Approximating the little Grothendieck problem over the orthogonal and unitary groups

Abstract

The little Grothendieck problem consists of maximizing $\sum _{ij}C_{ij}x_ix_j$ for a positive semidefinite matrix C, over binary variables $x_i\in \{\pm 1\}$. In this paper we focus on a natural generalization of this problem, the little Grothendieck problem over the orthogonal group. Given $C\in \mathbb {R}^{dn\times dn}$ a positive semidefinite matrix, the objective is to maximize $\sum _{ij}{\text {tr}}\left( C^T_{ij}O_iO_j^T\right)$ restricting $O_i$ to take values in the group of orthogonal matrices $\mathcal {O}_d$, where $C_{ij}$ denotes the (ij)-th $d\times d$ block of C. We propose an approximation algorithm, which we refer to as Orthogonal-Cut, to solve the little Grothendieck problem over the group of orthogonal matrices $\mathcal {O}_d$ and show a constant approximation ratio. Our method is based on semidefinite programming. For a given $d\ge 1$, we show a constant approximation ratio of $\alpha _{\mathbb {R}}(d)^2$, where $\alpha _{\mathbb {R}}(d)$ is the expected average singular value of a $d\times d$ matrix with random Gaussian $\mathcal {N}\left( 0,\frac{1}{d}\right)$ i.i.d. entries. For $d=1$ we recover the known $\alpha _{\mathbb {R}}(1)^2=2/\pi$ approximation guarantee for the classical little Grothendieck problem. Our algorithm and analysis naturally extends to the complex valued case also providing a constant approximation ratio for the analogous little Grothendieck problem over the Unitary Group $\mathcal {U}_d$. Orthogonal-Cut also serves as an approximation algorithm for several applications, including the Procrustes problem where it improves over the best previously known approximation ratio of $\frac{1}{2\sqrt{2}}$. The little Grothendieck problem falls under the larger class of problems approximated by a recent algorithm proposed in the context of the non-commutative Grothendieck inequality. Nonetheless, our approach is simpler and provides better approximation with matching integrality gaps. Finally, we also provide an improved approximation algorithm for the more general little Grothendieck problem over the orthogonal (or unitary) group with rank constraints, recovering, when $d=1$, the sharp, known ratios.

Appendices

Appendix 1: Technical proofs—analysis of algorithm for the Stiefel Manifold setting

Lemma 17

Let $r\ge d$. Let G be a $d\times r$ Gaussian random matrix with real valued i.i.d. $\mathcal {N}\left( 0,\frac{1}{r}\right)$ entries and let $\alpha _{\mathbb {R}}(d,r)$ as defined in Definition 11. Then,

$$\begin{aligned} \mathbb {E}\left( \mathcal {P}_{d,r}(G) G^T\right) = \mathbb {E}\left( G\mathcal {P}_{d,r}(G)^T\right) = \alpha _{\mathbb {R}}(d,r)I_{d\times d}. \end{aligned}$$

Furthermore, if G is a $d\times r$ Gaussian random matrix with complex valued i.i.d. $\mathcal {N}\left( 0,\frac{1}{r}\right)$ entries and $\alpha _{\mathbb {C}}(d,r)$ the analogous constant (Definition 11), then

$$\begin{aligned} \mathbb {E}\left( \mathcal {P}_{d,r}(G) G^H\right) = \mathbb {E}\left( G\mathcal {P}_{d,r}(G)^H\right) = \alpha _{\mathbb {C}}(d,r)I_{d\times d}. \end{aligned}$$

The proof of this Lemma is a simple adaptation of the proof of Lemma 6.

Proof

We restrict the presentation to the real case. As before, all the arguments are equivalent to the complex case, replacing all transposes with Hermitian adjoints and $\alpha _{\mathbb {R}}(d,r)$ with $\alpha _{\mathbb {C}}(d,r)$.

Let $G = U [\Sigma \ 0] V^T$ be the singular value decomposition of G. Since $GG^T = U\Sigma ^2 U^T$ is a Wishart matrix, it is well known that its eigenvalues and eigenvectors are independent and U is distributed according to the Haar measure in $\mathcal {O}_d$ (see e.g. Lemma 2.6 in [38]). To resolve ambiguities, we consider $\Sigma$ ordered such that $\Sigma _{11} \ge \Sigma _{22} \ge \cdots \ge \Sigma _{dd}$.

Let $Y = \mathcal {P}_{(d,r)}(G) G^T$. Since

$$\begin{aligned} \mathcal {P}_{(d,r)}(G) = \mathcal {P}_{(d,r)}(U [\Sigma \ 0] V^T) = U[I_{d\times d} \ 0]V^T, \end{aligned}$$

we have

$$\begin{aligned} Y = \mathcal {P}_{(d,r)}(U [\Sigma \ 0]V^T) (U [\Sigma \ 0] V^T)^T = U [I_{d\times d} \ 0]V^T V \Sigma U^T = U \Sigma U^T. \end{aligned}$$

Note that $G\mathcal {P}_{(d,r)}(G)^T = U \Sigma U^T = Y$.

Since $Y_{ij} = u_i \Sigma u_j^T$, where $u_1,\ldots ,u_d$ are the rows of U, and U is distributed according to the Haar measure, we have that $u_j$ and $-u_j$ have the same distribution conditioned on any $u_i$, for $i\ne j$, and $\Sigma$. This implies that, if $i\ne j, Y_{ij} = u_i \Sigma u_j^T$ is a symmetric random variable, and so $\mathbb {E}Y_{ij} = 0$. Also, $u_i\sim u_j$ implies that $Y_{ii}\sim Y_{jj}$. This means that $\mathbb {E}Y = c I_{d\times d}$ for some constant c. To obtain c,

$$\begin{aligned} c = c\frac{1}{d}{\text {tr}}(I_{d\times d}) = \frac{1}{d}\mathbb {E}{\text {tr}}(Y) = \frac{1}{d}\mathbb {E}{\text {tr}}(U \Sigma U^T) = \frac{1}{d}\mathbb {E}\sum _{k=1}^n\sigma _{k}(G) = \alpha _{\mathbb {R}}(d,r), \end{aligned}$$

which shows the lemma. $\square$

Lemma 18

Let $r\ge d$. Let $M,N\in \mathbb {R}^{d\times nd}$ such that $MM^T = NN^T = I_{d \times d}$. Let $R \in \mathbb {R}^{nd \times r}$ be a Gaussian random matrix with real valued i.i.d. entries $\mathcal {N}\left( 0, \frac{1}{r} \right)$. Then

$$\begin{aligned} \mathbb {E}\left[ \mathcal {P}_{(d,r)}(M R) (N R)^T\right] = \mathbb {E}\left[ (M R) \mathcal {P}_{(d,r)}(N R)^T\right] = \alpha _{\mathbb {R}}(d,r) MN^T, \end{aligned}$$

where $\alpha _{\mathbb {R}}(d,r)$ is the constant in Definition 11.

Analogously, if $M,N\in \mathbb {C}^{d\times nd}$ such that $MM^H = NN^H = I_{d \times d}$, and $R \in \mathbb {C}^{nd \times r}$ is a Gaussian random matrix with complex valued i.i.d. entries $\mathcal {N}\left( 0, \frac{1}{r} \right)$, then

$$\begin{aligned} \mathbb {E}\left[ \mathcal {P}_{(d,r)}(M R) (N R)^H\right] = \mathbb {E}\left[ (M R) \mathcal {P}_{(d,r)}(N R)^H\right] = \alpha _{\mathbb {C}}(d,r) MN^H, \end{aligned}$$

where $\alpha _{\mathbb {C}}(d,r)$ is the constant in Definition 11.

Similarly to above, the proof of this Lemma is a simple adaptation of the proof of Lemma 5.

Proof

We restrict the presentation of proof to the real case. Nevertheless, all the arguments trivially adapt to the complex case by, essentially, replacing all transposes with Hermitian adjoints and $\alpha _{\mathbb {R}}(d)$ and $\alpha _{\mathbb {R}}(d,r)$ with $\alpha _{\mathbb {C}}(d)$ and $\alpha _{\mathbb {C}}(d,r)$.

Let $A = \left[ M^T\text { } N^T\right] \in \mathbb {R}^{dn\times 2d}$ and $A=QB$ be the QR decomposition of A with $Q\in \mathbb {R}^{nd\times nd}$ an orthogonal matrix and $B \in \mathbb {R}^{nd \times 2d}$ upper triangular with non-negative diagonal entries; note that only the first 2d rows of B are nonzero. We can write

$$\begin{aligned} Q^TA = B = \begin{bmatrix} B_{11}&\quad B_{12}\\ 0_d&\quad B_{22}\\ 0_d&\quad 0_d\\\ \vdots&\quad \vdots \\ 0_d&\quad 0_d \end{bmatrix} \in \mathbb {R}^{dn \times 2d}, \end{aligned}$$

where $B_{11}\in \mathbb {R}^{d\times d}$ and $B_{22}\in \mathbb {R}^{d\times d}$ are upper triangular matrices with non-negative diagonal entries. Since

$$\begin{aligned} (Q^T M^T)^T_{11}(Q^T M^T)_{11}= & {} (Q^T M^T)^T(Q^T M^T) = M Q Q^T M^T = M I_{nd \times nd} M^T\\= & {} M M^T = I_{d \times d}, \end{aligned}$$

$B_{11} = (Q^T M^T)_{11}$ is an orthogonal matrix, which together with the non-negativity of the diagonal entries (and the fact that $B_{11}$ is upper-triangular) forces $B_{11}$ to be $B_{11} = I_{d\times d}$.

Since R is a Gaussian matrix and Q is an orthogonal matrix, $QR \sim R$ which implies

$$\begin{aligned} \mathbb {E}\left[ \mathcal {P}_{(d,r)}(M R) (N R)^T\right] = \mathbb {E}\left[ \mathcal {P}_{(d,r)}(M Q R)(NQ R)^T \right] . \end{aligned}$$

Since $MQ=[B_{11}^T, 0_{d\times d},\ldots ,0_{d\times d}] = [I_{d\times d}, 0_{d\times d},\ldots ,0_{d\times d}]$ and $NQ = [B_{12}^T, B_{22}^T, 0_{d\times d},\ldots ,0_{d\times d}]$,

$$\begin{aligned} \mathbb {E}\left[ \mathcal {P}_{(d,r)}(M R) (N R)^T\right] = \mathbb {E}\left[ \mathcal {P}_{(d,r)}(R_1)(B_{12}^T R_1 + B_{22}^T R_2)^T\right] , \end{aligned}$$

where $R_1$ and $R_2$ are the first two $d\times r$ blocks of R. Since these blocks are independent, the second term vanishes and we have

$$\begin{aligned} \mathbb {E}\left[ \mathcal {P}_{(d,r)}(M R) (N R)^T\right] = \mathbb {E}\left[ \mathcal {P}_{(d,r)}(R_1) R_1^T\right] B_{12}. \end{aligned}$$

The Lemma now follows from using Lemma 17 to obtain $\mathbb {E}\left[ \mathcal {P}_{(d,r)}(R_1) R_1^ T\right] = \alpha _{\mathbb {R}}(d,r)I_{d\times d}$ and noting that $B_{12} = (Q^TM^T)^T(Q^TN^T) = MN^T$.

The same argument, with $Q'B'$ the QR decomposition of $A' = \left[ N^T M^T\right] \in \mathbb {R}^{dn\times 2d}$ instead, shows

$$\begin{aligned} \mathbb {E}\left[ (M R) \mathcal {P}_{(d,r)}(N R)^T\right] = \mathbb {E}\left[ R_1\mathcal {P}_{(d,r)}(R_1)^T\right] MN^T = \alpha _{\mathbb {R}}(d,r)MN^T. \end{aligned}$$

$\square$

Appendix 2: Bounds for the average singular value

Lemma 19

Let $G_{\mathbb {C}} \in \mathbb {C}^{d\times d}$ be a Gaussian random matrix with i.i.d. complex valued $\mathcal {N}(0,d^{-1})$ entries and define $\alpha _{\mathbb {C}}(d):= \mathbb {E}\left[ \frac{1}{d} \sum _{j=1}^d \sigma _j(G_{\mathbb {C}})\right]$. We have the following bound

$$\begin{aligned} \alpha _{\mathbb {C}}(d) \ge \frac{8}{3\pi } - \frac{5.05}{d}. \end{aligned}$$

Proof

We express $\alpha _{\mathbb {C}}(d)$ as sums and products of Gamma functions and then use classical bounds to obtain our result.

Recall that from equation (16),

$$\begin{aligned} \alpha _{\mathbb {C}}(d) = d^{-3/2} \sum _{n=0}^{d-1} T_n, \end{aligned}$$

(26)

where

$$\begin{aligned} T_n = \int _0^{\infty } x^{1/2} e^{-x} L_n(x)^2 dx, \end{aligned}$$

and $L_n(x)$ is the nth Laguerre polynomial,

$$\begin{aligned} L_n(x) = \sum _{k=0}^n \left( {\begin{array}{c}n\\ k\end{array}}\right) \frac{(-1)^k}{k!} x^k. \end{aligned}$$

This integral can be expressed as [see [16] section 7.414 equation 4(1)]

$$\begin{aligned} T_n = \frac{ \Gamma (n+3/2)}{\Gamma (n+1)} \sum _{m=0}^n \frac{\left( \frac{-1}{2}\right) _m (-n)_m}{(m!)^2 (-n-\frac{1}{2})_m}, \end{aligned}$$

(27)

where $(x)_m$ is the Pochhammer symbol

$$\begin{aligned} (x)_m = \frac{\Gamma (x+m)}{\Gamma (x)}. \end{aligned}$$

The next lemma states a couple basic facts about the Gamma function that we will need in the subsequent computations. $\square$

Lemma 20

The Gamma function satisfies the following inequalities:

$$\begin{aligned} \frac{1}{\sqrt{n}} \le \frac{\Gamma (n)}{\Gamma (n+1/2)} \le \frac{1}{\sqrt{n-1/2}} \end{aligned}$$

$$\begin{aligned} \sqrt{n} \le \frac{\Gamma (n+1)}{\Gamma (n+1/2)} \le \sqrt{n+1/2}. \end{aligned}$$

Proof

See [4] page 255. $\square$

We want to bound the summation in (27), which we rewrite as

$$\begin{aligned} \sum _{m=0}^n \frac{\left( \frac{-1}{2}\right) _m (-n)_m}{(m!)^2 (-n-\frac{1}{2})_m}= & {} \sum _{m=0}^{\infty } \frac{(\frac{-1}{2})_m^2}{(m!)^2} - \sum _{m= n+1}^{\infty }\frac{(\frac{-1}{2})_m^2}{(m!)^2}\\&-\sum _{m=0}^n \frac{(\frac{-1}{2})_m^2}{(m!)^2} \left( 1 - \frac{(-n)_m}{(-n-\frac{1}{2})_m} \right) . \end{aligned}$$

For simplicity define

$$\begin{aligned} (I):= & {} \sum _{m=0}^{\infty } \frac{(\frac{-1}{2})^2_m}{(m!)^2}\\ (II):= & {} \sum _{m= n+1}^{\infty }\frac{(\frac{-1}{2})_m^2}{(m!)^2}\\ (III):= & {} \sum _{m=0}^n \frac{(\frac{-1}{2})_m^2}{(m!)^2} \left( 1 - \frac{(-n)_m}{(-n-\frac{1}{2})_m} \right) , \end{aligned}$$

so that (27) becomes

$$\begin{aligned} T_n = \frac{\Gamma (n+3/2)}{\Gamma (n+1)} ((I) + (II) + (III)). \end{aligned}$$

The first term we can compute explicitly (see [16]) as

$$\begin{aligned} (I) = \frac{4}{\pi }. \end{aligned}$$

For the second term we use the fact that $(\frac{-1}{2})_m = \Gamma (m-1/2)/\Gamma (-1/2)$ to get

$$\begin{aligned} (II)= \sum _{m = n+1}^{\infty } \frac{1}{\Gamma (-1/2)^2} \frac{\Gamma (m-1/2)^2}{\Gamma (m+1)^2} = \frac{1}{4\pi } \sum _{m=n+1}^{\infty } \frac{\Gamma (m-1/2)^2}{\Gamma (m+1)^2}. \end{aligned}$$

Using the first inequality in Lemma 20 and the multiplication formula for the Gamma function,

$$\begin{aligned} \frac{\Gamma (m-1/2)}{\Gamma (m+1)} = \frac{1}{m-1/2} \frac{\Gamma (m+1/2)}{\Gamma (m+1)} \le \frac{1}{(m-1/2)\sqrt{m}} \end{aligned}$$

so we have

$$\begin{aligned} (II) \le \frac{1}{4\pi } \sum _{m=n+1}^{\infty } \frac{1}{(m-1/2)^2 m} \le \frac{1}{4\pi } \int _{n-1/2}^{\infty } \frac{1}{x^3} dx = \frac{1}{2\pi (2n-1)^2}. \end{aligned}$$

For the third term, we use the formula $(x)_m = \frac{\Gamma (x+n)}{\Gamma (x)}$ to deduce

$$\begin{aligned} (III)&= \sum _{m=0}^n \frac{(\frac{-1}{2})_m^2}{(m!)^2} \left( 1 - \frac{(-n)_m}{(-n-\frac{1}{2})_m} \right) \\&= \frac{1}{4\pi } \sum _{m=0}^n \frac{\Gamma (m-1/2)^2}{\Gamma (m+1)^2} \left( 1 - \frac{\Gamma (n+1) \Gamma (n-m+3/2)}{\Gamma (n+3/2) \Gamma (n-m+1)}\right) \nonumber \\&= \frac{\Gamma (n+1)}{\Gamma (n+3/2)} \frac{1}{4\pi } \sum _{m=0}^{n} \frac{\Gamma (m-1/2)^2}{\Gamma (m+1)^2} \left( \frac{\Gamma (n+3/2)}{\Gamma (n+1)} - \frac{\Gamma (n-m+3/2)}{\Gamma (n-m+1)}\right) . \end{aligned}$$

Using the second bound in Lemma 20,

$$\begin{aligned} \frac{\Gamma (n-m+3/2)}{\Gamma (n-m+1)} \ge \sqrt{n-m+1/2}, \end{aligned}$$

and also

$$\begin{aligned} \frac{\Gamma (n+3/2)}{\Gamma (n+1)} \le \sqrt{n+1}, \end{aligned}$$

so that

$$\begin{aligned} (III) \le \frac{1}{4\pi \sqrt{n+1/2}} \sum _{m=0}^n \left( \frac{1}{(m-1/2)\sqrt{m+1/2}}\right) ^2 \left( \sqrt{n+1} - \sqrt{n-m+1/2}\right) . \end{aligned}$$

If we multiply top and bottom by $\sqrt{n+1} + \sqrt{n-m+1/2}$ and use the fact that

$$\begin{aligned} \frac{m+1/2}{\sqrt{n+1} + \sqrt{n-m+1/2}} \le \frac{m+1/2}{\sqrt{n+1}}, \end{aligned}$$

then

$$\begin{aligned} (III)&\le \frac{1}{4\pi \sqrt{n+1/2}} \sum _{m=0}^n \frac{1}{(m-1/2)^2 (m+1/2)} \frac{m+1/2}{\sqrt{n+1}}\\&\le \frac{1}{2\pi (n+1)} \sum _{m=0}^n \frac{1}{(m-1/2)^2}\\&\le \frac{1}{n+1} \frac{8+\pi ^2}{2\pi }\\&\le \frac{3}{n+1}. \end{aligned}$$

Combining our bounds for (I), (II) and (III),

$$\begin{aligned} T_n&= \frac{\Gamma (n+3/2)}{\Gamma (n+1)} \left[ (I) - (II)-(III)\right] \\&\ge \frac{\Gamma (n+3/2)}{\Gamma (n+1)}\left( \frac{4}{\pi } - \frac{1}{2\pi (2n-1)^2} - \frac{3}{n+1}\right) \\&\ge \sqrt{n+1/2} \left( \frac{4}{\pi } - \frac{1}{2\pi (2n-1)^2} - \frac{3}{n+1}\right) , \end{aligned}$$

and by (26),

$$\begin{aligned} \alpha _{\mathbb {C}}(d) \ge \frac{1}{d^{3/2}} \sum _{n=1}^{d-1} \sqrt{n+1/2} \left( \frac{4}{\pi } - \frac{1}{2\pi (2n-1)^2} - \frac{3}{n+1}\right) . \end{aligned}$$

The term $\frac{1}{d^{3/2}} \sum _{n=1}^{d-1} 4 \sqrt{n+1/2}/ \pi$ is the main term and can be bounded below by

$$\begin{aligned} \frac{1}{d^{3/2}} \sum _{n=1}^{d-1} \frac{4 \sqrt{n+1/2}}{ \pi }&\ge \frac{1}{d^{3/2}} \frac{8}{3\pi } \left( (d-1/2)^{3/2} - (1/2)^{3/2}\right) \\ {}&\ge (1 - (2d)^{-1}) \frac{8}{3\pi } - (2d)^{-3/2}\\&\ge \frac{8}{3\pi } - \left( \frac{8}{3\pi } + \frac{1}{2}\right) d^{-1} . \end{aligned}$$

The other error terms are at most

$$\begin{aligned} d^{-3/2} \sum _{n=1}^{d-1} \sqrt{n+1/2}\left( \frac{1}{2\pi (2n-1)^2} + \frac{3}{n+1}\right)&\le \frac{1}{d^{3/2}} \sum _{n=1}^{d-1}\frac{4}{\pi (n+1)} \sqrt{n+1/2}\\&\le \frac{1}{d^{3/2}} \sum _{n=1}^{d-1} \frac{4}{\pi (n+1)^{1/2}}\\&\le \frac{4}{\pi d^{3/2}}2\sqrt{d+1}. \end{aligned}$$

Combining the main and error term bounds, the lemma follows. $\square$

Lemma 21

For $G_{\mathbb {K}} \in \mathbb {K}^{d\times d}$ a Gaussian random matrix with i.i.d. $\mathbb {K}$ valued $\mathcal {N}(0,d^{-1})$ entries, define $\alpha _{\mathbb {K}}(d) := \mathbb {E}\left[ \frac{1}{d} \sum _{j=1}^d \sigma _j(G_{\mathbb {K}})\right]$. The following holds

$$\begin{aligned} \alpha _{\mathbb {C}}(d) - \alpha _{\mathbb {R}}(d) \le 4.02 d^{-1}. \end{aligned}$$

Proof

To find an explicit formula for $\alpha _{\mathbb {R}}(d)$, we need an expression for the spectral distribution of the wishart matrix $d G_{\mathbb {R}} G_{\mathbb {R}}^T$, which we call $p_d^{\mathbb {R}}(x)$, given by equation (16) in [24]:

$$\begin{aligned} p_d^{\mathbb {R}}(x) = \frac{1}{2d}\left( 2R_d(x) - \frac{\Gamma \left( \frac{d}{2}+\frac{1}{2} \right) }{\Gamma \left( \frac{d}{2} \right) }L_{d-1}(x) \left\{ \psi _1\left( x\right) - \psi _2\left( x\right) \right\} \right) , \end{aligned}$$

where

$$\begin{aligned} \psi _1(x)= & {} e^{-x} \sum _{k=0}^{(\kappa +d-2)/2} \delta _k L_{2k+1-\kappa }(x),\\ \psi _2(x)= & {} \left( \frac{x}{2}\right) ^{-1/2}e^{-\frac{x}{2}} \left[ (1-\kappa )\frac{2\Gamma \left( \frac{1}{2},\frac{x}{2} \right) }{\Gamma \left( \frac{1}{2}\right) } + 2\kappa -1 \right] ,\\ R_d(x)= & {} e^{-x}\sum _{m=0}^{d-1}\left( L_{m}(x) \right) ^2,\\ \delta _k= & {} \frac{\Gamma \left( k + 1 - \frac{\kappa }{2} \right) }{\Gamma \left( k + \frac{3}{2} - \frac{\kappa }{2}\right) }, \end{aligned}$$

$\kappa = d\mod 2$ and $\Gamma (a,y) = \int _y^\infty t^{a-1}e^{-t}dt$ is the incomplete Gamma function.

This means that

$$\begin{aligned} \alpha _{\mathbb {R}}(d)&= d^{-1/2} \int _0^{\infty } x^{1/2} p_d^{\mathbb {R}}(x) dx\\&= \frac{1}{d^{3/2}}\int _0^\infty x^{1/2}R_d(x)dx\\&\quad - \frac{1}{2d^{3/2}}\int _0^\infty x^{1/2} \frac{\Gamma \left( \frac{d}{2}+\frac{1}{2} \right) }{\Gamma \left( \frac{d}{2} \right) }L_{d-1}(x) \left\{ \psi _1\left( x\right) - \psi _2\left( x\right) \right\} dx. \end{aligned}$$

Recall that (see Sect. 5)

$$\begin{aligned} \alpha _{\mathbb {C}}(d) = d^{-3/2}\sum _{n=0}^{d-1}\int _{0}^\infty x^{1/2}e^{-x}L_n(x)^2 dx \end{aligned}$$

which implies

$$\begin{aligned} \alpha _{\mathbb {R}}(d) = \alpha _{\mathbb {C}}(d) - \frac{1}{2d^{3/2}}\int _0^\infty x^{1/2}\frac{\Gamma \left( \frac{d}{2}+\frac{1}{2} \right) }{\Gamma \left( \frac{d}{2} \right) }L_{d-1}(x) \left\{ \psi _1\left( x\right) - \psi _2\left( x\right) \right\} dx. \end{aligned}$$

(28)

We are especially interested in the following terms which appear in the full expression for $\alpha _{\mathbb {R}}(d)$:

$$\begin{aligned} Q(m,k)= \int _0^{\infty } x^{1/2} e^{-x} L_m(x) L_k(x) dx. \end{aligned}$$

(29)

From [16] section 7.414 equation 4(1), we have

$$\begin{aligned} Q(m,k) = \frac{1}{4\pi } \sum _{i=0}^{\min \{m,k\}} \frac{\Gamma (i+3/2)}{\Gamma (i+1)}\frac{\Gamma (m-i-1/2)}{\Gamma (m-i+1)}\frac{\Gamma (k-i-1/2)}{\Gamma (k-i+1)}. \end{aligned}$$

The following lemma deals with bounds on sums involving Q(m, k) terms. $\square$

Lemma 22

For Q(m, k) as defined in (29) we have the following bounds

$$\begin{aligned}&\sum _{k=0}^m \frac{\Gamma (k+1/2)}{\Gamma (k+1)} Q(2m,2k) \le 2.8\end{aligned}$$

(30)

$$\begin{aligned}&\sum _{k=1}^m \frac{\Gamma (k+3/2)}{\Gamma (k+1)} Q(2m-1,2k-1) \le 5.6 \end{aligned}$$

(31)

Proof

Note that in (30),

$$\begin{aligned} Q(2m,2k) = \frac{1}{4\pi } \sum _{i=0}^{2k} \frac{\Gamma (i+3/2)}{\Gamma (i+1)} \frac{ \Gamma (2m - i - 1/2)}{\Gamma (2m-i+1)}\frac{ \Gamma (2k - i - 1/2)}{ \Gamma (2k-i+1)} \end{aligned}$$

since $m\ge k$.

For $0<i<2k-1$, the ith term in the summation of Q(2m, 2k) can be bounded above by

$$\begin{aligned}&\frac{\Gamma (i+3/2)}{\Gamma (i+1)} \frac{ \Gamma (2m - i - 1/2)}{\Gamma (2m-i+1)}\frac{ \Gamma (2k - i - 1/2)}{ \Gamma (2k-i+1)}\\&\quad \le \sqrt{i+1} \frac{1}{(2k-i)\sqrt{2k-i-1}}\frac{1}{(2m-i)\sqrt{2m-i-1}}\\&\quad \le \sqrt{i+1} \frac{1}{(2k-i-1)^{3/2} (2m-i-1)^{3/2}}. \end{aligned}$$

This means that

$$\begin{aligned} Q(2m,2k)&\le \frac{1}{8 \sqrt{\pi }} \frac{\Gamma (2m-1/2) \Gamma (2k-1/2)}{\Gamma (2m+1) \Gamma (2k+1)}\\&\quad + \frac{1}{4\pi } \sum _{i=1}^{2k-1} \sqrt{i+1} \frac{1}{(2k-i-1)^{3/2} (2m-i+1)^{3/2}}\\&\quad + \frac{1}{4\pi } \sqrt{\pi } \frac{\Gamma (2k+1/2)}{\Gamma (2k)} \frac{\Gamma (2m-2k+1/2)}{\Gamma (2m-2k+2)}\\&\quad + \frac{1}{4\pi } \max \left( \frac{\Gamma (2k+3/2)}{\Gamma (2k+1)} \frac{\Gamma (2m-2k-1/2)}{\Gamma (2m-2k+1)}(-2\sqrt{\pi }) , 0\right) . \end{aligned}$$

We bound the sum from $i=1$ to $2k-3$ by

$$\begin{aligned}&\frac{1}{4\pi } \sum _{i=1}^{2k-3} \sqrt{i+1} \frac{1}{(2k-i-1)^{3/2} (2m-i+1)^{3/2}}\\&\quad \le \frac{1}{4\pi } \sum _{i=0}^{2k-3} \sqrt{i+1} \frac{1}{(2k-i-1)^{3/2}} \frac{1}{(2m-2k+1)^{3/2}}\\&\quad \le \frac{1}{4\pi (2m-2k+1)^{3/2}} \int _0^{2k-2} \frac{\sqrt{x+1}}{(2k-x-1)^{3/2}} dx\\&\quad \le \frac{1}{4\pi (2m-2k+1)^{3/2}} \left( \sqrt{8k} + \frac{\sqrt{4k}}{2k-1}\right) , \end{aligned}$$

so that for $k \ge 1$,

$$\begin{aligned} Q(2m,2k)&\le \frac{1}{8\sqrt{\pi }} \frac{\Gamma (2m-1/2) \Gamma (2k-1/2)}{\Gamma (2m+1) \Gamma (2k+1)}\\&\quad + \frac{1}{4\pi (2m-2k+1)^{3/2}} \left( \sqrt{8k} + \frac{\sqrt{4k}}{2k-1}\right) \\&\quad + \frac{1}{4\pi } \frac{\sqrt{2k-1}}{(2m-2k+3)^{3/2}}\\&\quad + \frac{1}{4\pi } \sqrt{\pi } \frac{\Gamma (2k+1/2)}{\Gamma (2k)} \frac{\Gamma (2m-2k+1/2)}{\Gamma (2m-2k+2)}\\&\quad + \frac{1}{4\pi } \max \left( \frac{\Gamma (2k+3/2)}{\Gamma (2k+1)} \frac{\Gamma (2m-2k-1/2)}{\Gamma (2m-2k+1)}(-2\sqrt{\pi }) , 0\right) . \end{aligned}$$

For $k=0, Q(2m,0) < 0$ except for the term $Q(0,0) = \sqrt{\pi }/2$ which also becomes negative in the full sum, so we ignore these terms.

We now turn our attention to the full sum $\sum _{k=0}^m \frac{\Gamma (k+1/2)}{\Gamma (k+1)} Q(2m,2k)$. As before, we define for clarity

$$\begin{aligned} (I)&:=\frac{1}{8\sqrt{\pi }} \sum _{k=1}^m \frac{\Gamma (k+1/2)}{\Gamma (k+1)} \frac{\Gamma (2m-1/2) \Gamma (2k-1/2)}{\Gamma (2m+1) \Gamma (2k+1)}\\ (II)&:=\frac{1}{4\pi } \sum _{k=1}^m \frac{\Gamma (k+1/2)}{\Gamma (k+1)} \frac{1}{(2m-2k+1)^{3/2}} \left( \sqrt{8k} + \frac{\sqrt{4k}}{2k-1}\right) \\ (III)&:=\frac{1}{4\pi } \sum _{k=1}^m \frac{\Gamma (k+1/2)}{\Gamma (k+1)} \frac{\sqrt{2k-1}}{(2m-2k+3)^{3/2}}\\ (IV)&:= \frac{1}{4\sqrt{\pi } } \sum _{k=1}^m \frac{\Gamma (k+1/2)}{\Gamma (k+1)} \frac{\Gamma (2k+1/2)}{\Gamma (2k)} \frac{\Gamma (2m-2k+1/2)}{\Gamma (2m-2k+2)}\\ (V)&:= \frac{1}{4\pi } \sum _{k=1}^m \frac{\Gamma (k+1/2)}{\Gamma (k+1)} \max \left( \frac{\Gamma (2k+3/2)}{\Gamma (2k+1)} \frac{\Gamma (2m-2k-1/2)}{\Gamma (2m-2k+1)}(-2\sqrt{\pi }) , 0\right) . \end{aligned}$$

Using the bounds in lemma 20,

$$\begin{aligned} (I)&\le \sum _{k=1}^m \frac{1}{32 \sqrt{\pi } k^{1/2}} \frac{1}{mk} \frac{1}{\sqrt{2m-1}\sqrt{2k-1}} \le \frac{1}{32 \sqrt{\pi }},\\ (II)&\le \frac{1}{4\pi } \sum _{k=1}^m \frac{1}{k^{1/2} (2m-2k+1)^{3/2}} (4 \sqrt{k})\\&\le \frac{1}{\pi } \left( 1 - \frac{1}{\sqrt{2m-1}}\right) \le \frac{1}{\pi },\\ (III)&\le \frac{1}{4\pi } \sum _{k=1}^m \frac{1}{k^{1/2}} \frac{\sqrt{2k-1}}{(2m -2k + 3)^{3/2}}\\&\le \frac{1}{\sqrt{24}\pi },\\ (IV)&\le \frac{1}{4\sqrt{\pi }} \left( \sum _{k=1}^m \frac{(2k)^{1/2}}{k^{1/2}} \frac{1}{(2m-2k+1)\sqrt{2m-2k}} + \sqrt{\pi }\right) \\&\le \frac{1}{2\pi } + \frac{1}{2},\\ (V)&= \frac{1}{4\pi } 4 \pi \frac{\Gamma (2m+3/2)}{\Gamma (2m+1)}\frac{\Gamma (m+1/2)}{\Gamma (m+1)}\\&\le \frac{\sqrt{2m+1}}{\sqrt{m}} \le \sqrt{3}. \end{aligned}$$

Finally,

$$\begin{aligned} \sum _{k=0}^m \frac{\Gamma (k+1/2)}{\Gamma (k)} Q(2m,2k)&\le (I) + (II) + (III) + (IV) + (V)\\&\le 2.8. \end{aligned}$$

To deduce the inequality (31), we use the previously derived bounds to show that

$$\begin{aligned} Q(2m-1,2k-1)&\le \frac{1}{4\pi } \sum _{i=1}^{2k-3} \sqrt{i+1} \frac{1}{(2k-i-2)^{3/2} (2m-i)^{3/2}}\\&\quad + \frac{1}{4\pi } \sqrt{\pi } \frac{\Gamma (2k-1/2)}{\Gamma (2k-1)} \frac{\Gamma (2m-2k+1/2)}{\Gamma (2m-2k+2)}\\&\quad + \frac{1}{4\pi } \max \left( \frac{\Gamma (2k+1/2)}{\Gamma (2k)} \frac{\Gamma (2m-2k+1/2)}{\Gamma (2m-2k+2)}(-2\sqrt{\pi }) , 0\right) , \end{aligned}$$

so that $Q(2m-1,2k-1) \le Q(2m,2k)$. Now it suffices to note that in the full sum, $\sum _{k=1}^m \frac{\Gamma (k+3/2)}{\Gamma (k+1)} Q(2m-1,2k-1) \le 2 \sum _{k=1}^m \frac{\Gamma (k+1/2)}{\Gamma (k)} Q(2m-1,2k-1)$ and we get

$$\begin{aligned} \sum _{k=1}^m \frac{\Gamma (k+3/2)}{\Gamma (k+1)} Q(2m-1,2k-1)\le 2 \sum _{k=1}^m\frac{\Gamma (k+1/2)}{\Gamma (k)} Q(2m,2k) \le 5.6. \end{aligned}$$

$\square$

We now return our focus to finding a bound on the expression for $\alpha _{\mathbb {R}}(d)$ given in (28). Since $\psi _1,\psi _2$ depend on the parity of d, we split in to two cases.

Odd $d=2m+1$

From (see [16] section 7.414 equation 6),

$$\begin{aligned} \int _0^{\infty } e^{-x/2} L_{2m}(x) dx = 2, \end{aligned}$$

thus Eq. (28) becomes

$$\begin{aligned}&\alpha _{\mathbb {C}}(2m+1) - \alpha _{\mathbb {R}}(2m+1)\\&\quad = \frac{1}{(2m+1)^{3/2}} \frac{\Gamma (m+1)}{\Gamma (m+1/2)} \left( \sum _{k=0}^{m} \frac{\Gamma (k+1/2)}{\Gamma (k)} Q(2m,2k) - 2^{1/2}\right) , \end{aligned}$$

and using the first bound in Lemma 22,

$$\begin{aligned} \alpha _{\mathbb {C}}(2m+1) - \alpha _{\mathbb {R}}(2m+1) \le 2.8 \sqrt{m+1/2} \frac{1}{(2m+1)^{3/2}} \le m^{-1}. \end{aligned}$$

Even $d=2m$

For $d=2m$, we have

$$\begin{aligned} \alpha _{\mathbb {R}}(2m) = \alpha _{\mathbb {C}}(2m) - \frac{1}{2(2m)^{3/2}} \int _0^{\infty } x^{1/2} \frac{\Gamma (m+1/2)}{\Gamma (m)} L_{2m-1}(x) \{ \psi _1(x) - \psi _2(x)\} dx. \end{aligned}$$

We split the integral into two parts,

$$\begin{aligned} (I):= & {} \frac{1}{2(2m)^{3/2}} \int _0^{\infty } x^{1/2} \frac{\Gamma (m+1/2)}{\Gamma (m)} L_{2m-1}(x) \psi _1(x) dx\\ (II):= & {} \frac{-1}{2(2m)^{3/2}} \int _0^{\infty } x^{1/2} \frac{\Gamma (m+1/2)}{\Gamma (m)} L_{2m-1}(x) \psi _2(x) dx. \end{aligned}$$

Expanding from the definition of $\psi _1$ above, we have

$$\begin{aligned} (I)&= \frac{1}{2(2m)^{3/2}} \int _0^{\infty } \frac{\Gamma (m+1/2)}{\Gamma (m)} x^{1/2} L_{2m-1}(x)e^{-x} \sum _{k=0}^{m-1} \frac{\Gamma (k)}{\Gamma (k+1/2)} L_{2k-1}(x) dx\\&= \frac{1}{2(2m)^{3/2}} \frac{\Gamma (m+1/2)}{\Gamma (m)} \sum _{k=1}^{m} \frac{\Gamma (k+3/2)}{\Gamma (k+1)} Q(2m-1,2k-1), \end{aligned}$$

so by Lemma 22,

$$\begin{aligned} (I)&\le \frac{1}{2(2m)^{3/2}} \frac{\Gamma (m+1/2)}{\Gamma (m)} 5.6 \le \frac{1}{ m^{1/2}}. \end{aligned}$$

The other part of the integral is

$$\begin{aligned} (II)&= \frac{-1}{2(2m)^{3/2}} \int _0^{\infty } x^{1/2} \frac{\Gamma (m+1/2)}{\Gamma (m)} L_{2m-1}(x) (x/2)^{-1/2}\\&\quad \times e^{-x/2} \left[ \frac{2 \Gamma (1/2,x/2)}{\Gamma (1/2)} - 1\right] dx\\&= \frac{-1}{4m^{1/2}} \int _0^{\infty } \frac{\Gamma (m+1/2)}{\Gamma (m)} L_{2m-1}(x) e^{-x/2} \frac{2 \Gamma (1/2,x/2)}{\Gamma (1/2)} dx \\&\quad + \frac{1}{2m^{3/2}} \frac{\Gamma (m+1/2)}{\Gamma (m)}, \end{aligned}$$

where we use the fact that for odd $2m-1$ (see [16] section 7.414 equation 6),

$$\begin{aligned} \int _0^{\infty } L_{2m-1}(x) e^{-x/2} dx = -2. \end{aligned}$$

We can bound the first integral in the expression of (II) by

$$\begin{aligned}&\left| \int _0^{\infty } L_{2m-1}(x) e^{-x/2} \Gamma (1/2,x/2) dx\right| \\&\quad \le \left( \int _0^{\infty } e^{-x} L_{2m-1}(x)^2 dx\right) ^{1/2} \left( \int _0^{\infty } \Gamma (1/2,x/2)^2 dx\right) ^{1/2}\\&\quad = \left[ \int _0^{\infty } \left( \int _x^{\infty } t^{-1/2} e^{-t} dt\right) ^2 dx\right] ^{1/2}\\&\quad = \left[ \int _0^1 \left( \int _x^{\infty } t^{-1/2} e^{-t} dt \right) ^2 dx + \int _1^{\infty } \left( \int _x^{\infty } t^{-1/2} e^{-t} dt\right) ^2 dx\right] ^{1/2}\\&\quad \le \left( \Gamma (1/2)^2 + \int _1^{\infty } (e^{-x})^2 dx\right) ^{1/2} \le (\pi + e^{-2}/2)^{1/2}, \end{aligned}$$

so finally

$$\begin{aligned} (II)&\le \frac{ (\pi + 1/2 e^{-2})^{1/2}}{2\sqrt{\pi } m^{3/2}} \frac{\Gamma (m+1/2)}{\Gamma (m)} + \frac{m^{1/2}}{2m^{3/2}}\\&\le 1.01 m^{-1}. \end{aligned}$$

Combining the above bounds we see that in the case of even $d=2m$,

$$\begin{aligned} \alpha _{\mathbb {C}}(2m) - \alpha _{\mathbb {R}}(2m) =(I) + (II) \le 2.01 m^{-1}. \end{aligned}$$

$\square$