Weil positivity and trace formula the archimedean place

Abstract

We provide a potential conceptual reason for the positivity of the Weil functional using the Hilbert space framework of the semi-local trace formula of Connes (Sel Math (NS) 5(1):29–106, 1999). We explore in great details the simplest case of the single archimedean place. The root of this result is the positivity of the trace of the scaling action compressed onto the orthogonal complement of the range of the cutoff projections associated to the cutoff in phase space, for $\Lambda =1$. We express the difference between the Weil distribution and the trace associated to the above compression of the scaling action, in terms of prolate spheroidal wave functions, and use, as a key device, the theory of hermitian Toeplitz matrices to control that difference. All the concepts and tools used in this paper make sense in the general semi-local case, where Weil positivity implies RH.

Appendices

Appendix A: Fourier versus Mellin transforms

We use the convolution algebra $C_c^\infty ({{\mathbb {R}}}_+^*)$ of smooth complex valued functions with compact support on the multiplicative group ${{\mathbb {R}}}_+^*$. Its convolution product and involution are given by

$$\begin{aligned} (f* g)(u):=\int f(v)g(u/v)d^*v, \qquad f^*(u):=\overline{f(u^{-1})}. \end{aligned}$$

(145)

The (multiplicative) Fourier transform of f (see (22))

$$\begin{aligned} {{\hat{f}}}(s)=\int _0^\infty f(u)u^{-is}d^*u \end{aligned}$$

transforms convolution into pointwise product and the involution into the pointwise complex conjugation, $s\in {{\mathbb {R}}}$. For complex values of s, the evaluation $f\mapsto {{\hat{f}}}(s)$ is still multiplicative but no longer compatible with the involution.

The translation to formulas using the Mellin transform is done in (41) and uses the isomorphism

$$\begin{aligned} k{\mathop {\mapsto }\limits ^{\sim }} \Delta ^{1/2}(k)=f, \qquad f(x):= x^{1/2}k(x) \end{aligned}$$

(146)

which respects the convolution product, and transforms $x^{-1}k(x^{-1})$ into $f(x^{-1})$. Hence, after taking complex conjugates, the natural involution $k\mapsto {{\bar{k}}}^\sharp$ becomes $f\mapsto f^*$ . The Mellin transform ${{\tilde{k}}}(z):=\int _0^\infty k(u)u^zd^*u$ is related to the (multiplicative) Fourier transform of f by (41) i.e.

$$\begin{aligned} {{\tilde{k}}}(\frac{1}{2}+is)=\int _0^\infty k(u)u^{\frac{1}{2}+is}d^*u =\int _0^\infty f(u)u^{is}d^*u={{\hat{f}}}(-s), \end{aligned}$$

(147)

where the sign in $-s$ is due to the convention for the multiplicative Fourier transform (22).

Appendix B: Explicit formula

In this appendix, we gather different sources on the normalization of the archimedean contribution to the explicit formula. Following [5], one defines the Mellin transform of a function $f\in C^\infty _c({{\mathbb {R}}}_+^*)$ as

$$\begin{aligned} {{\tilde{f}}}(s):=\int _0^\infty f(x)x^{s-1}dx. \end{aligned}$$

(148)

Then, with $f^\sharp (x):=x^{-1}f(x^{-1})$ the explicit formula takes the form

$$\begin{aligned} \sum _{\rho }{{\tilde{f}}}(\rho )=\int _0^\infty f(x)dx+\int _0^\infty f^\sharp (x)dx-\sum _v {{\mathcal {W}}}_v(f), \end{aligned}$$

(149)

where v runs over all places $\{{{\mathbb {R}}},2,3,5,\ldots \}$ of ${{\mathbb {Q}}}$, the sum on the left hand side is over all complex zeros $\rho$ of the Riemann zeta function, and for $v=p$

$$\begin{aligned} {{\mathcal {W}}}_p(f)=(\log p)\sum _{m=1}^\infty \left( f(p^m)+f^\sharp (p^m)\right) . \end{aligned}$$

(150)

The archimedean distribution is defined as

$$\begin{aligned} {{\mathcal {W}}}_{{\mathbb {R}}}(f):=(\log 4\pi +\gamma )f(1)+\int _{1}^\infty \left( f(x)+f^\sharp (x)-\frac{2}{x} f(1)\right) \frac{dx}{x-x^{-1}}.\nonumber \\ \end{aligned}$$

(151)

One then has

$$\begin{aligned} {{\mathcal {W}}}_{{\mathbb {R}}}(f)=(\log \pi )f(1)-\frac{1}{2\pi i}\int _{1/2+iw}\mathfrak {R}\left( \frac{\Gamma '}{\Gamma }\left( \frac{w}{2}\right) \right) {{\tilde{f}}}(w)dw. \end{aligned}$$

(152)

In [4, 6] a positivity result for the distribution ${{\mathcal {W}}}_\infty =-{{\mathcal {W}}}_{{\mathbb {R}}}$ is proven, (for test functions with support in a small enough interval around 1), by writing the distribution ${{\mathcal {W}}}_\infty$ in terms of the Mellin transform of the test function as follows

$$\begin{aligned} {{\mathcal {W}}}_\infty (f)=\int _{w=1/2+i\tau }h_+(\tau ){{\tilde{f}}}(w)\frac{d\tau }{2\pi }. \end{aligned}$$

(153)

The function $h_+(\tau )$ is

$$\begin{aligned} h_+(\tau )=-\log \pi + \mathfrak {R}(\lambda (1/4 +i\tau /2)), \qquad \lambda (z)=\Gamma '(z)/\Gamma (z). \end{aligned}$$

(154)

It is the derivative of $2\,\theta (\tau )$, where $\theta$ is the Riemann-Siegel angular function defined as

$$\begin{aligned} \theta (E) = - \frac{E}{2} \log \pi + \mathfrak {I}\log \Gamma \left( \frac{1}{4} + i \frac{E}{2} \right) , \end{aligned}$$

(155)

with $\log \Gamma (s)$, for $\mathfrak {R}(s)>0$, the branch of the $\log$ which is real for s real.

In order to reflect the unitarity of the scaling action it is convenient to use the automorphism of $C_c^\infty ({{\mathbb {R}}}_+^*)$, $f\mapsto \Delta ^{1/2}f$, $\Delta ^{1/2}f(x):=x^{1/2}f(x)$ which replaces the involution $f\mapsto {{\bar{f}}}^\sharp$ by the involution $f\mapsto f^*$ of the convolution $C^*$-algebra and the restriction of the Mellin transform to the critical line by the Fourier transform. One has, for any place v, $W_v(f):={{\mathcal {W}}}_v(\Delta ^{-1/2}f)$, $\forall f\in C_c^\infty ({{\mathbb {R}}}_+^*)$.

Appendix C: Positivity criterion

This appendix shows that one may impose finitely many vanishing conditions to test functions without altering the validity of Weil’s positivity criterion. We follow [31] and state the following equivalence, using the Mellin transform

Proposition 6.12

Let $Z\subset {{\mathbb {C}}}$ be the set of non-trivial zeros of the Riemann zeta function and $F \subset {{\mathbb {C}}}$ a finite set disjoint from Z and containing $\{0,1\}$, then

$$\begin{aligned} RH \iff \sum _v {{\mathcal {W}}}_v(g*{{\bar{g}}}^\sharp )\le 0, \quad \forall g\in C_c^\infty ({{\mathbb {R}}}_+^*)\mid {{\tilde{g}}}(z)=0,\ \forall z\in F. \end{aligned}$$

(156)

Proof

The implication “$\Rightarrow$” follows from the explicit formula (149) and the hypothesis $\{0,1\}\subset F$. Conversely, the proof of Proposition 1 of [31] applies verbatim, provided one first refines the proof of Lemma 1 of op.cit. by showing that, given $\epsilon >0$ and $\rho _0\in Z$, there exists $g_0 \in C_c^\infty ({{\mathbb {R}}}_+^*)$ such that

$$\begin{aligned} {{\tilde{g}}}_0(z)=0,\quad \forall z\in F; \quad {{\tilde{g}}}_0(\rho _0)=1; \quad \vert {{\tilde{g}}}_0(\rho )\vert \le \epsilon /\vert \rho -\rho _0\vert ^2,\quad \forall \rho \in Z, \ \rho \ne \rho _0. \end{aligned}$$

In order to fulfill the additional vanishing condition: ${{\tilde{g}}}_0(z)=0, \forall z\in F$, one adjoins F to the finite set of zeros fulfilling $\vert \rho -\rho _0\vert <R$ (same notation as in Proposition 1 of [31]), and one then proceeds exactly as in op.cit. $\square$

Appendix D: Quantized calculus redux

Let C be a locally compact abelian group endowed with the proper homomorphism

$$\begin{aligned}\mathrm{Mod}:C\rightarrow {{\mathbb {R}}}_+^*,\qquad \mathrm{Mod}(u)=\vert u\vert \quad u \in C.\end{aligned}$$

We let ${\widehat{C}}$ be the Pontrjagin dual of C endowed with its Haar measure. The elements $f\in L^\infty ({\widehat{C}})$ act as multiplication operators on the Hilbert space ${{\mathcal {H}}}:= L^2({\widehat{C}})$. We define the “quantized” differential of f to be the operator

$$\begin{aligned} ^-\!\!\!\!\!d\,f:= [H,\,f]= Hf-fH, \end{aligned}$$

(157)

where the operator H on ${{\mathcal {H}}}$ is

$$\begin{aligned} H:= 2 {{\mathbb {F}}}_C\,\mathbf{1}_P\,{{\mathbb {F}}}_C^{-1} -1, \end{aligned}$$

(158)

where ${{\mathbb {F}}}_C: L^2(C) \rightarrow {{\mathcal {H}}}$ is the Fourier transform, and $\mathbf{1}_P$ is the multiplication by the characteristic function of the set $P=\{u \in C\,| \, \vert u\vert \ge 1\}$.

We take the case $C={{\mathbb {R}}}$ with module $\exp : {{\mathbb {R}}}\rightarrow {{\mathbb {R}}}_+^*$ considered in this paper and identify the dual ${{\widehat{C}}} \sim {{\mathbb {R}}}$ using the bi-character $\nu (s,t):=\exp (-ist)$ which corresponds to (21) under the isomorphism given by the module. We give a “geometric” proof of the following Lemma (see [7] Chapter IV for the general theory, a compact operator has infinite order when its characteristic values form a sequence of rapid decay; this implies that it is of trace class).

Lemma 6.13

For $f\in {{\mathcal {S}}}({{\widehat{C}}})$ the quantized differential [H, f] is an infinitesimal of infinite order and in particular a trace class operator.

Proof

Let us work in the Hilbert space $L^2(C)$ so that the action of $K={{\mathbb {F}}}_C^{-1} f{{\mathbb {F}}}_C$ is a convolution operator with Schwartz kernel $k(x,y)={{\widehat{f}}}(x-y)$. The projection $\mathbf{1}_P$ is the multiplication by the characteristic function of the halfline $[0,\infty ]$. It is enough to show that the operator $PK(1-P)$ is of infinite order. After precomposition with the symmetry $\sigma (\xi )(x):=\xi (-x)$ the Schwartz kernel of $PK(1-P)\sigma$ is $h(x,y)=P(x){{\widehat{f}}}(x+y)P(y)$. Let $\phi \in C^\infty ({{\mathbb {R}}})$ be a smooth function which is identically 0 for $x\le -1$ and identically 1 for $x\ge 0$. Let $g={{\widehat{f}}}$ and T be the operator in $L^2({{\mathbb {R}}})$ given by

$$\begin{aligned} T\xi (x):=\int \phi (x)g(x+y)\phi (y)\xi (y)dy. \end{aligned}$$

One has $PTP=PK(1-P)\sigma$ and thus it is enough to show that T is of infinite order. Let $A:=-\partial _x^2 +x^2$ be the harmonic oscillator, it is enough to show that the operator $A^nT$ is bounded for any $n>0$. Indeed the eigenvalues of A are the positive integers with multiplicity 1 and the above boundedness ensures that the characteristic values of T are of rapid decay. Now the Schwartz kernel of $A^nT$ is a finite linear combination of products of the form

$$\begin{aligned} \phi (y)\phi (x)^{(\ell ')} x^k g^{(\ell )}(x+y). \end{aligned}$$

Since $g={{\widehat{f}}}\in {{\mathcal {S}}}({{\mathbb {R}}})$ the derivatives $g^{(\ell )}$ are of rapid decay and for any given $m>0$ one has an inequality of the form $\vert g^{(\ell )}(a)\vert \le C_m (3+a)^{-m}$ for all $a\ge -2$. it follows that one controls the Hilbert Schmidt norm by the square root of the integral

$$\begin{aligned} C_m^2\int _{-1}^\infty \int _{-1}^\infty \vert \phi (y)\phi (x)^{(\ell ')}\vert ^2\vert x\vert ^{2k}(3+x+y)^{-m}dx dy \end{aligned}$$

which is finite for m large enough. This shows as required that $A^nT$ is bounded for any $n>0$. $\square$

Remark 6.14

One can give two alternate proofs of Lemma 6.13. The first uses the conformal invariance of the quantized calculus to get a unitary operator $U:L^2(S^1)\rightarrow L^2({{\mathbb {R}}})$ of the form $(U\xi )(t):=\frac{\pi ^{-1/2}}{t+i}\ \xi (\frac{t-i}{t+i})$ which conjugates, up to sign, the Hilbert transform H (which acts in $L^2({{\widehat{C}}})$) by the operator $2P_{H^2}-1$ where $P_{H^2}$ is the orthogonal projection on boundary values of holomorphic functions. Moreover the conjugate of the multiplication operator by $f\in {{\mathcal {S}}}({{\mathbb {R}}})$ is the multiplication by the smooth function $g(z)=f(i(1+z)/(1-z))$. Then the result follows since the quantized differential of a smooth function $g\in C^\infty (S^1)$ is of the form $2\sum {{\widehat{g}}}(n)[P_{H^2},z^n]$ which is of infinite order because the ${{\widehat{g}}}(n)$ are of rapid decay. The fact that $g\in C^\infty (S^1)$ comes from the smoothness os the extension of a Schwartz function to $P^1({{\mathbb {R}}})$ by the value 0 at $\infty$.

Another instructive alternate proof is to estimate directly, for $f\in {{\mathcal {S}}}({{\mathbb {R}}})$, the Schwartz kernel given by

$$\begin{aligned} k(x,y)=\frac{f(x)-f(y)}{x-y}. \end{aligned}$$

Appendix E: Signs and normalizations

We follow [27] and use the classical formula expressing the Fourier transform as a composition of the inversion

$$\begin{aligned} I(f)(s):= f(s^{-1}) \end{aligned}$$

(159)

and a multiplicative convolution operator. In our framework the unitary u is given by the ratio of archimedean local factors on the critical line

$$\begin{aligned} u(s)=\,\frac{\pi ^{-z/2}\Gamma (z/2)}{\pi ^{-(1-z)/2}\Gamma ((1-z)/2)},\qquad z=1/2+is. \end{aligned}$$

(160)

In terms of the Riemann-Siegel angular function (155), one has

$$\begin{aligned} u(s)= e^{2\, i\, \theta (s)}, \end{aligned}$$

(161)

so that the function $Z(t):=e^{i\theta (t)}\zeta (\frac{1}{2}+it)$ is real valued. Indeed, this follows from the functional equation since the complete zeta function $\zeta _{{\mathbb {Q}}}(z):=\pi ^{-z/2}\Gamma (z/2)\zeta (z)$ is real valued on the critical line. The quantized differential $\ ^-\!\!\!\!\!df$ of f is given by the kernel

$$\begin{aligned} k(s,t)= \frac{i}{ \pi } \, \frac{f(s)-f(t)}{s-t}. \end{aligned}$$

(162)

Thus when one takes the logarithmic derivative of u one obtains on the diagonal

$$\begin{aligned} u^*\,^-\!\!\!\!\!du(s)=e^{-2\, i\, \theta (s)}\frac{i}{ \pi }\partial _s e^{2\, i\, \theta (s)}=-\frac{2}{ \pi }\partial _s\theta (s)~\Longrightarrow ~\frac{1}{2} u^*\,^-\!\!\!\!\!du(s)=\frac{-2\partial _s\theta (s)}{2 \pi }.\nonumber \\ \end{aligned}$$

(163)

One can then write

$$\begin{aligned} \mathrm{Tr}({{\hat{h}}}_1\left( \frac{1}{2} u^{-1}\, ^-\!\!\!\!\!d\,u\right) {{\hat{h}}}_2)=-\int {{\hat{h}}}_1(t){{\hat{h}}}_2(t)\frac{2\partial _t\theta (t)}{2 \pi }dt. \end{aligned}$$

(164)

This corresponds to (153) since ${{\mathcal {W}}}_{{\mathbb {R}}}=-{{\mathcal {W}}}_\infty$ and to the semi-local trace formula

$$\begin{aligned} \mathrm{Tr}({{\hat{h}}}_1\left( \frac{1}{2} u^{-1}\, ^-\!\!\!\!\!d\,u\right) {{\hat{h}}}_2) = \sum _{v \in S} \int '_{{{\mathbb {Q}}}^*_v} \frac{\vert w \vert ^{1/2} }{ \vert 1-w \vert } \,h(w) d^* w, \quad h=h_1*h_2, \end{aligned}$$

(165)

for the single archimedean place.

Appendix F: Issues of convergence

We gather several inequalities which ensure the convergence of the series (99) of Proposition 5.3. We first consider the terms

$$\begin{aligned} A_n(\rho ):= \frac{\lambda (n)}{\sqrt{1-\lambda (n)^2}}\ \rho ^{1/2}\int _{\rho ^{-1}}^1(D_u\xi _n)(x)(D_u\zeta _n)(\rho x)dx. \end{aligned}$$

We estimate the integral using Schwarz’s inequality

$$\begin{aligned} \bigg |\rho ^{1/2}\int _{\rho ^{-1}}^1(D_u\xi _n)(x)(D_u\zeta _n)(\rho x)dx\bigg |\le \left( \int _{\rho ^{-1}}^1(D_u\xi _n)(x)^2 dx\right) ^{\frac{1}{2}} \left( \int _{\rho ^{-1}}^1(D_u\zeta _n)(\rho x)^2 \rho dx\right) ^{\frac{1}{2}}. \end{aligned}$$

One has, using $D_u(f)(x)=x\partial _xf(x)$

$$\begin{aligned} \int _{\rho ^{-1}}^1(D_u\zeta _n)(\rho x)^2 \rho dx=\int _1^\rho (D_u\zeta _n)(y)^2dy\le \rho ^2 \int _1^\rho (\partial _y\zeta _n)(y)^2dy. \end{aligned}$$

With $\zeta _n(x)=\frac{1}{\sqrt{1-\lambda (n)^2}}\,\eta _n(x)$ and $\eta _n={{\mathbb {F}}}_{e_{{\mathbb {R}}}}\xi _n$ one thus obtains

$$\begin{aligned} \int _1^\rho (\partial _y\zeta _n)^2(y)dy=\frac{1}{1-\lambda (n)^2}\int _1^\rho (\partial _y\eta _n)^2(y)dy\le \frac{1}{1-\lambda (n)^2} (2 \pi )^2, \end{aligned}$$

since $\partial _y\eta _n$ is the Fourier transform of $2 \pi i x \xi _n(x)$ whose $L^2$-norm is bounded by $2\pi$. We thus get

$$\begin{aligned} \left( \int _{\rho ^{-1}}^1(D_u\zeta _n)^2(\rho x) \rho dx\right) ^{\frac{1}{2}}\le \rho \frac{2\pi }{\sqrt{1-\lambda (n)^2}}. \end{aligned}$$

To estimate $\int _{\rho ^{-1}}^1(D_u\xi _n)(x)^2 dx$, we rewrite the equality (67) as follows

$$\begin{aligned} (\mathbf{W}f)(x) = -\left( 1-x^2\right) f''(x)+2 x f'(x)+4 \pi ^2 x^2 f(x), \end{aligned}$$

(166)

so that since $\xi _n$ is an eigenvector of $\mathbf{W}$ (i.e. $\mathbf{W}\xi _n=\chi _{2n}^{2\pi }\xi _n$), using the notations of [28], we get

$$\begin{aligned} D_u(\xi _n)(x)=\frac{1}{2} \left( 1-x^2\right) \xi _n''(x)+(\chi _{2n}^{2\pi }-2 \pi ^2 x^2) \xi _n(x). \end{aligned}$$

Assuming $n\ge 3$ to ensure $\chi _{2n}^{2\pi }\ge 2 \pi ^2$ one then derives

$$\begin{aligned} \Vert D_u(\xi _n)\Vert \le \chi _{2n}^{2\pi }+ \frac{1}{2}\left( \int _{0}^1 (\xi _n''(x))^2(1-x^2)^2dx\right) ^{\frac{1}{2}}. \end{aligned}$$

By [28] (Theorem 3.6) one has (note the different normalization of inner product due to (16))

$$\begin{aligned} \left( \int _{0}^1 (\xi _n''(x))^2(1-x^2)^2dx\right) ^{\frac{1}{2}}\le (2n)^2 + (6 \pi + 1)2n + 3(2\pi + 1)^2, \end{aligned}$$

(167)

while the eigenvalues $\chi _{2n}^{2\pi }$ fulfill (see op.cit. ): $\chi _{2n}^{2\pi }\le 2n(2n+1)+(2\pi )^2$. Thus, one obtains the inequality

$$\begin{aligned} \Vert D_u(\xi _n)\Vert \le 8 n^2 +(6 \pi + 2)2n+ 16 \pi ^2+12 \pi +1 \end{aligned}$$

and the following uniform bound (take $\rho \le 2$)

$$\begin{aligned} \vert A_n(\rho )\vert \le \frac{\lambda (n)}{1-\lambda (n)^2}4 \pi (8 n^2 +(6 \pi + 2)2n+ 16 \pi ^2+12 \pi +1),\qquad \forall \rho , \ 1\le \rho \le 2.\nonumber \\ \end{aligned}$$

(168)

We then consider the terms

$$\begin{aligned} B_n(\rho ):=\frac{\lambda (n)}{\sqrt{1-\lambda (n)^2}}\ \left( \rho ^{-1/2}(D_u\xi _n)(\rho ^{-1})\zeta _n(1)-\rho ^{1/2}\xi _n(1)(D_u\zeta _n)(\rho )\right) . \end{aligned}$$

By (101), one has $\vert \xi _n(1)\vert \le \sqrt{2n+\frac{1}{2}}$. One also has $\zeta _n=\frac{1}{\sqrt{1-\lambda (n)^2}}\,\eta _n$ and $\eta _n={{\mathbb {F}}}_{e_{{\mathbb {R}}}}\xi _n$ thus

$$\begin{aligned} (D_u\zeta _n)(\rho )=\frac{1}{\sqrt{1-\lambda (n)^2}} \ \rho \eta _n'(\rho )~\Longrightarrow ~ \vert (D_u\zeta _n)(\rho )\vert \le \frac{8 \pi }{\sqrt{1-\lambda (n)^2}},\qquad \forall \rho , \ 1\le \rho \le 2 \end{aligned}$$

using the equality $\eta _n'(\rho )=-4\pi \int _0^1 \sin (2\pi \rho x) \xi _n(x)xdx$ and Schwarz’s inequality. Hence

$$\begin{aligned} \frac{\lambda (n)}{\sqrt{1-\lambda (n)^2}}\bigg |\rho ^{1/2}\xi _n(1)(D_u\zeta _n)(\rho )\bigg |\le \frac{\lambda (n)}{ 1-\lambda (n)^2}8\pi \sqrt{2(2n+\frac{1}{2})}, \qquad \forall \rho , \ 1\le \rho \le 2. \end{aligned}$$

By (75) one gets: $\eta _n(x)=\lambda (n)\xi _n(x)$, for $x\in [0,1]$ thus one obtains by proportionality

$$\begin{aligned} (D_u\xi _n)(\rho ^{-1})\zeta _n(1)=\frac{1}{\sqrt{1-\lambda (n)^2}}(D_u\xi _n)(\rho ^{-1})\eta _n(1)=\frac{1}{\sqrt{1-\lambda (n)^2}}(D_u\eta _n)(\rho ^{-1})\xi _n(1) \end{aligned}$$

and the above bound for $\eta _n'(y)=-4\pi \int _0^1 \sin (2\pi y x) \xi _n(x)xdx$, applied for $y=\rho ^{-1}$ thus gives

$$\begin{aligned} \vert (D_u\xi _n)(\rho ^{-1})\zeta _n(1)\vert \le \frac{4 \pi }{\sqrt{1-\lambda (n)^2}}\sqrt{2n+\frac{1}{2}} \end{aligned}$$

so that

$$\begin{aligned} \frac{\lambda (n)}{\sqrt{1-\lambda (n)^2}}\vert \rho ^{-1/2}(D_u\xi _n)(\rho ^{-1})\zeta _n(1) \vert \le \frac{\lambda (n)}{ 1-\lambda (n)^2}4\pi \sqrt{2n+\frac{1}{2}},\qquad \forall \rho , \ 1\le \rho \le 2. \end{aligned}$$

The above inequalities then give

$$\begin{aligned} \vert B_n(\rho )\vert \le \frac{\lambda (n)}{1-\lambda (n)^2}\left( 8\pi \sqrt{2}+4\pi \right) \sqrt{2n+\frac{1}{2}}. \end{aligned}$$

(169)

We thus obtain

Lemma 6.15

(i) The series (99) of Proposition 5.3 is convergent and the remainder (after replacing the infinite sum by the sum of the first N terms) is majored as follows

$$\begin{aligned} \vert Q\epsilon (\rho )-\sum _0^N \frac{\lambda (k)}{\sqrt{1-\lambda (k)^2}} T_k(\rho )\vert \le \sum _{N+1}^\infty \frac{2^{2 n+2} \pi ^{2 n+\frac{3}{2}} p(n)((2 n)!)^2}{(4 n)! \Gamma \left( 2 n+\frac{3}{2}\right) } \end{aligned}$$

(170)

where $p(n)=16 n^2+8 (1+3 \pi ) n+(4+\sqrt{2}) \sqrt{4 n+1}+32 \pi ^2+24 \pi +2$.

(ii) For $N=10$, the remainder is less than $2.366 \times 10^{-12}$ for any $\rho \in [1,2]$:

$$\begin{aligned} \vert Q\epsilon (\rho )-\sum _0^{10} \frac{\lambda (k)}{\sqrt{1-\lambda (k)^2}} T_k(\rho )\vert \le 2.366 \times 10^{-12},\qquad \forall \rho \in [1,2]. \end{aligned}$$

(171)

Proof

(i)  follows from (168) and (169) which, together with (72), combine to yield for $n\ge 3$,

$$\begin{aligned}&\bigg |\frac{\lambda (n)}{\sqrt{1-\lambda (n)^2}} T_n(\rho )\bigg |\le 2\lambda (n)\left( \vert A_n(\rho )\vert +\vert B_n(\rho )\vert \right) \le \\&\quad \le \frac{2^{2 n+2} \pi ^{2 n+\frac{3}{2}} \left( 16 n^2+8 (1+3 \pi ) n+(4+\sqrt{2}) \sqrt{4 n+1}+32 \pi ^2+24 \pi +2\right) ((2 n)!)^2}{(4 n)! \Gamma \left( 2 n+\frac{3}{2}\right) } \end{aligned}$$

which gives (170).

(ii)To compute the upper bound of the right hand side of (170) for $N=10$, one splits the sum in two, using the simple estimate $p(n)\le 120 n^2$ for $n\ge 35$:

$$\begin{aligned} \frac{2^{2 n+2} \pi ^{2 n+\frac{3}{2}} p(n)((2 n)!)^2}{(4 n)! \Gamma \left( 2 n+\frac{3}{2}\right) }\le \frac{15\ 2^{2 n+4} n^2 \pi ^{2 n+\frac{1}{2}} ((2 n)!)^2}{(4 n)! \Gamma \left( 2 n+\frac{3}{2}\right) },\qquad \forall n\ge 35. \end{aligned}$$

With $\nu _n$ the right hand side of this inequality, one obtains the relation

$$\begin{aligned} \nu _{n+1}/\nu _n=\frac{8 \pi ^2 (n+1)^3 (2 n+1)}{n^2 (4 n+1) (4 n+3)^2 (4 n+5)}=\frac{\pi ^2}{16 n^2}+\frac{\pi ^2}{32 n^3}+O\left( n^{-4}\right) \end{aligned}$$

and $n^2\nu _{n+1}/\nu _n<1$ for all $n\ge 35$. One has $\nu _{35}\le 5 \times 10^{-81}$, and thus using the trivial bound by the geometric series one gets

$$\begin{aligned} \sum _{35}^\infty \nu _n\le \frac{1225}{1224} \nu _{35}\le 10^{-80}. \end{aligned}$$

One then simply computes the missing terms and they give

$$\begin{aligned} \sum _{11}^{34}\frac{2^{2 n+2} \pi ^{2 n+\frac{3}{2}} p(n)((2 n)!)^2}{(4 n)! \Gamma \left( 2 n+\frac{3}{2}\right) }\sim 2.365 \times 10^{-12} \end{aligned}$$

Thus combining the above inequalities we obtain (171).$\square$

Remark 6.16

For completeness, we give a short proof of an improved form of (167). As in [28] (Equation (3.26)) one has, using integration by parts, the identity, for $f\in C^\infty ([-1,1],{{\mathbb {R}}})$, and $c=2 \pi$, using (67)

$$\begin{aligned} \int _{-1}^1(\mathbf{W}f)^2(x)dx&=\int _{-1}^1(1-x^2)^2\vert f''(x)\vert ^2dx+2\int _{-1}^1(1-x^2)(1+c^2x^2)\vert f'(x)\vert ^2dx \\&\quad +c^2\int _{-1}^1(c^2 x^4+6x^2-2)\vert f(x)\vert ^2dx. \end{aligned}$$

Applying this to $f=\xi _n$ and using $\mathbf{W}\xi _n=\chi _{2n}^{2\pi }\xi _n$ one gets

$$\begin{aligned} \int _{-1}^1 (\xi _n''(x))^2(1-x^2)^2dx\le \int _{-1}^1(\mathbf{W}\xi _n)^2(x)dx+2 c^2\int _{-1}^1\xi _n(x)^2dx \end{aligned}$$

providing the following improvement of (167)

$$\begin{aligned} \left( \int _{0}^1 (\xi _n''(x))^2(1-x^2)^2dx\right) ^{\frac{1}{2}}\le \sqrt{(\chi _{2n}^{2\pi })^2+2c^2}\le 2n(2n+1)+(2\pi )^2(1+\sqrt{2}). \end{aligned}$$