass term cancels because ${\sum }_i\Vert f_i{\Vert }_2^2=\Vert f{\Vert }_2^2$. The function $K_f$ is real, even, smooth, supported in $[-2L,2L]$, and $K_f(0)=K_f^{\prime }(0)=0$. It need not be nonnegative for complex or sign-changing $f$.

Keep the pole and primes together

The global pole does not split into independent local squares. Expanding it gives

$$P(f)-\sum _{i}P(f_i)=\iint e^{(x+y)/2}\sum _i({\chi }_i(x)-{\chi }_i(y){)}^2\Re (f(x)\overline{f(y)})dxdy.$$

Evenness of both $f$ and the multipliers permits $y\mapsto -y$, changing the exponential to $e^{(x-y)/2}$ without changing the remaining factor. Setting $t=x-y$ and pairing positive and negative $t$ yields

$$P(f)-\sum _{i}P(f_i)=2{\int }_0^{\infty }\cosh (t/2)K_f(t)dt.$$

Consequently the complete localization correction is

$$Q(f)-\sum _{i}Q(f_i)={\int }_0^{\infty }{e}^{t/2}{K}_f(t)dt-\sum _{2\le n\le e^{2L}}\frac{\Lambda (n)}{\sqrt{n}}{K}_f(\log n)-{\int }_0^{\infty }\frac{e^{-5t/2}}{1-e^{-2t}}{K}_f(t)dt.\text{\hspace{1em}(1)}$$

This uses the exact cancellation $2\cosh (t/2)-k_{\Gamma }(t)=e^{t/2}-e^{-5t/2}/(1-e^{-2t})$. Estimating the pole and prime terms separately would discard this common-source cancellation.

Let $\Psi (u)={\sum }_{n\le u}\Lambda (n)$ be the right-continuous Chebyshev function, distinct from the golden conjugate, and define $W_f(u)=u^{-1/2}{K}_f(\log u)$ for $u\ge 1$. Equation (1) becomes

$$Q(f)-\sum _{i}Q(f_i)={\int }_1^{e^{2L}}(\Psi (u)-u)W_f^{\prime }(u)du-{\int }_1^{e^{2L}}\frac{W_f(u)}{u(u^2-1)}du.\text{\hspace{1em}(2)}$$

Indeed the first two terms of (1) equal $-\int W_{f}d(\Psi -u)$. Integration by parts has no boundary charge: $W_f(1)=W_f(e^{2L})=0$. At $u=1$, $W_f(u)=O((u-1{)}^2)$, so the second integrand is integrable. Compact support makes this a finite-scale identity without any PNT or RH hypothesis. This calculation preserves signs; it is not a bound on the Chebyshev error.

For a joint Lipschitz bound $\Vert \chi (x)-\chi (y){\Vert }_{{\ell }^2}\le B|x-y|$ on $[-L,L]$, Cauchy–Schwarz gives $|K_f(t)|\le B^2{t}^2\Vert f{\Vert }_2^2$. Thus the last term of (1) has absolute value at most $C_{\ast }{B}^2\Vert f{\Vert }_2^2$, where

$$C_{\ast }:={\int }_0^{\infty }\frac{t^2{e}^{-5t/2}}{1-e^{-2t}}dt=2\sum _{m\ge 0}(2m+5/2{)}^{-3}<\infty .$$

A cutoff family with $B\le C/L$ would give an $O(L^{-2})$ bound for this Gamma remainder. The existence of such a family subordinate to increasingly fine FIB windows is an additional condition, not a consequence of the window count. The signed first integral in (2) remains the arithmetic obligation for the same actual $f$ and cutoffs. A large absolute-value envelope neither supplies that lower bound nor proves no better estimate is possible.

Ordinary windows require both poles

The preceding even-test calculation has a general paper-level extension. For an arbitrary complex smooth compactly supported $g$, use

$$A_{\pm }(g)=\int e^{\pm x/2}g(x)dx,P_{\mathrm{full}}(g)=2\Re (A_{+}(g)\overline{A_{-}(g)}).$$

This is the two-pole term from the general explicit formula, rather than an application of the even-only rank-one energy bundle to a non-even function. With this pole term, write $Q_{\mathrm{full}}$ for the same energy expression with the same ambient $L$ and $c_L$. It agrees with $Q$ on even tests. Its pole kernel is $2\cosh ((x-y)/2)$, so for arbitrary smooth $f$ and real smooth square multipliers on its ambient interval,

$$P_{\mathrm{full}}(f)-\sum _i{P}_{\mathrm{full}}({\chi }_{i}f)=2{\int }_0^{\infty }\cosh (t/2)K_f(t)dt.$$

Thus (1) and (2) apply to $Q_{\mathrm{full}}$ without requiring the individual pieces to be even. On odd pieces the pole term is negative, not a positive square. Translation of $g$ multiplies $A_{+}$ and $A_{-}$ by opposite real factors, which cancel in their product; the full form is translation invariant. The actual general explicit-form interface is reusable, but the complete non-even jump-energy and localization bridge displayed here has not been compiled as a new Lean theorem.

Reuse the stronger small-window spectral floor

For an interval of length $d<\log 2$, prime autocorrelations vanish. This does not make the whole-line Gamma energy vanish: its interaction with the exterior supplies an endpoint potential. The relevant existing result is Suzuki v3, Corollary 1.2 and Theorem 1.4, not a new small-window positivity claim. With ${\lambda }_a^W$ the source’s lowest Weil eigenvalue on $(-a,a)$, it gives

$${\lambda }_{d/2}^W=\log (1/d)-\gamma -\log \pi +{\mu }_1+O(d),{\mu }_1>0.\text{\hspace{1em}(3)}$$

Here ${\mu }_1$ is the lowest Rayleigh value of the closure of the form in its equation (4.4),

$$\mathcal{L}(v)=\frac{1}{4}{\iint }_{(-1,1{)}^2}\frac{|v(x)-v(y){|}^2}{|x-y|}dxdy-\frac{1}{2}{\int }_{-1}^1\log (1-x^2)|v(x){|}^{2}dx.$$

The initial domain is $H_0^1(-1,1)$; a constant is not in that initial domain. The smooth approximation in the source’s §3.2 and the logarithmic Fourier form norm in (4.6) put the constant in the closed form domain. Its Rayleigh quotient is $1-\log 2$, hence

$$0<{\mu }_1\le 1-\log 2.\text{\hspace{1em}(4)}$$

This upper comparison uses the source’s domain argument, rather than assuming that an arbitrary boundary value is admissible. A direct exterior-potential lower bound with asymptotic $\log (1/d)-\gamma -\log \pi +O(d)$ discards ${\mu }_1$ and is weaker than (3). Reproving that weaker local positivity would not close the present gap. The source theorem and this application have not been independently formalized here.

The cost of a fine square partition

Suppose now that the smooth cutoffs are nonnegative and their vector has Lipschitz constant $B>0$ on the common ambient interval. The two unit cutoff vectors give

$$0\le D_{\chi }(x,t)\le \min (B^2{t}^2,2),|K_f(t)|\le \min (B^2{t}^2,2)\Vert f{\Vert }_2^2$$

on pairs relevant to the correlation. Let $k_{\ast }(t)=e^{-5t/2}/(1-e^{-2t})$. A sharper scalar Gamma envelope than the earlier $C_{\ast }{B}^2$ is

$$I_2(B)={\int }_0^{\infty }{k}_{\ast }(t)\min (B^2{t}^2,2)dt=\log B+c_{\mathrm{IMS}}+R(B),c_{\mathrm{IMS}}=\frac{3}{2}\log 2+\frac{\pi }{2}-\frac{7}{2},\text{\hspace{1em}(5)}$$

where, writing $a=\sqrt{2}/B$,

$$R(B)={\int }_0^a(2-B^2{t}^2)\left(\frac{1}{2t}-k_{\ast }(t)\right)dt>0,R(B)=O(B^{-1}).$$

To obtain (5), split at $a$ and use ${\int }_s^{\infty }{k}_{\ast }(t)dt=\mathrm{atanh}(e^{-s/2})+\arctan (e^{-s/2})-2e^{-s/2}$. The subtracted kernel tends to $3/4$ at zero and is positive for $t>0$; this also proves the stated remainder properties. For the actual finite support cutoff, when $2L>a$,

$$I_2(B,2L)=I_2(B)-2{\int }_{2L}^{\infty }{k}_{\ast }(t)dt.\text{\hspace{1em}(6)}$$

The discarded tail tends to zero as $L\to \infty$. These formulas concern a uniform upper envelope, not the value or a lower bound of the signed Gamma correction for a particular $f$.

There is also a geometric cost. If every cutoff support has diameter at most $d$ and the ambient interval contains a segment longer than $d$, cutoff vectors at distance greater than $d$ are orthogonal. The square partition traces the unit sphere, so that segment’s image has length at least $\pi /2$. Its length is at most $B$ times the segment length. Taking distances down to $d$ gives

$$Bd\ge \pi /2.\text{\hspace{1em}(7)}$$

Consider cofinal supports $L_j\to \infty$ and fine partitions with $d_j\downarrow 0$, whose actual smooth cutoff support intervals have lengths $d_{j,i}\in [d_j/r,d_j]$ for a fixed $r\ge 1$. For nonzero tests $f_j$ supported in the corresponding ambient intervals, set

$${\mathcal{B}}_j=\frac{{\sum }_i{\lambda }_{d_{j,i}/2}^W\Vert {\chi }_{j,i}{f}_j{\Vert }_2^2}{\Vert f_j{\Vert }_2^2}-I_2(B_j,2L_j).$$

Equations (3)–(7) give the uniform paper-level comparison

$$\underset{j}{\mathrm{limsup}}{\mathcal{B}}_j\le \log r+\frac{9}{2}-\gamma -2\log \pi -\frac{3}{2}\log 2-\frac{\pi }{2}.\text{\hspace{1em}(8)}$$

The right side is approximately $-0.977193$ for $r=1$ and $-0.495981$ for $r=\phi$. The latter is a conditional width-ratio example: the ratio of bare FIB cylinder lengths does not construct a smooth square partition or prove this ratio for its support intervals. Smooth cutoffs must overlap at the seams, and their actual widths and Lipschitz constants must be checked.

For $1\le r\le \phi$, this rejects a particular sufficient budget even when the optimal local spectral constants are used: subtracting this independent Gamma envelope and merely using a zero lower bound for the signed arithmetic term in (2) cannot give positive fine-scale margins under these conditions. Equation (8) does not give this rejection for every fixed width ratio. It does not give a negative value of the actual Weil form, a lower bound on its actual Gamma cost, or an obstruction to stronger joint estimates. The local surpluses, localization defect and arithmetic term depend on the same $f$ and cutoffs; retaining those relations is the remaining route. The analytic asymptotics and comparison (8) are not a new compiled general theorem.

A smooth realization of the physical windows

The width-ratio hypothesis can be realized by standard mollification and normalization. This constructs windows in the chosen physical interval chart; it does not intertwine FIB ATOM operations with prime translation.

Take adjacent tiles $I_i=[b_i,b_{i+1})$ of lengths in $[a,\phi a]$, $a>0$, and a nonnegative smooth mollifier $\eta$ of integral one, positive on $(-1,1)$ with topological support $[-1,1]$. Put ${\eta }_{\epsilon }(x)={\epsilon }^{-1}\eta (x/\epsilon )$, $0<\epsilon <a/2$, and

$${\rho }_i={\eta }_{\epsilon }\ast 1_{I_i},s=(\sum _i{\rho }_i^2{)}^{1/2}.$$

Index the tiles by $0\le i<N$, include tiles $0,N-1$ as padding in this sum, and keep the source support compactly inside $(b_1+\epsilon ,b_{N-1}-\epsilon )$. On $U=(b_0+\epsilon ,b_N-\epsilon )$, the ${\rho }_i$ sum to one and at most two are nonzero, so $s\ge 1/\sqrt{2}$. For interior tiles define ${\chi }_i={\rho }_i/s$ on $U$ and zero outside. Each retained numerator’s support is compactly contained in $U$, making this extension smooth. The retained family has square unity near the source support and global squared norm at most one.

The complete cutoff support widths and a joint Lipschitz bound are

$$d_i=b_{i+1}-b_i+2\epsilon ,\frac{\max d_i}{\min d_i}\le \frac{\phi a+2\epsilon }{a+2\epsilon }<\phi ,B\le \frac{2\Vert \eta {\Vert }_{\infty }}{\epsilon }.$$

Indeed ${\rho }_i^{\prime }={\eta }_{\epsilon }(x-b_i)-{\eta }_{\epsilon }(x-b_{i+1})$. At most one seam contributes at a point, so $\Vert {\rho }^{\prime }{\Vert }_{{\ell }^2}\le \sqrt{2}\Vert \eta {\Vert }_{\infty }/\epsilon$. Differentiating $\rho /\Vert \rho \Vert$ is orthogonal projection followed by division by $s$, giving the bound. No additional endpoint cutoff is inserted. These widths belong to the complete ${\chi }_i$, not the possibly smaller supports of a particular ${\chi }_{i}f$. This uses standard smooth partition ingredients and supplies no new arithmetic positivity.

The actual residual requires a positive arithmetic contribution

The same-function comparison avoids replacing the Gamma defect by a scalar envelope. Let $f\in C_c^{\infty }(\mathbb{R};\mathbb{C})$ have support in $[-L,L]$, and let a finite real smooth square partition hold there. Suppose each $g_i={\chi }_{i}f$ has support in an interval of length at most $d<\log 2$. Keep the full two-pole form and the same ambient prime cutoff for all terms. Define

$$R_f(t)=\Re \int f(x)\overline{f(x-t)}dx,\mathcal{C}(g)=\iint e^{|x-y|/2}\Re (g(x)\overline{g(y)})dxdy.$$

Every local prime correlation vanishes. Since $K_f(t)=2(R_f(t)-{\sum }_i{R}_{g_i}(t))$, at every prime-power atom $K_f(\log n)=2R_f(\log n)$, independently of the partition. The signed arithmetic integral in (2), denoted $J_{\chi }(f)$, decomposes exactly as

$$J_{\chi }(f)=J_{0,L}(f)-A_{\chi }(f),A_{\chi }(f)=\sum _i\mathcal{C}(g_i),$$

$$J_{0,L}(f)=2{\int }_0^{2L}{e}^{t/2}{R}_f(t)dt-2\sum _{2\le n\le e^{2L}}\frac{\Lambda (n)}{\sqrt{n}}{R}_f(\log n).$$

The interval kernel gives a uniform bound without cutoff derivatives:

$$|A_{\chi }(f)|\le 2(e^{d/2}-1)\Vert f{\Vert }_2^2.\text{\hspace{1em}(9)}$$

For $x\in [a,b]$, $b-a\le d$, its row integral is $2(e^{(x-a)/2}+e^{(b-x)/2}-2)\le 2(e^{d/2}-1)$. Apply $2|uv|\le |u{|}^2+|v{|}^2$ and sum the local masses to obtain (9). Thus $J_{\chi }$ converges to $J_{0,L}$ as the maximum width tends to zero, uniformly on unit tests with an error bound independent of $L$. Refinement below the first prime shift changes only this controlled local continuum term.

Write $E_{\ast }(f)={\int }_0^{\infty }{k}_{\ast }(t)E_t(f)dt$ and $a_0=c_{\Gamma }+4$. The complete accounting is

$$Q_{\mathrm{full}}(f)=a_0\Vert f{\Vert }_2^2+E_{\ast }(f)+J_{0,L}(f),\text{\hspace{1em}(10)}$$

$$S_{\chi }(f):=\sum _i{Q}_{\mathrm{full}}(g_i)-{\int }_0^{\infty }{k}_{\ast }(t)K_f(t)dt=a_0\Vert f{\Vert }_2^2+E_{\ast }(f)+A_{\chi }(f).\text{\hspace{1em}(11)}$$

To check the cancellation, the full pole has kernel $e^{|x-y|/2}+e^{-|x-y|/2}$, while

$${\int }_0^{\infty }{e}^{-t/2}{E}_t(g)dt=4\Vert g{\Vert }_2^2-\iint e^{-|x-y|/2}\Re (g(x)\overline{g(y)})dxdy.$$

Using $k_{\Gamma }-k_{\ast }=e^{-t/2}$ cancels the decaying kernel and leaves $\mathcal{C}(g)$. The whole-line energy is essential: for shifts larger than the support width it is $2\Vert g{\Vert }_2^2$, not zero. The project’s pole-continuum and shifted-digamma identities and renormalized Weil multiplier already supply the completion for their even bundled tests. Equation (11) remains a paper application of the general explicit formula to the possibly non-even local pieces.

The baseline is strictly negative. Its series, supplied by GammaMu, gives

$$a_0=\mathrm{digamma}(5/4)-\log \pi =-\gamma -\log \pi +\sum _{m\ge 1}\left(\frac{1}{m}-\frac{1}{m+1/4}\right)<-1.$$

Indeed each summand is at most $\frac{1}{2}(1/m-1/(m+1))$, whose sum is $1/2$, while $\gamma >1/2$ and $\log \pi >1$. Numerically $a_0\approx -1.372183419$; the strict sign uses the series bound rather than this decimal.

For a fixed even $h\in C_c^{\infty }(\mathbb{R};\mathbb{C})$ with $\Vert h{\Vert }_2=1$, put $f_R(x)=R^{-1/2}h(x/R)$, $R>0$. The standard translation estimate $E_t(f_R)\le t^2\Vert h^{\prime }{\Vert }_2^2/R^2$ gives

$$|S_{\chi }(f_R)-a_0|\le 2(e^{d/2}-1)+\frac{C_{\ast }\Vert h^{\prime }{\Vert }_2^2}{R^2},0<C_{\ast }\le \frac{26}{125}.\text{\hspace{1em}(12)}$$

For completeness, the translation estimate follows by integrating $f_R^{\prime }$ along a segment, applying Cauchy–Schwarz and then Fubini. The moment bound follows from $1/(1-e^{-2t})=1+1/(e^{2t}-1)\le 1+1/(2t)$ for $t>0$ and ${\int }_0^{\infty }(t^2+t/2)e^{-5t/2}dt=26/125$. If $h$ and the cutoffs are nonnegative, then also $S_{\chi }(f_R)\ge a_0$.

As $R\to \infty$ and $d\to 0$, the actual residual therefore tends to $a_0$, uniformly over these partitions, without any width-ratio, window-count or cutoff-derivative bound. Even if the widths merely remain below $\log 2$, (12) gives

$$\underset{R\to \infty }{\mathrm{limsup}}{S}_{\chi }(f_R)\le a_0+2(\sqrt{2}-1)<2\sqrt{2}-3<0.\text{\hspace{1em}(13)}$$

The first bound is approximately $-0.543756294$. This uses actual local forms and the actual signed Gamma defect of the same test. It is not a negative-full-Weil example. Since $Q_{\mathrm{full}}(f_R)=S_{\chi }(f_R)+J_{\chi }(f_R)$, positivity for these tests requires

$$J_{\chi }(f_R)\ge -a_0-2(e^{d/2}-1)-\frac{C_{\ast }\Vert h^{\prime }{\Vert }_2^2}{R^2}.\text{\hspace{1em}(14)}$$

Equation (14) is a necessary condition under positivity, not a supplied prime-discrepancy estimate. In particular a zero lower bound for $J_{\chi }$ cannot alone certify these directions. The needed arithmetic estimate must provide this positive contribution and also control the other, oscillatory test directions. The identities and uniform estimates (9)–(14) are paper-level deductions; they do not settle RH or establish a new compiled localization theorem.

Separating higher prime powers with the same test

The weighted Mertens application controls the higher-prime-power contribution for every compact smooth complex test, including the non-even local pieces. In the notation of (10), define the primary-prime discrepancy

$$C_L(f)=2{\int }_0^{2L}{e}^{t/2}{R}_f(t)dt-2\sum _{p\le e^{2L}}\frac{\log p}{\sqrt{p}}{R}_f(\log p).$$

Keep the actual same $f$ throughout. Then $J_{0,L}(f)=C_L(f)-T(f)$, where $T$ is the higher-power sum in that note; compact support makes its infinite-index notation identical to the common finite cutoff. Writing

$$T(f)=\frac{1}{2}|\int f{|}^2+2\kappa \Vert f{\Vert }_2^2+ϵ(f),|ϵ(f)|\le (8M_1+D_3)\Vert f^{\prime }{\Vert }_2^2,$$

the remaining comparison is exactly

$$Q_{\mathrm{full}}(f)=C_L(f)-\frac{1}{2}|\int f{|}^2+(a_0-2\kappa )\Vert f{\Vert }_2^2+E_{\ast }(f)-ϵ(f).$$

The combined remainder has absolute value at most $(26/125+8M_1+D_3)\Vert f^{\prime }{\Vert }_2^2$. Its higher-power energy has a favorable sign: that note also gives $-ϵ(f)=4\int E(v)R_f^{\prime }(2v)dv+{\mathcal{D}}_{\ge 3}(f)$ with ${\mathcal{D}}_{\ge 3}(f)\ge 0$. Together with $E_{\ast }(f)\ge 0$, this yields the paper-level sufficient condition

$$C_L(f)\ge \frac{1}{2}|\int f{|}^2+(2\kappa -a_0)\Vert f{\Vert }_2^2+8M_1\Vert f^{\prime }{\Vert }_2^2⟹Q_{\mathrm{full}}(f)\ge 0.$$

No such lower bound for $C_L(f)$ has been supplied. The derivative budget is useful on fixed smooth slow dilations, where it decreases quadratically, and need not be affordable on oscillatory tests. The Mertens supplier closes one paper-level debit estimate while leaving the signed primary-prime comparison unresolved. It provides no FIB-to-prime-translation intertwiner, all-support positivity, or RH conclusion.

The complementary logarithmic budget applies the classical entire-xi Hadamard expansion to the actual even-power boundary pairing, with the pole retained. In that note’s equation (8), the same test satisfies

$$E_{\ast }(f)-ϵ(f)+d_{\xi }\Vert f{\Vert }_2^2=\mathcal{Z}(f)+{\mathcal{D}}_{\mathrm{odd}}(f)+\mathcal{J}(f),d_{\xi }=4+\gamma -\log (4\pi )-\frac{\pi }{2}.$$

The three terms on the right are nonnegative; $\mathcal{Z}$ is the actual zero-strip Poisson contribution, not a critical-line zero sum or an RH assumption. Reusing the classical digamma values gives $2\kappa -a_0+d_{\xi }=2C_{\mathrm{odd}}+\log 2$, where $C_{\mathrm{odd}}={\sum }_p\log p/(\sqrt{p}(p-1))$. The full same-test account becomes

$$Q_{\mathrm{full}}(f)=C_L(f)-\frac{1}{2}|\int f{|}^2-(2C_{\mathrm{odd}}+\log 2)\Vert f{\Vert }_2^2+\mathcal{Z}(f)+{\mathcal{D}}_{\mathrm{odd}}(f)+\mathcal{J}(f).\text{\hspace{1em}(15)}$$

Dropping all three positive terms yields a sufficient condition for an individual test,

$$C_L(f)\ge \frac{1}{2}|\int f{|}^2+(2C_{\mathrm{odd}}+\log 2)\Vert f{\Vert }_2^2.$$

This condition cannot hold for every compact smooth test. For a normalized test supported in an interval of width $d<\log 2$, all primary-prime correlations vanish and the continuum estimate (9) gives $|C_L(f)|\le 2(e^{d/2}-1)$. This tends to zero as $d\downarrow 0$, while the required scalar is at least $\log 2>0$. Such tests refute the universal sufficient condition, not positivity of their full Weil form. A useful all-test comparison must retain the actual positive contributions or a sharper joint estimate. In the basic budget (8), $E_{\ast }$ has been spent; it cannot also be added to (15) as an independent reserve. Equation (15) and this domain boundary remain paper-level deductions without a new Lean declaration, Robin estimate or RH conclusion.

The classical Stechkin application supplies a stronger paper-level bound for the actual $\mathcal{Z}$ on this same test:

$$\mathcal{Z}(f)\ge {\kappa }_S{E}_{\ast }(f)-2\Vert f{\Vert }_2^2,{\kappa }_S=1/\sqrt{5}.$$

Thus the independently justified remainder

$${\mathcal{R}}_S(f)=\mathcal{Z}(f)-{\kappa }_S{E}_{\ast }(f)+2\Vert f{\Vert }_2^2\ge 0$$

gives the exact account

$$\begin{array}{rl}{Q}_{\mathrm{full}}(f)=& C_L(f)-\frac{1}{2}|\int f{|}^2-(2C_{\mathrm{odd}}+\log 2+2)\Vert f{\Vert }_2^2\\ & +{\kappa }_S{E}_{\ast }(f)+{\mathcal{R}}_S(f)+{\mathcal{D}}_{\mathrm{odd}}(f)+\mathcal{J}(f).\end{array}\text{\hspace{1em}(16)}$$

The retained fraction is paid for by a lower bound on $\mathcal{Z}$, including the scalar cost $2\Vert f{\Vert }_2^2$. It is not a second use of the full $E_{\ast }$ and is not obtained by defining an unknown Robin or Weil margin to be positive. A sufficient joint comparison is now

$$C_L(f)+{\kappa }_S{E}_{\ast }(f)\ge \frac{1}{2}|\int f{|}^2+(2C_{\mathrm{odd}}+\log 2+2)\Vert f{\Vert }_2^2⟹Q_{\mathrm{full}}(f)\ge 0.\text{\hspace{1em}(17)}$$

Equation (17) is sufficient rather than necessary: retaining the actual ${\mathcal{R}}_S$, ${\mathcal{D}}_{\mathrm{odd}}$ and $\mathcal{J}$ can reduce the required margin. The construction below shows that its antecedent cannot hold for every admitted test, even with the retained Gamma fraction. Every term belongs to the same actual $f$; separate extrema for $C_L$ and $E_{\ast }$ do not prove their joint comparison. The source reuse preserves a positive Gamma fraction without an RH hypothesis, but does not solve the remaining signed primary-prime/pole estimate or identify its tests with the Robin configurations.

A smooth obstruction to the retained-fraction scalarization

The earlier scalarization obstruction discarded every positive contribution. Here the question is different: does keeping ${\kappa }_S{E}_{\ast }$ suffice after charging the scalar cost $2\Vert f{\Vert }_2^2$? A fixed smooth test refutes that stronger universal requirement. The construction uses standard convolution and translation estimates, not a new approximation or prime-distribution theorem.

The classical suppliers are Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (Springer, 2011), §4.4, Theorem 4.15, printed p.104 (convolution contraction), Propositions 4.18 and 4.20, pp.106–107 (support and smoothness), and §9.1, Lemma 9.1 and Proposition 9.3, pp.266–267 (weak-derivative commutation and translation). The parameter map is dimension one, $p=2$, the whole space $\mathbb{R}$, and the zero-extended seed below. Positive compact convolution and explicit norm control are used directly; no unspecified approximation sequence or extension of the Weil form to nonsmooth tests is needed.

Start with the compact absolutely continuous function

$$f_0(x)=\{\begin{array}{ll}2\cos (2\pi x),& |x|\le 1/4,\\ 0,& |x|>1/4.\end{array}\Vert f_0{\Vert }_2^2=1,\Vert f_0^{\prime }{\Vert }_2^2=4{\pi }^2<40.$$

Choose any nonnegative even $\eta \in C_c^{\infty }(-1,1)$ of integral one. Put $\epsilon =1/1000$, ${\eta }_{\epsilon }(x)={\epsilon }^{-1}\eta (x/\epsilon )$, $g={\eta }_{\epsilon }\ast f_0$, and $f=g/\Vert g{\Vert }_2$. Young’s convolution inequality gives $\Vert g^{\prime }{\Vert }_2\le \Vert f_0^{\prime }{\Vert }_2$. Reuse the segment-integral translation estimate above, valid also for this absolutely continuous $f_0$, to obtain

$$\Vert g-f_0{\Vert }_2\le \epsilon \Vert f_0^{\prime }{\Vert }_2<1/100,\Vert g{\Vert }_2>99/100,\Vert f^{\prime }{\Vert }_2^2<\frac{40}{(99/100{)}^2}<42.$$

This is an actual nonnegative even real compact smooth test with $\Vert f{\Vert }_2=1$, supported in $[-251/1000,251/1000]$. Its support width is less than $11/20<\log 2$, so every primary-prime correlation vanishes. The existing continuum bound therefore gives

$$0\le C_L(f)\le 2(e^{11/40}-1)<1.$$

For its shifted energy, nonnegativity makes $R_f(t)\ge 0$, so the same translation estimate and norm identity give

$$\Vert f-{\tau }_{t}f{\Vert }_2^2\le \min (42t^2,2).$$

Since $1-e^{-2t}=2e^{-t}\sinh t\ge 2t{e}^{-t}$ for $t>0$, its actual kernel satisfies

$$\frac{e^{-5t/2}}{1-e^{-2t}}\le \frac{e^{-3t/2}}{2t}.$$

Split the energy integral at $t=1/4$ and $t=1$. On the first interval use the derivative bound, on the second use the bound two, and on the last also use $1/t\le 1$. This yields

$$E_{\ast }(f)\le \frac{42}{64}+\log 4+\frac{2}{3}{e}^{-3/2}<\frac{21}{32}+\frac{7}{5}+\frac{1}{6}=\frac{1067}{480}<\frac{9}{4}.$$

The scalar comparisons use only ${\pi }^2<10$, $\log 4<7/5$, $e^{3/2}>4$, $\log 2>11/20$, and $\log (3/2)>11/40$. Since ${\kappa }_S<1/2$ and $C_{\mathrm{odd}}\ge 0$,

$$\begin{array}{rl}{C}_L(f)+{\kappa }_S{E}_{\ast }(f)& <1+\frac{9}{8}=\frac{17}{8},\\ \frac{1}{2}|\int f{|}^2+(2C_{\mathrm{odd}}+\log 2+2)\Vert f{\Vert }_2^2& >\frac{5}{2}.\end{array}$$

Thus the antecedent in (17) fails by more than $3/8$ on an admitted smooth even test. Enlarging the ambient support does not repair it for this same $f$: the added continuum integrand is zero, and all newly admitted prime correlations still vanish.

The discarded positive terms have concrete contributions on this very test. Every odd-power shift $k\log p$, $k\ge 3$, exceeds its support width, so the existing energy definition gives

$${\mathcal{D}}_{\mathrm{odd}}(f)=2C_{\mathrm{odd}},{\mathcal{R}}_S(f)=\mathcal{Z}(f)-{\kappa }_S{E}_{\ast }(f)+2\ge 2-{\kappa }_S{E}_{\ast }(f)>\frac{7}{8}.$$

These compensations do not by themselves determine the sign of the full form: the displayed estimates supply only a lower bound for the reduced deficit. They show why discarding independently positive terms is a substantive loss. This is a paper-level obstruction to that sufficient scalarization, not a negative-full-Weil example or a counterexample to RH or Robin. The actual values in (16) must be retained or bounded jointly if this account is used for an all-test proof. The construction and bounds have not been compiled in Lean.

The exact auxiliary line does not replace the discarded residual

Retain the auxiliary line itself, rather than replacing it by the scalar lower bound in (16). Use the classical notation $Q_{\sigma }(t)=\Re ({\xi }^{\prime }/\xi )(\sigma +it)$, ${\kappa }_S=1/\sqrt{5}$, and $\phi =(1+\sqrt{5})/2$, with the same entire $\xi$ as the Stechkin supplier. For the same even compact smooth test set

$${\mathcal{A}}_{\phi }(f)=\frac{1}{2\pi }{\int }_{\mathbb{R}}2{\kappa }_S{Q}_{\phi }(2\tau )|\hat{f}(\tau ){|}^{2}d\tau ,{\mathcal{R}}_{\phi }(f)=\mathcal{Z}(f)-{\mathcal{A}}_{\phi }(f)\ge 0.$$

Dropping only this exact Stechkin residual from (15), while keeping every other actual contribution, gives

$$\begin{array}{rl}{Q}_{\mathrm{cmp}}(f):=Q_{\mathrm{full}}(f)-{\mathcal{R}}_{\phi }(f)=& C_L(f)-\frac{1}{2}|\int f{|}^2-(2C_{\mathrm{odd}}+\log 2)\Vert f{\Vert }_2^2\\ & +{\mathcal{A}}_{\phi }(f)+{\mathcal{D}}_{\mathrm{odd}}(f)+\mathcal{J}(f).\end{array}\text{\hspace{1em}(18)}$$

Universal nonnegativity of (18) is still too strong. The applicable classical boundary is Lagarias, arXiv:math/0404394v4, §3, equation (3.1), printed p.12. His analytic class $\mathcal{A}$ consists of functions holomorphic on $0<\Re s<1$ with a uniform $O(1/|s|)$ bound away from $|\Im s|\le 1$. He explicitly gives the nonzero null vector $F(s)=\xi (s,\pi )$ for the Weil scalar product. For the trivial representation this is $2\xi (s)$; that factor affects neither its zeros nor the logarithmic derivative. This is an existing null-vector result, not a new construction. Its compact-test application requires the domain argument below.

Reuse the original positive even theta kernel $\Phi$ from Romik, equations (1.6)–(1.11), Online First pp.2–3:

$$\Phi (x)=\sum _{n\ge 1}(4{\pi }^2{n}^4{e}^{9x/2}-6\pi n^2{e}^{5x/2})e^{-\pi n^2{e}^{2x}},\hat{\Phi }(z)=\xi (1/2+iz).\text{\hspace{1em}(19)}$$

The Fourier convention here is $\hat{f}(z)=\int e^{-izx}f(x)dx$. Since $\Phi$ is even, Romik’s plus-sign convention agrees for every complex $z$, with no additional prefactor. The project’s theta transform, source_theta_fourier_eq_xi, already proves the all-complex identity for the identified kernel; source_theta_normalized supplies $\xi (1/2)>0$. Neither identity needs a new implementation.

For the compact-domain bridge use the original all-real expression in (19), which is smooth at zero. Extending individual positive-half-line summands with $|x|$ would not by itself prove smoothness there. On $x\ge 0$, termwise differentiation of the normally convergent theta series gives, for every derivative order $j$, constants $C_j,c_j$ such that

$$|{\Phi }^{(j)}(x)|\le C_j{e}^{c_{j}x}{e}^{-(\pi /2)e^{2x}}.$$

Indeed each derivative is a finite sum of exponential factors times $n^d{e}^{-\pi n^2{e}^{2x}}$; split the latter exponent using $n^2{e}^{2x}\ge (n^2+e^{2x})/2$ and sum $n^d{e}^{-\pi n^2/2}$. Evenness controls the negative half-line. In particular every derivative is integrable against $e^{a|x|}$ for each fixed $a>0$, and all tilted integration-by-parts boundary terms vanish. These derivative estimates and the cutoff passage are paper applications of the classical kernel, not consequences already bundled into the compiled Fourier identity.

The multiplier of ${\mathcal{R}}_{\phi }$ is

$$r(\tau )=2[Q_1(2\tau )-{\kappa }_S{Q}_{\phi }(2\tau )]\ge 0.$$

It is strictly positive at zero. To see the strictness without any low-zero computation, write an actual zero as $\rho =\beta +i\gamma$, let $u=\beta (1-\beta )>0$, $y={\gamma }^2$, and define the reflected kernel pair

$$F_{\sigma }(\rho )=\frac{\sigma -\beta }{(\sigma -\beta {)}^2+{\gamma }^2}+\frac{\sigma -1+\beta }{(\sigma -1+\beta {)}^2+{\gamma }^2}.$$

With positive denominators

$$D_1=y^2+(1-2u)y+u^2,D_{\phi }=y^2+(3-2u)y+(1+u{)}^2,$$

the exact kernel difference is

$$F_1(\rho )-{\kappa }_S{F}_{\phi }(\rho )=\frac{y^2+6uy+u(1+u)}{D_1{D}_{\phi }}>0.\text{\hspace{1em}(20)}$$

Reflection preserves the actual zero multiset and multiplicities $m_{\rho }$, so $Q_{\sigma }(0)=\frac{1}{2}{\sum }_{\rho }{m}_{\rho }{F}_{\sigma }(\rho )$. The multiplier’s factor two cancels this one-half, including for fixed critical-line zeros. The existing absolutely convergent real resolvent and nonempty zero set therefore give $r(0)>0$. Continuity and $\hat{\Phi }(0)=\xi (1/2)>0$ imply

$$0<{\mathcal{R}}_{\phi }(\Phi )<\infty .\text{\hspace{1em}(21)}$$

Finiteness uses the same Poisson-pair bound in the supplier’s equation (5): it is uniform over widths $0<a\le 2$ and summable over the actual zero ordinates. All relevant widths $1-\beta$ and $\phi -\beta$ lie in this interval. Weighted kernel derivatives give the necessary decay of $|\hat{\Phi }{|}^2$ on the real Fourier line; no pointwise bound on narrow Poisson peaks is assumed.

Take a real even $0\le \chi \le 1$ in $C_c^{\infty }$, equal to one on $[-1,1]$ and supported in $[-2,2]$, and put $f_N(x)=\chi (x/N)\Phi (x)$. These are actual nonnegative even compact smooth tests. Their weighted two-jet errors satisfy

$${ϵ}_N=\int e^{|x|/2}(|f_N-\Phi |+|(f_N-\Phi {)}^{\prime \prime }|)dx⟶0.$$

Product differentiation and the weighted derivative tails above justify this limit. Integration by parts, with its vanishing boundary terms, gives

$$|\widehat{f_N-\Phi }(v+iw)|\le \frac{{ϵ}_N}{1+v^2},|w|\le 1/2.\text{\hspace{1em}(22)}$$

This is the same weighted-jet mechanism as closed-strip decay; that module’s bundled test domain is compact, so its noncompact use here is an explicitly justified paper extension. The fixed-support rational approximation theorem is not substituted for this expanding-support cutoff.

For every actual zero use $z_{\rho }=\gamma -i(\beta -1/2)$ and retain the correct paired factors

$$Q_{\mathrm{full}}(f_N)=\sum _{\rho }{m}_{\rho }{\hat{f}}_N(z_{\rho })\overline{{\hat{f}}_N(\overline{z_{\rho }})}.$$

Equation (19) makes the limiting factor $\hat{\Phi }(z_{\rho })=\xi (\rho )$ vanish at every actual zero, without RH. A finite weighted two-jet budget $B$ for $\Phi$ and (22) bound the paired-product error by ${ϵ}_N(2B+{ϵ}_N)/(1+{\gamma }^2{)}^2$. The existing multiplicity-weighted inverse-fourth zero summability and compact-test explicit formula yield $Q_{\mathrm{full}}(f_N)\to 0$. This defines no new arithmetic form on noncompact functions. Off the critical line the paired expression is not $|{\hat{f}}_N(z_{\rho }){|}^2$.

On the real Fourier line the same product estimate, followed by the supplier’s Poisson-pair bound applied separately to $Q_1$ and $Q_{\phi }$, gives ${\mathcal{R}}_{\phi }(f_N)\to {\mathcal{R}}_{\phi }(\Phi )$. Consequently

$$Q_{\mathrm{cmp}}(f_N)⟶-{\mathcal{R}}_{\phi }(\Phi )<0.\text{\hspace{1em}(23)}$$

Thus for all sufficiently large $N$ an admitted nonnegative even smooth test makes (18) negative. Normalization by its nonzero $L^2$ norm preserves the sign. This applies the published null-vector boundary with an explicit compact-domain bridge; it makes no originality claim. It excludes discarding only the exact residual as an all-test positivity certificate, even when the auxiliary line and all other terms are kept. It supplies no sign for $Q_{\mathrm{full}}(f_N)$ itself, no Robin estimate, and no RH counterexample or proof. The theta Fourier identity is existing formalized mathematics; equations (18), (20)–(23) and their compact-domain application have not been compiled as a new Lean result.

A prime edge crossing an intermediate FIB window

The five first-level internal-coordinate intervals have geometric order $[3],[null],[5],[2],[25]$. Put $\phi =(1+\sqrt{5})/2$. In particular

$$I_3=[-1,\phi -2],I_{null}=[\phi -2,2\phi -3],I_5=[2\phi -3,\phi -1].$$

For this test construction, choose these numerical intervals as windows in the Weil physical $x$ coordinate. This choice does not identify prime translation with an operation on the original golden-coordinate source or prove an intertwining theorem. Translation by $\log 2$ sends the entire interval $[-41/100,-39/100]\subset \mathrm{int}{I}_3$ into $\mathrm{int}{I}_5$, across $I_{null}$. This is an edge of the actual prime translation, not a legal-digit seam transition.

For a smooth example take a nonzero even real $\eta \in C_c^{\infty }(-1/100,1/100)$, $a=2/5$, $b=\log 2-2/5$, and

$$f(x)=\eta (x-a)+\eta (x+a)+\eta (x-b)+\eta (x+b).$$

It is even with support in $[-41/100,41/100]$. At this radius $2<e^{2L}<3$, so the only active prime power is $2$. The four bump supports are disjoint. Exactly the ordered centre pairs $(a,-b)$ and $(b,-a)$ have separation $\log 2$, giving

$$\int f(x)f(x-\log 2)dx=2\Vert \eta {\Vert }_2^2>0.$$

The original prime contribution to $Q$ is therefore $-2\sqrt{2}\log 2\Vert \eta {\Vert }_2^2$. The pair $(b,-a)$ crosses $I_{null}$, whereas $(a,-b)$ connects $I_{null}$ to $I_5$. Dropping the nonadjacent pair alone loses $-\sqrt{2}\log 2\Vert \eta {\Vert }_2^2$, half of the displayed prime contribution, even for a legitimate even smooth test. This is not a negative-full-form example and does not refute RH. Pairing mirrored windows to keep evenness changes their support geometry and must retain the corresponding cross terms.

Reuse and remaining research target

The standard IMS identity is reusable mathematics, not a new FIB positivity theorem. The project’s smooth rational approximation controls the complete paired zero sum; its golden cofinal interface still requires positivity for every admitted test at every chosen scale. Neither supplies a lower bound for the signed integral in (2).

A sufficient next input would combine genuine lower margins for the localized tests with a lower bound for the same $W_f$ against $\Psi -u$, paying the displayed Gamma remainder uniformly over all allowed coefficients and growing supports. Equation (14) is a necessary benchmark on the slow dilations, not this all-test sufficient bound. The family $W_f$ is constrained by a common $f$ and square partition; it cannot be replaced by arbitrary independently optimized weights. The corresponding Robin research also retains a joint signed prime fluctuation, but identifying those two test families requires another explicit map.

The analytic localization, Stieltjes calculations and uniform comparisons above are paper-level applications with their hypotheses displayed, not a new Lean closure. Transient exact Lean checks cover only the stated rational interval placement, first-prime support threshold, pointwise four-bump pairing, scalar Gamma cancellation, negativity of the scalar expression in (8) at $r=\phi$, and the actual scalar inequalities $c_{\Gamma }+4<-1$ and $c_{\Gamma }+4+2(\sqrt{2}-1)<0$; no named wrapper is retained. The scalar checks do not prove the spectral-domain, asymptotic, integral or cutoff premises. The nonlocal source, these checks, and the missing all-scale arithmetic estimate have different evidentiary roles.