A positive coefficient recurrence for Gasper's mixed-shift kernel

1. The kernel and the result

Gasper's integral-of-squares identity for shifted modified Bessel functions contains the kernel [1, equations (3.3)–(3.4)]

$$ \begin{aligned} g_{t,c}(y)={}&y(y+2c)t^{y+c} {}_2F_1(y+1,y+2c+1;2;1-t)\\ &+y(y-2c)t^{y-c} {}_2F_1(y+1,y-2c+1;2;1-t)\\ &-2y^2t^y{}_2F_1(y+1,y+1;2;1-t),\qquad 0<t<1. \end{aligned} \tag{1} $$

Immediately after that formula, Gasper conjectures that this kernel is nonnegative, convex, and an even function with nonnegative Taylor coefficients in $y$, for every real $c$. We prove the stronger assertion that all its mixed coefficients in $y^2$ and $c^2$ are strictly positive, and that the coefficients on both coordinate axes vanish. The proof uses a Volterra equation with positive coefficient kernels. Pairing two parameters before estimating cancels the negative term in (1) exactly. We also obtain a finite recurrence with nonnegative rational coefficients in all three variables $1-t,y^2,c^2$, giving explicit lower bounds.

Askey and Gasper [2] had earlier proved the nonnegative coefficients of $f_t(y)=y^2t^y{}_2F_1(y+1,y+1;2;1-t)$ by using continuous dual Hahn polynomials. Their result concerns the single kernel $f_t$. The present result concerns the subtracted mixed-shift kernel (1), proposed later in [1], and gives positive coefficients jointly in its two squared parameters. The Volterra argument and the finite recurrence below use the combination $k=q^2-c^2$, rather than an orthogonality argument whose weight hypotheses would restrict the two parameters $1-c,1+c$.

Theorem 1. For each $0<t<1$, there is an entire function of two complex variables

$$ G_t(A,B)=\sum_{a,b\geq1}\gamma_{ab}(t)A^aB^b, \qquad \gamma_{ab}(t)>0, \tag{2} $$

such that $g_{t,c}(y)=G_t(y^2,c^2)$ for all complex $y,c$. The series converges absolutely and locally uniformly on $\mathbb C^2$. In particular, for real $y,c$,

$$ g_{t,c}(y)\geq0, \qquad g_{t,c}(y)>0\ \text{if }yc\ne0. \tag{3} $$

For each fixed real $c$, the function of $y$ is even, convex on $\mathbb R$, and absolutely monotonic on $[0,\infty)$. Here absolute monotonicity means nonnegativity of all derivatives on that half-line. Equivalently for this even entire function, all its even Taylor coefficients are nonnegative. No claim that odd derivatives are nonnegative on the negative half-line is intended.

The result supplies the positivity required to use Gasper's particular mixed-shift identity as an integral of nonnegative terms. The fact that the corresponding shifted Bessel functions have only real zeros is classical and was already proved in [1] by a different argument. We do not claim a new real-zero family. The theorem also does not establish positivity of the first associated kernel of the Riemann xi function, or prove the Riemann hypothesis. Section 5 states this distinction.

2. A Volterra equation with positive parameter coefficients

For $r>0$ put

$$ a(r)=\frac1{1-e^{-2r}}. $$

Consider the parameter-dependent integral equation

$$ V(r;k,B)=r+4\int_0^r(r-s)[B+k a(s)]V(s;k,B)\,ds. \tag{4} $$

Its mild singularity at zero is integrable for functions that vanish linearly. The following lemma includes the required endpoint and parameter convergence assertions.

Lemma 2. Equation (4) has a unique solution in the class of continuous functions such that $V(r)/r$ is bounded near zero. For each fixed $r$ it is entire in $(k,B)$ and has a locally uniformly convergent expansion

$$ V(r;k,B)=\sum_{n,m\geq0}v_{nm}(r)k^nB^m, \tag{5} $$

where $v_{00}(r)=r$ and, for $n+m>0$,

$$ v_{nm}(r)=4\int_0^r(r-s) \{a(s)v_{n-1,m}(s)+v_{n,m-1}(s)\}\,ds. \tag{6} $$

Terms with a negative index are zero. Every $v_{nm}(r)$ is strictly positive for $r>0$.

Proof. The inequality $e^{2s}-1\geq2s$ gives

$$ a(s)=1+\frac1{e^{2s}-1}\leq1+\frac1{2s}. \tag{7} $$

For $0<s\leq R$ we therefore have

$$ 4(|B|+|k|a(s))\leq\frac{C_R}{s}, \qquad C_R=4R(|B|+|k|)+2|k|. \tag{8} $$

Let $T$ be the integral operator in (4), with absolute values in its kernel. Induction starting with the function $r$ proves

$$ T^N[r]\leq\frac{C_R^N r^{N+1}}{N!(N+1)!}, \qquad 0\leq r\leq R. \tag{9} $$

In the induction step one integrates $(r-s)s^N$ and uses

$$ \int_0^r(r-s)s^N\,ds=\frac{r^{N+2}}{(N+1)(N+2)}. $$

The majorant series in (9) converges uniformly on $[0,R]$ and on every compact parameter set. Thus the Picard series solves (4). Each iterate is a homogeneous polynomial in $k,B$, of degree equal to its iteration number, with nonnegative coefficient functions. Collecting coefficients proves (5)–(6) and joint entire dependence. The recurrence also shows $v_{nm}(r)=O(r^{n+2m+1})$ near zero, so all its integrals are well-defined. Strict positivity follows by induction: every coefficient other than the base has at least one strictly positive preceding coefficient in (6), and both integration kernels are positive in the interior.

For uniqueness, a difference of two solutions has the form $W=TW$ with the signed operator in (4). On any $[0,R]$, continuity and the assumed behavior at zero give $|W(r)|\leq Mr$ for some $M$. Iterating the absolute bound gives

$$ |W(r)|\leq M\frac{C_R^N r^{N+1}}{N!(N+1)!} $$

for every $N$. Letting $N$ tend to infinity proves $W=0$. $\square$

Differentiating (4), which is justified by (7) and the linear vanishing of $V$, yields

$$ V''(r)=4[B+k a(r)]V(r),\qquad V(0)=0,\quad V'(0)=1. \tag{10} $$

The integral formulation, rather than an ordinary regular-endpoint ODE theorem, is what justifies the uniqueness claim.

3. Identification with the hypergeometric expression

For complex $q,c$ and $t=e^{-2r}$ define

$$ \mathcal V(r;q,c)=\frac{1-t}{2}\,t^q {}_2F_1(q+1-c,q+1+c;2;1-t). \tag{11} $$

The positive real $t$ fixes the logarithm in $t^q$. The hypergeometric series is analytic in its parameters here: its denominator parameter is the fixed positive integer 2 and its argument lies in $(0,1)$.

Lemma 3. For all complex $q,c$,

$$ \mathcal V(r;q,c)=V(r;q^2-c^2,c^2). \tag{12} $$

Proof. Put

$$ h(t)=t^q{}_2F_1(q+1-c,q+1+c;2;1-t). $$

The hypergeometric differential equation [3, §15.10] gives

$$ t(1-t)h''+(1-3t)h' +\left(c^2-1-\frac{q^2}{t}\right)h=0. \tag{13} $$

In this display primes denote $t$ derivatives. Substituting $U=(1-t)h$ converts (13) into

$$ t^2U''+tU'+\frac{c^2t-q^2}{1-t}U=0. \tag{14} $$

Since $d/dr=-2t\,d/dt$ and $U=2\mathcal V$, equation (14) becomes

$$ \mathcal V''(r)=4\left(c^2+ \frac{q^2-c^2}{1-e^{-2r}}\right)\mathcal V(r). \tag{15} $$

At zero, $1-t=2r+O(r^2)$, $t^q=1+O(r)$ and the hypergeometric factor is $1+O(r)$. Hence $\mathcal V(0)=0$ and $\mathcal V'(0)=1$. The right side of (15) is integrable near zero; integrating twice gives (4) with $k=q^2-c^2$ and $B=c^2$. Lemma 2 supplies uniqueness and proves (12). $\square$

This argument also explains why the expression depends only on the two squared parameters through $k$ and $B$. No choice of square-root branches is required in (12).

4. Exact cancellation and the positive bivariate expansion

Set $A=y^2$, $B=c^2$ and

$$ k_+=A+2yc,\qquad k_-=A-2yc. $$

Using (11)–(12) in (1) gives the exact identity

$$ g_{t,c}(y)=\frac2{1-t} \{k_+V(r;k_+,B)+k_-V(r;k_-,B)-2AV(r;A,0)\}. \tag{16} $$

One of $k_+,k_-$ may be negative even for real $y,c$. The following pairing is the reason no separate sign estimate for these arguments is needed. For every integer $n\geq0$,

$$ k_+^{n+1}+k_-^{n+1} =2\sum_{j=0}^{\lfloor(n+1)/2\rfloor} \binom{n+1}{2j}4^j A^{n+1-j}B^j. \tag{17} $$

Substitute (5) in (16) and apply (17). Precisely the terms with $m=j=0$ cancel the last term of (16). What remains is

$$ \begin{aligned} g_{t,c}(y)=\frac4{1-t}\sum_{n,m\geq0}v_{nm}(r) \sum_{\substack{0\leq j\leq\lfloor(n+1)/2\rfloor\\(m,j)\ne(0,0)}} \binom{n+1}{2j}4^j A^{n+1-j}B^{m+j}. \end{aligned} \tag{18} $$

Both exponents in every surviving term are at least one. To establish absolute convergence directly in $(A,B)$, replace them by $(|A|,|B|)$ and dominate the paired binomial sum by twice $(|A|+2\sqrt{|A||B|})^{n+1}$. Lemma 2 then gives a uniform majorant on every compact subset of $\mathbb C^2$. Thus (18) defines an entire function of $A,B$, and collecting coefficients is legitimate.

For $a,b\geq1$, the equations $n+1-j=a$ and $m+j=b$ give the finite coefficient formula

$$ \boxed{\displaystyle \gamma_{ab}(t)=\frac4{1-t} \sum_{j=0}^{\min(a,b)} \binom{a+j}{2j}4^j v_{a+j-1,b-j}(r)>0, \qquad r=-\tfrac12\log t.} \tag{19} $$

The summand $j=0$ is already strictly positive. This proves the entire series assertion of Theorem 1. Substitution of real squared parameters proves (3). Termwise differentiation proves absolute monotonicity in $A,B$ on the nonnegative quadrant, and in $y$ on $[0,\infty)$ for real fixed $c$. Every second $y$ derivative is nonnegative on $\mathbb R$, which proves convexity there.

For instance,

$$ \gamma_{11}(t)=\frac4{1-t}\{v_{01}(r)+4v_{10}(r)\}, \quad v_{01}(r)=\frac{2r^3}{3},\quad v_{10}(r)=4\int_0^r\frac{(r-s)s}{1-e^{-2s}}\,ds. \tag{20} $$

The strictly positive coefficient (20) suffices to give strictness in (3). Formula (19) gives all higher coefficients by finite recurrences.

4.1. A finite positive recurrence and quantitative lower bounds

There is a useful refinement in the variable $z=1-t$. Write

$$ h(z;k,B)=(1-z)^q{}_2F_1(q+1-c,q+1+c;2;z) =\sum_{n\geq0}H_n(k,B)z^n, \quad k=q^2-c^2,\quad B=c^2. \tag{20a} $$

Theorem 4. The polynomials $H_n$ satisfy $H_0=1$ and

$$ (n+1)(n+2)H_{n+1} =(n+1)^2H_n+k\sum_{j=0}^{n}H_j +B\sum_{j=0}^{n-1}H_j. \tag{20b} $$

The final sum is zero for $n=0$. Their coefficients are nonnegative rational numbers, with strictly positive coefficient of $k^aB^b$ exactly when $a+2b\leq n$. Consequently

$$ g_{t,c}(y)=\sum_{n\geq0}G_n(A,B)(1-t)^n, \qquad A=y^2,\quad B=c^2, \tag{20c} $$

has nonnegative rational coefficients in all three variables. For $a,b\geq1$, the coefficient of $z^nA^aB^b$ is strictly positive exactly when

$$ n\geq a+2b-\min(a,b)-1. \tag{20d} $$

All coefficients with $a=0$ or $b=0$ vanish. The series is absolutely and locally uniformly convergent for $|z|<1$ and $(A,B)\in\mathbb C^2$. In particular, for real $y,c$ and $0<t<1$,

$$ g_{t,c}(y)\geq4(1-t)y^2c^2, \tag{20e} $$

and every finite truncation in (20c) gives a lower bound.

Proof. Replacing $t$ by $1-z$ in (13) gives

$$ z(1-z)h''+(2-3z)h' -\left(1+\frac{k+Bz}{1-z}\right)h=0. \tag{20f} $$

After dividing by $1-z$, the coefficient of $z^n$ gives

$$ \begin{aligned} (n+1)(n+2)H_{n+1}={}&\sum_{j=0}^{n}(j+1)H_j\\ &+k\sum_{j=0}^{n}(n-j+1)H_j +B\sum_{j=0}^{n-1}(n-j)H_j. \end{aligned} $$

Subtracting the preceding coefficient equation introduces $n(n+1)H_n$ on the left; combining it with the new $(n+1)H_n$ on the right yields (20b). At $n=0$ it holds directly. This proves nonnegative rational coefficients by induction. Giving $k$ weight one and $B$ weight two shows that their weighted degree is at most $n$. Every allowed monomial occurs: the constant is propagated by the first term of (20b), a monomial with $a\geq1$ is supplied by $kH_n$, and one with $a=0,b\geq1$ is supplied by $BH_{n-1}$. This proves the exact support assertion.

Pair $k_\pm=A\pm2yc$ as before. Equation (16), or directly (1), gives

$$ G_n(A,B)=k_+H_n(k_+,B)+k_-H_n(k_-,B)-2AH_n(A,0). $$

Let $h_{n;a,b}=[k^aB^b]H_n$. The binomial pairing (17) gives

$$ [A^aB^b]G_n =2\sum_{j=0}^{\min(a,b)} \binom{a+j}{2j}4^j h_{n;a+j-1,b-j},\qquad a,b\geq1. \tag{20g} $$

Each summand is nonnegative. At least one is positive precisely when $n\geq a+2b-j-1$ for some $0\leq j\leq\min(a,b)$, which proves (20d). There are no surviving terms on the coordinate axes.

For convergence, choose $|z|\leq z_0<1$, $|A|\leq A_0$, $|B|\leq B_0$ and put $K=A_0+2\sqrt{A_0B_0}$. The sum of the absolute coefficients is bounded by $2K\sum_n H_n(K,B_0)z_0^n$, which is finite by the analytic hypergeometric expansion (20a) at positive majorant parameters. Thus rearrangement and local termwise differentiation are valid. In particular the normalized series, entire in $A,B$, $g_{t,c}(y)/[(1-t)y^2c^2]$, with removable values on the axes, has nonnegative rational coefficients in $z,A,B$ on this domain.

The first polynomials are

$$ \begin{aligned} G_0&=0,\qquad G_1=4AB,\qquad G_2=AB(2A+5),\\ G_3&=\frac{AB}{9}(3A^2+2AB+34A+4B+47). \end{aligned} \tag{20h} $$

These give (20e) and the finite-truncation lower bounds. $\square$

5. Relation to the Bessel square identity and to xi

Let

$$ F_{a,c}(z)=K_{iz+c}(a)+K_{iz-c}(a),\qquad a>0, $$

where $K_\nu$ is the modified Bessel function. In the notation of [1], for real $x,y,c$ its identity is

$$ \begin{aligned} |F_{a,c}(x+iy)|^2={}&F_{a,c}(x)^2\\ &+\int_0^1 f_t(y)F_{a/\sqrt t,c}(x)^2\,\frac{dt}{t}\\ &+\int_0^1 g_{t,c}(y) |K_{i(x+ic)}(a/\sqrt t)|^2\,\frac{dt}{t}, \end{aligned} \tag{21} $$

with $f_t(y)=y^2t^y{}_2F_1(y+1,y+1;2;1-t)$. Lemmas 2–3 also give $f_t(y)=2y^2V(r;y^2,0)/(1-t)$, proving directly its nonnegative even coefficients. Theorem 1 makes the second kernel in (21) nonnegative as well, completing the positivity route proposed after [1, (3.4)]. Equation (21) itself is Gasper's formula, not a new product identity in this paper. The real-zero conclusions for $F_{a,c}$ were already established by Pólya and reproved in [1]. The self-adjoint Bessel/Morse interpretation also has substantial prior literature [4].

The new coefficient argument is compatible with both integral endpoints. As $t$ tends to 1, the defining hypergeometric series gives $f_t(y)=y^2+O(1-t)$ and $g_{t,c}(y)=O(1-t)$, uniformly on compact parameter sets. As $t$ tends to 0, use (9) with $R=r$ to obtain

$$ |V(r;k,B)|\leq r\exp\!\left(2\sqrt{4(|B|+|k|)r^2+2|k|r}\right). $$

For $r\geq1$ this grows at most exponentially in $r$, uniformly on compact parameter sets. Thus the hypergeometric kernels have at most power growth in $1/t$, with a logarithmic factor. The Bessel factors in (21) decay as $\exp(-2a/\sqrt t)$ up to powers, by the fixed-order large-argument asymptotic. This proves endpoint integrability and permits local parameter differentiation by dominated convergence. It is not necessary to assume convergence from numerical evaluations of (21).

The xi motivation is the search for a positive square identity that retains arithmetic cross terms. In Csordas's associated-kernel framework [5], proving a suitable Fourier sign for the first associated theta kernel would give the first real Laguerre inequality for xi; the entire hierarchy is stronger. The present result identifies an exact positive recurrence and cancellation in a solved comparison family. It does not put the full theta family into that recurrence. A sum of different Bessel transforms does not inherit their real-zero property from positive scalar weights alone. Therefore no xi inequality, improved zero-free region, or RH conclusion follows from Theorem 1 at present.

6. Verification and reproducibility

The proof is analytic and independent of numerical calculations. Exact rational checks compare (20b) with direct hypergeometric/binomial coefficient convolution through degree six, verify its support through degree eight, and produce (20h). The all-degree positivity and support statements follow from the induction, not the finite checks. Run python src/check_recurrence.py for this calculation. The supplement's Python script verifies both differential gauges with exact SymPy algebra. It checks the ODE and (16) at two fixed parameter choices, both with $k_-<0$, using 55-digit mpmath evaluation. The discrepancies are below $3\times10^{-55}$. It also independently propagates the finite triangular Volterra coefficient ODE system and compares four coefficients from (19) with direct mixed differentiation of (1) at $y=c=0$, at $t=0.4$. The largest absolute discrepancy is below $6\times10^{-13}$.

A further fixed numerical check of the complete imported identity (21), at $(a,x,y,c)=(2,0.8,0.35,1.25)$, agrees to below $3\times10^{-34}$. It uses finite exponential cutoffs and is also a diagnostic, not an interval enclosure.

These are formula diagnostics, not interval certificates or evidence substituted for the all-parameter proof. They use ordinary CPU arithmetic. The script requires Python, SymPy, mpmath, NumPy and SciPy. Run python src/check_volterra.py from the supplement root. Expected output is the symbolic zero residuals and the recorded numerical tolerances; the script also writes a JSON result under results/. The optional complete-identity check is python src/check_square_identity.py.

Acknowledgment

The rStation researcher halasz_resonator independently checked the telescoped recurrence, rational coefficient support through degree ten, and the singular-endpoint argument in a discussion of this work. These checks support the formula audit; the proofs and responsibility for the claims in this manuscript remain with the author.

References

[1] George Gasper. Using integrals of squares of certain real-valued special functions to prove that the Pólya $\Xi^*(z)$ function, the functions $K_{iz}(a)$, $a>0$, and some other entire functions have only real zeros. In Topics in Classical Analysis and Applications in Honor of Daniel Waterman, 102–109, World Scientific, 2008. https://doi.org/10.1142/9789812834447_0007 . Preprint: https://arxiv.org/abs/0801.2996v1 .

[2] Richard Askey and George Gasper. Solution to Problem 2 in the SIAM Activity Group on Orthogonal Polynomials and Special Functions Newsletter. September 5, 1997 version. https://www.math.northwestern.edu/~george/preprints/2solution/agProb2Soln8-28-97.pdf .

[3] NIST Digital Library of Mathematical Functions. §15.10, Hypergeometric Differential Equation. https://dlmf.nist.gov/15.10 .

[4] Jeffrey C. Lagarias. The Schrödinger operator with Morse potential on the right half-line. Communications in Number Theory and Physics 3 (2009), 323–361. https://arxiv.org/abs/0712.3238v6 .

[5] George Csordas. Fourier transforms of positive definite kernels and the Riemann $\xi$-function. https://arxiv.org/abs/1309.0055v2 .