Explicit Examples of Eigenfunctions of the Hermite Operator Saturating Some L p Bounds

Abstract

The goal of the paper is to provide a sequence of eigenfunctions that saturates the $L^{p}$ bounds obtained by Koch and Tataru for the multidimensional Hermite operator. More precisely, several such sequences of eigenfunctions have already been identified by Koch and Tataru, and we present an example in another range of $p$ .

Keywords

Hermite operator quantum harmonic oscillator eigenfunctions Lebesgue spaces

Let $(E_{n})_{n \in N}$ denote the sequence of eigenspaces of the Hermite operator (also called the quantum harmonic oscillator) $- Δ + | x |^{2}$ on $L^{2} (R^{d})$ with $d \geq 2$ :

\begin{aligned} E_{n} = \ker (- Δ + | x |^{2} - 2 n - d) . \end{aligned}

Each eigenspace

E_{n}

is finite-dimensional (more precisely

\dim (E_{n})

grows like

\frac{n^{d - 1}}{(d - 1)!}

n \to + \infty

). The classical paper by Koch and Tataru (2005) gives an almost complete set of optimal

L^{2} \to L^{p}

inequalities for all

p \in [2, + \infty]

for the eigenspaces

E_{n}

. The result reads as follows:

\forall f \in E_{n} ‖ f ‖_{L^{p} (R^{d})} \leq C (d, p) n^{a (p)} ‖ f ‖_{L^{2} (R^{d})}

(1)

where

a (p)

is an explicit convex function of

\frac{1}{p}

, defined by

\begin{aligned} a (p) = {\begin{array}{rcl} - \frac{1}{2} (\frac{1}{2} - \frac{1}{p}) & with & 2 \leq p < \frac{2 (d + 3)}{d + 1}, \\ \frac{1}{2} (- \frac{1}{3} + \frac{d}{3} (\frac{1}{2} - \frac{1}{p})) & with & \frac{2 (d + 3)}{d + 1} < p \leq \frac{2 d}{d - 2}, \\ \frac{1}{2} (- 1 + d (\frac{1}{2} - \frac{1}{p})) & with & \frac{2 d}{d - 2} \leq p \leq + \infty . \end{array} \end{aligned}

Let us write a few words about the critical exponent

p_{0} = \frac{2 (d + 3)}{d + 1}

. The following bound is shown in Koch et al. (2007) (and is probably obtainable by the arguments of Koch and Tataru (2005)):

\begin{aligned} \forall f \in E_{n} ‖ f ‖_{L^{p_{0}} (R^{d})} \leq C (d) n^{a (p_{0})} (\log (n))^{\frac{1}{p_{0}}} ‖ f ‖_{L^{2} (R^{d})} . \end{aligned}

In the remarkable paper by Jeong et al. (2024b), it is shown that the logarithmic factor is indeed unnecessary for

d \geq 3

, while the case

d = 2

remains open. We also refer the reader to Jeong et al. (2024a) for a related topic.

The paper by Koch and Tataru (2005) also provides proofs of sharpness of the $L^{2} \to L^{p}$ bounds and it is interesting to see whether one could give explicit examples of sequence of eigenfunctions saturating such bounds. At the end of Koch and Tataru (2005), explicit examples are provided for

\begin{aligned} p \in [2, \frac{2 (d + 3)}{d + 1}) ⋃ [\frac{2 d}{d - 2}, + \infty] . \end{aligned}

The aim of this article is to extend this analysis in the middle range

(\frac{2 (d + 3)}{d + 1}, \frac{2 d}{d - 2})

Before going into details, we introduce some notations for the one-dimensional eigenfunctions, namely Hermite functions:

h_{n} (x) = \frac{H_{n} (x)}{\sqrt{n! 2^{n} \sqrt{π}}} e^{- x^{2} / 2}

(2)

where

H_{n}

denotes the Hermite polynomial of order

n

(physicists’ convention). For example, the ground state is given by

h_{0} (x) = π^{- \frac{1}{4}} e^{- x^{2} / 2}

. The following equalities are well-known:

- h_{n}^{″} (x) + x^{2} h_{n} (x) = (2 n + 1) h_{n} (x) and ‖ h_{n} ‖_{L^{2} (R)} = 1.

(3)

In the multidimensional case, the eigenspace

E_{n}

(corresponding to the eigenvalue

2 n + d

- Δ + | x |^{2}

) can be described using a simple orthonormal basis as follows:

\begin{aligned} E_{n} = Span ({h_{n_{1}} \otimes \dots \otimes h_{n_{d}}, with \sum_{i = 1}^{d} n_{i} = n}) . \end{aligned}

For the sequel, it is worthwhile to note, as at the end of Koch and Tataru (2005), that if

J \subset N^{d}

is a non-empty set consisting of tuples

(n_{1}, \dots, n_{d})

satisfying

n_{1} + \dots + n_{d} = n

, then the following function is an

L^{2} (R^{d})

-normalized eigenfunction corresponding to the eigenvalue

2 n + d

x \in R^{d} \mapsto \frac{1}{\sqrt{Card (J)}} \sum_{(n_{1}, \dots, n_{d}) \in J} h_{n_{1}} (x_{1}) \dots h_{n_{d}} (x_{d}) .

(4)

With such notations, we now briefly recall the known explicit examples:

- for $p \in [2, \frac{2 (d + 3)}{d + 1})$ , the $L^{p}$ norm of the eigenfunction $h_{n} \otimes h_{0} \otimes \dots \otimes h_{0}$ (which corresponds to the eigenvalue $2 n + d$ ) satisfies for $n ≫ 1$ :

\begin{aligned} ‖ h_{n} \otimes h_{0} \otimes \dots \otimes h_{0} ‖_{L^{p} (R^{d})} & = {(\int_{R^{d}} | \frac{1}{π^{\frac{d - 1}{4}}} h_{n} (x_{1}) e^{- \frac{x_{2}^{2} + \dots + x_{d}^{2}}{2}} |^{p} d x_{1} \dots d x_{d})}^{1 / p} \\ = \frac{1}{π^{\frac{d - 1}{4}}} {(\int_{R} | h_{n} (x) |^{p} d x)}^{\frac{1}{p}} {(\int_{R} e^{- p x^{2} / 2} d x)}^{\frac{d - 1}{p}} \\ ≃ ‖ h_{n} ‖_{L^{p}} . \end{aligned}

Moreover, it is known that

‖ h_{n} ‖_{L^{p} (R)} ≃ n^{- \frac{1}{2} (\frac{1}{2} - \frac{1}{p})}

for

p \in [2, 4)

(see Thangavelu, 1993, Lemma 1.5.2). This covers the relevant case since

\frac{2 (d + 3)}{d + 1} < 4

(this example is presented in Koch and Tataru (2005, Subsection 5.1)).

- for $p \geq \frac{2 d}{d - 2}$ if $d \geq 3$ (set $p = \infty$ if $d = 2$ ), as shown in Koch and Tataru (2005, Subsection 5.2), one may consider a sequence of eigenfunctions of the form (4) for saturating the $L^{2} \to L^{p}$ bounds. Another construction is given in Imekraz et al. (2016, Proposition 2.4) showing that the sequence of radial eigenfunctions of $- Δ + | x |^{2}$ is also convenient. Geometrically, both examples concentrate around the origin.

The result of this article shows that one may also construct examples similar to (4) in the middle range $p \in (\frac{2 (d + 3)}{d + 1}, \frac{2 d}{d - 2})$ .

Theorem 1

We assume $d \geq 2$ . For any $α \in (0, \frac{1}{6}]$ , $β \geq 0$ and $n \in N$ , define

J_{n}^{(α, β)} := {(n_{1}, \dots, n_{d}) \in N^{d}, \forall i \in {1, \dots, d} | n_{i} - \frac{n}{d} | \leq α (\frac{2 n}{d} + 1)^{\frac{1}{3}} + β and \sum_{i = 1}^{d} n_{i} = n}

and the following eigenfunction of

- Δ + | x |^{2}

corresponding to the eigenvalue

2 n + d

v_{n}^{(α, β)} (x_{1}, \dots, x_{d}) := \frac{1}{\sqrt{Card (J_{n}^{(α, β)})}} \sum_{(n_{1}, \dots, n_{d}) \in J_{n}^{(α, β)}} h_{n_{1}} (x_{1}) \dots h_{n_{d}} (x_{d}) .

(5)

In particular, we have

‖ v_{n}^{(α, β)} ‖_{L^{2} (R^{d})} = 1

. Let

p \geq 1

be fixed. For any

n ≫ 1

the norm

‖ v_{n}^{(α, β)} ‖_{L^{p} (R^{d})}

saturates the Koch-Tataru

L^{2} \to L^{p}

estimates in the middle range:

‖ v_{n}^{(α, β)} ‖_{L^{p} (R^{d})} ≳ n^{\frac{1}{2} (- \frac{1}{3} + \frac{d}{3} (\frac{1}{2} - \frac{1}{p}))} = n^{\frac{d}{12} - \frac{1}{6} - \frac{d}{6 p}} .

(6)

Due to (1), the bound from below (6) is actually an equivalence for $p \in (\frac{2 (d + 3)}{d + 1}, \frac{2 d}{d - 2}]$ .

We finish this introduction by outlining the organization of the paper and the structure of the proof of Theorem 1.

Section 1 is devoted to giving two graphical representations in dimension $2$ that illustrate the concentration of the eigenfunctions $v_{n}^{(α, β)}$ for $α = \frac{1}{6}$ and $β = 4$ . We emphasize that the parameter $β$ plays no essential role from a theoretical point of view in the statement of Theorem 1, it is only introduced to ensure $Card (J_{n}^{(α, β)}) ≫ 1$ for small values of $n$ in graphical representations.

In Section 2, we prove two results (Proposition 3 and Corollary 5) that provide a quantitative statement about the spatial concentration of the one-dimensional Hermite functions $(h_{n})$ . More precisely, for $n ≫ 1$ , Corollary 5 will imply that if $| n_{i} - \frac{n}{d} | = O (n^{\frac{1}{3}})$ then $h_{n_{i}} (x_{i})$ is bounded below by $n^{- \frac{1}{12}}$ (up to a multiplicative constant) for $x_{i}$ belonging on the following interval

[\sqrt{\frac{2 n}{d} + 1} - \frac{3.89}{{(\frac{2 n}{d} + 1)}^{\frac{1}{6}}}, \sqrt{\frac{2 n}{d} + 1} - \frac{3.72}{{(\frac{2 n}{d} + 1)}^{\frac{1}{6}}}] .

(7)

To prove such a concentration property, we use a well-known WKB approximation of the Hermite functions. More precisely, by setting

g_{n} (y) := \frac{(2 n + 1)}{2} (\arccos (y) - y \sqrt{1 - y^{2}}) - \frac{π}{4}, \forall y \in (- 1, 1),

the following holds uniformly for

y \in (- 1, 1)

h_{n} (y \sqrt{2 n + 1}) = \frac{\sqrt{2}}{\sqrt{π} (2 n + 1)^{\frac{1}{4}}} \frac{\cos (g_{n} (y))}{(1 - y^{2})^{\frac{1}{4}}} + O (\frac{1}{(2 n + 1)^{\frac{5}{4}} (1 - y^{2})^{\frac{7}{4}}}) .

(8)

Written in this form, this WKB approximation of

h_{n}

merely yields unknown constants in (7) (instead of

3.89

and

3.72

) and also an unknown upper bound for the constant

α

in Theorem 1. Without an explicit remainder in (8), the conclusion of Theorem 1 remains true for

α

small enough, which seems essentially useless for numerical applications. To obtain the explicit condition

α \in (0, \frac{1}{6}]

, we need to provide an explicit constant in the remainder term of (8). This is the purpose of Proposition 2, whose proof is postponed to Section 6.

In Section 3, we give an elementary proof of the asymptotics of the cardinality of the sets $J_{n}^{(α, β)}$ for $n ≫ 1$ . For any choice of $α > 0$ , Lemma 6 below ensures the following estimate

Card (J_{n}^{(α, β)}) ≃ n^{\frac{d - 1}{3}} .

However, let us give an intuition about why this asymptotics is the good one for our purpose (the complete argument is developed in Section 4). Once we have the idea to consider an eigenfunction as (4), we search a specific set

J

d

-tuples

(n_{1}, \dots, n_{d})

. Due to (7), if one can ensure that each

(n_{1}, \dots, n_{d}) \in J

satisfies

| n_{i} - \frac{n}{d} | = O (n^{\frac{1}{3}})

then each

h_{n_{1}} (x_{1}) \dots h_{n_{d}} (x_{d})

in (4) is bounded below by

n^{\frac{- d}{12}}

(up to a multiplicative constant) on the following hypercube

H_{n, d} := {[\sqrt{\frac{2 n}{d} + 1} - \frac{3.89}{{(\frac{2 n}{d} + 1)}^{\frac{1}{6}}}, \sqrt{\frac{2 n}{d} + 1} - \frac{3.72}{{(\frac{2 n}{d} + 1)}^{\frac{1}{6}}}]}^{d} .

(9)

Hence, the

L^{p}

norm of the function (4) is bounded below (up to a multiplicative constant) by

\begin{aligned} n^{- \frac{d}{12}} \times \frac{1}{\sqrt{Card (J)}} \times Card (J) \times Mes (H_{n, d})^{\frac{1}{p}} \end{aligned}

namely

\sqrt{Card (J)} n^{\frac{- d}{12} - \frac{d}{6 p}}

because the volume of the hypercube (9) is of order

n^{- \frac{d}{6}}

. In order to reach the lower bound (6), we require a bound like

\begin{aligned} Card (J) ≳ n^{\frac{d - 1}{3}} . \end{aligned}

Note now that each

(n_{1}, \dots, n_{d}) \in J

satisfies

n_{1} + \dots + n_{d} = n

(so that (4) is an eigenfunction corresponding to

2 n + d

). The technical condition

| n_{i} - \frac{n}{d} | = O (n^{\frac{1}{3}})

finally leads to the definition of

J_{n}^{(α, β)}

in Theorem 1.

As written above, the last point is to give an explicit constant in the remainder in (8). In Section 5, we establish several explicit inequalities for the Airy functions which do not appear explicitly in the literature but can be deduced from Olver (1997). These explicit inequalities are then used in Section 6, where the constants in the analysis of Olver (1997) are followed.

In all the paper, we use the notation $N = {0, 1, 2, 3, \dots}$ and the following classical notations: –

for any positive numbers $u$ and $w$ , we write $u ≲ w$ to mean that $u \leq C (d, p) w$ for some positive constant $C (d, p)$ (if the exponent $p$ is relevant in $u$ and $w$ ),

–

$u ≳ w$ means $w ≲ u$ ,

–

$u ≃ w$ means $u ≲ w$ and $w ≲ u$ .

1. Graphical Representations in Dimension

d = 2

In dimension $d = 2$ , the sets $J_{n}^{(α, β)}$ are easy to describe and we indeed have

\begin{aligned} J_{n}^{(α, β)} & = {(k, n - k), with ⌈ \frac{n}{2} - α \sqrt[3]{n + 1} - β ⌉ \leq k \leq ⌊ \frac{n}{2} + α \sqrt[3]{n + 1} + β ⌋} \\ Card (J_{n}^{(α, β)}) & = 1 - ⌈ \frac{n}{2} - α \sqrt[3]{n + 1} - β ⌉ + ⌊ \frac{n}{2} + α \sqrt[3]{n + 1} + β ⌋ \\ \underset{n \to + \infty}{\sim} \frac{\sqrt[3]{n}}{3} if α = \frac{1}{6} . \end{aligned}

n \to + \infty

, the number

Card (J_{n}^{(α, β)})

tends to

+ \infty

, and it appears graphically that the different eigenfunctions in (5) change signs except if

(x_{1}, x_{2})

lies in a concentration region. Since our choice of

n

is somewhat restricted if we want to use (2) with factorials, and since

\frac{1}{6} (n + 1)^{\frac{1}{3}} < 1

for

n \leq 200

, we will use the extra parameter

β

in order to make

Card (J_{n}^{(α, β)})

large enough to expect a cancellation phenomenon outside a concentration region. We choose

α = \frac{1}{6}

, which is the maximal value allowed by Theorem 1, and

β = 4

that provide quite satisfactory graphical representations. Here is the formula for the eigenfunction

v_{n}^{(\frac{1}{6}, 4)}

(corresponding to the eigenvalue

2 n + 2

\begin{aligned} v_{n}^{(\frac{1}{6}, 4)} : (x_{1}, x_{2}) \mapsto \frac{1}{\sqrt{Card (J_{n}^{(\frac{1}{6}, 4)})}} \sum_{k = ⌈ \frac{n}{2} - \frac{1}{6} \sqrt[3]{n + 1} - 4 ⌉}^{⌊ \frac{n}{2} + \frac{1}{6} \sqrt[3]{n + 1} + 4 ⌋} h_{k} (x_{1}) h_{n - k} (x_{2}), \end{aligned}

Here are two graphical representations on $[- \sqrt{2 n + 2}, \sqrt{2 n + 2}]^{2}$ for $n = 20$ and $n = 80$ of $v_{n}^{(\frac{1}{6}, 4)}$ and we check the equalities

\begin{aligned} J_{20}^{(\frac{1}{6}, 4)} & = {(6, 14), (7, 13), (8, 12), (9, 11), (10, 10), (11, 9), (12, 8), (13, 7), (14, 6)}, \\ J_{80}^{(\frac{1}{6}, 4)} & = {(36, 44), (37, 43), (38, 42), (39, 41), (40, 40), (41, 39), (42, 38), (43, 37), (44, 36)} . \end{aligned}

As mentioned in the introduction, these graphical representations confirm the concentration at least around the two points with coordinates

x_{1} = x_{2}

around

\pm \sqrt{n + 1}

2. Study of the One-dimensional Eigenfunctions

For any $y \in (- 1, 1)$ and any positive integer $n$ , we define:

g_{n} (y) = \frac{(2 n + 1)}{2} (\arccos (y) - y \sqrt{1 - y^{2}}) - \frac{π}{4} .

(10)

We require an explicit approximation of the one-dimensional Hermite functions in the allowed region

(- \sqrt{2 n + 1}, \sqrt{2 n + 1})

, as stated in the following result:

Proposition 2

For any $n ≫ 1$ , the following inequality holds uniformly for any $y \in (- 1, 1)$ :

| h_{n} (y \sqrt{2 n + 1}) - \frac{\sqrt{2}}{\sqrt{π} (2 n + 1)^{\frac{1}{4}}} \frac{\cos (g_{n} (y))}{(1 - y^{2})^{\frac{1}{4}}} | \leq \frac{0.5}{(2 n + 1)^{\frac{5}{4}} (1 - y^{2})^{\frac{7}{4}}} .

(11)

The previous result is obtained by carefully following the constants in a standard WKB method with explicit remainders (see Section 6 for details of the proof). The simple constant $0.5$ has been chosen because it is sufficient for what follows¹ but the proof in Section 6 provides a slightly better constant. Even without the explicit constant $0.5$ , such an approximation is very well-known (see Skovgaard, 1959 and Erdélyi (1960, pages 22 $-$ 23)). Proposition 2 is only meaningful when $y$ is not too close to $1$ and becomes useful precisely if the upper bound in (11) is controlled by the amplitude of the oscillating term, which gives the condition in Muckenhoupt (1970, (2.4)):

\begin{aligned} \frac{1}{(2 n + 1)^{\frac{1}{6}}} ≲ \sqrt{2 n + 1} - y \sqrt{2 n + 1} . \end{aligned}

For

y \to 1

, a more precise approximation of the eigenfunction

h_{n}

is obtained by comparing it with an Airy function (see Section 6). Proposition 3

Define the following set

C := {(\frac{3 π}{\sqrt{2}})^{\frac{2}{3}} k^{\frac{2}{3}}, with k \in N and k \geq 1} .

(12)

For any fixed constant

C \in C

, for any

n ≫ 1

and any

x

in the interval

[\sqrt{2 n + 1} - \frac{C + \frac{1}{\sqrt{C}}}{(2 n + 1)^{\frac{1}{6}}}, \sqrt{2 n + 1} - \frac{C}{(2 n + 1)^{\frac{1}{6}}}]

(13)

the following inequality holds uniformly

\begin{aligned} h_{n} (x) \geq \frac{0.3}{C^{\frac{1}{4}} (2 n + 1)^{\frac{1}{12}}} . \end{aligned}

Remark 4

In the sequel, we only need the previous result for a single choice of $C$ , for instance the smallest one $C = {(\frac{3 π}{\sqrt{2}})}^{\frac{2}{3}} \approx 3.541$ . In particular, we note that an immediate corollary is the bound

\begin{aligned} ‖ h_{n} ‖_{L^{p} (R)} ≳ n^{- \frac{1}{12} - \frac{1}{6 p}} \end{aligned}

which is known to be optimal for

p \in (4, + \infty]

(see Thangavelu, 1993, Lemma 1.5.2).

Proof.

Set $x = y \sqrt{2 n + 1}$ . Then the zone (13) becomes

y \in [1 - \frac{C + \frac{1}{\sqrt{C}}}{(2 n + 1)^{\frac{2}{3}}}, 1 - \frac{C}{(2 n + 1)^{\frac{2}{3}}}] .

(14)

Using Proposition 2, we may write

h_{n} (x) = P_{n} (y) \cos (g_{n} (y)) + R_{n} (y)

(15)

with

P_{n} (y) = \frac{\sqrt{2}}{\sqrt{π} (2 n + 1)^{\frac{1}{4}} (1 - y^{2})^{\frac{1}{4}}} and | R_{n} (y) | \leq \frac{0.5}{(2 n + 1)^{\frac{5}{4}} (1 - y^{2})^{\frac{7}{4}}} .

(16)

We temporarily admit that the condition

C \in C

implies the following inequality (see Step 2 for justification):

\cos (g_{n} (y)) \geq \frac{1}{2} .

(17)

Step 1. We now prove that (17) implies the expected inequality

h_{n} (x) \geq \frac{1}{4 C^{\frac{1}{4}} (2 n + 1)^{\frac{1}{12}}}

. We first observe that the following holds uniformly for any

y \in [0, 1 - \frac{C}{(2 n + 1)^{\frac{2}{3}}}]

n \to + \infty

\begin{aligned} | R_{n} (y) | \leq \frac{0.5}{(2 n + 1)^{\frac{5}{4}} {(1 - {(1 - \frac{C}{(2 n + 1)^{\frac{2}{3}}})}^{2})}^{\frac{7}{4}}} \underset{n \to + \infty}{\sim} \frac{0.5}{(2 C)^{\frac{7}{4}} (2 n + 1)^{\frac{1}{12}}} . \end{aligned}

For any

C \in C

, the inequality

C^{\frac{3}{2}} \geq \frac{3 π}{\sqrt{2}} = 6.66 \dots

yields the following rough bound

C + \frac{1}{\sqrt{C}} \leq \frac{7}{6} C

. We then provide a lower bound of the principal term

P_{n} (y)

under the condition

1 - \frac{\frac{7}{6} C}{(2 n + 1)^{\frac{2}{3}}} \leq y

\begin{aligned} \frac{P_{n} (y)}{2} \geq \frac{1}{\sqrt{2 π} (2 n + 1)^{\frac{1}{4}} {(1 - {(1 - \frac{\frac{7}{6} C}{(2 n + 1)^{\frac{2}{3}}})}^{2})}^{\frac{1}{4}}} \underset{n \to + \infty}{\sim} \frac{1}{\sqrt{2 π} {(\frac{14}{6})}^{\frac{1}{4}} C^{\frac{1}{4}} (2 n + 1)^{\frac{1}{12}}} . \end{aligned}

We again use the inequality

C^{\frac{3}{2}} \geq \frac{3 π}{\sqrt{2}}

to obtain the following numerical lower bound² :

\frac{1}{\sqrt{2 π} {(\frac{14}{6})}^{\frac{1}{4}} C^{\frac{1}{4}}} - \frac{0.5}{2^{\frac{7}{4}} C^{7 / 4}} \geq \frac{1}{C^{\frac{1}{4}}} (\frac{1}{\sqrt{2 π} {(\frac{14}{6})}^{\frac{1}{4}}} - \frac{0.5 \times \sqrt{2}}{2^{\frac{7}{4}} \times 3 π}) > \frac{0.3}{C^{\frac{1}{4}}} .

(18)

Combining the above considerations, we obtain the following uniform lower bound for

n ≫ 1

\begin{aligned} \frac{P_{n} (y)}{2} - | R_{n} (y) | \geq \frac{0.3}{C^{\frac{1}{4}} (2 n + 1)^{\frac{1}{12}}} . \end{aligned}

Now (15) and (17) finally show the bound from below:

\begin{aligned} h_{n} (x) \geq \frac{0.3}{C^{\frac{1}{4}} (2 n + 1)^{\frac{1}{12}}} . \end{aligned}

Step 2 (Use of (12) to Prove (17)). Since we aim to bound from below $\cos (g_{n} (y))$ , it suffices to ensure that $g_{n} (y)$ is near $2 π N$ . For $C \in C$ , we shall use the condition

\frac{2 \sqrt{2}}{3} C^{\frac{3}{2}} \in 2 π N .

(19)

We first integrate the Taylor expansion at $ω = 0$ as $a \to 0^{+}$ :

\begin{aligned} \arccos (1 - a^{2}) & = \sqrt{2} \int_{0}^{a} \frac{1}{\sqrt{1 - \frac{ω^{2}}{2}}} d ω = \sqrt{2} \int_{0}^{a} 1 + \frac{ω^{2}}{4} + O (ω^{4}) d ω \\ = \sqrt{2} a + \frac{\sqrt{2}}{12} a^{3} + O (a^{5}) . \end{aligned}

θ \to 0

, one also has

\begin{aligned} 2 θ - \sin (2 θ) = \frac{4}{3} θ^{3} - \frac{4}{15} θ^{5} + O (θ^{7}) . \end{aligned}

By composition with

θ = \arccos (1 - a^{2})

, we get the following expansion for

a \to 0^{+}

\begin{aligned} 2 \arccos (1 - a^{2}) - \sin (2 \arccos (1 - a^{2})) & = 2 θ - \sin (2 θ) \\ = \frac{8 \sqrt{2}}{3} a^{3} - \frac{2 \sqrt{2}}{\sqrt{5}} a^{5} + O (a^{7}) . \end{aligned}

Remembering the inequality

C^{\frac{3}{2}} \geq \frac{3 π}{\sqrt{2}}

, we may define the following positive constant:

\begin{aligned} K = \frac{8 \sqrt{2}}{3} - \frac{π}{3 C^{\frac{3}{2}}} . \end{aligned}

By using

K < \frac{8 \sqrt{2}}{3}

, we deduce the following two-sided inequalities for

a \to 0^{+}

\begin{aligned} K a^{3} \leq 2 \arccos (1 - a^{2}) - \sin (2 \arccos (1 - a^{2})) \leq \frac{8 \sqrt{2}}{3} a^{3} . \end{aligned}

We now introduce

\begin{aligned} G (y) := 2 \arccos (y) - 2 y \sqrt{1 - y^{2}} = 2 θ (y) - \sin (2 θ (y)) with θ (y) := \arccos (y) . \end{aligned}

For

y

in the zone (14), we set

a = \sqrt{1 - y} \in [\frac{\sqrt{C}}{(2 n + 1)^{\frac{1}{3}}}, \frac{\sqrt{C + \frac{1}{\sqrt{C}}}}{(2 n + 1)^{\frac{1}{3}}}]

. We then deduce the following inequalities for

n ≫ 1

\begin{aligned} K C^{\frac{3}{2}} & \leq (2 n + 1) G (y) \leq \frac{8 \sqrt{2}}{3} (C + \frac{1}{\sqrt{C}})^{\frac{3}{2}} \\ \frac{8 \sqrt{2}}{3} C^{\frac{3}{2}} - \frac{π}{3} & \leq (2 n + 1) G (y) \leq \frac{8 \sqrt{2}}{3} C^{\frac{3}{2}} + \frac{8 \sqrt{2}}{3} [(C + \frac{1}{\sqrt{C}})^{\frac{3}{2}} - C^{\frac{3}{2}}] . \end{aligned}

Since

C^{\frac{3}{2}}

is larger than

1

, we may bound³ :

\begin{aligned} {(C + \frac{1}{\sqrt{C}})}^{\frac{3}{2}} - C^{\frac{3}{2}} = C^{\frac{3}{2}} ({(1 + \frac{1}{C^{\frac{3}{2}}})}^{\frac{3}{2}} - 1) \leq \sqrt{8} - 1. \end{aligned}

Due to the relation

g_{n} (y) = \frac{(2 n + 1) G (y) - π}{4}

, we thus obtain

\begin{aligned} \frac{2 \sqrt{2}}{3} C^{\frac{3}{2}} - \frac{π}{3} \leq g_{n} (y) \leq \frac{2 \sqrt{2}}{3} C^{\frac{3}{2}} + \frac{\frac{8 \sqrt{2}}{3} (\sqrt{8} - 1) - π}{4} . \end{aligned}

A numerical computation shows that the upper bound is less than

\frac{2 \sqrt{2}}{3} C^{\frac{3}{2}} + \frac{π}{3}

. Finally, the condition (19) proves that

g_{n} (y)

belongs to

[\frac{- π}{3}, \frac{π}{3}] + 2 π Z

and hence

\cos (g_{n} (y)) \geq \frac{1}{2}

For the multidimensional case, the following corollary will be required. Corollary 5

Let $C \in C$ be a constant as in Proposition 3 and fix $α > 0$ satisfying $3 α < \frac{1}{\sqrt{C}}$ . Moreover, we consider $β \geq 0$ . Define the following intervals

I_{t} := [\sqrt{2 t + 1} - \frac{C + \frac{2}{3 \sqrt{C}}}{(2 t + 1)^{\frac{1}{6}}}, \sqrt{2 t + 1} - \frac{C + \frac{1}{3 \sqrt{C}}}{(2 t + 1)^{\frac{1}{6}}}] \forall t \geq 1.

(20)

For any

t ≫ 1

and

n \in N

satisfying

| n - t | \leq α (2 t + 1)^{\frac{1}{3}} + β

, the following inequality holds

min_{x \in I_{t}} h_{n} (x) \geq \frac{1}{4 C^{\frac{1}{4}} (2 t + 1)^{\frac{1}{12}}} .

(21)

Proof.

For simplicity, we use the following notations:

\begin{aligned} C_{1} = C, C_{1}^{'} = C + \frac{1}{3 \sqrt{C}} C_{2}^{'} = C + \frac{2}{3 \sqrt{C}}, C_{2} := C + \frac{1}{\sqrt{C}} . \end{aligned}

We will show that if

α

is chosen small enough then the inequality

| n - t | \leq α (2 t + 1)^{\frac{1}{3}} + β

implies that the interval

I_{t}

is included in the one appearing in (13):

\begin{aligned} [\sqrt{2 t + 1} - \frac{C_{2}^{'}}{(2 t + 1)^{\frac{1}{6}}}, \sqrt{2 t + 1} - \frac{C_{1}^{'}}{(2 t + 1)^{\frac{1}{6}}}] \subset [\sqrt{2 n + 1} - \frac{C_{2}}{(2 n + 1)^{\frac{1}{6}}}, \sqrt{2 n + 1} - \frac{C_{1}}{(2 n + 1)^{\frac{1}{6}}}] . \end{aligned}

By invoking Proposition 3, such an inclusion will yield the uniform bound

h_{n} (x) \geq \frac{0.3}{C^{\frac{1}{4}} (2 n + 1)^{\frac{1}{12}}}

for

x \in I_{t}

, which in turn implies the expected bound (21) because

n \sim t

for

t ≫ 1

We first compare the lower bounds of the two intervals and we have to show the inequality

\begin{aligned} \sqrt{2 n + 1} - \frac{C_{2}}{(2 n + 1)^{\frac{1}{6}}} \leq \sqrt{2 t + 1} - \frac{C_{2}^{'}}{(2 t + 1)^{\frac{1}{6}}} . \end{aligned}

The sequence

n \mapsto \sqrt{2 n + 1} - \frac{C_{2}}{(2 n + 1)^{\frac{1}{6}}}

is increasing, which allows us to apply the bound

\begin{aligned} n \leq t + α T^{\frac{1}{3}} + β with T = 2 t + 1, \end{aligned}

and

\begin{aligned} \sqrt{2 n + 1} - \frac{C_{2}}{(2 n + 1)^{\frac{1}{6}}} & \leq \sqrt{T + 2 α T^{\frac{1}{3}} + 2 β} - \frac{C_{2}}{(T + 2 α T^{\frac{1}{3}} + 2 β)^{\frac{1}{6}}} \\ \leq \sqrt{T} - \frac{C_{2} - α}{T^{\frac{1}{6}}} + \frac{β}{\sqrt{T}} + o (\frac{1}{\sqrt{T}}) . \end{aligned}

It suffices to show for

T ≫ 1

that the last upper bound is less or equal to

\begin{aligned} \sqrt{T} - \frac{C_{2}^{'}}{T^{\frac{1}{6}}} . \end{aligned}

It is clear that it is sufficient to set

0 < α < C_{2} - C_{2}^{'} = \frac{1}{3 \sqrt{C}}

For the comparison of the upper bounds of the two intervals, we briefly give the similar arguments. We use $t - α T^{\frac{1}{3}} - β \leq n$ so that it is sufficient to prove the inequality

\begin{aligned} \sqrt{T} - \frac{C_{1}^{'}}{T^{\frac{1}{6}}} \leq \sqrt{T - 2 α T^{\frac{1}{3}} - 2 β} - \frac{C_{1}}{(T - 2 α T^{\frac{1}{3}} - 2 β)^{\frac{1}{6}}} = \sqrt{T} - \frac{C_{1} + α}{T^{\frac{1}{6}}} - \frac{β}{\sqrt{T}} + o (\frac{1}{\sqrt{T}}) \end{aligned}

that would lead to the sufficient condition

0 < α < C_{1}^{'} - C_{1} = \frac{1}{3 \sqrt{C}}

3. Cardinality of the Set

J_{n}^{(α, β)}

We require a combinatorial lemma in which we will work under the assumptions $T \to + \infty$ and $T = o (n)$ for $n \to + \infty$ .

Lemma 6
For any integer $d \geq 2$ , there exists a constant $C (d) \geq 1$ , with the property that, for any $n \geq d (d - 1)$ and any real number $T$ satisfying $d - 1 \leq T \leq \frac{n}{d}$ , the cardinality of the set $Λ$ defined below is of order $T^{d - 1}$ :
$\begin{aligned} Λ & := {(n_{1}, \dots, n_{d}) \in N^{d}, max_{1 \leq i \leq d} | n_{i} - \frac{n}{d} | \leq T and \sum_{i = 1}^{d} n_{i} = n}, \end{aligned}$

$\begin{aligned} \frac{1}{C (d)} T^{d - 1} & \leq Card (Λ) \leq C (d) T^{d - 1} . \end{aligned}$

Proof.
The case $d = 2$ follows directly from
$\begin{aligned} Card (Λ) = Card (N \cap [\frac{n}{2} - T, \frac{n}{2} + T]) ≃ T . \end{aligned}$
In what follows, we assume $d \geq 3$ .

Step 1. The upper bound follows from the fact that each $n_{i}$ belongs to $[\frac{n}{d} - T, \frac{n}{d} + T]$ . Thus $(n_{1}, \dots, n_{d - 1})$ runs over a set of size $O (T^{d - 1})$ and $- n_{d} = n_{1} + \dots + n_{d - 1}$ is determined by the choice of the $d - 1$ first integers.

We now explain the lower bound of $Card (Λ)$ .

Step 2. Let $n_{1}^{⋆}, \dots, n_{d - 1}^{⋆}$ be $d - 1$ non-negative integers satisfying for each $i \in {1, \dots, d - 1}$
$| \frac{n}{d} - n_{i}^{⋆} | \leq \frac{1}{2} .$
(22)
The integer $n_{d}^{⋆} := n - (n_{1}^{⋆} + \dots + n_{d - 1}^{⋆}) = \frac{n}{d} + \sum_{i = 1}^{d - 1} (\frac{n}{d} - n_{i}^{⋆})$ satisfies
$| \frac{n}{d} - n_{d}^{⋆} | \leq \frac{d - 1}{2} and n_{d}^{⋆} \geq \frac{n}{d} - \frac{d - 1}{2} \geq \frac{d - 1}{2} > 0.$
(23)
We have just identified an element $(n_{1}^{⋆}, \dots, n_{d}^{⋆})$ of $Λ$ .

Step 3. In order to construct other elements of $Λ$ , consider $(μ_{1}, \dots, μ_{d - 1}) \in Z^{d - 1}$ satisfying
$\begin{aligned} max_{1 \leq i \leq d - 1} | μ_{i} | \leq \frac{T}{2 (d - 1)} . \end{aligned}$
The number of tuples $(μ_{1}, \dots, μ_{d - 1})$ is ${(1 + 2 ⌊ \frac{T}{2 (d - 1)} ⌋)}^{d - 1} \geq {(\frac{T}{2 (d - 1)})}^{d}$ . For $T \geq d - 1$ , we will use the following consequence:
$| μ_{1} | + \dots + | μ_{d - 1} | \leq \frac{T}{2} \leq T - \frac{d - 1}{2} .$
(24)
We now introduce the following linearly independent vectors:
$\begin{aligned} e_{1} := (\begin{matrix} - 1 \\ 0 \\ ⋮ \\ ⋮ \\ 0 \\ 1 \end{matrix}), e_{2} := (\begin{matrix} 0 \\ - 1 \\ 0 \\ ⋮ \\ 0 \\ 1 \end{matrix}), \dots, e_{d - 1} := (\begin{matrix} 0 \\ ⋮ \\ ⋮ \\ 0 \\ - 1 \\ 1 \end{matrix}) . \end{aligned}$
We claim that all the above conditions ensure that the following point belongs to $Λ$ :
$(\begin{matrix} n_{1}^{⋆} \\ ⋮ \\ n_{d - 1}^{⋆} \\ n_{d}^{⋆} \end{matrix}) + \sum_{k = 1}^{d - 1} μ_{k} e_{k} = (\begin{matrix} n_{1}^{⋆} - μ_{1} \\ ⋮ \\ n_{d - 1}^{⋆} - μ_{d - 1} \\ n_{d}^{⋆} + μ_{1} + \dots + μ_{d - 1} \end{matrix}) .$
(25)

$∙$ It is clear that the sum of the coordinates equals $n$ .

$∙$ We now check
$max_{1 \leq i \leq d - 1} | n_{i}^{⋆} - μ_{i} - \frac{n}{d} | \leq T and | n_{d}^{⋆} + μ_{1} + \dots + μ_{d - 1} - \frac{n}{d} | \leq T .$
(26)
Such inequalities come directly from (22), (23) and (24).

$∙$ We finally note that (26) and the assumption $T \leq \frac{n}{d}$ show that each coordinate of (25) is non-negative.

4. Proof of Theorem 1

Let $C = min (C) = {(\frac{3 π}{\sqrt{2}})}^{\frac{2}{3}}$ as in Proposition 3. In particular, $\frac{1}{3 \sqrt{C}} > 0.17 > \frac{1}{6}$ . We fix $α \in (0, \frac{1}{6}]$ so that we can apply Corollary 5. We then set

\begin{aligned} T = α (\frac{2 n}{d} + 1)^{\frac{1}{3}} + β . \end{aligned}

In particular, one has

T = o (n)

. Lemma 6 then ensures that, for

n ≫ 1

, the set

J_{n}^{(α, β)}

has size of order

n^{\frac{d - 1}{3}}

. Consider the interval

I_{t}

defined in (20) for

t = \frac{n}{d}

. For any

(x_{1}, \dots, x_{d})

belonging to the subset

I_{\frac{n}{d}} \times \dots \times I_{\frac{n}{d}}

, the following holds:

\begin{aligned} v_{n}^{(α, β)} (x_{1}, \dots, x_{n}) & ≳ \frac{1}{\sqrt{Card (J_{n}^{(α, β)}})} \sum_{(n_{1}, \dots, n_{d}) \in J_{n}^{(α, β)}} n^{- \frac{d}{12}} \\ ≳ \sqrt{Card (J_{n}^{(α, β)})} n^{- \frac{d}{12}} \\ ≳ n^{\frac{d}{12} - \frac{1}{6}} . \end{aligned}

A numerical computation allows us to restrict

I_{\frac{n}{d}} \times \dots \times I_{\frac{n}{d}}

to the following hypercube:

\begin{aligned} {[\sqrt{\frac{2 n}{d} + 1} - \frac{3.89}{{(\frac{2 n}{d} + 1)}^{\frac{1}{6}}}, \sqrt{\frac{2 n}{d} + 1} - \frac{3.72}{{(\frac{2 n}{d} + 1)}^{\frac{1}{6}}}]}^{d} \end{aligned}

whose volume is of order

n^{- \frac{d}{6}}

. We finally integrate the norm in

L^{p} (R^{d})

of the function equalling

1

on this hypercube to get the conclusion:

\begin{aligned} ‖ v_{n}^{(α, β)} ‖_{L^{p} (R^{d})} ≳ n^{\frac{d}{12} - \frac{1}{6}} \times n^{- \frac{d}{6 p}} . \end{aligned}

5. Appendix: Explicit Inequalities About Airy Functions

We recall several notations and write explicit inequalities which are consequences of Olver (1997). The Airy function is defined as the improper integral

\begin{aligned} Ai (X) = \frac{1}{π} \int_{0}^{+ \infty} \cos (\frac{t^{3}}{3} + X t) d t, \forall X \in R . \end{aligned}

X \to + \infty

, the following asymptotic is known:

Ai (X) \underset{X \to + \infty}{\sim} \frac{1}{2 \sqrt{π} X^{\frac{1}{4}}} e^{- \frac{2}{3} X^{\frac{3}{2}}} .

(27)

We require the following result that gives a quantitative statement of the oscillating behaviour on

(- \infty, 0)

. Lemma 7

The following holds true for any $X > 0$ :

Ai (- X) = \frac{1}{\sqrt{π} X^{\frac{1}{4}}} \cos (\frac{2}{3} X^{\frac{3}{2}} - \frac{π}{4}) + \tilde{ε} (X) with | \tilde{ε} (X) | \leq \frac{5}{48 \sqrt{π} X^{\frac{7}{4}}} .

(28)

Moreover the constant

\frac{5}{48 \sqrt{π}}

is optimal.

Proof.

Fix $ν > \frac{- 1}{2}$ , we shall consider the first Hankel function $H_{ν}^{(1)}$ which may be defined by the following formula (see Watson, 1945, page 168, line (3) or Olver, 1997, page 268, lines (13.01) and (13.07)) for positive $x$ :

\begin{aligned} H_{ν}^{(1)} (x) = \sqrt{\frac{2}{π x}} \frac{e^{i (x - \frac{π ν}{2} - \frac{π}{4})}}{Γ (ν + \frac{1}{2})} \int_{0}^{+ \infty} e^{- τ} τ^{ν - \frac{1}{2}} {(1 + \frac{i τ}{2 x})}^{ν - \frac{1}{2}} d τ . \end{aligned}

Following Olver (1997, page 269, Ex 13.1), we introduce two real terms⁴

P (ν, x)

and

Q (ν, x)

as follows:

\begin{aligned} P (ν, x) + i Q (ν, x) & = H_{ν}^{(1)} (x) \sqrt{\frac{π x}{2}} e^{- i (x - \frac{π ν}{2} - \frac{π}{4})} \\ = \frac{1}{Γ (ν + \frac{1}{2})} \int_{0}^{+ \infty} e^{- τ} τ^{ν - \frac{1}{2}} {(1 + \frac{i τ}{2 x})}^{ν - \frac{1}{2}} d τ . \end{aligned}

(29)

Since the following inequality will be needed at the end of this section, we first require

ν \in (\frac{- 1}{2}, \frac{1}{2}]

\begin{aligned} \sqrt{P (ν, x)^{2} + Q (ν, x)^{2}} & \leq \frac{1}{Γ (ν + \frac{1}{2})} \int_{0}^{+ \infty} e^{- τ} τ^{ν - \frac{1}{2}} {| 1 + \frac{i τ}{2 x} |}^{ν - \frac{1}{2}} d τ \\ \leq \frac{1}{Γ (ν + \frac{1}{2})} \int_{0}^{+ \infty} e^{- τ} τ^{ν - \frac{1}{2}} {(1 + \frac{τ^{2}}{4 x^{2}})}^{\frac{ν}{2} - \frac{1}{4}} d τ \\ \leq \frac{1}{Γ (ν + \frac{1}{2})} \int_{0}^{+ \infty} e^{- τ} τ^{ν - \frac{1}{2}} d τ \\ \leq 1. \end{aligned}

(30)

For

ν \in (\frac{- 1}{2}, \frac{3}{2}]

, we now claim that the following inequality holds true for any

x > 0

\sqrt{(P (ν, x) - 1)^{2} + Q (ν, x)^{2}} \leq \frac{| ν - \frac{1}{2} | (ν + \frac{1}{2})}{2 x} .

(31)

The proof follows directly from the previous integral representations:

\begin{aligned} P (ν, x) - 1 + i Q (ν, x) & = \frac{1}{Γ (ν + \frac{1}{2})} \int_{0}^{+ \infty} e^{- τ} τ^{ν - \frac{1}{2}} [{(1 + \frac{i τ}{2 x})}^{ν - \frac{1}{2}} - 1] d τ \\ \sqrt{(P (ν, x) - 1)^{2} + Q (ν, x)^{2}} & \leq \frac{1}{Γ (ν + \frac{1}{2})} \int_{0}^{+ \infty} e^{- τ} τ^{ν - \frac{1}{2}} | {(1 + \frac{i τ}{2 x})}^{ν - \frac{1}{2}} - 1 | d τ . \end{aligned}

By using

ν \leq \frac{3}{2}

, we may invoke the inequality

\begin{aligned} | {(1 + \frac{i τ}{2 x})}^{ν - \frac{1}{2}} - 1 | = | \int_{0}^{\frac{τ}{2 x}} (ν - \frac{1}{2}) (1 + i ω)^{ν - \frac{3}{2}} d ω | \leq \frac{| ν - \frac{1}{2} | τ}{2 x} . \end{aligned}

As a consequence, we continue our computations to prove (31):

\begin{aligned} \sqrt{(P (ν, x) - 1)^{2} + Q (ν, x)^{2}} & \leq \frac{| ν - \frac{1}{2} |}{2 x Γ (ν + \frac{1}{2})} \int_{0}^{+ \infty} e^{- τ} τ^{ν + \frac{1}{2}} d τ \\ \leq \frac{| ν - \frac{1}{2} | Γ (ν + \frac{3}{2})}{2 x Γ (ν + \frac{1}{2})} \\ \leq \frac{| ν - \frac{1}{2} | (ν + \frac{1}{2})}{2 x} . \end{aligned}

For

ν = \frac{1}{3}

, this inequality reads

\sqrt{(P (\frac{1}{3}, x) - 1)^{2} + Q (\frac{1}{3}, x)^{2}} \leq \frac{5}{72 x} .

(32)

We use the following identity, which relates the Airy function to the Hankel function

H_{\frac{1}{3}}^{(1)}

and, consequently, to

P (\frac{1}{3}, \cdot)

and

Q (\frac{1}{3}, \cdot)

(see Olver, 1997, page 392, line (1.05) and then use (29)):

\begin{aligned} Ai (- X) & = \sqrt{\frac{X}{3}} ℜ (e^{i \frac{π}{6}} H_{\frac{1}{3}}^{(1)} (ξ)) with ξ = \frac{2}{3} X^{\frac{3}{2}} \\ = \frac{1}{\sqrt{π} X^{\frac{1}{4}}} ℜ (e^{i (ξ - \frac{π}{4})} (P (\frac{1}{3}, ξ) + i Q (\frac{1}{3}, ξ))) \\ = \frac{1}{\sqrt{π} X^{\frac{1}{4}}} (\cos (ξ - \frac{π}{4}) P (\frac{1}{3}, ξ) - \sin (ξ - \frac{π}{4}) Q (\frac{1}{3}, ξ)) . \end{aligned}

(33)

Then (28) is a straightforward consequence of (32) and the last identity.

The sharpness of (28) for $X \to + \infty$ is due to the direct appearance of the constant $\frac{5}{48 \sqrt{π}}$ in the classical asymptotics of the Airy function⁵ :

Ai (- X) = \frac{1}{\sqrt{π} X^{\frac{1}{4}}} \cos (\frac{2}{3} X^{\frac{3}{2}} - \frac{π}{4}) + \frac{5}{48 \sqrt{π} X^{\frac{7}{4}}} \sin (\frac{2}{3} X^{\frac{3}{2}} - \frac{π}{4}) + O (\frac{1}{X^{\frac{13}{4}}}) .

(34)

It is sufficient to make

X

tend to

+ \infty

under the condition

| \sin (\frac{2}{3} X^{\frac{3}{2}} - \frac{π}{4}) | = 1

Following Olver (1997, pages 394 $-$ 395), one may define three functions as follows.

$∙$ The Airy function $Bi$ of the second kind is defined by

\begin{aligned} Bi (X) = \frac{1}{π} \int_{0}^{+ \infty} e^{\frac{- 1}{3} t^{3} + X t} + \sin (\frac{t^{3}}{3} + X t) dt \forall X \in R . \end{aligned}

Moreover, a formula similar to (33) also holds (see Olver (1997, page 394) by setting

ξ = \frac{2}{3} X^{\frac{3}{2}}

or Olver (1997, page 393, line (1.14)) and (29)):

Bi (- X) = \frac{- 1}{\sqrt{π} X^{\frac{1}{4}}} (\sin (ξ - \frac{π}{4}) P (\frac{1}{3}, ξ) + \cos (ξ - \frac{π}{4}) Q (\frac{1}{3}, ξ)) for X > 0.

(35)

$∙$ A weight function $E$ has a definition involving

\begin{aligned} c = - 0.366 \dots \end{aligned}

the largest negative solution of

Ai (c) = Bi (c)

. We then define

\begin{aligned} E (X) = {\begin{array}{rcl} 1 & for & X < c, \\ \sqrt{\frac{Bi (X)}{Ai (X)}} & for & X \geq c . \end{array} \end{aligned}

As explained in Olver (1997), the function

E

is non-decreasing and satisfies

E (X) \geq 1, \forall X \in R .

(36)

$∙$ A modulus function $M$ is finally given by

\begin{aligned} M (X) = {\begin{array}{rcl} \sqrt{Ai {(X)}^{2} + Bi {(X)}^{2}} & for & X < c, \\ \sqrt{2 Ai (X) Bi (X)} & for & X \geq c . \end{array} \end{aligned}

Note that for any

X > 0

, we may write

\begin{aligned} M (- X) \leq \sqrt{Ai (- X)^{2} + Bi (- X)^{2}} . \end{aligned}

From (33) and (35) and then (30), we infer the following

\begin{aligned} M (- X) \leq & \frac{1}{\sqrt{π} X^{\frac{1}{4}}} \sqrt{P (\frac{1}{3}, ξ)^{2} + Q (\frac{1}{3}, ξ)^{2}} \\ \leq & \frac{1}{\sqrt{π} X^{\frac{1}{4}}} . \end{aligned}

(37)

We conclude this section with a remark that will be used below. For any $X > 0$ (and thus $X \geq c$ ), the following identity holds:

\frac{M (X)}{E (X)} = \sqrt{2} Ai (X) .

(38)

6. Appendix: Proof of Proposition 2

Instead of following the constants in the proof of Skovgaard (1959), we prefer to use the results presented in Olver (1997, page 403, Ex 4.2 & 4.3) since some explicit computations are already provided.

We aim to study the following function for $n ≫ 1$ :

\begin{aligned} R_{n} : y \in [0, 1) \mapsto (2 n + 1)^{\frac{5}{4}} (1 - y^{2})^{\frac{7}{4}} (h_{n} (y \sqrt{2 n + 1}) - \frac{\sqrt{2}}{\sqrt{π} (2 n + 1)^{\frac{1}{4}}} \frac{\cos (g_{n} (y))}{(1 - y^{2})^{\frac{1}{4}}}) . \end{aligned}

While precise statements are given below, we first note that, as

y \to 1

h_{n}

is approximated by an Airy-type function, in its oscillatory regime, that roughly reads

h_{n} (y \sqrt{2 n + 1}) \approx \frac{2^{\frac{1}{3}}}{(2 n + 1)^{\frac{1}{12}}} Ai ((2 n + 1)^{\frac{2}{3}} ζ (y)) with ζ (y) \underset{y \to 1^{-}}{\sim} - 2^{\frac{1}{3}} (1 - y) .

(39)

The figure below shows two graphical representations of

R_{n}

for

n \in {20, 50}

, compared with approximations

R_{n}

in which an Airy-type function is used instead of

h_{n}

(in fact, the principal term of (41) was employed because it is more precise than (39)).

In particular, we note the following observations: –

the two curves are essentially superposed for $y$ near $1$ , indicating that $h_{n}$ is well approximated by an Airy-type function near the turning point $\sqrt{2 n + 1}$ ,

–

if the Airy function is replaced with its classical asymptotics at $- \infty$ , see (34), one conjectures that a natural expected bound of $R_{n}$ would be $\frac{5 \sqrt{8}}{48 \sqrt{π}} = 0.166..$ .

However, a residual term in the proof yields a larger upper bound $0.5$ (see (49)), which is sufficient for our purpose (see (18)). Proposition 2 constitutes a special case of the following result.

Proposition 8

Given any $K > \frac{5 \sqrt{2}}{24 \sqrt{π}}$ , there exists a constant $y_{K} \in [0, 1)$ such that, for all $n ≫ 1$ , the following inequality holds uniformly for $y \in (- 1, y_{K}] \cup [y_{K}, 1)$ :

| h_{n} (y \sqrt{2 n + 1}) - \frac{\sqrt{2}}{\sqrt{π} (2 n + 1)^{\frac{1}{4}}} \frac{\cos (g_{n} (y))}{(1 - y^{2})^{\frac{1}{4}}} | \leq \frac{K}{(2 n + 1)^{\frac{5}{4}} (1 - y^{2})^{\frac{7}{4}}} .

(40)

Moreover, for the constant

K = 0.5

, the value

y_{0.5} = 0

is convenient.

We now outline the main ideas for obtaining an explicit constant in the upper bound of (40).

From (10), the formulas $g_{n} (- y) = n π - g_{n} (y)$ and $h_{n} (- x) = (- 1)^{n} h_{n} (x)$ ensure that it is sufficient to prove (40) for $y \in [0, 1)$ .

From the differential equation (3) satisfied by $h_{n}$ , we deduce that the function $y \mapsto h_{n} (y \sqrt{2 n + 1})$ satisfies the following one:

\begin{aligned} - \frac{d^{2} w}{d y^{2}} + (2 n + 1)^{2} (y^{2} - 1) w (y) = 0 \forall y \in [0, + \infty) . \end{aligned}

The set of such

w

forms a two-dimensional vector space. We now invoke Olver (1997, page 399, Theorem 3.1) that provides two solutions expressed with the Airy functions

Ai

and

Bi

: one solution diverges to

+ \infty

as its argument tend to

+ \infty

whereas the other tends to

0

. It follows that these two solutions are linearly independent and constitute a basis of the vector space of all solutions. In particular, there is a unique solution, up to a multiplicative constant, that tends to

0

+ \infty

. Since

y \mapsto h_{n} (y \sqrt{2 n + 1})

tends to

0

+ \infty

, we may use the asymptotic form in Olver (1997, page 399, Theorem 3.1 and page 403, Ex 4.2 and 4.3) and claim that there exists a constant

C_{n}

satisfying:

h_{n} (y \sqrt{2 n + 1}) = C_{n} (\frac{- ζ (y)}{1 - y^{2}})^{\frac{1}{4}} (Ai ((2 n + 1)^{\frac{2}{3}} ζ (y)) + ε (n, y)) \forall y \geq 0

(41)

in which the function

ζ

and the remainder

ε

are now given. It should be noted that the factor of the Airy function is essentially constant for

n ≫ 1

(see (47)) and

y \to 1

(see (42)):

\begin{aligned} C_{n} (\frac{- ζ (y)}{1 - y^{2}})^{\frac{1}{4}} \underset{n \to + \infty}{\sim} \frac{\sqrt{2}}{(2 n + 1)^{\frac{1}{12}}} \times \frac{1}{2^{\frac{1}{6}}} . \end{aligned}

The function

ζ

is a continuously differentiable solution of

ζ (y) ζ^{'} {(y)}^{2} = y^{2} - 1

, yielding the classical formulas:

\begin{array}{ccccl} 0 \leq y \leq 1 & \Rightarrow & ζ (y) = - Ω {(y)}^{\frac{2}{3}} & with & Ω (y) = \frac{3}{4} (\arccos (y) - y \sqrt{1 - y^{2}}) \\ = \frac{3}{2} \int_{y}^{1} \sqrt{1 - ω^{2}} d ω . \\ 1 \leq y & \Rightarrow & ζ (y) = Φ {(y)}^{\frac{2}{3}} & with & Φ (y) = \frac{3}{4} (y \sqrt{y^{2} - 1} - \ln (y + \sqrt{y^{2} - 1})) \\ = \frac{3}{2} \int_{1}^{y} \sqrt{ω^{2} - 1} d ω . \end{array}

(42)

Here is a remark useful for the end of the proof:

\begin{aligned} 2 Ω (y) - (1 - y^{2})^{\frac{3}{2}} = \int_{y}^{1} 3 (1 - ω) \sqrt{1 - ω^{2}} d ω \geq 0, \forall y \in [0, 1] \end{aligned}

which indeed reads

(- ζ (y))^{\frac{3}{2}} \geq \frac{(1 - y^{2})^{\frac{3}{2}}}{2}, \forall y \in [0, 1] .

(43)

Explicit bounds of the remainder $ε$ for $y \in [0, 1)$ . Once numerical computations have been performed in the result presented in Olver (1997, Theorem 3.1), the following explicit bounds are reported in Olver (1997, page 403, Ex 4.3) for $y \in [0, 1]$ :

\begin{aligned} | ε (n, y) | \leq 0.97 (e^{\frac{0.28}{2 n + 1}} - 1) \frac{M ((2 n + 1)^{\frac{2}{3}} ζ (y))}{E ((2 n + 1)^{\frac{2}{3}} ζ (y))} . \end{aligned}

Using (36) and (37) from the previous section, the following inequality holds uniformly in the region

y \in [0, 1)

for

n ≫ 1

| ε (n, y) | \leq \frac{0.28}{2 n + 1} \times \frac{1}{\sqrt{π} (2 n + 1)^{\frac{1}{6}} (- ζ (y))^{\frac{1}{4}}} \leq \frac{0.16}{(2 n + 1)^{\frac{7}{6}} (- ζ (y))^{\frac{1}{4}}} .

(44)

Negligible contribution of the remainder $ε$ as $y \to + \infty$ . We use Olver (1997, page 397, line (2.15) and page 403, Ex 4.2) and (38) to obtain

| ε (n, y) | \leq \frac{Ai ((2 n + 1)^{\frac{2}{3}} ζ (y))}{1.04} [\exp (\frac{1.05}{2 n + 1} \int_{y}^{+ \infty} | H^{'} (ω) | d ω) - 1]

(45)

for a suitable explicit function

H : (- 1, + \infty) \to [0, + \infty)

given by

\begin{aligned} H (y) = \frac{5}{24 | ζ (y) |^{\frac{3}{2}}} + \frac{y^{3} - 6 y}{12 | y^{2} - 1 |^{\frac{3}{2}}} . \end{aligned}

In particular, the derivative of

H

(1, + \infty)

can be expressed in terms of

Φ (y)

(see (42)) and satisfies, as

y \to + \infty

\begin{aligned} H^{'} (y) = \frac{- 5 Φ^{'} (y)}{24 Φ {(y)}^{2}} + \frac{3 y^{2} + 2}{4 (y^{2} - 1)^{5 / 2}} = O (\frac{1}{y^{3}}) . \end{aligned}

Hence, for a fixed

n

, (45) shows

| ε (n, y) | = o (Ai ((2 n + 1)^{\frac{2}{3}} ζ (y))), y \to + \infty .

(46)

Computation and asymptotics of $C_{n}$ in (41). We fix $n$ and study the asymptotics, as $y \to + \infty$ , of the two sides of (41). For the left-hand side of (41), we refer to (2) and recall that $H_{n}$ is a polynomial with leading term $2^{n} X^{n}$ . For the right-hand side of (41), we refer to (46), (27) and use $ζ {(y)}^{\frac{3}{2}} = Φ (y) = \frac{3}{4} y^{2} - \frac{3}{4} \ln (2 y) - \frac{3}{8} + o (1)$ . We then write

\begin{aligned} \frac{2^{\frac{n}{2}} (2 n + 1)^{\frac{n}{2}}}{π^{\frac{1}{4}} \sqrt{n!}} y^{n} e^{- \frac{(2 n + 1)}{2} y^{2}} & \underset{y \to + \infty}{\sim} \frac{C_{n}}{2 \sqrt{π} (2 n + 1)^{\frac{1}{6}} \sqrt{y}} e^{- \frac{2}{3} (2 n + 1) ζ {(y)}^{\frac{3}{2}}} \\ \underset{y \to + \infty}{\sim} \frac{C_{n}}{2 \sqrt{π} (2 n + 1)^{\frac{1}{6}} \sqrt{y}} e^{\frac{2 n + 1}{4}} (2 y)^{\frac{2 n + 1}{2}} e^{- \frac{(2 n + 1)}{2} y^{2}} . \end{aligned}

The previous comparison leads to the following computation:

\begin{aligned} C_{n} = π^{\frac{1}{4}} \frac{(2 n + 1)^{\frac{n}{2} + \frac{1}{6}} e^{- \frac{2 n + 1}{4}}}{2^{\frac{n - 1}{2}} \sqrt{n!}} . \end{aligned}

In what follows, we require a two-term asymptotic expansion of the sequence

(C_{n})

for

n \to + \infty

C_{n} = \frac{\sqrt{2}}{(2 n + 1)^{\frac{1}{12}}} + \frac{\sqrt{2}}{24 (2 n + 1)^{\frac{13}{12}}} + o (\frac{1}{(2 n + 1)^{\frac{13}{12}}})

(47)

and also an approximation of

\frac{C_{n}}{\sqrt{π} (2 n + 1)^{\frac{1}{6}}}

. A numerical computation yields

\frac{\sqrt{2}}{24 \sqrt{π}} < 0.1

(at this point, we prefer a much larger bound to get a simple final constant

0.5

in (11)). Consequently, for

n ≫ 1

, the following inequality holds:

| \frac{C_{n}}{\sqrt{π} (2 n + 1)^{\frac{1}{6}}} - \frac{\sqrt{2}}{\sqrt{π} (2 n + 1)^{\frac{1}{4}}} | \leq \frac{0.1}{(2 n + 1)^{\frac{5}{4}}} .

(48)

Conclusion. We now have all the ingredients to prove (40). We first exploit (48):

\begin{aligned} | h_{n} (y \sqrt{2 n + 1}) - \frac{\sqrt{2}}{\sqrt{π} (2 n + 1)^{\frac{1}{4}}} \frac{\cos (g_{n} (y))}{(1 - y^{2})^{\frac{1}{4}}} | \\ \leq | h_{n} (y \sqrt{2 n + 1}) - \frac{C_{n}}{\sqrt{π} (2 n + 1)^{\frac{1}{6}}} \frac{\cos (g_{n} (y))}{(1 - y^{2})^{\frac{1}{4}}} | + \frac{0.1}{(2 n + 1)^{\frac{5}{4}} (1 - y^{2})^{\frac{1}{4}}} . \end{aligned}

By the triangle inequality, the last upper bound can be majorized by

\begin{aligned} | h_{n} (y \sqrt{2 n + 1}) - C_{n} (\frac{- ζ (y)}{1 - y^{2}})^{\frac{1}{4}} Ai ((2 n + 1)^{\frac{2}{3}} ζ (y)) | \\ + | C_{n} (\frac{- ζ (y)}{1 - y^{2}})^{\frac{1}{4}} Ai ((2 n + 1)^{\frac{2}{3}} ζ (y)) - \frac{C_{n}}{\sqrt{π} (2 n + 1)^{\frac{1}{6}}} \frac{\cos (g_{n} (y))}{(1 - y^{2})^{\frac{1}{4}}} | + \frac{0.1}{(2 n + 1)^{\frac{5}{4}} (1 - y^{2})^{\frac{1}{4}}} . \end{aligned}

We note that

g_{n} (y)

equals

\frac{2}{3} X^{\frac{3}{2}} - \frac{π}{4}

with

X = - (2 n + 1)^{\frac{2}{3}} ζ (y)

(see (10) and (42)). We may then use the Airy-function approximation (41) and (28) to bound the last term by

\begin{aligned} C_{n} (\frac{- ζ (y)}{1 - y^{2}})^{\frac{1}{4}} (| ε (n, y) | + | \tilde{ε} (- (2 n + 1)^{\frac{2}{3}} ζ (y)) |) + \frac{0.1}{(2 n + 1)^{\frac{5}{4}} (1 - y^{2})^{\frac{1}{4}}} . \end{aligned}

Then (44) and the bound of the remainder in (28) give the following upper bound:

\begin{aligned} \frac{C_{n} \times 0.16}{(2 n + 1)^{\frac{7}{6}} (1 - y^{2})^{\frac{1}{4}}} + \frac{C_{n} \times 5}{(2 n + 1)^{\frac{14}{12}} \times 48 \sqrt{π} (1 - y^{2})^{\frac{1}{4}} (- ζ (y))^{\frac{6}{4}}} + \frac{0.1}{(2 n + 1)^{\frac{5}{4}} (1 - y^{2})^{\frac{1}{4}}} . \end{aligned}

We are now in a position to conclude. We write

\begin{aligned} K = \frac{5 \sqrt{2}}{24 \sqrt{π}} + 2 δ with δ > 0 \end{aligned}

and we use (47) and (43) so that we obtain the following upper bound for

n ≫ 1

\frac{\sqrt{2} \times 0.16 + δ}{(2 n + 1)^{\frac{15}{12}} (1 - y^{2})^{\frac{1}{4}}} + \frac{\frac{\sqrt{2} \times 5}{48 \sqrt{π}} \times 2 + δ}{(2 n + 1)^{\frac{15}{12}} (1 - y^{2})^{\frac{7}{4}}} + \frac{0.1}{(2 n + 1)^{\frac{5}{4}} (1 - y^{2})^{\frac{1}{4}}}

(49)

which is less than

\frac{1}{(2 n + 1)^{\frac{5}{4}} (1 - y^{2})^{\frac{7}{4}}} ((0.33 + δ) (1 - y^{2})^{\frac{6}{4}} + \frac{5 \sqrt{2}}{24 \sqrt{π}} + δ) .

(50)

The term

(0.33 + δ) (1 - y^{2})^{\frac{6}{4}}

is the one we referred to as residual before the statement of Proposition 8. Such an upper bound finally proves Proposition 8 because

–

the last factor is less than $K = \frac{5 \sqrt{2}}{24 \sqrt{π}} + 2 δ$ for $y \in [y_{K}, 1)$ for a suitable $y_{K}$ near $1^{-}$ ,

–

one has $0.33 + \frac{5 \sqrt{2}}{24 \sqrt{π}} < 0.5$ and $(1 - y^{2})^{\frac{6}{4}} \leq 1$ . We deduce that, for $δ > 0$ small enough, the factor in (50) is less than $0.5$ uniformly for $y \in [0, 1)$ .

Footnotes

Acknowledgements

The author would like to thank Marie Landeiro Dos Reis for motivating the presentation of the results in this article.

Ethical Considerations

Not applicable.

Funding Statements

The author received no financial support for the research, authorship, and/or publication of this article.

Declaration of Conflicting Interests

The author declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.

Notes

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