On hyperplane sections and projections in $l_p^n$

Abstract

For $2< p < p_0 \simeq 26.265$, the hyperplane section of the $l_p^n$-unit ball $B_p^n$ perpendicular to $a^{(n)} = \frac{1}{\sqrt{n}} (1 ,\ldots ,1)$ for large n has larger volume than the one orthogonal to $a^{(2)} = \frac{1}{\sqrt{2}} (1,1,0 ,\ldots ,0)$, as shown by Oleszkiewicz. This is different from the case of $l_\infty ^n$ considered by Ball. We give a quantitative estimate for which dimensions n this happens, namely for $n > c (\frac{1}{p_0-p} + \frac{1}{p-2})$ for some absolute constant $c>0$. Correspondingly for projections of $B_q^n$ onto hyperplanes, Barthe and Naor showed that projections onto hyperplanes perpendicular to $a^{(n)}$ have smaller volume for large n than onto the one orthogonal to $a^{(2)}$, if $\frac{4}{3}< q < 2$, different from the case $q=1$. We show that this happens for all $n > 5 (\frac{1}{q-\frac{4}{3}} + \frac{1}{2-q})$.

1 Introduction and main results

In a well-known paper Ball [2] proved that the hyperplane section of the n-cube $B_\infty ^n:= [-1,1]^n$ perpendicular to $a^{(2)} = \frac{1}{\sqrt{2}} (1,1,0 ,\ldots ,0) \in S^{n-1} \subset {\mathbb {R}}^n$ has maximal volume among all hyperplane sections. Earlier Hadwiger [6] and Hensley [7] had shown independently of one another that coordinate hyperplanes, e.g. orthogonal to $a^{(1)} = (1,0 ,\ldots ,0) \in S^{n-1}$, yield the minimal $(n-1)$-dimensional cubic sections.

For $0< p < \infty$, let $B_p^n:= \{ x = (x_j)_{j=1}^n \in {\mathbb {R}}^n \ | \ \Vert x \Vert _p:= (\sum _{j=1}^n |x_j|^p )^{\frac{1}{p}} \le 1 \ \}$ be the unit ball of $l_p^n = ({\mathbb {R}}^n, \Vert \cdot \Vert _p)$. Meyer and Pajor [13] found extremal sections of the $l_p^n$ balls $B_p^n$: They proved that the normalized volume of sections of $B_p^n$ by a fixed hyperplane is monotone increasing in p. This implies that coordinate planes provide the minimal sections for $2 \le p < \infty$, as for $p=\infty$, and the maximal sections for $1 \le p \le 2$. The minimal hyperplane sections of $B_1^n$ are those orthogonal to a main diagonal, e.g. $a^{(n)} = \frac{1}{\sqrt{n}}(1 ,\ldots ,1) \in S^{n-1}$, see also [13]. Koldobsky [10] extended this to the full range $1\le p \le 2$.

This left open the case of the maximal hyperplane section of $B_p^n$ for $2< p < \infty$. The situation there is more complicated, since the maximal hyperplane may depend as well on p as on the dimension n: Oleszkiewicz [14] proved that Ball’s result does not transfer to the balls $B_p^n$ if $2< p < p_0 \simeq 26.265$: then the intersection of the hyperplane perpendicular to $a^{(n)}$ has larger volume than the one orthogonal to $a^{(2)}$, for sufficiently large dimensions n. The value $p_0$ is the unique value in $(2,\infty )$ such that $\lim _{n \rightarrow \infty } \frac{\textrm{vol}_{n-1}(B_p^n \cap (a^{(n)})^\perp )}{\textrm{vol}_{n-1}(B_p^n \cap (a^{(2)})^\perp )} > 1$ for all $2< p < p_0$. Oleszkiewicz’s result is an asymptotic one, not determining dimensions n for which this happens. We derive a quantitative estimate for dimensions n such that this holds, namely for $n > c \left( \frac{1}{p_0-p} + \frac{1}{p-2} \right)$. On the other hand, recently Eskenazis, Nayar and Tkocz [5] proved that Ball’s result is stable for $l_p^n$ and very large p: $(a^{(2)})^\perp \cap B_p^n$ is the maximal hyperplane section of $B_p^n$ for all dimensions, provided that $p_1:= 10^{15} \le p < \infty$. They call it “resilience of cubic sections".

Dual to hyperplane sections of convex bodies are projections of convex bodies onto hyperplanes. The known results for $l_p^n$-balls show a duality between sections and projections, when maximal and minimal directions a and p and the conjugate index $q = \frac{p}{p-1}$ are interchanged. Nevertheless the proofs in both situations are different, since volume does not behave well under duality. Barthe and Naor [3] determined the extremal hyperplane projections of $l_q^n$-balls except for the minimal hyperplane projections when $1< q < 2$, corresponding to the dual maximal section case mentioned above when $2< p < \infty$. For $q=1$, the projection of $B_1^n$ onto the hyperplane perpendicular to $a^{(1)} = (1,0 ,\ldots ,0)$ is maximal, the projection onto the hyperplane orthogonal to $a^{(2)}$ is minimal, which essentially is a consequence of Szarek’s result [16] on the best constants in the Khintchine inequality for $q=1$. Barthe and Naor [3] proved that this does no transfer to $\frac{4}{3}< q < 2$, namely that the projection onto $a^{(n) \perp }$ has smaller volume than the one onto $a^{(2) \perp }$ for large dimensions n. Here $q_0:= \frac{4}{3}$ is the unique value in (1, 2) such that $\lim _{n \rightarrow \infty } \frac{\textrm{vol}_{n-1}(P_{(a^{(n)})^\perp } (B_q^n))}{\textrm{vol}_{n-1}(P_{(a^{(2)})^\perp } (B_q^n))} < 1$ for all $q_0< q < 2$, where $P_{a^\perp }$ denotes the orthogonal projection onto the hyperplane $a^\perp$. In this case, we also give a quantitative estimate for dimensions n when this happens, namely when $n > 5 \left( \frac{1}{p-\frac{4}{3}} + \frac{1}{2-p} \right)$. Note that there is no complete duality here, since $\frac{4}{3}$ is not the dual index of $p_0 \simeq 26.265$.

Let $a \in S^{n-1}:= \{ \ x \in {\mathbb {R}}^n \ | \ \Vert x \Vert _2 =1 \ \}$ be a direction vector. We introduce the normalized section function

$$\begin{aligned} A_{n,p}(a):= \frac{\textrm{vol}_{n-1}(a^\perp \cap B_p^n)}{\textrm{vol}_{n-1}(B_p^{n-1})} \, \end{aligned}$$

and the normalized projection function

$$\begin{aligned} P_{n,p}(a):= \frac{\textrm{vol}_{n-1}(P_{a^\perp } (B_p^n))}{\textrm{vol}_{n-1}(B_p^{n-1})} . \end{aligned}$$

In terms of this notation, Ball’s result states $A_{n,\infty }(a) \le A_{n,\infty }(a^{(2)}) = \sqrt{2}$ for all $a \in S^{n-1}$ and Eskenazis, Nayar and Tkocz’s result reads $A_{n,p}(a) \le A_{n,p}(a^{(2)}) = 2^{\frac{1}{2} - \frac{1}{p}}$ for all $a \in S^{n-1}$ and $10^{15} \le p < \infty$. We note that $A_{n,p}(a^{(2)}) = 2^{\frac{1}{2} - \frac{1}{p}}$ is independent of $n \ge 2$. But as shown by Oleszkiewicz, $\lim _{n \rightarrow \infty } A_{n,p}(a^{(n)}) > A_{n,p}(a^{(2)}) = 2^{\frac{1}{2} - \frac{1}{p}}$ for $2< p < p_0$. In the projection case, $P_{n,1}(a^{(2)}) \le P_{n,1}(a)$ for all $a \in S^{n-1}$, which Eskenazis, Nayar and Tkocz [5] extended to $P_{n,q}(a^{(2)}) \le P_{n,q}(a)$ for all $a \in S^{n-1}$ and $1 < q \le 1 + 10^{-12}$. Again $P_{n,q}(a^{(2)}) = 2^{\frac{1}{2} - \frac{1}{q}}$ is independent of $n \ge 2$. However, by Barthe and Naor [3], $P_{n,q}(a^{(2)}) > \lim _{n \rightarrow \infty } P_{n,q}(a^{(n)})$ for $\frac{4}{3}< q < 2$.

Our two main results study these limits in more detail.

Theorem 1.1

Let $2< p < \infty$ and $n \in {\mathbb {N}}$. Then for all $2< p < p_0 \simeq 26.265$

$$\begin{aligned} \lim _{n \rightarrow \infty } \frac{A_{n,p}(a^{(n)})}{A_{n,p}(a^{(2)})} = \sqrt{\frac{3}{\pi }} \sqrt{\frac{2^{\frac{2}{p}} \Gamma \left( 1+\frac{1}{p} \right) ^3}{\Gamma \left( 1+\frac{3}{p} \right) }} > 1 . \end{aligned}$$

We have the following quantitative estimate: $A_{n,p}(a^{(n)}) > A_{n,p}(a^{(2)})$ holds if

• (a) either $5 \le p < p_0$ and $n \ge \frac{650}{p_0-p}$   or  

• (b) $2< p < 5$ and $n > \frac{65}{p-2}$ is satisfied.

Remarks. (a) The constant 650 in the statement for $5 \le p < p_0$ is not optimal, but by necessity fairly large since the p-derivative of $f(p):=\sqrt{\frac{3}{\pi }} \sqrt{\frac{2^{\frac{2}{p}} \Gamma \left( 1+\frac{1}{p} \right) ^3}{\Gamma \left( 1+\frac{3}{p} \right) }}$ at $p_0$ with $f(p_0)=1$ is small, $f'(p_0) \simeq - \frac{1}{1316}$. The derivative at 2 is positive and larger in modulus, namely $f'(2) = \frac{1}{4} (1 - \ln 2) \simeq \frac{1}{13}$.

(b) The case of complex hyperplane sections of $l_p^n({\mathbb {C}})$ is considered in [8].

Theorem 1.2

Let $1< q < 2$ and $n \in {\mathbb {N}}$. Then for all $\frac{4}{3}< q < 2$

$$\begin{aligned} \lim _{n \rightarrow \infty } \frac{P_{n,q}(a^{(n)})}{P_{n,q}(a^{(2)})} = \sqrt{\frac{1}{\pi }} \sqrt{2^{\frac{2}{q}} \Gamma \left( \frac{1}{q} \right) \Gamma \left( 2-\frac{1}{q} \right) } < 1 . \end{aligned}$$

We have the following quantitative estimate: $P_{n,q}(a^{(n)}) < P_{n,q}(a^{(2)})$ holds if

$$\begin{aligned} n > \frac{\frac{32}{15}}{q-\frac{4}{3}} + \frac{\frac{24}{5}}{2-q} . \end{aligned}$$

Remark. For the derivative of $g(q):= \sqrt{\frac{1}{\pi }} \sqrt{2^{\frac{2}{q}} \Gamma \left( \frac{1}{q} \right) \Gamma \left( 2-\frac{1}{q} \right) }$ we have $g'(\frac{4}{3}) = \frac{9}{32} \left( 4 - \pi - 2 \ln 2 \right) \simeq - \frac{1}{6.73}$ and $g'(2) = \frac{1}{4} \left( 1 - \ln 2 \right) \simeq \frac{1}{13}$.

The limits in Theorems 1.1 and 1.2 were already determined by Oleszkiewicz [14] and Barthe, Naor [3]. Meyer and Pajor [13] showed that $A_{n,p}(a)$ is monotone increasing in p for any fixed n and a. Barthe and Naor proved that $P_{n,q}(a)$ is monotone increasing in q for any fixed n and a.

2 Formulas

Eskenazis, Nayar and Tkocz [5], Proposition 6, proved the following formula for the normalized volume of hyperplane sections.

Proposition 2.1

Let $1 \le p < \infty$, $n \in {\mathbb {N}}$ and $a = (a_j)_{j=1}^n \in S^{n-1} \subset {\mathbb {R}}^n$. Then

$$\begin{aligned} A_{n,p}(a) = \Gamma \left( 1+\frac{1}{p} \right) \ {\mathbb {E}}_{\xi ,R} \frac{1}{ ||\sum _{j=1}^n a_j R_j \xi _j||_2 } \, \end{aligned}$$

(2.1)

where $(\xi _j)_{j=1}^n$ are i.i.d. random vectors uniformly distributed on the sphere $S^2 \subset {\mathbb {R}}^3$ and $(R_j)_{j=1}^n$ are i.i.d. random variables with density $c_p^{-1} t^p \exp (-t^p)$ on $[0,\infty )$, $c_p:= \frac{1}{p} \Gamma \left( 1+\frac{1}{p} \right)$, independent of the $(\xi _j)_{j=1}^n$.

For $p = \infty$ with $R_j=1$ one has $A_{n,\infty }(a) = {\mathbb {E}}_{\xi } \frac{1}{ ||\sum _{j=1}^n a_j \xi _j||_2 }$, see König, Koldobsky [11]. We will use another formula for $A_{n,p}(a)$ derived from (2.1).

Proposition 2.2

Let $1 \le p < \infty$, $n \in {\mathbb {N}}$ and $a = (a_j)_{j=1}^n \in S^{n-1} \subset {\mathbb {R}}^n$. Then

$$\begin{aligned} A_{n,p}(a)&= \Gamma \left( 1+\frac{1}{p} \right) \ \frac{2}{\pi }\int _0^\infty \prod _{j=1}^n \gamma _p(a_j s) \ ds \ , \nonumber \\ \gamma _p(s)&:= \frac{1}{\Gamma \left( 1+\frac{1}{p} \right) } \int _0^\infty \cos (sr) \exp (-r^p) \ dr \ . \end{aligned}$$

(2.2)

Proof

Define $sinc (x):= \frac{\sin x}{x}$ for $x \ne 0$, $sinc (0):= 1$. Let $t>0$, $e \in S^2$ be fixed and m denote the normalized Haar surface measure on $S^2$. Then

$$\begin{aligned} sinc (t) = \int _{S^2} \exp (i t \langle e, u \rangle ) \ dm(u). \end{aligned}$$

(2.3)

Here $\langle \ \cdot , \cdot \ \rangle$ denotes the standard inner product on ${\mathbb {R}}^n$. This implies for $(b_j)_{j=1}^n \in {\mathbb {R}}^n$

$$\begin{aligned} \prod _{j=1}^n sinc (b_j s)&= \int _{(S^2)^n} \exp \left( i s \left\langle e, \sum _{j=1}^n b_j u_j \right\rangle \right) \prod _{j=1}^n dm(u_j) \nonumber \\&= \int _{(S^2)^n} sinc \left( \left| \left| \sum _{j=1}^n b_j u_j\right| \right| _2 s \right) \prod _{j=1}^n dm(u_j) = {\mathbb {E}}_{\xi } sinc \left( \left| \left| \sum _{j=1}^n b_j \xi _j\right| \right| _2 s \right) \ , \end{aligned}$$

(2.4)

where the second equality follows from (2.3) by integration over dm(e). Note that the first equality holds for all $e \in S^2$.

For all $t>0$ we have $\frac{2}{\pi }\int _0^\infty sinc (ts) ds = \frac{1}{t}$ and (2.1) may be rewritten

$$\begin{aligned} A_{n,p}(a)&= \Gamma \left( 1+\frac{1}{p} \right) \frac{2}{\pi }{\mathbb {E}}_{\xi ,R} \int _0^\infty sinc \left( \left| \left| \sum _{j=1}^n a_j R_j \xi _j\right| \right| _2 s \right) \ ds \\&= \Gamma \left( 1+\frac{1}{p} \right) \frac{2}{\pi }\int _0^\infty {\mathbb {E}}_{\xi ,R} \ sinc \left( \left| \left| \sum _{j=1}^n a_j R_j \xi _j\right| \right| _2 s \right) \ ds \ . \end{aligned}$$

The sinc-integral is only a conditionally convergent Riemann integral. The verification that ${\mathbb {E}}_{\xi ,R}$ and $\int _0^\infty$ may be interchanged is the same as in the proof of Proposition 3.2 (a) of König, Rudelson [12]. Using (2.4) and the independence of the $(R_j)_{j=1}^n$, we get

$$\begin{aligned} A_{n,p}(a)&= \Gamma \left( 1 + \frac{1}{p} \right) \frac{2}{\pi }\int _0^\infty {\mathbb {E}}_R \left( \prod _{j=1}^n sinc (a_j R_j s) \right) \ ds \\&= \Gamma \left( 1 + \frac{1}{p} \right) \frac{2}{\pi }\int _0^\infty \prod _{j=1}^n {\mathbb {E}}_{R_j} sinc (a_j R_j s) \ ds \ . \end{aligned}$$

Denoting $\gamma _p(s):= {\mathbb {E}}_{R_1} sinc (R_1 s)$, integration by parts gives

$$\begin{aligned} \gamma _p(s)&= c_p^{-1} \int _0^\infty sinc (sr) r^p \exp (-r^p) dr \\&= c_p^{-1} \frac{1}{p} \int _0^\infty \cos (sr) \exp (-r^p) dr = \frac{1}{\Gamma \left( 1+\frac{1}{p} \right) } \int _0^\infty \cos (sr) \exp (-r^p) dr \ . \end{aligned}$$

$\square$

Equation (2.1) yields $A_{n,p}(a^{(2)}) = 2^{\frac{1}{2}-\frac{1}{p}}$, see [5, Section 3.2].

Remarks. (a) Proposition 2.2 is also found in Koldobsky [9, Theorem 3.2], with a different proof.

(b) For $1 \le p \le 2$ the $\gamma _p$ are just the densities of the (positive) p-stable random variables. In the case interesting for us, namely $2< p < \infty$, the functions $\gamma _p$ take positive and negative values. For $p \notin 2 {\mathbb {N}}$, $\gamma _p$ has only finitely many real zeros, see Pólya [15], whereas for $p \in 2 {\mathbb {N}}$, $\gamma _p$ has infinitely many real zeros, see Boyd [4].

Barthe and Naor [3] proved the following formula for the volume of the orthogonal projection of $B_q^n$ onto hyperplanes.

Proposition 2.3

Let $1 \le q < \infty$, $p:= \frac{q}{q-1}$ be the conjugate index, $n \in {\mathbb {N}}$ and $a = (a_j)_{j=1}^n \in S^{n-1} \subset {\mathbb {R}}^n$. Then

$$\begin{aligned} P_{n,q}(a) = \Gamma \left( \frac{1}{q} \right) \ {\mathbb {E}}\left| \sum _{j=1}^n a_j X_j \right| \, \end{aligned}$$

(2.5)

where the $X_j$ are i.i.d. symmetric random variables with density function $d_q^{-1} |t|^{p-2} \exp (-|t|^p)$, $t \in {\mathbb {R}}$, $d_q = \frac{2}{p} \Gamma \left( \frac{1}{q} \right)$. A second formula for $P_{n,q}(a)$ is

$$\begin{aligned} P_{n,q}(a)&= \Gamma \left( \frac{1}{q} \right) \frac{2}{\pi }\int _0^\infty \frac{1 - \prod _{j=1}^n \delta _q(a_j s)}{s^2} ds \ , \nonumber \\ \delta _q(s)&:= \frac{p}{\Gamma \left( \frac{1}{q} \right) } \int _0^\infty \cos (s r) \ r^{p-2} \exp (-r^p) dr \ . \end{aligned}$$

(2.6)

Note that ${\mathbb {E}}|X_1| = \frac{1}{\Gamma \left( \frac{1}{q} \right) }$. To deduce (2.6) from (2.5), apply the usual formula $|x|= \frac{1}{\pi }\int _{\mathbb {R}}\frac{1 - Re(\exp (ixs))}{s^2} ds$ to find

$$\begin{aligned} {\mathbb {E}}\left| \sum _{j=1}^n a_j X_j \right|&= \frac{1}{\pi }\int _{\mathbb {R}}\frac{1 - {\mathbb {E}}\exp (i (\sum _{j=1}^n a_j X_j) s)}{s^2} ds = \frac{1}{\pi }\int _{\mathbb {R}}\frac{1 - \prod _{j=1}^n {\mathbb {E}}\exp (i a_j X_j s))}{s^2} ds \\&= \frac{2}{\pi }\int _0^\infty \frac{1 - \prod _{j=1}^n {\mathbb {E}}\cos (a_j X_j s))}{s^2} ds = \frac{2}{\pi }\int _0^\infty \frac{1 - \prod _{j=1}^n \delta _q(a_j s)}{s^2} ds \ , \\ \delta _q(s)&= \frac{p}{\Gamma \left( \frac{1}{q} \right) } \int _0^\infty \cos (sr) \ r^{p-2} \exp (-r^p) dr \ . \end{aligned}$$

Differentiation and integration by parts yields a relation between the functions $\delta _q$ and $\gamma _p$ in (2.6) and (2.2):

$$\begin{aligned} \delta _q'(s)&= - \frac{p}{\Gamma \left( \frac{1}{q} \right) } \int _0^\infty \sin (sr) \ r^{p-1} \exp (-r^p) dr \nonumber \\&= - \frac{s}{\Gamma \left( \frac{1}{q} \right) } \int _0^\infty \cos (sr) \exp (-r^p) dr = - \frac{\Gamma \left( 1+ \frac{1}{p} \right) }{\Gamma \left( 1-\frac{1}{p} \right) } \ s \ \gamma _p(s) \ . \end{aligned}$$

(2.7)

Since $\gamma _4''(s) = - \frac{\Gamma (\frac{3}{4})}{4 \Gamma (\frac{5}{4})} \ \delta _{\frac{4}{3}}(s)$, we have $\gamma _4'''(s) = \frac{1}{4}\,s \gamma _4(s)$. Similarly, for all $k \in {\mathbb {N}}$, $\gamma _{2k}^{(2k-1)}(s)= (-1)^k \frac{1}{2 k} s \gamma _{2 k}(s)$. Therefore the functions $\gamma _{2k}$ studied by Boyd [4] satisfy a linear differential equation.

For $q \searrow 1$, the variables $X_j$ tend to the Rademacher variables with $\delta _q(s) \rightarrow \delta _1(s) = \cos (s)$, and the best constants in the Khintchine inequality, which were determined by Szarek [16], yield the extrema of $P_{n,1}$: $a^{(2)}$ for the minimum and $a^{(1)}$ for the maximum.

3 Prerequisites for the proof of Theorem 1.1

For the proof of Theorem 1.1 we need two lemmas on $\Gamma$-functions.

Lemma 3.1

• (a) Let $f(p):= \frac{\Gamma \left( 1+\frac{3}{p} \right) }{\Gamma \left( 1+\frac{1}{p} \right) }$. Then $f(p) \ge 0.9429$ for all $3 \le p < \infty$.

• (b) Let $g(p):= \Big ( \frac{3}{\pi }\frac{ 2^{\frac{2}{p}} \Gamma \left( 1+\frac{1}{p} \right) ^3 }{ \Gamma \left( 1+\frac{3}{p} \right) } \Big )^{\frac{1}{2}}$. Then there is exactly one solution $p_0 \in (2,\infty )$ of $g(p) = 1$, $p_0 \simeq 26.265$. For all $2< p < p_0$ we have $g(p) > 1$. The function $g'$ has exactly one zero $p_1 \in [2,\infty )$, $p_1 \simeq 4.192$. For $2 \le p < p_1$, g is strictly increasing, for $p_1< p < \infty$, g is strictly decreasing. The following lower estimates hold:

  $$\begin{aligned}{} & {} g(p) \ge 1 + \frac{p_0-p}{1317} \, \ p \in [5,p_0] \;, \; g(p) \\{} & {} \quad > \frac{25}{24} \, \ p \in [4,5] \;, \; g(p) \ge 1 + \frac{p-2}{44} \, \ p \in [2,4] . \end{aligned}$$

Proof

(a) In terms of the Digamma function $\Psi := (\ln \Gamma )'$ we have

$$\begin{aligned} f'(p) = \frac{f(p)}{p^2} \left( \Psi \left( 1+\frac{1}{p} \right) - 3 \Psi \left( 1+\frac{3}{p} \right) \right) . \end{aligned}$$

For $F(p):= \Psi \left( 1+\frac{1}{p} \right) - 3 \Psi \left( 1+\frac{3}{p} \right)$ one has $F'(p) = \frac{1}{p^2} \left( 9 \Psi ' \left( 1+\frac{3}{p} \right) - \Psi ' \left( 1+\frac{1}{p} \right) \right)$. By Abramowitz, Stegun [1, 6.3.16 and 6.4.10] for all $x >0$

$$\begin{aligned} \Psi (1+x) = -\gamma + \sum _{n=1}^\infty \frac{x}{n(n+x)} \;, \; \Psi '(1+x) = \sum _{n=1}^\infty \frac{1}{(n+x)^2} \, \end{aligned}$$

(3.1)

where $\gamma \simeq 0.5772$ denotes the Euler–Mascheroni constant. Therefore $\Psi '$ is decreasing, and we conclude for all $0 \le x \le 1$ that $\frac{\pi ^2}{6} -1 = \Psi '(2) \le \Psi '(1+x) \le \Psi '(1) = \frac{\pi ^2}{6}$. Hence $F'(p) \ge \frac{1}{p^2} \left( \frac{4 \pi ^2}{3} -9 \right) > 0$ for all $p \ge 3$. Thus F is increasing. Since $F(9) \simeq -0.012$, $F(10) \simeq 0.084$, F has exactly one zero $p_1 \in (3,\infty )$, $p_1 \simeq 9.115$. Hence f is decreasing in $(3,p_1)$ and increasing in $(p_1,\infty )$. For all $p \ge 3$, $f(p) \ge f(p_1) > 0.9429$.

(b) Let $h(p):= \frac{ 2^{\frac{2}{p}} \Gamma \left( 1+\frac{1}{p} \right) ^3 }{ \Gamma \left( 1+\frac{3}{p} \right) }$. Then

$$\begin{aligned} h'(p) = \frac{h(p)}{p^2} \left( 3 \Psi \left( 1+\frac{3}{p} \right) - 3 \Psi \left( 1+\frac{1}{p} \right) - 2 \ln 2 \right) . \end{aligned}$$

By (3.1) and the geometric series we find for $p > 3$

$$\begin{aligned} (\ln h)'(p)&= \frac{h'(p)}{h(p)} = \frac{1}{p^2} \left( 3 \sum _{n=1}^\infty \left( \frac{\frac{3}{p}}{n(n+\frac{3}{p})} - \frac{\frac{1}{p}}{n(n+\frac{1}{p})} \right) - 2 \ln 2 \right) \\&= \frac{1}{p^2} \left( 3 \sum _{n=1}^\infty \sum _{k=0}^\infty \frac{(-1)^k}{n^{k+2}} \frac{3^{k+1}-1}{p^{k+1}} - 2 \ln 2 \right) \\&= \frac{1}{p^2} \left( 3 \sum _{k=0}^\infty (-1)^k \zeta (k+2) \frac{3^{k+1}-1}{p^{k+1}} - 2 \ln 2 \right) \ . \end{aligned}$$

The sum is an alternating series with decreasing coefficients. We find that

$$\begin{aligned} (\ln h)'(p) \le -\frac{2 \ln 2}{p^2} + \frac{\pi ^2}{p^3} - \frac{24 \zeta (3)}{p^4} + \frac{13}{15} \frac{\pi ^4}{p^5} - \frac{240 \zeta (5)}{p^6} + \frac{242}{315} \frac{\pi ^6}{p^7} < 0\nonumber \\ \end{aligned}$$

(3.2)

holds for all $5 \le p < \infty$. Therefore $\ln h$, h and $g(p) = \sqrt{\frac{3}{\pi }h(p)}$ are strictly decreasing in $[5,\infty )$. We have $\lim _{p \rightarrow \infty } g(p) = \sqrt{\frac{3}{\pi }} < 1$, $g(26) \simeq 1.00020$, $g(27) \simeq 0.99945$: There is exactly one $p_0 \in [5,\infty )$ with $g(p_0)=1$, $p_0 \simeq 26.265$, and for $5 \le p < p_0$ we have $g(p) > 1$. Inequality (3.2) yields for $5 \le p \le p_0$ that $(\ln h)'(p) \le - \frac{1.04768}{p^2}$. Hence for these p

$$\begin{aligned} g'(p) = g(p) (\ln g)'(p) = \frac{1}{2} g(p) (\ln h)'(p) \le - \frac{1}{2} \frac{1.04768}{p^2} \le - \frac{1}{2} \frac{1.04768}{p_0^2} < - \frac{1}{1317} . \end{aligned}$$

This implies $g(p) \ge 1 + \frac{1}{1317} (p_0-p)$ for all $5 \le p \le p_0$.

To show $g(p) > 1$ also for $2< p < 5$, note that $g(2)=1$ and

$$\begin{aligned} g'(p) = \frac{1}{2} g(p) (\ln h)'(p) = \frac{3}{2} \frac{g(p)}{p^2} \left( \Psi \left( 1+\frac{3}{p} \right) - \Psi \left( 1+\frac{1}{p} \right) - \frac{2}{3} \ln 2 \right) . \end{aligned}$$

Again by (3.1)

$$\begin{aligned} \Psi \left( 1+\frac{3}{p} \right) - \Psi \left( 1+\frac{1}{p} \right) = \sum _{n=1}^\infty \left( \frac{\frac{3}{p}}{n (n + \frac{3}{p})} - \frac{\frac{1}{p}}{n (n + \frac{1}{p})} \right) = \sum _{n=1}^\infty \frac{2 p}{(np+1)(np+3)} . \end{aligned}$$

All summands are decreasing in p. Thus $k(p):= \Psi (1+\frac{3}{p}) - \Psi (1+\frac{1}{p}) - \frac{2}{3} \ln 2$ is strictly decreasing in p, with $k(4) = \pi - \frac{8}{3} - \frac{2}{3} \ln 2 \simeq 0.0128 >0$ and $k(5) \simeq -0.0470 <0$. Thus $g'$ has exactly one zero $p_1 \in (2,\infty )$, $p_1 \simeq 4.193$, and g is strictly increasing in $(2,p_1)$ and strictly decreasing in $(p_1,\infty )$. We know already that $g(5) > 1$ and hence $g(p)>1$ for all $2 < p \le 5$. For $p \in [4,5]$, $g(p) \ge \min (g(4),g(5)) = g(5) > \frac{25}{24}$. Further

$$\begin{aligned} \left( \frac{g(p)}{p^2} \right) ' = \frac{3}{2} \frac{g(p)}{p^4} \left( k(p) - \frac{4}{3} p \right) < 0 \, \end{aligned}$$

since $k(p)- \frac{4}{3} p \le k(2)-\frac{8}{3} = - \left( 2+ \frac{2}{3} \ln (2) \right) < 0$. Therefore $\frac{g(p)}{p^2}$ and k(p) are both strictly decreasing and positive for $p \in [2,p_1)$, and with $g'(p) = \frac{g(p)}{p^2} k(p)$, $g'$ is decreasing and hence g is concave in $[2,p_1]$. Therefore for $2 \le p \le 4$

$$\begin{aligned} g(p) \ge 1 + \frac{g(4)-1}{2} (p-2) > 1 + \frac{p-2}{44} \, \end{aligned}$$

which proves all lower estimates stated in Lemma 3.1. $\square$

For $p \rightarrow \infty$, the functions $\gamma _p$ in (2.2) tend to $\gamma _\infty$, $\gamma _\infty (s) = sinc (s)$. We estimate their difference for $p \ge 2$.

Lemma 3.2

Let $2 \le p < \infty$. Then for all $s > 0$

$$\begin{aligned} \left| sinc (s) - \int _0^\infty \cos (s r) \exp (-r^p) dr \right| \le 0.3926 . \end{aligned}$$

This implies $\gamma _p(s) > 0$ for all $0 \le s \le \frac{2}{3} \pi$.

Proof

We have $\int _0^\infty \exp (-s^p) ds = \Gamma \left( 1 + \frac{1}{p} \right) < 1$. Since $sinc (s) = \int _0^1 \cos (sr) dr$, we find

$$\begin{aligned}&\left| sinc (s) - \int _0^\infty \cos (s r) \exp (-r^p) dr \right| \\&\quad = \left| \int _0^1 \cos (s r) (1-\exp (-r^p)) dr - \int _1^\infty \cos (s r) \exp (-r^p) dr \right| \\&\quad \le \left( 1 - \Gamma \left( 1 + \frac{1}{p} \right) \right) + 2 \int _1^\infty \exp (-r^p) dr \\&\quad = \left( 1 - \Gamma \left( 1 + \frac{1}{p} \right) \right) + \frac{2}{p} \int _1^\infty u^{\frac{1}{p} -1} \exp (-u) du =: \phi (p) \ . \end{aligned}$$

Then $\phi ' < 0$, since for $p \ge 2$

$$\begin{aligned} \phi '(p) = - \frac{1}{p^2} \left( 2 \int _1^\infty u^{\frac{1}{p} -1} \left( 1+\frac{\ln (u)}{p} \right) \exp (-u) du - \Gamma ' \left( 1 + \frac{1}{p} \right) \right) . \end{aligned}$$

Since $\Gamma (x)$ has a (positive) minimum at ${\tilde{x}} \simeq 1.4616$, $\Gamma '(1+ \frac{1}{{\tilde{p}}}) = 0$ for ${\tilde{p}} \simeq 2.1662$. For $p > {\tilde{p}}$, $\Gamma '(1 + \frac{1}{p}) < 0$ and for $2 \le p < {\tilde{p}}$, $0 \le \Gamma '(1+ \frac{1}{p}) < \frac{3}{100}$. The last inequality holds since $\Gamma$ is log-convex and positive in (1, 2), thus convex, and for $2 \le p < {\tilde{p}}$, $\Gamma '(1 + \frac{1}{p})$ is decreasing, with $\Gamma ' \left( \frac{3}{2}\right) = \Gamma \left( \frac{3}{2} \right) \Psi \left( \frac{3}{2} \right) = \sqrt{\pi }(1 - \ln 2 - \frac{\gamma }{2}) < \frac{3}{100}$, see Abramowitz, Stegun [1, 6.3.4]. Therefore for all $p \ge 2$

$$\begin{aligned} \phi '(p) \le - \frac{1}{p^2} \left( 2 \int _1^\infty u^{-1} \exp (-u) du - \frac{3}{100} \right)< - \frac{2}{5} \frac{1}{p^2} < 0 \, \end{aligned}$$

using that $\int _1^\infty u^{-1} \exp (-u) du \simeq 0.219$. We conclude $\phi (p) \le \phi (2) < 0.3926$.

This yields for all $0 \le s \le \frac{2}{3} \pi$ and $p \ge 2$

$$\begin{aligned} \Gamma \left( 1+\frac{1}{p} \right) \gamma _p(s)= & {} \int _0^\infty \cos (sr) \exp (-r^p) dr \ge sinc (s) \\ {}{} & {} - 0.3926 \ge \frac{3 \sqrt{3}}{4 \pi } -0.3926> \frac{1}{50} > 0 . \end{aligned}$$

$\square$

Remark. The derivative of $p \phi (p)$ is increasing with

$$\begin{aligned} \lim _{p \rightarrow \infty } \left( p \phi (p) \right) ' = \gamma + 2 \int _1^\infty \frac{1}{u} \exp (-u) du \le 1.016 . \end{aligned}$$

Thus for all $p \ge 1$, $\left| sinc (s) - \int _0^\infty \cos (s r) \exp (-r^p) dr \right| \le \frac{1.016}{p}$. Actually, $\gamma _p(s) > 0$ for all $p \ge 1$ and $s \in [0, \pi ]$. However, we do not need this.

4 Proof of Theorem 1.1

Proof of Theorem 1.1.

(i) To estimate $A_{n,p}(a^{(n)})$ from below, we first find a lower bound for $\gamma _p \left( \frac{s}{\sqrt{n}} \right)$ for all $s \le \frac{3}{2} \sqrt{n}$. By the series representation for $\cos x$ we have for $0 \le x \le \frac{3}{2}$, by pairing up terms in the alternating series,

$$\begin{aligned} \cos x - \left( 1 - \frac{x^2}{2} + \frac{x^4}{26} \right) = \frac{x^4}{312} - \frac{x^6}{720} + \sum _{k=4}^\infty (-1)^m \frac{x^{2m}}{(2m)!} > 0 \, \end{aligned}$$

$1 - \frac{x^2}{2} + \frac{x^4}{26} > 0$. Therefore for $s \le \frac{3}{2} \sqrt{n}$

$$\begin{aligned} \gamma _p \left( \frac{s}{\sqrt{n}} \right)&\ge \frac{1}{\Gamma \left( 1+\frac{1}{p} \right) } \Big [ \int _0^{\frac{3}{2} \frac{\sqrt{n}}{s}} \left( 1 - \frac{s^2 r^2}{2n} + \frac{s^4 r^4}{26 n^2} \right) \exp (-r^p) dr \\&\quad + \int _{\frac{3}{2} \frac{\sqrt{n}}{s}}^\infty \cos \left( \frac{sr}{\sqrt{n}} \right) \exp (-r^p) dr \Big ] \ . \end{aligned}$$

To estimate this from below, write $\int _0^{\frac{3}{2} \frac{\sqrt{n}}{s}} = \int _0^\infty - \int _{\frac{3}{2} \frac{\sqrt{n}}{s}}^\infty$ and use that $|\cos (x)| \le 1$,

$$\begin{aligned} \gamma _p \left( \frac{s}{\sqrt{n}} \right)&\ge \frac{1}{\Gamma \left( 1+\frac{1}{p} \right) } \left[ \int _0^\infty \left( 1 - \frac{s^2 r^2}{2n} + \frac{s^4 r^4}{26 n^2} \right) \exp (-r^p) dr - R \right] \nonumber \\&= \frac{1}{\Gamma \left( 1+\frac{1}{p} \right) } \left[ \Gamma \left( 1+\frac{1}{p} \right) - \frac{\Gamma \left( 1+\frac{3}{p} \right) s^2}{6 n} + \frac{\Gamma \left( 1+\frac{5}{p} \right) s^4}{130 n^2} - R \right] \ , \end{aligned}$$

(4.1)

where

$$\begin{aligned} R&:= \int _{\frac{3}{2} \frac{\sqrt{n}}{s}}^\infty \left( 2 - \frac{s^2 r^2}{2n} + \frac{s^4 r^4}{26 n^2} \right) \exp (-r^p) dr \nonumber \\&= \frac{1}{p} \int _{\left( \frac{3}{2} \frac{\sqrt{n}}{s} \right) ^p}^\infty \left( 2 u^{\frac{1}{p} - 1} - \frac{s^2}{2n} u^{\frac{3}{p} - 1} + \frac{s^4}{26 n^2} u^{\frac{5}{p} - 1} \right) \exp (-u) du \; ; \; u \ge 1 \ . \end{aligned}$$

(4.2)

(ii) We first consider the case $p \ge 5$. Since $u \ge 1$ in the above integral, $u^{\frac{5}{p} -1} \le 1$, $2 u^{\frac{1}{p} -1} \le 2 \left( \frac{2}{3} \frac{s}{\sqrt{n}} \right) ^{p-1}$ and $\int _{\left( \frac{3}{2} \frac{\sqrt{n}}{s} \right) ^p} \exp (-u) du = \exp \left( -\left( \frac{3}{2} \frac{\sqrt{n}}{s} \right) ^p \right)$. The remainder term R will be smaller than the fourth order term $\frac{\Gamma \left( 1+\frac{5}{p} \right) s^4}{130 n^2}$ provided that

$$\begin{aligned} \frac{1}{p \Gamma \left( 1+\frac{5}{p} \right) } \left( 2 \left( \frac{2}{3} \frac{s}{\sqrt{n}} \right) ^{p-1} + \frac{s^4}{26 n^2} \right) \exp \left( - \left( \frac{3}{2} \frac{\sqrt{n}}{s} \right) ^p \right) < \frac{s^4}{130 n^2} . \end{aligned}$$

For $s \le \frac{3}{2} \sqrt{n}$ the left side is decreasing in p, and therefore this condition is the strongest for $p=5$. Writing $y:=\frac{s}{\sqrt{n}}$, it means

$$\begin{aligned} \frac{1}{5} \left( \frac{32}{81} y^4 + \frac{1}{26} y^4 \right) \exp \left( -\left( \frac{3}{2} \frac{1}{y} \right) ^5 \right) < \frac{1}{130} y^4 \end{aligned}$$

or $\frac{913}{81} < \exp \left( \left( \frac{3}{2} \frac{1}{y} \right) ^5 \right)$, $y < \frac{3}{2} \frac{1}{\ln \left( \frac{913}{81} \right) ^{\frac{1}{5}}} \simeq 1.2567$. Choosing $s \le \frac{7}{6} \sqrt{n}$, $R \le \frac{\Gamma \left( 1+\frac{5}{p} \right) s^4}{130 n^2}$ is satisfied and therefore

$$\begin{aligned} \gamma _p \left( \frac{s}{\sqrt{n}} \right) \ge 1 - c \frac{s^2}{n}, \quad c:= \frac{1}{6} \frac{\Gamma \left( 1+\frac{3}{p} \right) }{\Gamma \left( 1+\frac{1}{p} \right) } . \end{aligned}$$

By the proof of Lemma 3.1 (a), $\frac{\Gamma \left( 1+\frac{3}{p} \right) }{\Gamma \left( 1+\frac{1}{p} \right) }$ is decreasing for $5 \le p \le p_1 \simeq 9.115$ and increasing for $p > p_1$. Its value at $p_0$ is less than the one at 5, so that $c \le 0.1622 < \frac{8}{49}$, for its value at $p=5$. Then for $s \le \frac{7}{6} \sqrt{n}$, $x:= c \frac{s^2}{n} \le \frac{8}{49} \frac{49}{36} = \frac{2}{9}$. We have

$$\begin{aligned} \ln (1-x) = - \sum _{j=1}^\infty \frac{x^j}{j} \ge -x - \frac{1}{2} x^2 \sum _{k=0}^\infty x^k = -x - \frac{1}{2} \frac{x^2}{1-x} \ge -x-\frac{9}{14} x^2 \, \end{aligned}$$

and hence

$$\begin{aligned} \gamma _p \left( \frac{s}{\sqrt{n}} \right) ^n \ge \left( 1 - c \frac{s^2}{n} \right) ^n \ge \exp \left( -c s^2 - \frac{9}{14} c^2 \frac{s^4}{n} \right) \ge \exp (-c s^2) \left( 1 - \frac{9}{14} c^2 \frac{s^4}{n} \right) . \end{aligned}$$

By Lemma 3.1 (a), $\frac{\Gamma \left( 1+\frac{3}{p} \right) }{\Gamma \left( 1+\frac{1}{p} \right) } \ge 0.9429$ and hence $c \ge 0.1571$ and for $s \le \frac{7}{6} \sqrt{n}$

$$\begin{aligned} \int _0^{\frac{7}{6} \sqrt{n}}&\gamma _p \left( \frac{s}{\sqrt{n}} \right) ^n \ ds \ge \int _0^{\frac{7}{6} \sqrt{n}} \exp (-c s^2) \left( 1 - \frac{9}{14} c^2 \frac{s^4}{n} \right) \ ds \\&= \int _0^\infty \exp (-c s^2) \left( 1 - \frac{9}{14} c^2 \frac{s^4}{n} \right) \ ds - \int _{\frac{7}{6} \sqrt{n}}^\infty \exp (-c s^2) \left( 1 - \frac{9}{14} c^2 \frac{s^4}{n} \right) \ ds \end{aligned}$$

For $s \ge \frac{7}{6} \sqrt{n}$ and $c \ge 0.1571$ we have $1 - \frac{9}{14} c^2 \frac{s^4}{n} <0$ for all $n \ge 35$. Actually, evaluating the last integral in terms of the error function shows that the integral is already negative for $n \ge 24$. Thus for $n \ge 24$

$$\begin{aligned} \int _0^{\frac{7}{6} \sqrt{n}} \gamma _p \left( \frac{s}{\sqrt{n}} \right) ^n \ ds&\ge \int _0^\infty \exp (-c s^2) \left( 1 - \frac{9}{14} c^2 \frac{s^4}{n} \right) \ ds \\&= \frac{1}{2} \sqrt{\frac{\pi }{c}} \left( 1 - \frac{27}{56} \frac{1}{n} \right) = \sqrt{\frac{3 \pi }{2}} \sqrt{\frac{\Gamma \left( 1+\frac{1}{p} \right) }{\Gamma \left( 1+\frac{3}{p} \right) }} \left( 1 - \frac{27}{56} \frac{1}{n} \right) \ , \end{aligned}$$

where we used that $\int _0^\infty \exp (-c s^2) ds = \frac{1}{2} \sqrt{\frac{\pi }{c}}$ and $\int _0^\infty \exp (-c s^2) c^2\,s^4 ds = \frac{3}{8} \sqrt{\frac{\pi }{c}}$. By Lemma 3.2 we have $\gamma _p(s) > 0$ for all $0 \le s \le 2$. Hence

$$\begin{aligned} 0 < \sqrt{n} \int _{\frac{7}{6}}^2 \gamma _p(s)^n du = \int _{\frac{7}{6} \sqrt{n}}^{2 \sqrt{n}} \gamma _p \left( \frac{s}{\sqrt{n}} \right) ^n ds . \end{aligned}$$

By the proof of Proposition 2.2

$$\begin{aligned} |\gamma _p(s)|&= \left| \frac{p}{\Gamma \left( 1+\frac{1}{p} \right) } \int _0^\infty sinc (sr) r^p \exp (-r^p) dr \right| \\&= \left| \frac{1}{s \Gamma \left( 1+\frac{1}{p} \right) } \int _0^\infty \sin (sr) \ p r^{p-1} \exp (-r^p) dr \right| \ , \ |\sin (sr)| \le 1 \\&\le \frac{1}{s \Gamma \left( 1+\frac{1}{p} \right) } \int _0^\infty \exp (-u) du = \frac{1}{s \Gamma (1+\frac{1}{p})} \ . \end{aligned}$$

This yields the tail estimate for $p \ge 5$

$$\begin{aligned} \int _{2 \sqrt{n}}^\infty \Big |\gamma _p \left( \frac{s}{\sqrt{n}} \right) \Big |^n ds&= \sqrt{n} \int _2^\infty |\gamma _p(s)|^n ds \le \frac{\sqrt{n}}{\Gamma \left( 1+\frac{1}{p} \right) ^n} \int _2^\infty s^{-n} ds \\&= \frac{2 \sqrt{n}}{n-1} \left( \frac{1}{2 \Gamma \left( 1+\frac{1}{p} \right) } \right) ^n < \frac{2 \sqrt{n}}{n-1} 0.5446^n \ . \end{aligned}$$

We conclude for $p \ge 5$ and $n \ge 24$ that

$$\begin{aligned} \int _0^\infty \gamma _p \left( \frac{s}{\sqrt{n}} \right) ^n \ ds&\ge \int _0^{\frac{7}{6} \sqrt{n}} \gamma _p \left( \frac{s}{\sqrt{n}} \right) ^n ds - \int _{2 \sqrt{n}}^\infty \left| \gamma _p \left( \frac{s}{\sqrt{n}} \right) \right| ^n \ ds \\&\ge \sqrt{\frac{3 \pi }{2}} \sqrt{\frac{\Gamma \left( 1+\frac{1}{p} \right) }{\Gamma \left( 1+\frac{3}{p} \right) }} \left( 1 - \frac{27}{56} \frac{1}{n} \right) - \frac{2 \sqrt{n}}{n-1} 0.5446^n \\&\ge \sqrt{\frac{3 \pi }{2}} \sqrt{\frac{\Gamma \left( 1+\frac{1}{p} \right) }{\Gamma \left( 1+\frac{3}{p} \right) }} \left( 1 - \frac{27}{56} \frac{1}{n} - \frac{\sqrt{n}}{n-1} 0.5446^n \right) \ , \end{aligned}$$

using $\sqrt{\frac{3 \pi }{2}} \sqrt{\frac{\Gamma \left( 1+\frac{1}{p} \right) }{\Gamma \left( 1+\frac{3}{p} \right) }} > 2$. For $n \ge 24$, $\frac{\sqrt{n}}{n-1} 0.5446^n < \frac{10^{-5}}{n}$ and $\frac{27}{56} + 10^{-5} < \frac{193}{400}$, so that

$$\begin{aligned} A_{n,p}(a^{(n)})&= \Gamma \left( 1+\frac{1}{p} \right) \frac{2}{\pi }\int _0^\infty \gamma _p \left( \frac{s}{\sqrt{n}} \right) ^n \ ds \\&\ge \sqrt{\frac{6}{\pi }} \sqrt{\frac{\Gamma \left( 1+\frac{1}{p} \right) ^3}{\Gamma \left( 1+\frac{1}{p} \right) }} \left( 1 - \frac{193}{400} \frac{1}{n} \right) \ . \end{aligned}$$

This is $> A_{n,p}(a^{(2)}) = 2^{\frac{1}{2} - \frac{1}{p}}$, provided that

$$\begin{aligned} \frac{A_{n,p}(a^{(n)})}{A_{n,p}(a^{(2)})} \ge \sqrt{\frac{3}{\pi }} \sqrt{\frac{2^{\frac{2}{p}} \Gamma \left( 1+\frac{1}{p} \right) ^3}{\Gamma \left( 1+\frac{3}{p} \right) }} \left( 1 - \frac{193}{400} \frac{1}{n} \right) > 1 . \end{aligned}$$

By Lemma 3.1 (b), the quotient $g(p):= \sqrt{\frac{3}{\pi }} \sqrt{\frac{2^{\frac{2}{p}} \Gamma \left( 1+\frac{1}{p} \right) ^3}{\Gamma \left( 1+\frac{3}{p} \right) }}$ is $> 1$ for all $2< p < p_0 \simeq 26.265$, with $g(p) \ge 1 + \frac{1}{1317} (p_0-p)$ for all $5 \le p \le p_0$. We find for $p \ge 5$ and $n \ge 24$,

$$\begin{aligned} \frac{A_{n,p}(a^{(n)})}{A_{n,p}(a^{(2)})} \ge \left( 1 + \frac{1}{1317} (p_0-p) \right) \left( 1-\frac{193}{400} \frac{1}{n} \right) . \end{aligned}$$

This is $>1$ provided that $5 \le p \le p_0$ and $n \ge \frac{650}{p_0-p}$; $n \ge 24$ being automatically satisfied.

(iii) Secondly we consider the case $2 < p \le 5$. To estimate the remainder R in (4.2), we use that in this case $u^{\frac{5}{p} -1} \le u^{\frac{3}{2}}$. For $x > 0$

$$\begin{aligned} \int _x^\infty u^{\frac{3}{2}} \exp (-u) \ du&\le \left( \left( \int _x^\infty u \exp (-u) \ du \right) \left( \int _x^\infty u^2 \exp (-u) \ du \right) \right) ^{\frac{1}{2}} \\&= \Big ( (1+x) (2+2 x +x^2) \Big )^{\frac{1}{2}} \exp (-x) \ , \end{aligned}$$

Since the function $\phi (x):= \frac{(1+x)(2+2x+x^2)}{x^4} = \left( \frac{1}{x} + \frac{1}{x^2} \right) \left( 2 + \frac{2}{x} + \frac{1}{x^2} \right)$ is decreasing in x with $\phi \left( \frac{9}{2} \right) = \left( \frac{5 \sqrt{110}}{81} \right) ^2 < \left( \frac{13}{20} \right) ^2$, we conclude that $\int _x^\infty u^{\frac{3}{2}} \exp (-u) \ du \le \frac{13}{20} x^2 \exp (-x)$ for all $x \ge \frac{9}{2}$. Now choose $s \le \sqrt{\frac{n}{2}}$ for $2 \le p \le 5$. Then $x:= \left( \frac{3}{2} \frac{\sqrt{n}}{s} \right) ^p \ge \left( \frac{3}{\sqrt{2}} \right) ^p \ge \frac{9}{2}$ and

$$\begin{aligned} R \le \frac{1}{p} \left( 2 \left( \frac{2}{3} \frac{s}{\sqrt{n}} \right) ^{p-1} + \frac{s^4}{26 n^2} \frac{13}{20} \left( \frac{3}{2} \frac{\sqrt{n}}{s} \right) ^{2p} \right) \exp \left( - \left( \frac{3}{2} \frac{\sqrt{n}}{s} \right) ^p \right) . \end{aligned}$$

Again we want this to be smaller than the fourth order term $\Gamma (1+\frac{5}{p}) \frac{s^4}{130 n^2}$, a condition which is strongest for $p=2$. We then require for $y:= \frac{s}{\sqrt{n}}$

$$\begin{aligned} \frac{1}{2} \left( \frac{4}{3} y + \frac{81}{640} \right) \exp \left( -\left( \frac{3}{2} \frac{1}{y} \right) ^2 \right) < \Gamma \left( \frac{7}{2} \right) \frac{y^4}{130} . \end{aligned}$$

This is equivalent to $g(y):= \frac{1}{\sqrt{\pi }} \left( 26 \frac{16}{9} \frac{1}{y^3} + 13 \frac{27}{80} \frac{1}{y^4} \right) \exp \left( -\left( \frac{3}{2} \frac{1}{y} \right) ^2 \right) < 1$. The function g is increasing and positive in $y>0$, with $g(0.7161) < 1$. Hence the condition is satisfied for all $0 < y \le 0.7161$, and in particular for our choice $y = \frac{s}{\sqrt{n}} \le \frac{1}{\sqrt{2}}$. Therefore for $s \le \sqrt{\frac{n}{2}}$, as in part (ii),

$$\begin{aligned} \gamma _p \left( \frac{s}{\sqrt{n}} \right) \ge 1 - c \frac{s^2}{n}, \quad c:= \frac{1}{6} \frac{\Gamma \left( 1+\frac{3}{p} \right) }{\Gamma \left( 1+\frac{1}{p} \right) } . \end{aligned}$$

We have $c \le \frac{1}{4}$ for $2 \le p \le 5$ and $x:= c \frac{s^2}{n} \le \frac{1}{8}$. Similarly as in (ii),

$\ln (1-x) \ge -x-\frac{x^2}{1-x} \ge -x-\frac{4}{7} x^2$ and

$$\begin{aligned} \gamma _p \left( \frac{s}{\sqrt{n}} \right)&\ge \left( 1-c \frac{s^2}{n} \right) ^n =\exp \left( n \ln \left( 1-c \frac{s^2}{n} \right) \right) \\&\ge \exp \left( -c s^2 - \frac{4}{7} c^2 \frac{s^4}{n} \right) \ge \exp (-c s^2) \left( 1- \frac{4}{7} c^2 \frac{s^4}{n} \right) \end{aligned}$$

and

$$\begin{aligned} \int _0^{\sqrt{\frac{n}{2}}}&\gamma _p \left( \frac{s}{\sqrt{n}} \right) ^n \ ds \ge \int _0^{\sqrt{\frac{n}{2}}} \exp (-c s^2) \left( 1 - \frac{4}{7} c^2 \frac{s^4}{n} \right) \ ds \\&\ge \int _0^\infty \exp (-c s^2) \left( 1 - \frac{4}{7} c^2 \frac{s^4}{n} \right) ds - \int _{\sqrt{\frac{n}{2}}}^\infty \exp (-c s^2) \ ds \\&= \frac{1}{2} \sqrt{\frac{\pi }{c}} \left( 1 - \frac{3}{7} \frac{1}{n} \right) - \int _{\sqrt{\frac{n}{2}}}^\infty \exp (-c s^2) \ ds \ . \end{aligned}$$

To estimate the error term, note that $c \ge .16219$, its approximate value for $p=5$, and

$$\begin{aligned} \int _{\sqrt{\frac{n}{2}}}^\infty \exp (-c s^2) ds&= \frac{1}{2 \sqrt{c}} \int _{\frac{c}{2} n}^\infty \frac{1}{\sqrt{u}} \exp (-u) du \\&\le \frac{1}{c \sqrt{2 n}} \int _{\frac{c}{2} n}^\infty \exp (-u) du = \frac{1}{c \sqrt{2 n}} \exp (-\frac{c}{2} n) \le \frac{4.36}{\sqrt{n}} \ 0.9222^n \ . \end{aligned}$$

Again $\int _{\sqrt{\frac{n}{2}}}^{2 \sqrt{n}} \gamma _p \left( \frac{s}{\sqrt{n}} \right) ^n \ ds \ge 0$, since by Lemma 3.2$\gamma _p(x) >0$ for all $0 \le x \le 2$ and, as in (ii), $\int _{2 \sqrt{n}}^\infty \left| \gamma _p \left( \frac{s}{\sqrt{n}} \right) \right| ^n ds \le \frac{2 \sqrt{n}}{n-1} \ 0.5446^n$, so that for $2 \le p \le 5$

$$\begin{aligned} \int _0^\infty \gamma _p \left( \frac{s}{\sqrt{n}} \right) ^n ds \ge \frac{1}{2} \sqrt{\frac{\pi }{c}} \left( 1 - \frac{3}{7} \frac{1}{n} \right) - \left( \frac{4.36}{\sqrt{n}} \ 0.9222^n + \frac{2 \sqrt{n}}{n-1} \ 0.5446^n \right) . \end{aligned}$$

Since $c \le \frac{1}{4}$, $\frac{1}{2} \sqrt{\frac{\pi }{c}} \ge \sqrt{\pi }$. Further for $n \ge 33$

$$\begin{aligned} \frac{3}{7} + \frac{1}{\sqrt{\pi }} \left( 4.36 \sqrt{n} \ 0.9222^n + \frac{2 n \sqrt{n}}{n-1} \ 0.5446^n \right) \le 1.405 \, \end{aligned}$$

so that $\int _0^\infty \gamma _p \left( \frac{s}{\sqrt{n}} \right) ^n \ ds \ge \frac{1}{2} \sqrt{\frac{\pi }{c}} \left( 1- \frac{1.405}{n} \right)$. For $p \in [2,4]$, using Lemma 3.1 (b),

$$\begin{aligned} \frac{A_{n,p}(a^{(n)})}{A_{n,p}(a^{(2)})}&\ge \sqrt{\frac{3}{\pi }} \sqrt{\frac{2^{\frac{2}{p}} \Gamma \left( 1+\frac{1}{p} \right) ^3}{\Gamma \left( 1+\frac{3}{p} \right) }} \left( 1 - \frac{1.405}{n} \right) \\&\ge \left( 1 + \frac{p-2}{44} \right) \left( 1 - \frac{1.405}{n} \right) \ . \end{aligned}$$

For $n \ge \frac{65}{p-2}$ this is $>1$, with $n \ge 33$ being automatically satisfied. For $p \in [4,5]$, again by Lemma 3.1 (b), $\sqrt{\frac{3}{\pi }} \sqrt{\frac{2^{\frac{2}{p}} \Gamma \left( 1+\frac{1}{p} \right) ^3}{\Gamma \left( 1+\frac{3}{p} \right) }} > \frac{25}{24}$ and $\frac{A_{n,p}(a^{(n)})}{A_{n,p}(a^{(2)})} > 1$ is satisfied for $n \ge \frac{65}{p-2}$, too. $\square$

5 Prerequisites for the proof of Theorem 1.2

We need two lemmas for the proof of Theorem 1.2.

Lemma 5.1

• (a) For $q \in [1,2]$, let $f(q):= \frac{\Gamma \left( 2-\frac{1}{q} \right) }{\Gamma \left( \frac{1}{q} \right) }$. Then f is decreasing, with $f(1)=1$, $f(2)=\frac{1}{2}$ and $f \left( \frac{4}{3} \right) \le 0.7397$.

• (b) For $q \in [1,2]$, let $g(q):= \sqrt{\frac{2^{\frac{2}{q}}}{\pi }\Gamma \left( \frac{1}{q} \right) \Gamma \left( 2-\frac{1}{q} \right) }$. Then $g'$ has exactly one zero in $q_1 \in (1,2)$, $q_1 \simeq 1.612$, and g is strictly decreasing in $[1,q_1)$ and strictly increasing in $(q_1,2]$, with $g \left( \frac{4}{3} \right) = g(2) =1$. For $q \in (\frac{4}{3},2)$, we have

  $$\begin{aligned} g(q) \le 1 - M \left( \frac{1}{q} - \frac{1}{2} \right) \left( \frac{3}{4} - \frac{1}{q} \right) , \quad M= 0.86326 . \end{aligned}$$

Proof

(a) Differentiation gives $f'(q) = \frac{f(q)}{q^2} \left( \Psi \left( \frac{1}{q} \right) + \Psi \left( 2-\frac{1}{q} \right) \right)$. Since $\Gamma$ is logarithmic convex,$\Psi$ is increasing. Hence $\Psi \left( \frac{1}{q} \right) \le \Psi (1) = - \gamma$ and $\Psi \left( 2-\frac{1}{q} \right) \le \Psi \left( \frac{3}{2} \right) = 2 \left( 1- \ln (2) \right) - \gamma$, $\Psi \left( \frac{1}{q} \right) +\Psi \left( 2-\frac{1}{q} \right) \le -2 \left( \gamma + \ln (2) -1 \right) < 0$. Therefore f is decreasing in [1, 2]. Moreover, $f \left( \frac{4}{3} \right) \le 0.7397$.

(b) For g we find $(\ln (g))'(q) = \frac{1}{2 q^2} \left( \Psi \left( 2-\frac{1}{q} \right) - \Psi \left( \frac{1}{q} \right) - 2 \ln (2) \right)$. The function $h(q):=\Psi \left( 2-\frac{1}{q} \right) - \Psi \left( \frac{1}{q} \right) - 2 \ln (2)$ is strictly increasing, since with $\frac{1}{q} + \frac{1}{p} =1$ we have, using (3.1),

$$\begin{aligned} h'(q) = \frac{1}{q^2} \left( \Psi ' \left( 2-\frac{1}{q} \right) + \Psi ' \left( \frac{1}{q} \right) \right) = \frac{1}{q^2} \sum _{n=1}^\infty \left( \frac{1}{(n+\frac{1}{p})^2} + \frac{1}{(n-\frac{1}{p})^2} \right) > 0 . \end{aligned}$$

Moreover $h(1) = - 2 \ \ln (2) < 0$, $h(2) = 2 \left( 1 - \ln (2) \right) > 0$. Thus h has exactly one zero $q_1 \in (1,2)$, $q_1 \simeq 1.612$. We get that g is decreasing in $[1,q_1)$ and increasing in $(q_1,2]$. We have $g(1) = \frac{2}{\sqrt{\pi }} > 1$, $g \left( \frac{4}{3} \right) = g(2) = 1$ and $g(q) < 1$ for $q \in (\frac{4}{3}, 2)$.

For $\frac{4}{3}< q < 2$, choose $\theta \in (0,1)$ with $\frac{1}{q} = (1-\theta ) \frac{1}{2} + \theta \frac{3}{4}$, $\theta = \frac{4}{q} -2$, $1-\theta = 3 - \frac{4}{q}$. Since $\Gamma$ is logarithmic convex, $F:= \ln \Gamma$ satisfies $F'' = \Psi ' > 0$ and by Taylor’s formula with second degree remainder we have for some $\eta \in (\frac{1}{2}, \frac{3}{4})$

$$\begin{aligned} F \left( \frac{1}{q} \right) \le (1-\theta ) F \left( \frac{1}{2} \right) + \theta F \left( \frac{3}{4} \right) - \frac{\Psi '(\eta )}{2} \left( \frac{1}{q} - \frac{1}{2} \right) \left( \frac{3}{4} - \frac{1}{q} \right) . \end{aligned}$$

Since by (3.1) $\Psi '$ is decreasing, $\min _{\eta \in [\frac{1}{2}, \frac{3}{4}]} \Psi '(\eta ) = \Psi ' \left( \frac{3}{4} \right)$ and

$$\begin{aligned} F \left( \frac{1}{q} \right) \le \left( 3 - \frac{4}{q} \right) F \left( \frac{1}{2} \right) + \left( \frac{4}{q} -2 \right) F \left( \frac{3}{4} \right) - \frac{\Psi ' \left( \frac{3}{4} \right) }{2} \left( \frac{1}{q} - \frac{1}{2} \right) \left( \frac{3}{4} - \frac{1}{q} \right) . \end{aligned}$$

Similarly, for $\frac{5}{4}< 2 - \frac{1}{q} < \frac{3}{2}$, choose $\theta \in (0,1)$ with $2 - \frac{1}{q} = (1-\theta ) {\frac{5}{4}} + \theta {\frac{3}{2}}$, $\theta = 3 - \frac{4}{q}$, $1 - \theta = \frac{4}{q} -2$, such that with $\min _{\eta \in [\frac{5}{4}, \frac{3}{2}]} \Psi '(\eta ) = \Psi ' \left( \frac{3}{2} \right)$

$$\begin{aligned} F \left( 2 - \frac{1}{q} \right) \le \left( \frac{4}{q}-2 \right) F \left( \frac{5}{4} \right) + \left( 3 - \frac{4}{q} \right) F \left( \frac{3}{2} \right) - \frac{\Psi ' \left( \frac{3}{2} \right) }{2} \left( \frac{1}{q} - \frac{1}{2} \right) \left( \frac{3}{4} - \frac{1}{q} \right) . \end{aligned}$$

This yields after exponentiation with $c:=\frac{\Psi ' \left( \frac{3}{4} \right) +\Psi ' \left( \frac{3}{2} \right) }{4}$

$$\begin{aligned} \Gamma \left( \frac{1}{q} \right) \Gamma \left( 2-\frac{1}{q} \right)&\le \left( \Gamma \left( \frac{1}{2} \right) \Gamma \left( \frac{3}{2} \right) \right) ^{3-\frac{4}{q}} \\&\quad \times \left( \Gamma \left( \frac{3}{4} \right) \Gamma \left( \frac{5}{4} \right) \right) ^{\frac{4}{q} -2} \exp \left( - 2 c \left( \frac{1}{q} - \frac{1}{2} \right) \left( \frac{3}{4} - \frac{1}{q} \right) \right) \ . \end{aligned}$$

Clearly $\Gamma \left( \frac{1}{2} \right) \Gamma \left( \frac{3}{2} \right) = \frac{\pi }{2}$, and by the complement formula for the $\Gamma$-function, see Abramowitz, Stegun [1], $\Gamma \left( \frac{3}{4} \right) \Gamma \left( \frac{5}{4} \right) = \frac{\pi }{4 \sin \left( \frac{\pi }{4} \right) } = \frac{\pi }{2 \sqrt{2}}$, so that

$$\begin{aligned} g(q) = \sqrt{\frac{2^{\frac{2}{q}}}{\pi }\Gamma \left( \frac{1}{q} \right) \Gamma \left( 2-\frac{1}{q} \right) } \le \exp \left( - c \left( \frac{1}{q} - \frac{1}{2} \right) \left( \frac{3}{4} - \frac{1}{q}\right) \right) =: k(q) . \end{aligned}$$

Let $\varepsilon := \left( \frac{1}{q} - \frac{1}{2} \right) \left( \frac{3}{4} - \frac{1}{q} \right)$. Then $\varepsilon \le \frac{1}{64}$. Numerical evaluation yields $c \ge 0.86917$. By Taylor expansion $k(q) \le 1 - c \varepsilon + \frac{c^2 \varepsilon ^2}{2} = 1 - c \left( 1-\frac{c \varepsilon }{2} \right) \varepsilon \le 1 - d \varepsilon$, $d:= c - \frac{c^2}{128} \ge 0.86326$ =: M. Therefore $g(q) \le 1 - M \left( \frac{1}{q} - \frac{1}{2} \right) \left( \frac{3}{4} - \frac{1}{q} \right)$. $\square$

Lemma 5.2

For all $\frac{4}{3} \le q \le 2$ and $0 \le s \le \frac{16}{5}$, the functions $\delta _q$ of (2.6) satisfy $\delta _{\frac{4}{3}}(s) \le \delta _q(s) \le \delta _2(s) = \exp \left( -\frac{s^2}{4} \right)$. Further $\delta _{\frac{4}{3}} \left( \frac{48}{25} \right) > 0$ and $\delta _{\frac{4}{3}} \left( \frac{16}{5} \right) > - 0.588$.

Proof

Let $2 \le p = \frac{q}{q-1} \le 4$ be the conjugate index of q and $s \in [\frac{48}{25},\frac{16}{5}]$. We will show that $\frac{d}{dq} \delta _q(s) > 0$, or equivalently $\frac{d}{dp} \delta _q(s) < 0$. For $m > -1$ we have

$$\begin{aligned} \int _0^\infty r^m \exp (-r^p) dr = \frac{1}{p} \int _0^\infty u^{\frac{m+1}{p} -1} \exp (-u) du = \frac{1}{p} \Gamma \left( \frac{m+1}{p} \right) . \end{aligned}$$

(5.1)

Expanding $\cos (s r)$ into its Taylor series at zero, we find using (5.1)

$$\begin{aligned} \delta _q(s) = \sum _{n=0}^\infty \left( f_{2n}(p) \frac{s^{4n}}{(4n)!} - f_{2n+1}(p) \frac{s^{4n+2}}{(4n+2)!} \right) =: \sum _{n=0}^\infty F_n(p,s) \, \end{aligned}$$

(5.2)

where $f_{2n}(p):= \frac{\Gamma \left( 1+\frac{4 n -1}{p} \right) }{\Gamma \left( 1-\frac{1}{p} \right) }$, $f_{2n+1}(p):= \frac{\Gamma \left( 1+\frac{4 n + 1}{p} \right) }{\Gamma \left( 1-\frac{1}{p} \right) }$. Since $\Gamma$ is logarithmic convex, we have for $x>0$ and $0 \le \theta \le 1$ that $\Gamma (x+\theta ) \le \Gamma (x)^{1-\theta } \Gamma (x+1)^\theta = x^\theta \Gamma (x)$. Let $n \ge 2$, $x:= 1 + \frac{4n-1}{p} \ge 1$ and $\theta := \frac{2}{p}$. We claim that $x^\theta = \left( 1+ \frac{4n-1}{p} \right) ^\theta \le \frac{4n+1}{p}$. This is equivalent to $(4n+p-1)^{\frac{2}{p}} p^{1-\frac{2}{p}} \le 4n+1$. Applying the inequality $a b \le \frac{a^r}{r} + \frac{b^{r'}}{r'}$ with $r:= \frac{p}{2}$ and $r' = \frac{p}{p-2}$, we get $(4n+p-1)^{\frac{2}{p}} p^{1-\frac{2}{p}} \le \frac{2}{p} (4n+p-1) + p - 2 = \frac{2}{p} (4n-1) + p$ which is $\le 4n+1$ if and only if $n \ge \frac{p+1}{4}$, which is satisfied, since $p \le 4$ and $n \ge 2$. Therefore

$$\begin{aligned} \frac{f_{2n+1}(p)}{f_{2n}(p)} = \frac{\Gamma \left( 1+\frac{4 n + 1}{p} \right) }{\Gamma \left( 1+\frac{4 n -1}{p} \right) } \le \frac{4n+1}{p} \end{aligned}$$

(5.3)

and

$$\begin{aligned} F_n(p,s)&= f_{2n}(p) \frac{s^{4n}}{(4n)!} \left( 1 - \frac{f_{2n+1}(p)}{f_{2n}(p)} \frac{s^2}{(4n+1)(4n+2)} \right) \\&\ge f_{2n}(p) \frac{s^{4n}}{(4n)!} \left( 1 - \frac{s^2}{2 p (2n+1)} \right) > 0 \end{aligned}$$

for $n \ge 2$ and $s \le \frac{16}{5} < \sqrt{20}$. Hence for all $m \ge 1$, $\delta _q(s) \ge \sum _{n=0}^m F_n(p,s)$. In particular, for $q=\frac{4}{3}$, we find by numerical evaluation $\delta _{\frac{4}{3}} \left( \frac{48}{25} \right)> 0.0026 > 0$, choosing $m=2$, and $\delta _{\frac{4}{3}} \left( \frac{16}{5} \right) > - 0.588$, choosing $m=4$. Formula (5.2) implies

$$\begin{aligned} \frac{d}{dp} \delta _q(s)&= \sum _{n=0}^\infty \frac{d}{dp} F_n(p,s) =: - \frac{1}{p^2} \sum _{n=0}^\infty G_n(p,s) \nonumber \\&=: - \frac{1}{p^2} \sum _{n=0}^\infty \left( f_{2n}(p)g_{2n}(p) \frac{s^{4n}}{(4n)!} - f_{2n+1}(p)g_{2n+1}(p) \frac{s^{4n+2}}{(4n+2)!} \right) \ , \end{aligned}$$

(5.4)

where in terms of the Digamma function $\Psi$, $g_{2n}(p)= (4 n -1) \Psi \left( 1+\frac{4n-1}{p} \right) + \Psi \left( 1-\frac{1}{p} \right)$, $g_{2n+1}(p)= (4 n + 1) \Psi \left( 1+\frac{4n+1}{p} \right) + \Psi \left( 1-\frac{1}{p} \right)$. By Abramowitz, Stegun [1], 6.3, $\Psi '$ is positive, decreasing and $\Psi '(1+x) \le \frac{1}{x+\frac{1}{2}}$. Therefore

$$\begin{aligned} \Psi \left( 1+\frac{4n+1}{p} \right)\le & {} \Psi \left( 1+\frac{4n-1}{p} \right) + \frac{2}{p} \Psi ' \left( 1+\frac{4n-1}{p} \right) \\ {}\le & {} \Psi \left( 1+\frac{4n-1}{p} \right) + \frac{1}{2n} (4n+1) \Psi \left( 1+\frac{4n+1}{p} \right) \\ {}\le & {} (4n-1) \Psi \left( 1+\frac{4n-1}{p} \right) + 2 \Psi \left( 1+\frac{4n-1}{p} \right) + \frac{4n+1}{2n} \, \end{aligned}$$

so that

$$\begin{aligned} \frac{g_{2n+1}(p)}{g_{2n}(p)} \le 1 + \frac{2 \Psi \left( 1+\frac{4n-1}{p} \right) + 2 + \frac{1}{2n}}{g_{2n}(p)} = 1 + \frac{2}{4n-1} + C_n(p) \, \end{aligned}$$

(5.5)

where $C_n(p):= \frac{2 - \frac{2 \Psi \left( 1-\frac{1}{p} \right) }{4n-1} + \frac{1}{2n}}{g_{2n}(p)}$. The function $k(x):= \Psi (1+x)- \ln (1+x)$ is increasing, since using (3.1) we find

$$\begin{aligned} k'(x) = \sum _{n=1}^\infty \frac{1}{(n+x)^2} - \frac{1}{1+x} > \int _1^\infty \frac{dy}{(y+x)^2} - \frac{1}{1+x} = 0 . \end{aligned}$$

Further, by Abramowitz, Stegun [1, 6.3.18], $\lim _{x \rightarrow \infty } \Psi (1+x) - \ln (1+x) = 0$. Therefore for any $x>0$, $k(x) < 0$. For $x \ge \frac{7}{4}$, $k(x) \ge k \left( \frac{7}{4} \right) \simeq - 0.193 > - \frac{1}{5}$. Hence for $n \ge 2$, $\Psi \left( 1 + \frac{4 n -1}{p} \right) \ge \ln \left( 1 + \frac{4 n -1}{p} \right) - \frac{1}{5}$ and

$$\begin{aligned} C_n(p) \le \frac{2 - \frac{2 \Psi \left( 1-\frac{1}{p} \right) }{4n-1} + \frac{1}{2n}}{(4n-1) \ln \left( 1 + \frac{4 n -1}{p} \right) - \frac{4n-1}{5} + \Psi \left( 1 - \frac{1}{p} \right) } =: R . \end{aligned}$$

We have that $\Psi \left( 1- \frac{1}{p} \right) \in [-1.964,-1.085]$. The right side is maximal for $p=4$, with $\Psi \left( \frac{3}{4} \right) \simeq -1.086$, the logarithmic term in the denominator being minimal then,

$$\begin{aligned} C_n(p) \le \frac{2 - \frac{2 \Psi \left( \frac{3}{4} \right) }{4n-1} + \frac{1}{2n}}{(4n-1) \ln \left( n + \frac{3}{4} \right) - \frac{4n-1}{5} + \Psi \left( \frac{3}{4} \right) } . \end{aligned}$$

A tedious calculation yields that $C_n(p) \le c_n(p):= \frac{2}{3 n \ln (n)}$ and $\frac{2}{4n-1} + c_n(2) \le \frac{1}{n + \frac{1}{4}}$ for all $n \ge 5$. For $n = 2, 3, 4$, keeping a dependence on p, we may estimate $C_n(p) \le R \le c_n(p)$, where

$$\begin{aligned} c_2(p) = \frac{1}{6} + \frac{p}{10}, c_3(p) = \frac{1}{11} + \frac{p}{36}, c_4(p) = \frac{1}{18} + \frac{p}{72} . \end{aligned}$$

Using this and Eqs. (5.5) and (5.3) we find for $n \ge 2$

$$\begin{aligned} \frac{f_{2n+1}(p) g_{2n+1}(p)}{f_{2n}(p) g_{2n}(p)}&\le \frac{4n+1}{p} \left( 1+\frac{2}{4n-1} + c_n(p) \right) \le \frac{4n+1}{2} \left( 1+\frac{2}{4n-1} + c_n(2) \right) \\&=: q_n \le \frac{4n+1}{2} \left( 1 + \frac{1}{n+ \frac{1}{4}} \right) = \frac{4n+5}{2} \ , \end{aligned}$$

the last estimate for $q_n$ holding for $n \ge 5$, whereas $q_2 \le \frac{15}{2}$, $q_3 \le \frac{25}{3}$ and $q_4 \le \frac{21}{2}$. This implies

$$\begin{aligned} G_n(p,s) \ge f_{2n}(p) g_{2n}(p) \frac{s^{4n}}{(4n)!} \left( 1 - \frac{q_n s^2}{(4n+1)(4n+2)} \right) =: {\tilde{G}}_n(p,s) > 0 \end{aligned}$$

for all $n \ge 2$ and $s \le \frac{16}{5} < \sqrt{12}$. Hence by (5.4) for all $m \ge 1$

$$\begin{aligned} \frac{d}{dp} \delta _q(s) \le - \frac{1}{p^2} \sum _{n=0}^m G_n(p,s) . \end{aligned}$$

For $m=1$, with $g_0(s) = 0$, we have

$$\begin{aligned} \frac{d}{dp} \delta _q(s) \le + \frac{s^2}{2 p^2} \left[ a(p) - b(p) \frac{s^2}{12} + c(p) \frac{s^4}{360} \right] =: \phi (p,s) \end{aligned}$$

with $-0.972 \le a(p):= f_1(p)g_1(p) \le -0.954$ varying very little, $-0.255 \le b(p):= f_2(p)g_2(p) \le 0.114$, $0 < b(p)$ for $p \le 2.83$ and $c(p):= f_3(p) g_3(p)$ decreasing in $p \in [2,4]$, with value 6.66 at $p=2$ and 1.64 at $p=4$. Therefore $360 |a(p)| \ge 343.5$ and $\phi (p,s) < 0$ will be satisfied if

$$\begin{aligned} s^2 < 15 \frac{b(p)}{c(p)} + \sqrt{ \left( 15 \frac{b(p)}{c(p)} \right) ^2 + \frac{343.5}{c(p)} } . \end{aligned}$$

This holds for all $0 \le s \le \frac{16}{5}$, if $c(p) \le 3.275$, i.e. $p \ge 2.81$. For $0 \le p \le 2$, the right side is minimal for $p=2$ and we require $s \le 2.72$. If $p < 2.81$ and $s > 2.72$ one needs two more terms, $m=3$, to show $\frac{d}{dp} \delta _q(s) < 0$,

$$\begin{aligned} \frac{d}{dp} \delta _q(s) \le - \frac{1}{p^2} \left[ G_0(p,s) + G_1(p,s) + {\tilde{G}}_2(p,s) + {\tilde{G}}_3(p,s) \right] < 0 . \end{aligned}$$

$\square$

Corollary 5.3

For all $\frac{4}{3} \le q \le 2$ and $\frac{48}{25} \le s \le \frac{16}{5}$, $\left| \delta _q(s) \right| \le 0.588$.

Proof

By Lemma 5.2, $\delta _{\frac{4}{3}}(s) \le \delta _q(s) \le \delta _2(s) = \exp \left( -\frac{s^2}{4} \right) \le \exp \left( - \left( \frac{24}{25} \right) ^2 \right) < \frac{2}{5}$ for all $s \in [\frac{48}{25},\frac{16}{5}]$. By (2.7), $\delta '_{\frac{4}{3}}(s) = - \frac{\Gamma (\frac{5}{4})}{\Gamma (\frac{3}{4})} s \gamma _4(s)$. According to Boyd [4], page 83, $\gamma _4(s) > 0$ for all $0 \le s \le 3.45$, the first positive zero of $\gamma _4$ being at $s_1 \simeq 3.4535$. Therefore $\delta _{\frac{4}{3}}$ is strictly decreasing in $[\frac{48}{25},\frac{16}{5}]$, with $\frac{2}{5}> \delta _{\frac{4}{3}}(s) \ge \delta _{\frac{4}{3}}\left( \frac{16}{5} \right) > - 0.588$ by Lemma 5.2. We conclude that

$$\begin{aligned} \max \left\{ \left| \delta _q(s) \right| \ | \ q \in \left[ \frac{4}{3}, 2\right] \, \ s \in \left[ \frac{48}{25},\frac{16}{5}\right] \right\} \le 0.588 . \end{aligned}$$

$\square$

Remark. For $q \searrow 1$, $\delta _q(s) \rightarrow \cos (s)$, so that $|\delta _q(\pi )| \rightarrow 1$. Corollary 5.3 does not extend to the range $1 \le q < \frac{4}{3}$.

6 Proof of Theorem 1.2

Proof of Theorem 1.1.

Barthe and Naor [3] showed for $1 \le q < \infty$ that

$$\begin{aligned} \lim _{n \rightarrow \infty } \frac{P_{n,q}(a^{(n)})}{P_{n,q}(a^{(2)})} = \sqrt{\frac{2^{\frac{2}{q}}}{\pi }\Gamma \left( \frac{1}{q} \right) \Gamma \left( 2-\frac{1}{q} \right) } \, \end{aligned}$$

and this is $< 1$ if and only if $\frac{4}{3}< q < 2$, see Lemma 5.1 (b). Now consider $\frac{4}{3}< q < 2$ and let $p = \frac{q}{q-1}$ be the conjugate index, $2< p < 4$. As in the proof of Theorem 1.1, we use $\cos (x) \ge 1 - \frac{x^2}{2} + \frac{x^4}{26} >0$ for all $0 \le x \le \frac{3}{2}$, so that by (2.6) for all $s \le \frac{3}{2} \sqrt{n}$

$$\begin{aligned} \delta _q \left( \frac{s}{\sqrt{n}} \right)&\ge \frac{p}{\Gamma \left( \frac{1}{q} \right) } \Big ( \int _0^{\frac{3}{2} \frac{\sqrt{n}}{s}} \left( 1 - \frac{s^2 r^2}{2n} + \frac{s^4 r^4}{26 n^2} \right) \ r^{p-2} \exp (-r^p) \ dr \\&\quad + \int _{\frac{3}{2} \frac{\sqrt{n}}{s}}^\infty \cos \left( \frac{sr}{\sqrt{n}} \right) \ r^{p-2} \exp (-r^p) \ dr \Big ) \\&\ge \frac{p}{\Gamma \left( \frac{1}{q} \right) } \left( \int _0^\infty \left( 1 - \frac{s^2 r^2}{2n} + \frac{s^4 r^4}{26 n^2} \right) \ r^{p-2} \exp (-r^p) \ dr - R \right) \\&= \frac{1}{\Gamma \left( 1 - \frac{1}{p} \right) } \left( \Gamma \left( 1 - \frac{1}{p} \right) - \Gamma \left( 1 + \frac{1}{p} \right) \frac{s^2}{2 n} + \Gamma \left( 1 + \frac{3}{p} \right) \frac{s^4}{26 n^2} - R \right) \ , \\ R&:= \int _{\frac{3}{2} \frac{\sqrt{n}}{s}}^\infty \left( 2-\frac{s^2}{2 n} r^2 + \frac{s^4}{26 n^2} r^4 \right) \ r^{p-2} \exp (-r^p) \ dr \\&= \frac{1}{p} \int _{\left( \frac{3}{2} \frac{\sqrt{n}}{s} \right) ^p}^\infty \left( 2 u^{-\frac{1}{p}} - \frac{s^2}{2 n} u^{\frac{1}{p}} + \frac{s^4}{26 n^2} u^{\frac{3}{p}} \right) \exp (-u) \ du \ . \end{aligned}$$

Then $u^{-\frac{1}{p}} \le \frac{2}{3} \frac{s}{\sqrt{n}}$ and $u^{\frac{3}{p}} \le u^{\frac{3}{2}}$. As in the proof of Theorem 1 (iii), for $x \ge \frac{9}{2}$

$$\begin{aligned} \int _x^\infty u^{\frac{3}{2}} \exp (-u) du \le \Big ( (1+x) (2+2 x +x^2) \Big )^{\frac{1}{2}} \exp (-x) \le \frac{13}{20} x^2 \exp (-x) . \end{aligned}$$

Choose again $s \le \sqrt{\frac{n}{2}}$. Then with $x:= \left( \frac{3}{2} \frac{\sqrt{n}}{s} \right) ^p \ge \left( \frac{3}{\sqrt{2}} \right) ^2 = \frac{9}{2}$ and $y:= \frac{s}{\sqrt{n}}$

$$\begin{aligned} R \le \frac{1}{p} \left( 2 \frac{2}{3} y + \frac{y^4}{26} \frac{13}{20} \left( \frac{3}{2} \frac{1}{y} \right) ^{2p} \right) \ \exp \left( - \left( \frac{3}{2} \frac{1}{y} \right) ^p \right) . \end{aligned}$$

We want $R \le \Gamma \left( 1+\frac{3}{p} \right) \frac{y^4}{26}$, a condition which is strongest for $p=2$ when it means

$$\begin{aligned} \frac{1}{2} \left( \frac{4}{3} y + \frac{81}{640} \right) < \frac{\Gamma \left( \frac{5}{2} \right) }{26} y^4 \ \exp \left( \frac{9}{4} \frac{1}{y^2} \right) \, \end{aligned}$$

which is the same requirement as in (iii) of the proof of Theorem 1.1, being valid for $0 \le y \le 0.7161$. Thus the choice of $s \le \sqrt{\frac{n}{2}}$ is allowed and then

$$\begin{aligned} \delta _q \left( \frac{s}{\sqrt{n}} \right) \ge 1 - c \frac{s^2}{n}, \quad c:= \frac{1}{2} \frac{\Gamma \left( 1+\frac{1}{p} \right) }{\Gamma \left( 1-\frac{1}{p} \right) } = \frac{1}{2} \frac{\Gamma \left( 2-\frac{1}{q} \right) }{\Gamma \left( \frac{1}{q} \right) } . \end{aligned}$$

We have by Lemma 5.1 (a) $\frac{1}{4} \le c \le 0.3699 < \frac{37}{100}$, the lower estimate attained for $p=q=2$, the maximum for c attained for $p=4$, $q= \frac{4}{3}$. Therefore for $x:= c \frac{s^2}{n} \le \frac{37}{200}$, $\ln (1-x) \ge -x - \frac{1}{2} \frac{x^2}{1-x} \ge -x - \frac{100}{163} x^2$ and $\left( 1-c \frac{s^2}{n} \right) ^n \ge \exp (-c s^2) \left( 1 - \frac{100}{163} c^2 \frac{s^4}{n} \right)$. This yields the estimate

$$\begin{aligned}&\int _0^{\sqrt{\frac{n}{2}}} \frac{1-\delta _q \left( \frac{s}{\sqrt{n}} \right) ^n}{s^2} \ ds \le \int _0^{\sqrt{\frac{n}{2}}} \frac{1-\exp (-c s^2) \left( 1-\frac{100}{163} c^2 \frac{s^4}{n} \right) }{s^2} \ ds \\&\quad \le \int _0^\infty \frac{1-\exp (-c s^2)}{s^2} ds - \int _{\sqrt{\frac{n}{2}}}^\infty \frac{1-\exp (-c s^2)}{s^2} ds + \frac{100}{163} \frac{c^2}{n} \int _0^\infty s^2 \exp (-c s^2) \ ds \\&\quad = \sqrt{\pi c} \left( 1+ \frac{25}{163} \frac{1}{n} \right) - \int _{\sqrt{\frac{n}{2}}}^\infty \frac{1-\exp (-c s^2)}{s^2} \ ds \ . \end{aligned}$$

Therefore, using (2.6),

$$\begin{aligned} P_{n,q}(a^{(n)})&= \Gamma \left( \frac{1}{q} \right) \frac{2}{\pi }\int _0^\infty \frac{1-\delta _q \left( \frac{s}{\sqrt{n}} \right) ^n}{s^2} \ ds \\&= \Gamma \left( \frac{1}{q} \right) \frac{2}{\pi }\left( \int _0^{\sqrt{\frac{n}{2}}} \frac{1-\delta _q \left( \frac{s}{\sqrt{n}} \right) ^n}{s^2} \ ds + \int _{\sqrt{\frac{n}{2}}}^\infty \frac{1-\delta _q \left( \frac{s}{\sqrt{n}} \right) ^n}{s^2} \ ds \right) \\&\le \Gamma \left( \frac{1}{q} \right) \frac{2}{\pi }\left( \sqrt{\pi c} \left( 1+ \frac{25}{163} \frac{1}{n} \right) + \int _{\sqrt{\frac{n}{2}}}^\infty \frac{\exp (-c s^2) - \delta _q \left( \frac{s}{\sqrt{n}} \right) ^n}{s^2} \ ds \right) \\&= \sqrt{\frac{2}{\pi }\Gamma \left( \frac{1}{q} \right) \Gamma \left( 2-\frac{1}{q} \right) } \left( 1+\frac{25}{163} \frac{1}{n} \right) + \Gamma \left( \frac{1}{q} \right) \frac{2}{\pi }S \ , \end{aligned}$$

$S:= \frac{1}{\sqrt{n}} \int _{\frac{1}{\sqrt{2}}}^\infty \frac{\exp (-c u^2)^n - \delta _q(u)^n}{u^2} du$. Since $c \ge \frac{1}{4}$ and $\delta _q(s) >0$ for all $0 \le u \le \frac{48}{25}$ by Lemma 5.2, we find

$$\begin{aligned} S \le \frac{1}{\sqrt{n}} \int _{\frac{1}{\sqrt{2}}}^\infty \frac{\exp \left( - \frac{u^2}{4} n \right) }{u^2} \ du + \frac{1}{\sqrt{n}} \int _{\frac{48}{25}}^\infty \frac{|\delta _q(u)|^n}{u^2} \ du . \end{aligned}$$

For $n \ge 8$, $v = \frac{u^2}{4} n \ge 1$ and

$$\begin{aligned} \frac{1}{\sqrt{n}} \int _{\frac{1}{\sqrt{2}}}^\infty&\frac{\exp \left( - \frac{u^2}{4} n \right) }{u^2} \ du = \frac{1}{4} \int _{\frac{n}{8}}^\infty \frac{\exp (-v)}{v^{\frac{3}{2}}} \ dv \\&\le \frac{1}{4} \left( \frac{8}{n} \right) ^{\frac{3}{2}} \int _{\frac{n}{8}}^\infty \exp (-v) \ dv = \frac{4 \sqrt{2}}{n^{\frac{3}{2}}} \exp \left( -\frac{n}{8} \right) \le \frac{4 \sqrt{2}}{n^{\frac{3}{2}}} \ 0.8825^n \ . \end{aligned}$$

By Corollary 5.3 we have $|\delta _q(s)| \le 0.588$ for all $\frac{48}{25} \le u \le \frac{16}{5}$. Therefore

$$\begin{aligned} \frac{1}{\sqrt{n}} \int _{\frac{48}{25}}^{\frac{16}{5}} \frac{|\delta _q(u)|^n}{u^2} \ du \le \frac{1}{\sqrt{n}} \ 0.588^n \int _{\frac{48}{25}}^{\frac{16}{5}} \frac{du}{u^2} = \frac{5}{24} \frac{1}{\sqrt{n}} \ 0.588^n . \end{aligned}$$

Integration by parts shows for $\frac{4}{3} \le q < 2$, $2 < p \le 4$ that

$$\begin{aligned} \left| \delta _q(u) \right|&= \left| \frac{p}{\Gamma \left( \frac{1}{q} \right) } \int _0^\infty \frac{\sin (ur)}{u} (p-2-p r^p) \ r^{p-3} \exp (-r^p) \ dr \right| \\&\le \frac{1}{u} \frac{p}{\Gamma \left( \frac{1}{q} \right) } \int _0^\infty |p-2-p r^p| \ r^{p-3} \ \exp (-r^p) \ dr \\&= \frac{1}{u} \frac{2 p}{\Gamma \left( 1 - \frac{1}{p} \right) } \left( \frac{1-\frac{2}{p}}{e} \right) ^{1-\frac{2}{p}} =: \phi (p) \frac{1}{u} , \end{aligned}$$

using that $\int (p-2-p r^p) \ r^{p-3} \ \exp (-r^p) \ dr = r^{p-2} \exp (-r^p)$. For $\phi$, we have that $(\ln \phi )'(p) = \frac{1}{p^2} \left( 2 \ln \left( 1 - \frac{2}{p} \right) + p - \Psi \left( 1- \frac{1}{p} \right) \right)$ and $\lambda (p):= 2 \ln \left( 1 - \frac{2}{p} \right) + p - \Psi \left( 1- \frac{1}{p} \right)$ is increasing, since $\lambda '(p) = \frac{p(3+(p-1)^2) - (p-2) \Psi ' \left( 1- \frac{1}{p} \right) }{p^2(p-2)}$ and $\Psi ' \left( 1- \frac{1}{p} \right) \le \Psi ' \left( \frac{1}{2} \right) = \frac{\pi ^2}{2}$, so that $\lambda '(p) \ge \frac{p(3+(p-1)^2) - (p-2) \frac{\pi ^2}{2}}{p^2(p-2)} > 0$. We have $\lambda (2) < 0$, $\lambda (4) > 0$, so that $\phi$ is first decreasing and then increasing in [2, 4]. Since $\phi (2)< \phi (4) < \frac{14}{5}$, we conclude that $\left| \delta _q(u) \right| < \frac{14}{5} \frac{1}{u}$. Then

$$\begin{aligned} \frac{1}{\sqrt{n}} \int _{\frac{16}{5}}^\infty \frac{|\delta _q(u)|^n}{u^2} \ du&\le \frac{1}{\sqrt{n}} \left( \frac{14}{5} \right) ^n \int _{\frac{16}{5}}^\infty \frac{du}{u^{n+2}} \\&= \frac{5}{16} \frac{1}{\sqrt{n} (n+1)} \left( \frac{7}{8} \right) ^n \le \frac{5}{16} \frac{1}{n^{\frac{3}{2}}} \ 0.875^n \ . \end{aligned}$$

We finally get that

$$\begin{aligned} S \le \frac{4 \sqrt{2}}{n^{\frac{3}{2}}} \ 0.8825^n + \frac{5}{24} \frac{1}{\sqrt{n}} \ 0.588^n + \frac{5}{16} \frac{1}{n^{\frac{3}{2}}} \ 0.875^n . \end{aligned}$$

By Lemma 5.1 (a), $\sqrt{\frac{2}{\pi }\frac{\Gamma \left( \frac{1}{q} \right) }{\Gamma \left( 2 - \frac{1}{q} \right) }} \le \frac{2}{\sqrt{\pi }}$, with equality for $p=q=2$, and $\frac{2}{\sqrt{\pi }} S \le \frac{0.10559}{n}$ for all $n > 20$. We conclude with $\frac{25}{163} + 0.10559 < 0.25896$ that

$$\begin{aligned} P_{n,q}(a^{(n)}) \le \sqrt{\frac{2}{\pi }\Gamma \left( \frac{1}{q} \right) \Gamma \left( 2-\frac{1}{q} \right) } \left( 1+ \frac{0.25896}{n} \right) . \end{aligned}$$

By Barthe, Naor [3] $P_{n,q}(a^{(2)}) = 2^{\frac{1}{2} -\frac{1}{q}}$, so that together with Lemma 5.2 (b)

$$\begin{aligned} \frac{P_{n,q}(a^{(n)})}{P_{n,q}(a^{(2)})}&\le \sqrt{\frac{2^{\frac{2}{q}}}{\pi }\Gamma \left( \frac{1}{q} \right) \Gamma \left( 2-\frac{1}{q} \right) } \left( 1+ \frac{0.25896}{n} \right) \\&\le \left( 1- M \left( \frac{1}{q} - \frac{1}{2} \right) \left( \frac{3}{4} - \frac{1}{q} \right) \right) \left( 1+ \frac{0.25896}{n} \right) , \; M= 0.86326. \end{aligned}$$

Suppose that n satisfies $n \ge \frac{4}{5} \frac{q^2}{(q-\frac{4}{3})(2-q)}$. Then the last product is $< 1$, since $\frac{0.25896}{n} \le M \left( \frac{1}{q} - \frac{1}{2} \right) \left( \frac{3}{4} - \frac{1}{q} \right)$ suffices and this requires $n > \frac{\frac{8}{3} \frac{0.25896}{M} q^2}{(q-\frac{4}{3})(2-q)}$, with $\frac{8}{3} \frac{0.25896}{M} < \frac{4}{5}$. The condition $n \ge \frac{4}{5} \frac{q^2}{(q-\frac{4}{3})(2-q)}$ will be satisfied if $n > \frac{\frac{32}{15}}{q- \frac{4}{3}} + \frac{\frac{24}{5}}{2-q} = \frac{\frac{8}{3} q - \frac{32}{15}}{(q-\frac{4}{3})(2-q)} \ge \frac{\frac{4}{5} q^2}{(q-\frac{4}{3})(2-q)}$, where the last inequality is an equality for $q=\frac{4}{3}$ and $q=2$. Hence for $n > \frac{\frac{32}{15}}{q- \frac{4}{3}} + \frac{\frac{24}{5}}{2-q} =: \phi (q)$ we have $P_{n,q}(a^{(n)}) < P_{n,q}(a^{(2)})$. The restriction $n>20$ is automatically satisfied since the minimum of $\phi$ is $\phi \left( \frac{8}{5} \right) = 20$. We have the equality $\lim _{n \rightarrow \infty } \frac{P_{n,q}(a^{(n)})}{P_{n,q}(a^{(2)})} = \sqrt{\frac{2^{\frac{2}{q}}}{\pi }\Gamma \left( \frac{1}{q} \right) \Gamma \left( 2-\frac{1}{q} \right) }$. $\square$

Remark. Barthe and Naor [3] show that $P_{n,q}(a) \le P_{n,q}(a^{(n)})$ for all $a \in S^{n-1}$ and $q \ge 2$. Since the quantities are differentiable and coincide for $q=2$, they conclude that $\frac{d}{dq} P_{n,q}(a)|_{q=2} \le \frac{d}{dq} P_{n,q}(a^{(n)})|_{q=2}$ holds. If this inequality were strict, $P_{n,q}(a) \ge P_{n,q}(a^{(n)})$ might hold for some $2- \varepsilon _n< q < 2$. In this case, for q close to 2, $a^{(n)}$ might always yield the minimum, and no singularity $O(\frac{1}{2-q})$ would occur in Theorem 1.2.

Similarly, in the case of sections, Koldobsky [10] showed that $A_{n,p}(a^{(n)}) \le A_{n,p}(a)$ holds for all $1< p < 2$ and $a \in S^{n-1}$, which together with $A_{n,2}(a^{(n)})=A_{n,2}(a)$ implies $\frac{d}{dp} A_{n,p}(a)|_{p=2} \le \frac{d}{dp} A_{n,p}(a^{(n)})|_{p=2}$. If this inequality were strict, it might be that $A_{n,p}(a) \le A_{n,p}(a^{(n)})$ for p close to 2, $2< p < 2 + \varepsilon _n$.

However, numerical evaluation of (2.3) yields for $p = 6, p = 8$ and $n=3, 4$ that $a^{(n)}$ is not maximal, since

$$\begin{aligned} A_{3,6}(a^{(2)})= & {} 2^{\frac{1}{3}} \simeq 1.260> A_{3,6}(a^{(3)}) \simeq 1.250 \, \\ A_{4,8}(a^{(2)})= & {} 2^{\frac{3}{8}} \simeq 1.297> A_{4,8}(a^{(4)}) \simeq 1.295 > A_{4,8}(a^{(3)}) \simeq 1.270 . \end{aligned}$$