Fluctuations of the Magnetization in the p-Spin Curie–Weiss Model

Abstract

In this paper we study the fluctuations of the magnetization in the p-spin Curie–Weiss model, for $p \geqslant 3$. We provide a complete description of the asymptotic distribution of the magnetization in the p-spin Curie–Weiss model, complementing the well-known results in the 2-spin case. Our results unearth various new phase transitions, such as the existence of a certain ‘critical’ curve in the parameter space, where the limiting distribution of the magnetization is a discrete mixture, with local Gaussian fluctuations around each of the atoms. The number of atoms (mixture components) is either two or three depending on the sign of one of the parameters and the parity of p. Another interesting revelation is the existence of certain ‘special’ points in the parameter space where the magnetization converges to a non-Gaussian limiting distribution at rate $N^{\frac{1}{4}}$.

Appendices

Appendix A. Properties of Special Functions and Approximation Lemmas

1.1 Special functions and their properties

In this section, we state few important properties of some special mathematical functions which arise in our analysis.

Definition 2

The gamma function $\Gamma : (0,\infty ) \mapsto \mathbb {R}$ is defined as:

$$\begin{aligned} \Gamma (x) := \int _0^\infty u^{x-1} e^{-u}~du. \end{aligned}$$

Definition 3

The digamma function $\Gamma : (0,\infty ) \mapsto \mathbb {R}$ is defined as:

$$\begin{aligned} \psi (x) := \frac{{\mathrm {d}}}{\mathrm{{d}} x} \log \Gamma (x) = \frac{\Gamma '(x)}{\Gamma (x)}. \end{aligned}$$

The following standard expansion of the digamma function will be very helpful in our analysis: As $x \rightarrow \infty$,

$$\begin{aligned} \psi (1+x) = \log x + \frac{1}{2x} + O(x^{-2}). \end{aligned}$$

(A.1)

Definition 4

For real numbers $x\geqslant y>0$, the binomial coefficient x choose y is defined as

$$\begin{aligned} \left( {\begin{array}{c}x\\ y\end{array}}\right) := \frac{\Gamma (x+1)}{\Gamma (y+1)\Gamma (x-y+1)}. \end{aligned}$$

Lemma A.1

Fix $u>0$. Then, for every $x \in (0,u)$, we have

$$\begin{aligned} \frac{{\mathrm {d}}}{\mathrm{{d}} x}\left( {\begin{array}{c}u\\ x\end{array}}\right) = \left( {\begin{array}{c}u\\ x\end{array}}\right) \left[ \psi (u-x+1) - \psi (x+1)\right] . \end{aligned}$$

Proof

Let $\iota (x) = \left( {\begin{array}{c}u\\ x\end{array}}\right)$. Then, $\log \iota (x) = \log \Gamma (u+1) - \log \Gamma (x+1) - \log \Gamma (u-x+1)$ and hence,

$$\begin{aligned} \frac{\iota '(x)}{\iota (x)} = \frac{{\mathrm {d}}}{\mathrm{{d}} x}\log \iota (x)= -\psi (x+1) + \psi (u-x+1). \end{aligned}$$

(A.2)

Lemma A.1 now follows from (A.2). $\square$

1.2 Mathematical approximations

In this section, we give three different types of standard mathematical approximations, which play crucial roles in our analysis.

Lemma A.2

(Riemann Approximation). Let $f:[a,b]\rightarrow \mathbb {R}$ be a differentiable function, and let $a=x_0<x_1<\ldots <x_n=b$. Let $x_s^* \in [x_{s-1},x_s]$ for each $1\leqslant k\leqslant n$. Then, we have:

$$\begin{aligned} \left| \int _a^b f - \sum _{k=1}^n (x_s - x_{s-1})f(x_s^*)\right| \leqslant \frac{1}{2}(b-a)\max _{1\leqslant k\leqslant n}(x_s - x_{s-1})\sup _{x\in [a,b]}|f'(x)|. \end{aligned}$$

Proof

Lemma A.2 follows from the following string of inequalities:

$$\begin{aligned} \left| \int _a^b f - \sum _{s=1}^n (x_s - x_{s-1})f(x_s^*) \right|&= \left| \sum _{s=1}^n \int _{x_{s-1}}^{x_s} (f(x) - f(x_s^*)){\mathrm {d}} x \right| \nonumber \\&\leqslant \sum _{s=1}^n \int _{x_{s-1}}^{x_s} \left| f(x) - f(x_s^*)\right| {\mathrm {d}} x \end{aligned}$$

(A.3)

$$\begin{aligned}&\leqslant \sup _{x\in [a,b]}|f'(x)| \sum _{s=1}^n \int _{x_{s-1}}^{x_s} \left| x-x_s^*\right| {\mathrm {d}} x \\&= \frac{1}{2}\sup _{x\in [a,b]}|f'(x)| \sum _{s=1}^n \left[ (x_s^*-x_{s-1})^2 + (x_s-x_s^*)^2\right] \nonumber \\&\leqslant \frac{1}{2}\sup _{x\in [a,b]}|f'(x)| \sum _{s=1}^n (x_s-x_{s-1})^2 \nonumber \\&\leqslant \frac{1}{2}(b-a)\max _{1\leqslant s \leqslant n}(x_s - x_{s-1})\sup _{x\in [a,b]}|f'(x)|. \nonumber \end{aligned}$$

(A.4)

Note that, in going from (A.3) to (A.4), we used the mean value theorem. $\square$

The following lemma gives a Laplace-type approximation of an integral over a shrinking interval. For the classical Laplace approximation, which approximates integrals over fixed intervals, refer to [13, 35]. Even though the proof of Lemma A.3 below is exactly similar to that of the classical Laplace approximation, we provide the proof here for the sake of completeness. To this end, for positive sequences $\{a_n\}_{n\geqslant 1}$ and $\{b_n\}_{n\geqslant 1}$, $a_n = O_\square (b_n)$ denotes $a_n \leqslant C_1(\square ) b_n$ and $a_n = \Omega _\square (b_n)$ denotes $a_n \geqslant C_2(\square ) b_n$, for all n large enough and positive constants $C_1(\square ), C_2(\square )$, which may depend on the subscripted parameters.

Lemma A.3

(Laplace-Type Approximation-I) . Let $a<b$ be fixed real numbers, $g: [a,b]\mapsto \mathbb {R}$ be a differentiable function on (a, b), and $h_n : [a,b]\mapsto \mathbb {R}$ be a sequence of thrice differentiable functions on (a, b). Suppose that $\{x_n\}$ is a sequence in (a, b) that is bounded away from both a and b, satisfying $h_n'(x_n) = 0$ and $h_n''(x_n) < 0$ for all n. Suppose further, that for every $a< u<v<b$, $\sup _{x\in [u,v]} |g'(x)| = O_{u,v}(1)$, $\sup _{n\geqslant 1}\sup _{x\in [u,v]} |h_n^{(3)}(x)|= O_{u,v}(1)$ and $\inf _{x \in [u,v]} |g(x)| = \Omega _{u,v}(1)$. Also, suppose that $\inf _{n\geqslant 1} |h_n''(x_n)| > 0$. Then, for all $\alpha \in \left( 0,\frac{1}{6}\right)$, we have as $n \rightarrow \infty$,

$$\begin{aligned} \int _{x_n-n^{-\frac{1}{2}+\alpha }}^{x_n+n^{-\frac{1}{2}+\alpha }} g(x) e^{n h_n(x)}{\mathrm {d}} x=\sqrt{\dfrac{2\pi }{n\left| h_n''(x_n)\right| }}g(x_n) e^{n h_n(x_n)}\left( 1+ O\left( n^{-\frac{1}{2}+3\alpha }\right) \right) . \end{aligned}$$

Proof

If we make the change of variables $y = \sqrt{n}(x-x_n)$, we have

$$\begin{aligned} \int _{x_n-n^{-\frac{1}{2}+\alpha }}^{x_n+n^{-\frac{1}{2}+\alpha }} g(x) e^{n h_n(x)}{\mathrm {d}} x = n^{-\frac{1}{2}} \int _{-n^{\alpha }}^{n^{\alpha }} g(yn^{-\frac{1}{2}}+x_n) e^{n h_n\left( yn^{-\frac{1}{2}}+x_n\right) }~{\mathrm {d}} y . \end{aligned}$$

(A.5)

By a Taylor expansion, we have for any sequence $y \in [-n^\alpha , n^\alpha ]$,

$$\begin{aligned} e^{n h_n\left( yn^{-\frac{1}{2}}+x_n\right) }= & {} \left( 1+O\left( n^{3\alpha - \frac{1}{2}}\right) \right) e^{nh_n(x_n) + \frac{y^2}{2}h_n''(x_n)}\nonumber \\ {\text {and }} g(yn^{-\frac{1}{2}}+x_n)= & {} \left( 1+O\left( n^{\alpha - \frac{1}{2}}\right) \right) g(x_n). \end{aligned}$$

(A.6)

Using (A.6), the right side of (A.5) becomes

$$\begin{aligned}&n^{-\frac{1}{2}}\left( 1+O\left( n^{3\alpha - \frac{1}{2}}\right) \right) g(x_n)e^{n h_n(x_n)} \int _{-n^\alpha }^{n^\alpha } e^{\frac{y^2}{2}h_n''(x_n)}~{\mathrm {d}} y \nonumber \\&= \left( 1+O\left( n^{3\alpha - \frac{1}{2}}\right) \right) \sqrt{\dfrac{2\pi }{n\left| h_n''(x_n)\right| }}g(x_n) e^{n h_n(x_n)} \mathbb {P}\left( \left| N\left( 0,\frac{1}{|h_n''(x_n)|}\right) \right| \leqslant n^\alpha \right) \\ {}&= \left( 1+O\left( n^{3\alpha - \frac{1}{2}}\right) \right) \sqrt{\dfrac{2\pi }{n\left| h_n''(x_n)\right| }}g(x_n) e^{n h_n(x_n)}\left( 1-O\left( e^{-n^\alpha }\right) \right) \\ {}&= \left( 1+O\left( n^{3\alpha - \frac{1}{2}}\right) \right) \sqrt{\dfrac{2\pi }{n\left| h_n''(x_n)\right| }}g(x_n) e^{n h_n(x_n)}. \end{aligned}$$

The proof of Lemma A.3 is now complete. $\square$

Lemma A.4

(Laplace-Type Approximation-II). Let $a<b$ be fixed real numbers, $g: [a,b]\mapsto \mathbb {R}$ be a differentiable function on (a, b), and $h_n : [a,b]\mapsto \mathbb {R}$ be a sequence of 5-times differentiable functions on (a, b). Suppose that $\{x_n\}$ is a sequence in (a, b) that is bounded away from both a and b, satisfying $h_n'(x_n) = 0$ for all $n\geqslant 1$. Also, assume that $n^\frac{1}{2}h''_n(x_n)=C_1 + O(n^{-\frac{1}{4}})$, $n^{\frac{1}{4}} h^{(3)}_n(x_n)=C_2 + O(n^{-\frac{1}{4}})$, and $h_n^{(4)}(x_n) = C_3 + O(n^{-\frac{1}{4}})$, where $C_1, C_2$ and $C_3$ are real constants. Suppose further, that for every $a< u<v<b$, $\sup _{x\in [u,v]} |g'(x)| = O_{u,v}(1)$, $\sup _{n\geqslant 1}\sup _{x\in [u,v]} |h_n^{(5)}(x)|= O_{u,v}(1)$ and $\inf _{x \in [u,v]} |g(x)| = \Omega _{u,v}(1)$. Then, for all $\alpha \in \left( 0,\frac{1}{20}\right)$, as $n \rightarrow \infty$,

$$\begin{aligned}&\int \limits _{x_n-n^{-\frac{1}{4}+\alpha }}^{x_n+n^{-\frac{1}{4}+\alpha }} g(x) e^{n h_n(x)}{\mathrm {d}} x=n^{-\frac{1}{4}} g(x_n)e^{n h_n(x_n)}\nonumber \\&\quad \int _{-n^\alpha }^{n^\alpha } e^{\frac{y^2}{2}C_1+\frac{y^3}{6}C_2 + \frac{y^4}{24}C_3}~{\mathrm {d}} y \left( 1+O\left( n^{5\alpha - \frac{1}{4}}\right) \right) . \end{aligned}$$

Proof

To begin with, by a change of variables $y = n^{\frac{1}{4}}(x-x_n)$, we have

$$\begin{aligned} \int _{x_n-n^{-\frac{1}{4}+\alpha }}^{x_n+n^{-\frac{1}{4}+\alpha }} g(x) e^{n h_n(x)}{\mathrm {d}} x = n^{-\frac{1}{4}} \int _{-n^{\alpha }}^{n^{\alpha }} g(yn^{-\frac{1}{4}}+x_n) e^{n h_n\left( yn^{-\frac{1}{4}}+x_n\right) }~{\mathrm {d}} y ~ \end{aligned}$$

(A.7)

Now, by a Taylor expansion of $n h_n\left( yn^{-\frac{1}{4}}+x_n\right)$ around $x_n$, we have for any sequence $y \in [n^{-\alpha },n^\alpha ]$,

$$\begin{aligned} n h_n\left( yn^{-\frac{1}{4}}+x_n\right)&= nh_n(x_n)+\frac{n^\frac{1}{2}y^2}{2}h_n''(x_n) \nonumber \\&\quad +\frac{n^\frac{1}{4}y^3}{6}h_n^{(3)}(x_n)+\frac{y^4}{24}h^{(4)}_n(x_n)+O\left( n^{-\frac{1}{4}}y^5\right) \nonumber \\&= nh_n(x_n)+\frac{y^2}{2}C_1+\frac{y^3}{6}C_2+\frac{y^4}{24}C_3+O\left( n^{5\alpha - \frac{1}{4}}\right) . \end{aligned}$$

(A.8)

It follows from (A.8), that

$$\begin{aligned} e^{n h_n\left( yn^{-\frac{1}{4}}+x_n\right) }&= \left( 1+O\left( n^{5\alpha - \frac{1}{4}}\right) \right) e^{nh_n(x_n) + \frac{y^2}{2}C_1+\frac{y^3}{6}C_2+\frac{y^4}{24}C_3}. \end{aligned}$$

(A.9)

Similarly, for any sequence $y \in [-n^\alpha ,n^\alpha ]$, we have

$$\begin{aligned} g(yn^{-\frac{1}{4}}+x_n) = \left( 1+O\left( n^{\alpha - \frac{1}{4}}\right) \right) g(x_n). \end{aligned}$$

(A.10)

Using (A.9) and (A.10), the right side of (A.7) becomes

$$\begin{aligned} n^{-\frac{1}{4}} g(x_n)e^{n h_n(x_n)} \int _{-n^\alpha }^{n^\alpha } e^{\frac{y^2}{2}C_1+\frac{y^3}{6}C_2 + \frac{y^4}{24}C_3}~{\mathrm {d}} y \left( 1+O\left( n^{5\alpha - \frac{1}{4}}\right) \right) . \end{aligned}$$

The proof of Lemma A.4 is now complete. $\square$

Lemma A.5

(Stirling’s Approximation of the Binomial Coefficient). Suppose that $x=x_N$ is a sequence in $(-1,1)$ that is bounded away from both 1 and $-1$. Then, as $N\rightarrow \infty$,

$$\begin{aligned} \left( {\begin{array}{c}N\\ N(1+x)/2\end{array}}\right) = 2^N\sqrt{\dfrac{2}{\pi N(1-x^2)}} \exp \left( -NI(x)\right) \left( 1+O(N^{-1})\right) . \end{aligned}$$

Proof

First, note that by the usual Stirling approximation for the gamma function, we have the following as all of u, v and $u-v \rightarrow \infty$,

$$\begin{aligned} \left( {\begin{array}{c}u\\ v\end{array}}\right)&= \frac{\sqrt{2\pi u}\left( \frac{u}{e}\right) ^u\left( 1+O\left( \frac{1}{u}\right) \right) }{\sqrt{2\pi v}\left( \frac{v}{e}\right) ^v\left( 1+O\left( \frac{1}{v}\right) \right) \sqrt{2\pi (u-v)}\left( \frac{u-v}{e}\right) ^{(u-v)}\left( 1+O\left( \frac{1}{ u -v }\right) \right) }\\&= \sqrt{\dfrac{u}{2\pi v (u-v)}} \cdot \dfrac{u^u}{v^v(u-v)^{u-v}}\left( 1+O\left( \frac{1}{u}\right) +O\left( \frac{1}{v}\right) +O\left( \frac{1}{u-v}\right) \right) . \end{aligned}$$

Substituting $u = N$ and $v = N(1+x)/2$ (the hypothesis of the lemma indeed implies that u, v and $u-v \rightarrow \infty$), we have

$$\begin{aligned} \left( {\begin{array}{c}N\\ N(1+x)/2\end{array}}\right)&= \sqrt{\dfrac{N}{2\pi \frac{N(1+x)}{2}\cdot \frac{N(1-x)}{2}}}\\&\quad \cdot \dfrac{N^N}{\left( \dfrac{N(1+x)}{2}\right) ^{N(1+x)/2} \left( \dfrac{N(1-x)}{2}\right) ^{N(1-x)/2}}\left( 1+O(N^{-1})\right) \\&=2^N\sqrt{\dfrac{2}{\pi N(1-x^2)}}\exp \left( -\frac{N(1+x)}{2} \log (1+x)\right. \\&\quad \left. -\frac{N(1-x)}{2}\log (1-x) \right) \left( 1+O(N^{-1})\right) \\&=2^N\sqrt{\dfrac{2}{\pi N(1-x^2)}} \exp \left( -NI(x)\right) \left( 1+O(N^{-1})\right) . \end{aligned}$$

This completes the proof of Lemma A.5. $\square$

Appendix B. Properties of the Function H and Other Technical Lemmas

This section is devoted to proving several technical lemmas that are used throughout the proofs of our main results. In “Appendix B.1”, we will prove several important properties of the function H. In “Appendix B.2” we collect the proofs of some other technical lemmas.

1.1 Properties of the function H

We start by showing that a p-strongly critical point arises if and only if $p \geqslant 4$ is even, and in that case, the only such point is $(\tilde{\beta }_p,0)$ (recall (2.2)).

Lemma B.1

(Basic properties of the function H ). The function $H_{\beta ,h,p}$ has the following properties.

• (1) $\sup _{x \in [-1,1]} H_{\beta ,h,p}(x) \geqslant 0$ and equality holds if and only if $(\beta ,h) \in [0,\tilde{\beta }_p]\times \{0\}$.

• (2) Every local maximizer of $H_{\beta ,h,p}$ lies in $(-1,1)$.

• (3) $H_{\beta ,h,p}$ can have at most two local maximizers for $p = 3$ and at most three local maximizers for $p \geqslant 4$. Further, it has three global maximizers if and only if $p\geqslant 4$ is even, $h=0$ and $\beta = \tilde{\beta }_p$.

Proof of (1)

First note that $\sup _{x \in [-1,1]} H_{\beta ,h,p}(x) \geqslant H_{\beta ,h,p}(0) = 0$. Now, it follows from first principles, that $\lim _{\varepsilon \rightarrow 0} H_{\beta ,h,p}(\varepsilon )/\varepsilon = H_{\beta ,h,p}'(0) = h$. If $h > 0$, then there exists $0<\varepsilon < 1$ such that $H_{\beta ,h,p}(\varepsilon )/\varepsilon > h/2$, and if $h < 0$, then there exists $-1<\varepsilon < 0$ such that $H_{\beta ,h,p}(\varepsilon )/\varepsilon < h/2$. In either case, $\sup _{x \in [-1,1]} H_{\beta ,h,p}(x) \geqslant H_{\beta ,h,p}(\varepsilon )> \varepsilon h /2 > 0$. Therefore, equality in (1) implies that $h = 0$, and hence, by the definition in (2.2), we must have $\beta \leqslant \tilde{\beta }_p$. This proves the “only if” direction. For the “if” direction, suppose that $(\beta ,h) \in [0,\tilde{\beta }_p]\times \{0\}$. Consider the case $\beta < \tilde{\beta }_p$ first, so that by the definition in (2.2), there exists $\beta ' > \beta$ such that $\sup _{x \in [-1,1]} H_{\beta ' ,0,p}(x) = 0$. Equality in (1) now follows from:

$$\begin{aligned}&0\leqslant \sup _{x \in [-1,1]} H_{\beta ,0,p}(x) = \sup _{x \in [-1,1]} H_{\beta ,0,p}(|x|)\\&\leqslant \sup _{x \in [-1,1]} H_{\beta ' ,0,p}(|x|) = \sup _{x \in [-1,1]} H_{\beta ' ,0,p}(x) = 0. \end{aligned}$$

Finally, let $\beta = \tilde{\beta }_p$, and suppose towards a contradiction, that $H_{\beta ,0,p}(x) > 0$ for some $x \in [-1,1]$. Then, $H_{\beta ,0,p}(|x|) \geqslant H_{\beta ,0,p}(x) > 0$, and hence, there exists $\beta ' < \beta$ such that

$$\begin{aligned} H_{\beta ' ,0,p}(|x|) = H_{\beta ,0,p}(|x|) + (\beta ' -\beta )|x|^p > 0. \end{aligned}$$

This contradicts our previous finding that $\sup _{x \in [-1,1]} H_{\underline{\beta } ,0,p}(x) = 0$ for all $\beta < \tilde{\beta }_p$. The proof of (1) is now complete. $\square$

Proof of (2)

Note that $\lim _{x \rightarrow -1^+} H_{\beta ,h,p}'(x) = +\infty$ and $\lim _{x \rightarrow 1^-} H_{\beta ,h,p}'(x) = -\infty$. Hence, there exists $\varepsilon > 0$, such that $H_{\beta ,h,p}$ is strictly increasing on $[-1,-1+\varepsilon ]$ and strictly decreasing on $[1-\varepsilon ,1]$, showing that none of $-1$ and 1 can be a local maximizer of $H_{\beta ,h,p}$. $\square$

Proof of (3)

Define

$$\begin{aligned} N_{\beta ,h,p}(x) := (1-x^2)H_{\beta ,h,p}''(x) = \beta p(p-1)x^{p-2}(1-x^2) - 1, \end{aligned}$$

for $x \in (-1,1)$. Note that on $(-1,1)$, $N_{\beta ,h,p}'(x) = \beta p(p-1)x^{p-3}(p-2-px^2)$ has exactly two roots $\pm \sqrt{1-2/p}$, for $p=3$, and an additional root 0 for $p\geqslant 4$. Define:

$$\begin{aligned} K_p := 2\varvec{1}\{p =3\} + 3\varvec{1}\{p \geqslant 4\}. \end{aligned}$$

Then, by Rolle’s theorem, $N_{\beta ,h,p}$, and hence, $H_{\beta ,h,p}''$ can have at most $K_p+1$ roots on $(-1,1)$. This shows that $H_{\beta ,h,p}'$ can have at most $K_p+2$ roots on $(-1,1)$, which by part (2), include all the local maximizers of $H_{\beta ,h,p}$. We now claim that for any two local maximizers $a< b$ of $H_{\beta ,h,p}$, there exists a root of $H_{\beta ,h,p}'$ in (a, b). To see this, note that since a and b are local maximizers of $H_{\beta ,h,p}$, by the mean value theorem, there must exist $a_1 < b_1 \in (a,b)$ such that $H_{\beta ,h,p}'(a_1) \leqslant 0$ and $H_{\beta ,h,p}'(b_1) \geqslant 0$. Now, by the intermediate value theorem applied on the continuous function $H_{\beta ,h,p}'$, we conclude that there is a $\zeta \in (a_1,b_1)$ such that $H_{\beta ,h,p}'(\zeta ) = 0$. Hence, if there are $\ell$ local maximizers of $H_{\beta ,h,p}$ on $(-1,1)$, then there are at least $2\ell -1$ roots of $H_{\beta ,h,p}'$ on $(-1,1)$. Thus,

$$\begin{aligned} 2\ell -1 \leqslant K_p+2,\quad \text {i.e.}\quad \ell \leqslant (K_p+3)/2, \end{aligned}$$

which proves the first part of (3).

To prove the second part of (3), first suppose that $H_{\beta ,h,p}$ has three global maximizers. By the first part, p must be at least 4. We will now show that p is even, by contradiction. If p is odd, then $H_{\beta ,h,p}''(x) < 0$ for all $x \leqslant 0$, and hence, by Rolle’s theorem, there can be at most one non-positive root of $H_{\beta ,h,p}'$. Now, if $H_{\beta ,h,p}'$ has at least four positive roots, then by repeated application of Rolle’s theorem, $N_{\beta ,h,p}'$ has at least two positive roots. This is a contradiction, since $\sqrt{1-2/p}$ is the only positive root of $N_{\beta ,h,p}'$. Hence, $H_{\beta ,h,p}'$ can have at most three positive roots. Thus, $H_{\beta ,h,p}'$ can have at most four roots, and hence, $H_{\beta ,h,p}$ can have at most two local maximizers, a contradiction. Hence, p must be even.

Next, we show that h must be 0. If $h > 0$, then $H_{\beta ,h,p}(x) < H_{\beta ,h,p}(-x)$ for all $x < 0$, and hence, all the three global maximizers of $H_{\beta ,h,p}$ must be positive. Thus, $H_{\beta ,h,p}'$ has at least 5 positive roots, which implies that $N_{\beta ,h,p}'$ has at least three positive roots, a contradiction. Similarly, if $h<0$, then all the three global maximizers of $H_{\beta ,h,p}$ must be negative, and thus, $H_{\beta ,h,p}'$ has at least 5 negative roots, which implies that $N_{\beta ,h,p}'$ has at least three negative roots, once again a contradiction. This shows that $h=0$.

Finally, we show that $\beta = \tilde{\beta }_p$. If $\beta > \tilde{\beta }_p$, then by the definition in (2.2), 0 is not a global maximizer of $H_{\beta ,h,p}$ and hence, $H_{\beta ,h,p}$ being an even function, must have an even number of global maximizers, a contradiction. Therefore, it suffices to assume that $\beta <\tilde{\beta }_p$. We will show that 0 is the only global maximizer of $H_{\beta ,h,p}$, which is enough to complete the proof of the only if implication. Towards this, suppose that there is a non-zero global maximizer $x^*$ of $H_{\beta ,h,p}$. Since $\beta < \tilde{\beta }_p$, we must have $H_{\beta ,h,p}(x^*) = 0$, and hence, for every $\beta ' \in (\beta ,\tilde{\beta }_p)$, we must have $H_{\beta ' ,h,p}(x^*) > 0$, a contradiction to the definition in (2.2). This completes the proof of the only if implication.

For the if implication, let $\beta := \tilde{\beta }_p + \frac{1}{N}$, whence by part (1), $\sup _{x\in [-1,1]} H_{\beta ,0,p}(x) > 0$ for all $N \geqslant 1$. Since $H_{\beta ,0,p}(0) = 0$, for each N there exists $x_N \ne 0$ such that $H_{\beta ,0,p}(x_N) > 0$. Let $x_{N_k}$ be a convergent subsequence of $x_N$, converging to a point $x^*$. Then,

$$\begin{aligned} \lim _{k \rightarrow \infty } H_{\beta _{N_k},0,p}(x_{N_k}) = H_{\tilde{\beta }_p,0,p}(x^*), \end{aligned}$$

and hence, $H_{\tilde{\beta }_p,0,p}(x^*) \geqslant 0$. However, by part (1), the reverse inequality is true, and hence, $H_{\tilde{\beta }_p,0,p}(x^*) = 0$, and hence, $0, x^*$ and $-x^*$ are all global maximizers of $H_{\tilde{\beta }_p,0,p}$. We will be done, if we can show that $x^* \ne 0$. Towards this, note that since $\lim _{\varepsilon \rightarrow 0} H_{\tilde{\beta }_p,0,p}(\varepsilon )/\varepsilon ^2 = -\frac{1}{2}$, there exists $\delta > 0$ such that $H_{\tilde{\beta }_p,0,p}(\varepsilon ) < -\varepsilon ^2/4$ whenever $|\varepsilon | < \delta$. Suppose that $x^* =0$, i.e. $x_{N_k} \rightarrow 0$ as $k \rightarrow \infty$. Then for all k large enough, we must have

$$\begin{aligned} H_{\beta _{N_k},0,p}(x_{N_k}) = H_{\tilde{\beta }_p,0,p}(x_{N_k}) + \frac{x_{N_k}^p}{N_k}< -\frac{x_{N_k}^2}{4} + \frac{x_{N_k}^p}{N_k} < 0, \end{aligned}$$

a contradiction. This shows that $x^* \ne 0$. The proof of (3) and Lemma B.1 is now complete. $\square$

Remark B.1

The argument in the last paragraph of the proof of Lemma B.1 can be adopted to show that for odd p, $H_{\tilde{\beta }_p,0,p}$ has exactly two global maximizers, one at 0 and the other one positive.

We now proceed to describe p-special points. To begin with, for convenience in the proof, we introduce the following notation.

Definition 5

A point $(\beta ,h) \in [0,\infty )\times \mathbb {R}$ is said to be p-locally special, if the function $H_{\beta ,h,p}$ has a local maximizer m satisfying $H_{\beta ,h,p}''(m) = 0$.

We will see that every p-locally special point is actually p-special, and hence, the two notions are identical. In the following lemma, we give exact expressions for p-special points.

Lemma B.2

(Description of p-special points). Define

$$\begin{aligned} \check{\beta }_p := \frac{1}{2(p-1)} \left( \frac{p}{p-2}\right) ^{\frac{p-2}{2}}\quad \text {and}\quad \check{h}_p := \tanh ^{-1}\left( \sqrt{\frac{p-2}{p}}\right) - p\check{\beta }_p \left( \frac{p-2}{p}\right) ^{\frac{p-1}{2}}. \end{aligned}$$

Then, we have the following:

• (1) If $p\geqslant 3$ is odd, then $\left( \check{\beta }_p, \check{h}_p\right)$ is the only p-locally special point in $[0,\infty )\times \mathbb {R}$. In this case, $m_*:= \sqrt{1-2/p}$ is the only solution to the equation $H_{\check{\beta }_p, \check{h}_p,p}''(x) =0$. In fact, $m_*$ is a global maximizer of $H_{\check{\beta }_p, \check{h}_p,p}$ satisfying $H_{\check{\beta }_p, \check{h}_p,p}^{(3)}(m_*) =0$ and $H_{\check{\beta }_p, \check{h}_p,p}^{(4)}(m_*) <0$. Further, $m_*$ is the unique stationary point of $H_{\check{\beta }_p, \check{h}_p,p}$.

• (2) If $p\geqslant 4$ is even, then $\left( \check{\beta }_p, \check{h}_p\right)$ and $\left( \check{\beta }_p, -\check{h}_p\right)$ are the only p-locally special points in $[0,\infty )\times \mathbb {R}$. In this case, $m_*(1) := \sqrt{1-2/p}$ and $m_*(-1) := -m_*(1)$ are the only solutions to each of the equations $H_{\check{\beta }_p, i\check{h}_p,p}''(x) =0$ for $i \in \{-1,1\}$. In fact, $m_*(i)$ is a global maximizer of $H_{\check{\beta }_p, i\check{h}_p,p}$ for $i \in \{-1,1\}$ satisfying

  $$\begin{aligned} H_{\check{\beta }_p, i\check{h}_p,p}^{(3)}(m_*(i)) =0 \text { and } H_{\check{\beta }_p, i\check{h}_p,p}^{(4)}(m_*(i)) <0, \quad \text { for } i \in \{-1,1\}. \end{aligned}$$

  Further, $m^*(i)$ is the unique global maximizer of $H_{\check{\beta }_p, i\check{h}_p,p}$ for $i \in \{-1,1\}$.

Hence, a point $(\beta ,h)$ is p-locally special if and only if it is p-special.

Proof of Lemma B.2

We start with the following proposition:$\square$

Proposition 1

Let $\beta := \check{\beta }_p$, $h \in \mathbb {R}$, and let $y \in (0,1)$ be a local maximum of $H_{\beta ,h,p}$, satisfying $H_{\beta ,h,p}''(y)=H_{\beta ,h,p}^{(3)}(y) = 0$. Then $H_{\beta ,h,p}^{(4)}(y) < 0$.

Proof

For convenience, we will denote $N_{\beta ,h,p} := (1-x^2)H_{\beta ,h,p}''(x)$ by N and $H_{\beta ,h,p}$ by H. Note that

$$\begin{aligned} N''(x) = (1-x^2) H^{(4)}(x) - 4x H^{(3)}(x) - 2 H''(x). \end{aligned}$$

By hypothesis, $N''(y) = (1-y^2)H^{(4)}(y)$. Now,

$$\begin{aligned} N''(x) = \beta p(p-1)(p-2)(p-3)x^{p-4} - \beta p^2(p-1)^2 x^{p-2} \end{aligned}$$

cannot have any root other than 0 and $\pm \sqrt{\frac{(p-2)(p-3)}{p(p-1)}}$. But we know from the proof of Lemma B.2 that $H_{\beta ,h,p}''$ cannot have any root other than $\pm \sqrt{1-2/p}$ (note that Proposition 1 is not needed to reach this conclusion, and hence, there is no circularity in the argument), and for $p\geqslant 3$, we have $\frac{(p-2)(p-3)}{p(p-1)} < \frac{p-2}{p}$. Therefore, y is not a root of $N''$, and hence, not a root of $H^{(4)}$. Proposition 1 now follows from the standard higher derivative test. $\square$

We are now proceed with the proof of Lemma B.2. We start by proving that the first coordinate of every p-locally special point in $[0,\infty )\times \mathbb {R}$ must be equal to $\check{\beta }_p$. Towards this, we first claim that $H_{\beta ,h,p}''(x) < 0$, or equivalently, $N_{\beta ,h,p}(x) < 0$ for all $x \in (-1,1)$, if $\beta < \check{\beta }_p$. This will rule out the possibility of $(\beta ,h)$ being a candidate for a p-locally special point, for $\beta <\check{\beta }_p$. Towards proving this claim, we can assume that

$$\begin{aligned} \sup _{x \in (-1,1)} N_{\beta ,h,p}(x) > -1, \end{aligned}$$

since otherwise we would be done. Since $N_{\beta ,h,p}(-1) = N_{\beta ,h,p}(0) = N_{\beta ,h,p}(1) = -1$, the function $N_{\beta ,h,p}$ attains maximum at some $m \in (-1,1){\setminus }\{0\}$, and hence, m is a non-zero solution to the equation $N_{\beta ,h,p}'(x) = 0$. Therefore, from the proof of (3) in Lemma B.1, that $m \in \{-q,q\}$, where $q := \sqrt{1-2/p}$. Since $N_{\beta ,h,p}(q) \geqslant N_{\beta ,h,p}(-q)$, we know for sure that q is a global maximizer of $N_{\beta ,h,p}$. Our claim now follows from the observation that $\beta< \check{\beta }_p \implies N_{\beta ,h,p}(q) < 0$.

Now, we are going to rule out the possibility $\beta > \check{\beta }_p$, as well. Suppose that $\beta > \check{\beta }_p$, and let $m_*$ be a local maximizer of $H_{\beta ,h,p}$ satisfying $H_{\beta ,h,p}''(m_*) = 0$, i.e. $N_{\beta ,h,p}(m_*) = 0$. Now, $N_{\beta ,h,p}(0) = -1 \implies m_* \ne 0$. Next, since $\beta > \check{\beta }_p$, it follows that $N_{\beta ,h,p}(q) > 0$, and hence, $m_* \ne q$. If p is even, then $N_{\beta ,h,p}(-q) = N_{\beta ,h,p}(q) > 0$, and if p is odd, then $N_{\beta ,h,p}(x) < -1$ for all $x < 0$. Thus, in either case, $m_* \ne -q$. All these show that $N_{\beta ,h,p}'(m_*) \ne 0$. Suppose that $N_{\beta ,h,p}'(m_*) > 0$. Since $N_{\beta ,h,p}(m_*) = 0$, there exists $\varepsilon > 0$ such that $N_{\beta ,h,p}(x) > 0$ for all $x \in (m_*,m_*+\varepsilon )$ and $N_{\beta ,h,p}(x) < 0$ for all $x \in (m_*,m_*-\varepsilon )$. Thus, $H_{\beta ,h,p}''(x) > 0$ for all $x \in (m_*,m_*+\varepsilon )$ and $H_{\beta ,h,p}''(x) < 0$ for all $x \in (m_*-\varepsilon ,m_*)$. Since $H_{\beta ,h,p}'(m_*) = 0$, we must have

$$\begin{aligned} H_{\beta ,h,p}'(x) > 0\quad \text {for all}~x \in (m_*-\varepsilon ,m_*+\varepsilon ){\setminus } \{m_*\}. \end{aligned}$$

This implies that $H_{\beta ,h,p}$ is strictly increasing on $[m_*,m_*+\varepsilon )$, contradicting that $m_*$ is a local maximizer of $H_{\beta ,h,p}$. Similarly, if $N_{\beta ,h,p}'(m_*) < 0$, then there exists $\varepsilon > 0$ such that $H_{\beta ,h,p}'(x) < 0$ for all $x \in (m_*-\varepsilon ,m_*+\varepsilon ){\setminus } \{m_*\}$, and so, $H_{\beta ,h,p}(x)$ is strictly decreasing on $(m_*-\varepsilon ,m_*]$, contradicting once again, that $m_*$ is a local maximizer of $H_{\beta ,h,p}$. We have thus proved our claim, that the first coordinate of every p-special point in $[0,\infty )\times \mathbb {R}$ must be equal to $\check{\beta }_p$. In what follows, let $\beta := \check{\beta }_p$.

Proof of (1)

Let $p \geqslant 3$ be odd and let $m_*$ be any solution to the equation $H_{\beta ,h,p}''(x) = 0$, or equivalently, to the equation $N_{\beta ,h,p}(x) = 0$. Since $N_{\beta ,h,p}(x) \leqslant -1$ for all $x \leqslant 0$, it follows that $m_* \in (0,1)$. Now, we already know that the only positive root of $N_{\beta ,h,p}'$ is $q:=\sqrt{1-2/p}$, and since $N_{\beta ,h,p}(q) = 0$, by Rolle’s theorem, $N_{\beta ,h,p}$ cannot have any positive root other than q. Thus, $m_* = q$ is the only root of $H_{\beta ,h,p}''$. Since $N_{\beta ,h,p}(m_*) = N_{\beta ,h,p}'(m_*) = 0$, we have

$$\begin{aligned} H_{\beta ,h,p}^{(3)}(m_*) = \frac{N_{\beta ,h,p}'(m_*)(1-m_*^2) + 2m_*N_{\beta ,h,p}(m_*)}{(1-m_*^2)^2} = 0. \end{aligned}$$

Now, $m_*$ is a stationary point of $H_{\beta ,h,p}$, i.e. $H_{\beta ,h,p}'(m_*) = 0$ if and only if $h = \check{h}_p$. Hence, $(\check{\beta }_p,\check{h}_p)$ is the only candidate for being a p-locally special point in $[0,\infty )\times \mathbb {R}$. Let $h := \check{h}_p$ throughout the rest of the proof of (a). Since $H_{\beta ,h,p}'(m_*) = 0$ and $m_*$ is the only root of $H_{\beta ,h,p}''$, by Rolle’s theorem, $H_{\beta ,h,p}'$ cannot have any root other than $m_*$. This implies that the sign of $H_{\beta ,h,p}'$ remains constant on each of the intervals $(-1,m^*)$ and $(m^*,1)$. Since

$$\begin{aligned} \lim _{x \rightarrow -1^+} H_{\beta ,h,p}' (x) = +\infty \quad \text {and}\quad \lim _{x \rightarrow 1^-} H_{\beta ,h,p}' (x) = -\infty , \end{aligned}$$

we conclude that $H_{\beta ,h,p}' > 0$ on $(-1,m^*)$ and $H_{\beta ,h,p}' < 0$ on $(m^*,1)$, thereby showing that $m^*$ is a global maximizer, and also the unique stationary point of $H_{\beta ,h,p}$, and verifying that $(\check{\beta }_p,\check{h}_p)$ is actually a p-special point. The result in part (1) now follows from Proposition 1. $\square$

Proof of (2)

Let $p \geqslant 4$ be even. Since $m_*(1)$ and $m_*(-1)$ are the only non-zero roots of $N_{\beta ,h,p}'$, and they are also roots of $N_{\beta ,h,p}$, by Rolle’s theorem, they are the only roots of $N_{\beta ,h,p}$, as well. Hence, the only roots of $H_{\beta ,h,p}''$ are $m_*(1)$ and $m_*(-1)$, and so, $H_{\beta ,h,p}^{(3)}(m_*(1)) = H_{\beta ,h,p}^{(3)}(m_*(-1)) = 0$.

For $i \in \{-1,1\}$, note that $m_*(i)$ is a stationary point of $H_{\beta ,h,p}$, i.e. $H_{\beta ,h,p}'(m_*(i)) = 0$, if and only if $h = i\check{h}_p$. Hence, $(\check{\beta }_p,\check{h}_p)$ and $(\check{\beta }_p,-\check{h}_p)$ are the only candidates for being p-locally special points in $[0,\infty )\times \mathbb {R}$. Let $h := \check{h}_p$ throughout the rest of the proof of (2). Since $H_{\beta ,ih,p}'(m_*(i)) = 0$ and $m_*(i)$ is the only root of $H_{\beta ,ih,p}''$ with sign i, by Rolle’s theorem, $H_{\beta ,ih,p}'$ cannot have 0 or any point with sign i as a root, other than $m_*(i)$. This implies that the sign of $H_{\beta ,h,p}'$ remains constant on each of the intervals $[0,m_*(1))$ and $(m_*(1),1)$, and the sign of $H_{\beta ,-h,p}'$ remains constant on each of the intervals $(-1,m_*(-1))$ and $(m_*(-1),0]$. Since

$$\begin{aligned} \lim _{x \rightarrow -1^+} H_{\beta ,\pm h,p}' (x) = +\infty \quad \text {and}\quad \lim _{x \rightarrow 1^-} H_{\beta ,\pm h,p}' (x) = -\infty , \end{aligned}$$

we conclude that $H_{\beta ,h,p}' < 0$ on $(m_*(1),1)$ and $H_{\beta ,-h,p}' > 0$ on $(-1,m_*(-1))$. Now, note that

$$\begin{aligned}&h = \tanh ^{-1}\left( \sqrt{\frac{p-2}{p}}\right) - \check{\beta }_p p \left( \frac{p-2}{p}\right) ^{\frac{(p-1)}{2}}\\&\quad =\left[ \sum _{k=0}^\infty \frac{\left( \sqrt{\frac{p-2}{p}}\right) ^{2k+1}}{2k+1}\right] - \frac{p}{2(p-1)} \sqrt{\frac{p-2}{p}}\\&\quad \geqslant \sqrt{\frac{p-2}{p}} - \frac{p}{2(p-1)} \sqrt{\frac{p-2}{p}} \\&\quad = \frac{p-2}{2(p-1)}\sqrt{\frac{p-2}{p}} > 0. \end{aligned}$$

Hence, $H_{\beta ,h,p}'(0) = h > 0$ and $H_{\beta ,-h,p}'(0) = -h < 0$. Consequently, $H_{\beta ,h,p}' > 0$ on $[0,m_*(1))$ and $H_{\beta ,-h,p}' < 0$ on $(m_*(-1),0]$. Thus, $m_*(i)$ is the unique global maximizer of $H_{\beta ,ih,p}$ over the interval $\mathscr {J}_i := \{ix: x \in [0,1]\}$. (Note that $\mathscr {J}_1=[0, 1]$ and $\mathscr {J}_{-1}=[-1, 0]$.) Now, it is easy to see that $H_{\beta ,ih,p}(x) < H_{\beta ,ih,p}(-x)$, for all $x \in [-1,1]{\setminus } \mathscr {J}_i$. This shows that $m_*(i)$ is the unique global maximizer of $H_{\beta ,ih,p}$ over $[-1,1]$. Part (2) now follows from Proposition 1, and the proof of Lemma B.2 is now complete. $\square$

Next, we give a description of p-weakly critical points that is, points $(\beta ,h)$ for which the function $H_{\beta ,h,p}$ has exactly two global maximizers). Note that we already have a full characterization of p-strongly critical points (that is, points $(\beta ,h)$ for which the function $H_{\beta ,h,p}$ has exactly three global maximizers) by part (3) of Lemma B.1. To elaborate, we know that there cannot be any p-strongly critical point if p is odd, and if $p\geqslant 4$ is even, then $(\tilde{\beta }_p,0)$ is the only p-strongly critical point. In the following lemma, we show that the set of all p-critical points is a one-dimensional continuous curve in the plane $[0,\infty ) \times \mathbb {R}$. We also prove some other interesting properties of this curve, for instance, the only limit point(s) of the curve which is (are) outside it, is (are) the p-special point(s).

Lemma B.3

(Description of p-weakly critical points). For every $p \geqslant 3$, $\check{\beta }_p < \tilde{\beta }_p$, and the set ${\mathscr {C}_p}^+$ can be characterized as follows.

• (1) For every even $p \geqslant 4$, there exists a continuous function $\varphi _p: (\check{\beta }_p,\infty )\mapsto [0,\infty )$ which is strictly decreasing on $(\check{\beta }_p,\tilde{\beta }_p)$ and vanishing on $[\tilde{\beta }_p,\infty )$, such that

  $$\begin{aligned} \mathscr {C}_p^+ = \left\{ (\beta , \pm \varphi _p(\beta )): \beta \in (\check{\beta }_p,\infty ){\setminus } \{\tilde{\beta }_p\}\right\} . \end{aligned}$$

• (2) For every odd $p \geqslant 3$, there exists a strictly decreasing, continuous function $\varphi _p: (\check{\beta }_p,\infty )\mapsto \mathbb {R}$ satisfying $\varphi _p(\tilde{\beta }_p) = 0$ and $\lim _{\beta \rightarrow \infty }\varphi _p(\beta ) = -\infty$, such that

  $$\begin{aligned} \mathscr {C}_p^+ = \left\{ (\beta , \varphi _p(\beta )): \beta \in (\check{\beta }_p,\infty )\right\} . \end{aligned}$$

In both cases, $\lim _{\beta \rightarrow \check{\beta }_p^+} \varphi _p(\beta ) = \tanh ^{-1}(m_*) - p\check{\beta }_p m_*^{p-1}$, where $m_* := \sqrt{\frac{p-2}{p}}$.

Proof

First, we prove that $\check{\beta }_p < \tilde{\beta }_p$ for all $p \geqslant 3$. Since

$$\begin{aligned} \sup _{x\in [-1,1]} H_{\beta ,0,p+1}(x) = \sup _{x\in [0,1]} H_{\beta ,0,p+1}(x) \leqslant \sup _{x\in [0,1]} H_{\beta ,0,p}(x) = \sup _{x\in [-1,1]} H_{\beta ,0,p}(x), \end{aligned}$$

it follows that $\tilde{\beta }_{p+1} \geqslant \tilde{\beta }_p$, i.e. $\tilde{\beta }_p$ is increasing in p. Therefore, $\tilde{\beta }_p \geqslant \tilde{\beta }_2 = \frac{1}{2}$ for all $p \geqslant 3$. First note that $\check{\beta }_3 = \frac{\sqrt{3}}{4} < \frac{1}{2}$. Next, note that for $p \geqslant 4$,

$$\begin{aligned} \check{\beta }_p = \frac{1}{2(p-1)} \left( 1+\frac{2}{p-2}\right) ^{\frac{p-2}{2} } \leqslant \frac{e}{2(p-1)} \leqslant \frac{e}{6} < \frac{1}{2}. \end{aligned}$$

Hence, $\check{\beta }_p < \frac{1}{2} \leqslant \tilde{\beta }_p$ for all $p \geqslant 3$.

Next, we show that $\mathscr {C}_p^+ \subseteq (\check{\beta }_p,\infty ) \times \mathbb {R}$. Towards this, first let $\beta < \check{\beta }_p$ and $h\in \mathbb {R}$. It follows from the proof of Lemma B.2, that $H_{\beta ,h,p}'' < 0$ on $[-1,1]$, so $H_{\beta ,h,p}$ is strictly concave on $[-1,1]$, and hence, can have at most one global maximum. Therefore, $(\beta , h)\notin \mathscr {C}_p^+$. Now, let $\beta = \check{\beta }_p$ and $h \in \mathbb {R}$. From the proof of Lemma B.2, we know that $H_{\beta ,h,p}''$ cannot have any root on $[-1,1]$ other than possibly $\pm \sqrt{1-2/p}$. Since $H_{\beta ,h,p}''(-1) = H_{\beta ,h,p}''(1)=-\infty$, $H_{\beta ,h,p}''(0) = -1$ and $H_{\beta ,h,p}''$ is continuous, $H_{\beta ,h,p}''(x) < 0$ for all $x \in [-1,1]{\setminus } \{\pm \sqrt{1-2/p} \}$. This shows that $H_{\beta ,h,p}'$ is strictly decreasing on $[-1,1]$, and hence, $H_{\beta ,h,p}$ can have at most one stationary point. Consequently, $(\beta , h)\notin \mathscr {C}_p^+$, proving our claim that $\mathscr {C}_p^+ \subseteq (\check{\beta }_p,\infty ) \times \mathbb {R}$. We now consider the cases of even and odd p separately.

Proof of (1)

Let $p \geqslant 4$ be even. Since $x \mapsto \beta x^p - I(x)$ is an even function, the set $\mathscr {C}_p^+$ is symmetric about the line $h=0$, i.e. $(\beta ,h) \in {\mathscr {C}_p}^+ \implies (\beta ,-h) \in {\mathscr {C}_p}^+$. Next, we show that for every $\beta > \check{\beta }_p$, there exists at most one $h \geqslant 0$ such that $(\beta ,h) \in {\mathscr {C}_p}^+$. Suppose towards a contradiction, that there exists $\beta > \check{\beta }_p$ and $h_2>h_1\geqslant 0$, such that both $(\beta ,h_1)$ and $(\beta ,h_2) \in {\mathscr {C}_p}^+$. Letting $m_* := \sqrt{1-2/p}$, it follows that $H_{\beta ,h,p}''(m_*) > 0$ for all $h \in \mathbb {R}$. Recalling that $H_{\beta ,h,p}''$ can have at most two roots in [0, 1], and using the facts

$$\begin{aligned} H_{\beta ,h,p}''(0)=-1, H_{\beta ,h,p}''(1) = -\infty , \end{aligned}$$

it follows that there exist $0<a_1<m_*<a_2<1$, such that $H_{\beta ,h,p}''<0$ on $[0,a_1)$, $H_{\beta ,h,p}''(a_1)=0$, $H_{\beta ,h,p}''>0$ on $(a_1,a_2)$, $H_{\beta ,h,p}''(a_2) = 0$ and $H_{\beta ,h,p}''< 0$ on $(a_2,1]$. This shows that $H_{\beta ,h,p}'$ is strictly decreasing on $[0,a_1]$, strictly increasing on $[a_1,a_2]$ and strictly decreasing on $[a_2,1]$.

First assume that $h_1 > 0$, whence the two global maximizers $m_1(h_i) < m_2(h_i)$ of $H_{\beta ,h_i,p}$ must be positive roots of $H_{\beta ,h_i,p}'$ for $i \in \{1,2\}$. Note that the monotonicity pattern of the function $H_{\beta ,h_i,p}'$ implies that $m_1(h_i) \in (0,a_1)$ and $m_2(h_i) \in (a_2,1)$. Hence, $H_{\beta ,h_i,p}'(a_1) < 0$ and $H_{\beta ,h_i,p}'(a_2) > 0$, and by the intermediate value theorem, there exists $m(h_i) \in (a_1,a_2)$ such that

$$\begin{aligned} H_{\beta ,h_i,p}'(m(h_i)) = 0. \end{aligned}$$

Observe that $H_{\beta ,h_i,p}'$ is positive on $[0,m_1(h_i))$, negative on $(m_1(h_i),m(h_i))$, positive on $(m(h_i),m_2(h_i))$ and negative on $(m_2(h_i),1]$. Since $h_2 > h_1$, it follows that $H_{\beta ,h_2,p}'>0$ on $[0,m_1(h_1)]$ and on $[m(h_1),m_2(h_1)]$. However, since $m_1(h_2), m(h_2)$ and $m_2(h_2)$ are roots of $H_{\beta ,h_2,p}'$ on $(0,a_1), (a_1,a_2)$ and $(a_2,1)$ respectively, it follows that $m_1(h_1)<m_1(h_2)$, $m(h_2)<m(h_1)$ and $m_2(h_1)<m_2(h_2)$. Combining all these, gives

$$\begin{aligned} \int _{m_1(h_1)}^{m(h_1)} H_{\beta ,h_1,p}'(t) {\mathrm {d}} t< \int _{m_1(h_2)}^{m(h_2)} H_{\beta ,h_1,p}' (t) {\mathrm {d}} t < \int _{m_1(h_2)}^{m(h_2)} H_{\beta ,h_2,p}' (t) {\mathrm {d}} t \end{aligned}$$

(B.1)

and

$$\begin{aligned} \int _{m(h_1)}^{m_2(h_1)} H_{\beta ,h_1,p}' (t) {\mathrm {d}} t< \int _{m(h_1)}^{m_2(h_1)} H_{\beta ,h_2,p}' (t) {\mathrm {d}} t < \int _{m(h_2)}^{m_2(h_2)} H_{\beta ,h_2,p}' (t) {\mathrm {d}} t \end{aligned}$$

(B.2)

Adding (B.1) and (B.2), we have

$$\begin{aligned} \int _{m_1(h_1)}^{m_2(h_1)} H_{\beta ,h_1,p}' (t) {\mathrm {d}} t < \int _{m_1(h_2)}^{m_2(h_2)} H_{\beta ,h_2,p}' (t) {\mathrm {d}} t . \end{aligned}$$

(B.3)

This is a contradiction, since both sides of (B.3) are 0.

Therefore, it must be that $h_1 = 0$. In this case, the global maximizers $m_1(h_1) < m_2(h_1)$ of $H_{\beta ,h_1,p}$ satisfy $m_1(h_1) = - m_2(h_1)$. Since $H_{\beta ,h_1,p}'$ vanishes at 0, it must be negative on $(0,a_1]$. Hence, $m_2(h_1) \in (a_2,1)$. This shows that $H_{\beta ,h_1,p}'(a_2) > 0$, and hence, there exists $m(h_1) \in (a_1,a_2)$ such that $H_{\beta ,h_1,p}'(m(h_1)) = 0$. Observe that $H_{\beta ,h_1,p}'$ is negative on $(0,m(h_1))$, positive on $(m(h_1),m_2(h_1))$ and negative on $(m_2(h_1),1)$. Therefore, since $h_2>h_1$, $H_{\beta ,h_2,p}'>0$ on $[m(h_1),m_2(h_1)]$. Since $m(h_2)$ and $m_2(h_2)$ are roots of $H_{\beta ,h_2,p}'$ on $(a_1,a_2)$ and $(a_2,1)$ respectively, we must have $m(h_2)<m(h_1)$ and $m_2(h_1) < m_2(h_2)$. Hence, we have

$$\begin{aligned} \int _{0}^{m(h_1)} H_{\beta ,h_1,p}' (t) {\mathrm {d}} t< \int _{m_1(h_2)}^{m(h_2)} H_{\beta ,h_1,p}' (t) {\mathrm {d}} t < \int _{m_1(h_2)}^{m(h_2)} H_{\beta ,h_2,p}' (t) {\mathrm {d}} t \end{aligned}$$

(B.4)

and

$$\begin{aligned} \int _{m(h_1)}^{m_2(h_1)} H_{\beta ,h_1,p}' (t) {\mathrm {d}} t< \int _{m(h_1)}^{m_2(h_1)} H_{\beta ,h_2,p}' (t) {\mathrm {d}} t < \int _{m(h_2)}^{m_2(h_2)} H_{\beta ,h_2,p}' (t) {\mathrm {d}} t \end{aligned}$$

(B.5)

Adding (B.4) and (B.5), gives

$$\begin{aligned} \int _{0}^{m_2(h_1)} H_{\beta ,h_1,p}' (t) {\mathrm {d}} t < \int _{m_1(h_2)}^{m_2(h_2)} H_{\beta ,h_2,p}'(t) {\mathrm {d}} t . \end{aligned}$$

(B.6)

Once again, this is a contradiction, since the right side of (B.6) is 0, whereas the left side of (B.6) is non-negative. This completes the proof of our claim that for every $\beta > \check{\beta }_p$, there exists at most one $h \geqslant 0$ such that $(\beta ,h) \in {\mathscr {C}_p}^+$.

We now show that for all $\beta \in (\check{\beta }_p,\infty ){\setminus } \{\tilde{\beta }_p\}$, there exists at least one $h \geqslant 0$ such that $(\beta ,h) \in {\mathscr {C}_p}^+$. First, suppose that $\beta > \tilde{\beta }_p$. In this case, $\sup _{x\in [-1,1]} H_{\beta ,0,p}(x) > 0$ by the definition in (2.2), and hence, $H_{\beta ,0,p}$ has a non-zero global maximizer $m_*$. Since $H_{\beta ,0,p}$ is an even function, $-m_*$ is also a global maximizer. It now follows from part (3) of Lemma B.1, that $H_{\beta ,0,p}$ has exactly two global maximizers, and hence, $(\beta ,0) \in {\mathscr {C}_p}^+$.

Next, let $\beta \in (\check{\beta }_p,\tilde{\beta }_p)$. Recall that the function $H_{\beta ,0,p}'$ is continuous and strictly decreasing on each of the intervals $[0,a_1]$ and $[a_2,1)$. Hence, the functions

$$\begin{aligned} \psi _1 := H_{\beta ,0,p}'\Big |_{[0,a_1]} \quad \text {and} \quad \psi _2 := H_{\beta ,0,p}'\Big |_{[a_2,1)} \end{aligned}$$

are invertible, and by Proposition 2.1 in [22], the functions $\psi _1^{-1}$ and $\psi _2^{-1}$ are continuous. Hence, the function $\Lambda : [H_{\beta ,0,p}'(a_1), \min \{0,H_{\beta ,0,p}'(a_2) \} ] \rightarrow \mathbb {R}$ defined as:

$$\begin{aligned} \Lambda (h) := \int _{\psi _1^{-1}(h)}^{\psi _2^{-1}(h)} H_{\beta ,-h,p}' (t) {\mathrm {d}} t = \int _{\psi _1^{-1}(h)}^{\psi _2^{-1}(h)} H_{\beta ,0,p}' (t) {\mathrm {d}} t + h\left( \psi _1^{-1}(h)-\psi _2^{-1}(h)\right) \end{aligned}$$

is continuous. Since the function $t \mapsto H_{\beta ,0,p}'(t) - H_{\beta ,0,p}'(a_1)$ is strictly positive on the interval $( a_1, \psi _2^{-1}(H_{\beta ,0,p}'(a_1)) )$ (because it is strictly increasing on $[a_1,a_2]$, strictly decreasing on $[a_2,1)$, and vanishes at the endpoints $a_1$ and $\psi _2^{-1}(H_{\beta ,0,p}'(a_1))$ of the interval),

$$\begin{aligned} \Lambda (H_{\beta ,0,p}'(a_1)) = \int _{a_1}^{\psi _2^{-1}(H_{\beta ,0,p}'(a_1))} \left( H_{\beta ,0,p}'(t) - H_{\beta ,0,p}'(a_1)\right) {\mathrm {d}}t > 0. \end{aligned}$$

(B.7)

Next, suppose that $H_{\beta ,0,p}'(a_2) \leqslant 0$. Since the function $t \mapsto H_{\beta ,0,p}'(t) - H_{\beta ,0,p}'(a_2)$ is strictly negative on the interval $(\psi _1^{-1}(H_{\beta ,0,p}'(a_2)), a_2 )$ (because it is strictly decreasing on $[0,a_1]$, strictly increasing on $[a_1,a_2]$, and vanishes at the endpoints $\psi _1^{-1}(H_{\beta ,0,p}'(a_2))$ and $a_2$ of the interval),

$$\begin{aligned} \Lambda (H_{\beta ,0,p}'(a_2)) = \int _{\psi _1^{-1}(H_{\beta ,0,p}'(a_2))}^{a_2} \left( H_{\beta ,0,p}'(t) - H_{\beta ,0,p}'(a_2)\right) {\mathrm {d}}t < 0. \end{aligned}$$

(B.8)

Finally, suppose that $H_{\beta ,0,p}'(a_2) > 0$. Then we have

$$\begin{aligned} \Lambda (0) = \int _{0}^{\psi _2^{-1}(0)} H_{\beta ,0,p}' (t) {\mathrm {d}} t = H_{\beta ,0,p}(\psi _2^{-1}(0)) < 0. \end{aligned}$$

(B.9)

The last inequality in (B.9) follows from the facts that $\psi _2^{-1}(0) > 0$ and $\beta < \tilde{\beta }_p$.

Using (B.7), (B.8), (B.9) and the intermediate value theorem, we conclude that there exists $h(\beta ) \in (H_{\beta ,0,p}'(a_1), \min \{0,H_{\beta ,0,p}'(a_2) \})$ such that $\Lambda (h(\beta )) = 0$, i.e.

$$\begin{aligned} H_{\beta ,-h(\beta ),p}(\psi _1^{-1}(h(\beta ))) = H_{\beta ,-h(\beta ),p}(\psi _2^{-1}(h(\beta ))). \end{aligned}$$

(B.10)

Now, $\psi _1^{-1}(h(\beta )) \in (0,a_1)$ and $\psi _2^{-1}(h(\beta )) \in (a_2,1)$, and hence, $H_{\beta ,-h(\beta ),p}'$ is strictly decreasing on some open neighborhoods of $\psi _1^{-1}(h(\beta ))$ and $\psi _2^{-1}(h(\beta ))$. Since $H_{\beta ,-h(\beta ),p}'(\psi _1^{-1}(h(\beta )))$ $= H_{\beta ,-h(\beta ),p}'(\psi _2^{-1}(h(\beta ))) = 0$, the points $\psi _1^{-1}(h(\beta ))$ and $\psi _2^{-1}(h(\beta ))$ are local maximizers of $H_{\beta ,-h(\beta ),p}$. Since $-h(\beta )>0$, any global maximizer of $H_{\beta ,-h(\beta ),p}$ must be a positive root of $H_{\beta ,-h(\beta ),p}'$, and further, it cannot lie on the interval $[a_1,a_2]$, since $H_{\beta ,-h(\beta ),p}'$ is strictly increasing on this interval. Hence, one of $\psi _1^{-1}(h(\beta ))$ and $\psi _2^{-1}(h(\beta ))$ must be a global maximizer of $H_{\beta ,-h(\beta ),p}$, and by (B.10), both must be global maximizers of $H_{\beta ,-h(\beta ),p}$. By part (3) of Lemma B.1, these are the only global maximizers of $H_{\beta ,-h(\beta ),p}$, and hence, $(\beta ,-h(\beta )) \in {\mathscr {C}_p}^+$.

Next, if $\beta = \tilde{\beta }_p$, then $H_{\beta ,0,p}$ has three global maximizers, so $(\beta ,0) \notin {\mathscr {C}_p}^+$. One of these global maximizers is 0 and the other two are negative of one another. It follows from the argument used in proving the uniqueness of h under the case $h_1 = 0$, that

$$\begin{aligned} \int _{m_1(h)}^{m_2(h)} H_{\beta ,h,p}'(t) {\mathrm {d}} t > 0, \end{aligned}$$

for every $h > 0$, where $m_2(h)> m_1(h) > 0$ are possible global maximizers of $H_{\beta ,h,p}$ (see inequality (B.6)), which is a contradiction. Hence,

$$\begin{aligned} {\mathscr {C}_p}^+ \subseteq \left( \{\tilde{\beta }_p\} \times \mathbb {R}\right) ^c. \end{aligned}$$

At this point, we completed proving that for every $\beta \in (\check{\beta }_p,\infty ){\setminus } \{\tilde{\beta }_p\}$, there exists unique $h \geqslant 0$ such that $(\beta ,h) \in {\mathscr {C}_p}^+$, and further, there exists no such h for $\beta = \tilde{\beta }_p$. Denote by $\varphi _p(\beta )$, this unique h corresponding to $\beta \in (\check{\beta }_p,\infty ){\setminus } \{\tilde{\beta }_p\}$. Our proof so far, also reveals that $\varphi _p(\beta ) = 0$ for $\beta > \tilde{\beta }_p$ and $\varphi _p(\beta ) > 0$ for $\beta \in (\check{\beta }_p,\tilde{\beta }_p)$. Define $\varphi _p(\tilde{\beta }_p) = 0$ for the sake of completing its definition on the whole of $(\check{\beta }_p,\infty )$.

We now show that $\varphi _p$ is strictly decreasing on $(\check{\beta }_p,\tilde{\beta }_p)$. Towards this, take $\check{\beta }_p< \beta _1<\beta _2<\tilde{\beta }_p$. Let $h_1 := \varphi _p(\beta _1)$ and $h_2 := \varphi _p(\beta _2)$ (we already know from the proof of the existence part, that $h_1$ and $h_2$ are positive), and suppose towards a contradiction, that $h_1 \leqslant h_2$. Then, $H_{\beta _1,h_1,p}' < H_{\beta _2,h_2,p}'$ on (0, 1]. Let $m_{11}<m_{13}$ be the global maximizers of $H_{\beta _1,h_1,p}$ and $m_{21}<m_{23}$ be the global maximizers of $H_{\beta _2,h_2,p}$. Also, let $m_{12} \in (m_{11},m_{13})$ and $m_{22}\in (m_{21},m_{23})$ be local minimizers of $H_{\beta _1,h_1,p}$ and $H_{\beta _2,h_2,p}$, respectively. We have already shown that for $i \in \{1,2\}$, the function $H_{\beta _i,h_i,p}'$ is positive on $[0,m_{i1})$, negative on $(m_{i1},m_{i2})$, positive on $(m_{i2},m_{i3})$ and negative on $(m_{i3},1)$. Since $H_{\beta _2,h_2,p}' > 0$ on $[0,m_{11}]$, we must have $m_{21} > m_{11}$. On the other hand, we have $m_{21}< m_* := \sqrt{1-2/p} < m_{13}$. This, combined with the fact that $H_{\beta _2,h_2,p}' > 0$ on $[m_{12},m_{13}]$, implies that $m_{21}<m_{12}$. Next, since $H_{\beta _1,h_1,p}' < 0$ on $[m_{21},m_{22}]$ and $H_{\beta _1,h_1,p}'(m_{12}) = 0$, it follows that $m_{22} < m_{12}$. Finally, since $H_{\beta _1,h_1,p}' <0$ on $[m_{23},1)$, we must have $m_{13}<m_{23}$. Hence, we have

$$\begin{aligned} m_{11}< m_{21}<m_{22}<m_{12}<m_{13}<m_{23}. \end{aligned}$$

Using this and proceeding exactly as in the proof of the uniqueness of h, we have

$$\begin{aligned}&\int _{m_{11}}^{m_{12}} H_{\beta _1,h_1,p}' (t) {\mathrm {d}} t< \int _{m_{21}}^{m_{22}} H_{\beta _2,h_2,p}' (t) {\mathrm {d}} t \\&\quad \text {and}\quad \int _{m_{12}}^{m_{13}} H_{\beta _1,h_1,p}' (t) {\mathrm {d}} t < \int _{m_{22}}^{m_{23}} H_{\beta _2,h_2,p}' (t) \mathrm d t . \end{aligned}$$

Adding the above two inequalities, we have

$$\begin{aligned} \int _{m_{11}}^{m_{13}} H_{\beta _1,h_1,p}' (t) {\mathrm {d}} t < \int _{m_{21}}^{m_{23}} H_{\beta _2,h_2,p}' (t) {\mathrm {d}} t , \end{aligned}$$

which is a contradiction once again, since both sides of the above inequality are 0. Hence, we must have $h_1 > h_2$, showing that $\varphi _p$ is strictly decreasing on $(\check{\beta }_p,\tilde{\beta }_p)$.

Next, we show that $\varphi _p$ is continuous on $(\check{\beta }_p,\tilde{\beta }_p]$. Towards this, first take $\beta \in (\check{\beta }_p,\tilde{\beta }_p)$, and let $\{\beta _n\}_{n\geqslant 1}$ be a monotonic sequence in $(\check{\beta }_p,\tilde{\beta }_p)$ converging to $\beta$. Since $\varphi _p$ is decreasing on $(\check{\beta }_p,\tilde{\beta }_p)$, it follows that $\varphi _p(\beta _n)$ is monotonic as well (the direction of monotonicity being opposite to that of $\beta _n$). Moreover, $\varphi _p(\beta _n)$ is bounded between $\varphi _p(\beta _1)$ and $\varphi _p(\beta )$. Hence, $\lim _{n \rightarrow \infty } \varphi _p(\beta _n)$ exists, which we call h. Let $m_1(n) <m_2(n)$ denote the global maximizers of $H_{\beta _n,\varphi _p(\beta _n),p}$. Choose a subsequence $n_k$ such that $m_1(n_k) \rightarrow m_1$ and $m_2(n_k)\rightarrow m_2$ for some $m_1,m_2 \in [-1,1]$. Since

$$\begin{aligned} H_{\beta _{n_k},\varphi _p(\beta _{n_k}),p}(m_i(n_k)) \geqslant H_{\beta _{n_k},\varphi _p(\beta _{n_k}),p}(x)\quad \text {for all}~ x \in [-1,1]~\text {and}~i \in \{1,2\}, \end{aligned}$$

taking limit as $k \rightarrow \infty$ on both sides, we have $H_{\beta ,h,p}(m_i) \geqslant H_{\beta ,h,p}(x)$ for all $x \in [-1,1]$ and $i \in \{1,2\}$, showing that $m_1$ and $m_2$ are global maximizers of $H_{\beta ,h,p}$. We now show that $m_1 < m_2$. Since $\beta _n \rightarrow \beta > \check{\beta }_p$, there exists $\underline{\beta } > \check{\beta }_p$ such that $\beta _n > \underline{\beta }$ for all large n. If $a_1(\underline{\beta }) < a_2(\underline{\beta })$ are the positive roots of $H_{\underline{\beta },0,p}''$, then $H_{\beta _n,0,p}'' > 0$ on $[a_1(\underline{\beta }), a_2(\underline{\beta })]$ for all large n, and hence, $m_1(n) < a_1(\underline{\beta })$ and $m_2(n) > a_2(\underline{\beta })$ for all large n. This shows that

$$\begin{aligned} m_1\leqslant a_1(\underline{\beta }) < a_2(\underline{\beta })\leqslant m_2 \end{aligned}$$

and hence, $m_1 < m_2$. Thus $H_{\beta ,h,p}$ has at least two global maximizers. But $\beta \ne \tilde{\beta }_p$, and $H_{\beta ,h,p}$ must therefore have exactly two global maximizers, showing that $(\beta ,h) \in {\mathscr {C}_p}^+$. Since $h \geqslant 0$, by the uniqueness property, we must have $h = \varphi _p(\beta )$. Hence, $\lim _{n\rightarrow \infty } \varphi _p(\beta _n) = \varphi _p(\beta )$, showing that $\varphi _p$ is continuous on $(\check{\beta }_p,\tilde{\beta }_p)$.

To show that $\lim _{\beta \rightarrow (\tilde{\beta }_p)^{-}} \varphi _p(\beta ) = 0$, take a sequence $\beta _n \in (\check{\beta }_p,\tilde{\beta }_p)$ increasing to $\tilde{\beta }_p$, whence $\varphi _p(\beta _n)$ decreases to some $h \geqslant 0$. By the same arguments as before, it follows that $H_{\tilde{\beta }_p,h,p}$ has at least two global maximizers. If $h > 0$, then $H_{\tilde{\beta }_p,h,p}$ will have exactly two global maximizers. Therefore $(\tilde{\beta }_p,h) \in {\mathscr {C}_p}^+$, contradicting our finding that ${\mathscr {C}_p}^+ \subseteq (\{\tilde{\beta }_p\} \times \mathbb {R})^c$. This shows that $h=0$, completing the proof of (1). $\square$

Proof of (2)

Let $p \geqslant 3$ be odd. In this case, $H_{\beta ,0,p}''<0$ on $[-1,0]$ for all $\beta \geqslant 0$. Let $\beta > \check{\beta }_p.$ Once again, $H_{\beta ,0,p}''$ can have at most two positive roots, which, together with the facts $H_{\beta ,0,p}''(m_*) > 0$ and $H_{\beta ,0,p}''(1) = -\infty$, imply the existence of $0<a_1<m_*<a_2<1$, such that $H_{\beta ,0,p}'' < 0$ on $[-1,a_1)\bigcup (a_2,1]$ and $H_{\beta ,0,p}''> 0$ on $(a_1,a_2)$. One can now follow the proof of (a) modulo obvious modifications, to show that there exists at most one $h \in \mathbb {R}$ such that $(\beta ,h) \in {\mathscr {C}_p}^+$.

To show the existence of at least one such $h \in \mathbb {R}$, one can once again essentially follow the proof of (a) modulo a couple of minor modifications. To be specific, if we modify the definition of $\psi _1$ to $H_{\beta ,0,p}'\big |_{(-1,a_1]}$, and change the domain of $\Lambda$ to $[H_{\beta ,0,p}'(a_1), H_{\beta ,0,p}'(a_2)]$, then by following the proof of (a), we can show the existence of $h(\beta ) \in (H_{\beta ,0,p}'(a_1), H_{\beta ,0,p}'(a_2))$ such that $(\beta ,-h(\beta )) \in {\mathscr {C}_p}^+$. If we denote the unique h corresponding to each $\beta > \check{\beta }_p$ such that $(\beta ,h) \in {\mathscr {C}_p}^+$ by $\varphi _p(\beta )$, then continuity and the strict decreasing nature of $\varphi _p$ once again follow from the proof of (a).

Next, it follows from Remark B.1, that $\varphi _p(\tilde{\beta }_p) = 0$. We now show that $\lim _{\beta \rightarrow \infty } \varphi _p(\beta ) = -\infty$. Towards this, note that the monotonicity pattern of $H_{\beta ,\varphi _p(\beta ),p}'$ for $\beta > \check{\beta }_p$ implies that $H_{\beta ,\varphi _p(\beta ),p}$ has exactly two local maximizers $m_1(\beta ) \in (-1,a_1(\beta ))$ and $m_2(\beta ) \in (a_2(\beta ),1)$, where $a_1(\beta )$ and $a_2(\beta )$ are the inflection points of $H_{\beta ,\varphi _p(\beta ),p}$, satisfying $0<a_1(\beta )< m_*< a_2(\beta )<1$ for all $\beta > \check{\beta }_p$. Hence, $m_1(\beta )$ and $m_2(\beta )$ are global maximizers of $H_{\beta ,\varphi _p(\beta ),p}$. Let $\beta > \tilde{\beta }_p$, whence the strictly decreasing nature of $\varphi _p$ implies that $\varphi _p(\beta ) < 0$. Since $H_{\beta ,\varphi _p(\beta ),p}'(-1) = \infty$ and $H_{\beta ,\varphi _p(\beta ),p}'(0) = \varphi _p(\beta ) < 0$, the intermediate value theorem implies that $m_1(\beta ) < 0$. Hence,

$$\begin{aligned} \beta (m_1(\beta ))^p - I(m_1(\beta ))< 0,\quad \text {that is,}\quad H_{\beta ,\varphi _p(\beta ),p}(m_1(\beta )) < \varphi _p(\beta ) m_1(\beta ). \end{aligned}$$

Now, since

$$\begin{aligned} H_{\beta ,\varphi _p(\beta ),p}(m_1(\beta )) = H_{\beta ,\varphi _p(\beta ),p}(m_2(\beta )) = \beta (m_2(\beta ))^p + \varphi _p(\beta )m_2(\beta )- I(m_2(\beta )), \end{aligned}$$

we have $\beta (m_2(\beta ))^p + \varphi _p(\beta )m_2(\beta )- I(m_2(\beta )) < \varphi _p(\beta )m_1(\beta )$. This implies,

$$\begin{aligned} -2\varphi _p(\beta )> \varphi _p(\beta )(m_1(\beta ) - m_2(\beta )) > \beta (m_2(\beta ))^p - I(m_2(\beta )) \geqslant \beta m_*^p - I(m_2(\beta )).\nonumber \\ \end{aligned}$$

(B.11)

The proof of our claim now follows from (B.11) since $\lim _{\beta \rightarrow \infty } \beta m_*^p - I(m_2(\beta )) = \infty$. This completes the proof of part (2).

Finally, we prove that $\lim _{\beta \rightarrow \check{\beta }_p^+} \varphi _p(\beta ) = \tanh ^{-1}(m_*) - p\check{\beta }_p m_*^{p-1}$, where $m_* := \sqrt{1-2/p}$. Towards this, let $0< \varepsilon < \tilde{\beta }_p-\check{\beta }_p$ be given, and take any

$$\begin{aligned} \beta \in \left( \check{\beta }_p,\check{\beta }_p + \frac{\varepsilon }{2p(p-1)}\right) . \end{aligned}$$

As before, let $0<a_1<a_2<1$ be the points such that $H_{\beta ,0,p}''<0$ on $[0,a_1)\bigcup (a_2,1]$ and $H_{\beta ,0,p}''>0$ on $(a_1,a_2)$. Since $H_{\check{\beta }_p,0,p}'' \leqslant 0$ on [0, 1], it follows that $H_{\beta ,0,p}'' \leqslant (\beta -\check{\beta }_p)p(p-1) < \varepsilon /2$ on [0, 1]. Hence, for every $h \in \mathbb {R}$, we have

$$\begin{aligned} H_{\beta ,h,p}'(a_2) - H_{\beta ,h,p}'(a_1) = \int _{a_1}^{a_2} H_{\beta ,0,p}'' (t) {\mathrm {d}} t \leqslant \varepsilon (a_2-a_1)/2 <\varepsilon /2. \end{aligned}$$

(B.12)

Since $H_{\beta ,0,p}''(m_*) > 0$, we must have $m_* \in (a_1,a_2)$. If $m_1 < m_2$ are the two global maximizers of $H_{\beta ,\varphi _p(\beta ),p}$, then $m_1 \in (0,a_1)$ and $m_2 \in (a_2,1)$. Since $H_{\beta ,\varphi _p(\beta ),p}'$ is strictly decreasing on each of the intervals $[0,a_1]$ and $[a_2,1)$, we must have $H_{\beta ,\varphi _p(\beta ),p}'(a_1) < 0$ and $H_{\beta ,\varphi _p(\beta ),p}'(a_2) > 0$. Hence, there exists $a_3 \in (a_1,a_2)$ such that $H_{\beta ,\varphi _p(\beta ),p}'(a_3) = 0$. Now, since $H_{\beta ,\varphi _p(\beta ),p}'$ is increasing on $[a_1,a_2]$, we have from (B.12),

$$\begin{aligned} \big |H_{\beta ,\varphi _p(\beta ),p}'(a_3) - H_{\beta ,\varphi _p(\beta ),p}'(m_*)\big | \leqslant H_{\beta ,\varphi _p(\beta ),p}'(a_2) - H_{\beta ,\varphi _p(\beta ),p}'(a_1) < \varepsilon /2, \end{aligned}$$

and hence, $\big |H_{\beta ,\varphi _p(\beta ),p}'(m_*)\big | = \big |\tanh ^{-1}(m_*) - p\beta m_*^{p-1} - \varphi _p(\beta )\big | < \varepsilon /2$. Now, $\big |p\beta m_*^{p-1} - p\check{\beta }_p m_*^{p-1}\big | \leqslant p(\beta - \check{\beta }_p) < \varepsilon /2$. By triangle inequality, we thus have

$$\begin{aligned} \big |\tanh ^{-1}(m_*) - p\check{\beta }_p m_*^{p-1} - \varphi _p(\beta )\big |&\leqslant \big |\tanh ^{-1}(m_*) - p\beta m_*^{p-1} - \varphi _p(\beta )\big | \nonumber \\&\quad + \big |p\beta m_*^{p-1} - p\check{\beta }_p m_*^{p-1}\big | \nonumber \\&< \varepsilon . \end{aligned}$$

(B.13)

Our claim now follows from (B.13). The proof of (2) and Lemma B.3 is now complete. $\square$

Now, we will prove some properties of the function H, when the underlying parameter $(\beta ,h)$ is perturbed to $(\beta ,h_N)$, where $(\beta ,h_N) \rightarrow (\beta ,h)$, as $N \rightarrow \infty$. Investigating the properties of the function $H_{\beta ,h_N,p}$ is especially important, since our analysis hinges more upon these perturbed functions, rather than the original function $H_{\beta ,h,p}$.

Lemma B.4

Suppose that $(\beta ,h_N) \in [0,\infty )\times \mathbb {R}$ is a sequence converging to a point $(\beta ,h) \in [0,\infty )\times \mathbb {R}$. Then, we have the following:

• (1) Suppose that $(\beta ,h)$ is a p-regular point, and let $m_*$ be the global maximizer of $H_{\beta ,h,p}$. Then, for any sequence $(\beta ,h_N) \in [0,\infty )\times \mathbb {R}$ converging to $(\beta ,h)$, the function $H_{\beta ,h_N,p}$ will have unique global maximizer $m_*(N)$ for all large N, and $m_*(N) \rightarrow m_*$ as $N \rightarrow \infty$.

• (2) Let m be a local maximizer of the function $H_{\beta ,h,p}$, where the point $(\beta ,h)$ is not p-special. Suppose that $(\beta ,h_N) \in [0,\infty )\times \mathbb {R}$ is a sequence converging to $(\beta ,h)$. Then for all large N, the function $H_{\beta ,h_N,p}$ will have a local maximizer m(N), such that $m(N) \rightarrow m$ as $N \rightarrow \infty$. Further, if $A\subseteq [-1,1]$ is a closed interval such that $m \in \text {int}(A)$ and $H_{\beta ,h,p}(m) > H_{\beta ,h,p}(x)$ for all $x \in A{\setminus } \{m\}$, then there exists $N_0\geqslant 1$, such that for all $N \geqslant N_0$, we have $H_N(m(N)) > H_N(x)$ for all $x \in A {\setminus } \{m(N)\}$.

Proof of (1)

The set $\mathcal {R}_p$ of all p-regular points is an open subset of $[0,\infty )\times \mathbb {R}$. To see this, note that $\mathcal {R}_p^c$ is given by ${\mathscr {C}_p}\bigcup \{(\check{\beta }_p,\check{h}_p)\}$ if p is odd, and by ${\mathscr {C}_p}\bigcup \{(\check{\beta }_p,\check{h}_p), (\check{\beta }_p,-\check{h}_p)\}$ if p is even. By Lemma B.3, $\mathcal {R}_p^c$ is a closed set in either case. Hence, the function $H_{\beta ,h_N,p}$ will have unique global maximizer $m_*(N)$ for all large N.

To show that $m_*(N) \rightarrow m_*$, let $\{N_k\}_{k \geqslant 1}$ be a subsequence of the natural numbers. Then, $\{N_k\}_{k \geqslant 1}$ will have a further subsequence $\{N_{k_\ell }\}_{\ell \geqslant 1}$, such that $m_*(N_{k_\ell })$ converges to some $m' \in [-1,1]$. Since $H_{\beta _{N_{k_\ell }},h_{N_{k_\ell }},p}\left( m_*(N_{k_\ell })\right) \geqslant H_{\beta _{N_{k_\ell }},h_{N_{k_\ell }},p}(x)$ for all $x \in [-1,1]$, by taking limit as $\ell \rightarrow \infty$ on both sides, we have $H_{\beta ,h,p}(m') \geqslant H_{\beta ,h,p}(x)$ for all $x \in [-1,1]$, showing that $m'$ is a global maximizer of $H_{\beta ,h,p}$. Since $m_*$ is the unique global maximizer of $H_{\beta ,h,p}$, it follows that $m' = m_*$, completing the proof of (1). $\square$

Proof of (2)

Let us denote $H_{\beta ,h,p}$ by H and $H_{\beta ,h_N,p}$ by $H_N$. It is easy to show that there exists $M \geqslant 1$ odd, and points $-1=a_0<a_1<\ldots <a_M=1$, such that $H'$ is strictly decreasing on $[a_{2i},a_{2i+1}]$ and strictly increasing on $[a_{2i+1},a_{2i+2}]$ for all $0\leqslant i \leqslant \frac{M-1}{2}$. Hence, the local maximizer m of H lies in $(a_{2i},a_{2i+1})$ for some $0\leqslant i \leqslant \frac{M-1}{2}$. Since $H'(a_{2i}) > 0$ and $H'(a_{2i+1})<0$, we also have $H_N'(a_{2i}) > 0$ and $H_N'(a_{2i+1})<0$ for all large N, and hence $H_N'$ has a root $m(N) \in (a_{2i},a_{2i+1})$ for all large N.

Let us now show that $m(N) \rightarrow m$. Towards this, let $\{N_k\}_{k\geqslant 1}$ be a subsequence of the natural numbers, whence there is a further subsequence $\{N_{k_\ell }\}_{\ell \geqslant 1}$ of $\{N_k\}_{k\geqslant 1}$, such that $m(N_{k_\ell }) \rightarrow m'$ for some $m' \in [a_{2i},a_{2i+1}]$. Since $H_{N_{k_\ell }}'(m(N_{k_\ell })) = 0$ for all $\ell \geqslant 1$, we have $H'(m') = 0$. But the strict decreasing nature of $H'$ on $[a_{2i},a_{2i+1}]$ implies that m is the only root of $H'$ on this interval, and hence, $m'=m$. This shows that $m(N) \rightarrow m$.

Next, we show that m(N) is a local maximizer of $H_N$ for all N sufficiently large. For this, we prove something stronger than needed, because this will be useful in proving the last statement of (2). Since $H''(m) < 0$, there exists $\varepsilon > 0$ such that $[m-\varepsilon ,m+\varepsilon ] \subset (a_{2i},a_{2i+1})$ and $H''<0$ on $[m-\varepsilon ,m+\varepsilon ]$. If $m_0 \in [m-\varepsilon ,m+\varepsilon ]$ is such that $H''(m_0) = \sup _{x \in [m-\varepsilon ,m+\varepsilon ]} H''(x) < 0$, then since $H_N''$ converges to $H''$ uniformly on $(-1,1)$,

$$\begin{aligned} \sup _{x\in [m-\varepsilon ,m+\varepsilon ]}H_N''(x) < H''(m_0)/2\quad \text {for all large}~N. \end{aligned}$$

In particular, since $m(N) \in [m-\varepsilon ,m+\varepsilon ]$ for all large N, we have $H_N''(m(N)) < 0$ for all large N, showing that m(N) is a local maximizer of $H_N$ for all large N. Also, since $H_N'(m(N)) = 0$ and $\sup _{x\in [m-\varepsilon ,m+\varepsilon ]}H_N''(x) < 0$ for all large N, we must have

$$\begin{aligned} H_N(m(N)) > H_N(x)\quad \text {for all}~ x \in [m-\varepsilon ,m+\varepsilon ]{\setminus }\{m(N)\},\quad \text {for all large} N. \end{aligned}$$

Finally, suppose that $A \subseteq [-1,1]$ is a closed interval such that $m \in \text {int}(A)$ and $H(m) > H(x)$ for all $x \in A{\setminus } \{m\}$. By Lemma B.11, there exists $\varepsilon ' >0$ such that for all $0 < \delta \leqslant \varepsilon '$, $\sup _{x \in A{\setminus } (m-\delta ,m+\delta )} H(x) = H(m\pm \delta )$. Let $\alpha = \min \{\varepsilon ,\varepsilon '\}$. Then,

$$\begin{aligned} H_N(m(N)) > H_N(x) \quad \text {for all} ~x \in [m-\alpha ,m+\alpha ]{\setminus } \{m(N)\},\quad \text {for all large}~ N,\nonumber \\ \end{aligned}$$

(B.14)

and $\sup _{x \in A{\setminus } (m-\alpha ,m+\alpha )} H(x) = H(m\pm \alpha ) < H(m)$ (since $H'(m) = 0$ and $H''< 0$ on $[m-\alpha ,m+\alpha ]$). Hence,

$$\begin{aligned} \sup _{x \in A{\setminus } (m-\alpha ,m+\alpha )} H_N(x) < H_N(m(N))\quad \text {for all large}~ N. \end{aligned}$$

(B.15)

The proof of (2) now follows from (B.14) and (B.15), and the proof of Lemma B.4 is now complete. $\square$

1.2 Other technical lemmas

In this section, we collect the proofs of the remaining technical lemmas, which are used in the proofs of the main results in various places. We start with a result that gives implicit expressions for the partial derivatives of any stationary point of $H_{\beta ,h,p}$ with respect to $\beta$ and h.

Lemma B.5

Let $m = m(\beta ,h,p)$ satisfy the implicit relation $H_{\beta ,h,p}'(m) = 0$, and suppose that $H_{\beta ,h,p}''(m) \ne 0$. Then, the partial derivatives of m with respect to $\beta$ and h are given by:

$$\begin{aligned} \frac{\partial m}{\partial \beta } = -\frac{pm^{p-1}}{H_{\beta ,h,p}''(m)}\quad \quad \text {and}\quad \quad \frac{\partial m}{\partial h} = -\frac{1}{H_{\beta ,h,p}''(m)}. \end{aligned}$$

(B.16)

Moreover, $\big |\frac{\partial ^2 m}{\partial \beta ^2}\big | < \infty$ and $\big |\frac{\partial ^2 m}{\partial h^2}\big | < \infty$, if $H_{\beta ,h,p}''(m) \ne 0$.

Proof

Differentiating both sides of the identity $\beta p m^{p-1} +h - \tanh ^{-1}(m) = 0$ with respect to $\beta$ and h separately, we get the following two first order partial differential equations, respectively:

$$\begin{aligned}&pm^{p-1} + \beta p(p-1)m^{p-2}\frac{\partial m}{\partial \beta } - \frac{1}{1-m^2} \frac{\partial m}{\partial \beta } = 0, \nonumber \\&\quad \text {that is,}\quad pm^{p-1} + H_{\beta ,h,p}''(m)\frac{\partial m}{\partial \beta } = 0~; \end{aligned}$$

(B.17)

$$\begin{aligned}&\beta p(p-1)m^{p-2}\frac{\partial m}{\partial h} + 1 - \frac{1}{1-m^2}\nonumber \\&\frac{\partial m}{\partial h} = 0,\quad \text {that is,}\quad 1+H_{\beta ,h,p}''(m)\frac{\partial m}{\partial h} = 0~; \end{aligned}$$

(B.18)

The expressions in (B.16) follow from (B.17) and (B.18). Another implicit differentiation of (B.17) with respect to $\beta$ and (B.18) with respect to h yields the following two second order partial differential equations, respectively:

$$\begin{aligned} 2p(p-1) m^{p-2} \frac{\partial m}{\partial \beta } + H_{\beta ,h,p}^{(3)}(m)\left( \frac{\partial m}{\partial \beta }\right) ^2 + H_{\beta ,h,p}''(m)\frac{\partial ^2 m}{\partial \beta ^2} = 0; \end{aligned}$$

(B.19)

$$\begin{aligned} H_{\beta ,h,p}^{(3)}(m)\left( \frac{\partial m}{\partial h}\right) ^2 + H_{\beta ,h,p}''(m)\frac{\partial ^2 m}{\partial h^2} = 0; \end{aligned}$$

(B.20)

The finiteness of the second order partial derivatives of m as long as $H_{\beta ,h,p}''(m) \ne 0$, now follow from the fact that $H_{\beta ,h,p}''(m)$ is the coefficient of $\frac{\partial ^2 m}{\partial \beta ^2}$ and $\frac{\partial ^2 m}{\partial h^2}$ in the differential equations (B.19) and (B.20). $\square$

We now derive some important properties of the function $\zeta$ defined in (3.10). The following lemma is used in the proof of Lemma 3.2.

Lemma B.6

For any sequence $x \in (-1,1)$ that is bounded away from both 1 and $-1$, we have

$$\begin{aligned} \zeta (x) = \sqrt{\frac{2}{\pi N (1-x^2)}} e^{NH_N(x)}\left( 1+O(N^{-1})\right) . \end{aligned}$$

Proof

The proof of Lemma B.6 follows immediately from Lemma A.5. $\square$

Now, we bound the derivative of the function $\zeta$ in a neighborhood of the point $m_*(N)$. This result appears in the proof of Lemma 3.2.

Lemma B.7

For every $\alpha \geqslant 0$ and p-regular point $(\beta ,h)$, we have the following bound:

$$\begin{aligned} \sup _{x\in A_{N,\alpha }} |\zeta ^\prime (x)| = \zeta (m_*(N))O\left( N^{\frac{1}{2}+\alpha }\right) , \end{aligned}$$

where $m_*(N)$ is the global maximizer of $H_N$ and $A_{N,\alpha } := \Big (m_*(N)-N^{-\frac{1}{2} + \alpha },m_*(N) +N^{-\frac{1}{2} + \alpha }\Big )$.

Proof of Lemma B.7

We begin with the following lemma: $\square$

Lemma B.8

For any sequence $x \in (-1,1)$ that is bounded away from both 1 and $-1$, we have

$$\begin{aligned} \zeta '(x) =\zeta (x)\left( NH_N'(x) + \frac{x}{1-x^2} + O(N^{-1})\right) . \end{aligned}$$

Proof

By Lemma A.1 and (A.1), we have

$$\begin{aligned}&\dfrac{{\mathrm {d}}}{{\mathrm {d}} x} \left( {\begin{array}{c}N\\ N(1+x)/2\end{array}}\right) \nonumber \\&= \frac{N}{2}\left( {\begin{array}{c}N\\ N(1+x)/2\end{array}}\right) \left[ \psi \left( 1-\frac{Nx}{2}+\frac{N}{2}\right) -\psi \left( 1+\frac{Nx}{2}+\frac{N}{2}\right) \right] \nonumber \\&= \frac{N}{2}\left( {\begin{array}{c}N\\ N(1+x)/2\end{array}}\right) \left( \log \left( \frac{ N}{2}(1-x)\right) -\log \left( \frac{N}{2}(1+x)\right) \right. \nonumber \\&\quad \left. +\frac{1}{N(1-x)}-\frac{1}{N(1+x)} + O(N^{-2})\right) \nonumber \\&= \left( {\begin{array}{c}N\\ N(1+x)/2\end{array}}\right) \left[ -N \tanh ^{-1}(x)+\frac{x}{1-x^2}+O(N^{-1})\right] . \end{aligned}$$

(B.21)

We thus have by the product rule of differential calculus and (B.21),

$$\begin{aligned} \zeta '(x)&= \zeta (x)(N\beta px^{p-1} + Nh_N) + \exp \left\{ N(\beta x^p + h_Nx - \log 2)\right\} \frac{{\mathrm {d}}}{\mathrm{{d}} x} \left( {\begin{array}{c}N\\ N(1+x)/2\end{array}}\right) \\ {}&= \zeta (x)(N\beta px^{p-1} + Nh_N) + \zeta (x)\left[ -N \tanh ^{-1}(x)+\frac{x}{1-x^2}+O(N^{-1})\right] \\ {}&= \zeta (x)\left( NH_N'(x) + \frac{x}{1-x^2} + O(N^{-1})\right) , \end{aligned}$$

completing the proof of Lemma B.8. $\square$

Now, we proceed with the proof of Lemma B.7. First note that, since $H_N'(m_*(N)) = 0$, we have by the mean value theorem,

$$\begin{aligned} \sup _{x \in A_{N,\alpha }}\big |H_N'(x)\big | \leqslant \sup _{x \in A_{N,\alpha }}\big |x-m_*(N)\big |\sup _{x\in A_{N,\alpha }}|H_N''(x)|=O\left( N^{-\frac{1}{2}+\alpha }\right) . \end{aligned}$$

(B.22)

It follows from (B.22) and Lemma B.8 that

$$\begin{aligned} \sup _{x \in A_{N,\alpha }}|\zeta '(x)| \leqslant O\left( N^{\frac{1}{2}+\alpha }\right) \sup _{x \in A_{N,\alpha }}\zeta (x). \end{aligned}$$

(B.23)

Now, Lemma B.6 implies that

$$\begin{aligned} \sup _{x \in A_{N,\alpha }} \zeta (x) \leqslant \left( 1+O(N^{-1})\right) \zeta (m_*(N))\sup _{x\in A_{N,\alpha }} \sqrt{\frac{1-m_*(N)^2}{1-x^2}} = \zeta (m_*(N))O(1).\nonumber \\ \end{aligned}$$

(B.24)

Lemma B.7 now follows from (B.23) and (B.24). $\square$

Lemma B.7 has an analogous version for p-special points $(\beta ,h)$, which is stated below. In this case, the bound on $\zeta '$ is better, and holds on a slightly larger region, too.

Lemma B.9

Let $m_*(N)$ be the unique global maximizer of $H_N := H_{\beta ,h_N,p}$, where $h_N := h +\bar{h}N^{-3/4}$ for some $\bar{h}\in \mathbb {R}$, and $(\beta ,h)$ is a p-special point. Then, for all $\alpha \geqslant 0$,

$$\begin{aligned} \sup _{x\in \mathcal {A}_{N,\alpha }} |\zeta '(x)| = \zeta (m_*(N))O\left( N^{\frac{1}{4}+3\alpha }\right) \end{aligned}$$

where $\mathcal {A}_{N,\alpha } := \left( m_*(N) - N^{-\frac{1}{4}+\alpha }, m_*(N) + N^{-\frac{1}{4} + \alpha }\right)$.

Proof

The proof of Lemma B.9 is similar to that of Lemma B.7, the only difference being a change in the estimate of $\sup _{x \in \mathcal {A}_{N,\alpha }} |H_N'(x)|$ from the estimate in (B.22). Note that

$$\begin{aligned} \sup _{x\in \mathcal {A}_{N,\alpha }} |H_N''(x)| \leqslant \sup _{x \in \mathcal {A}_{N,\alpha }} \tfrac{1}{2}(x-m_*)^2\sup _{x\in \mathcal {I}(\mathcal {A}_{N,\alpha }\cup \{m_*\})}H^{(4)} (x) = O\left( N^{-\frac{1}{2}+ 2\alpha }\right) , \end{aligned}$$

where $m_*$ denotes the global maximizer of $H_{\beta ,h,p}$ and for a set $A \subseteq \mathbb {R}$, $\mathcal {I}(A)$ denotes the smallest interval containing A. The last equality follows from the observation

$$\begin{aligned}&\sup _{x\in \mathcal {A}_{N,\alpha }} |x-m_*| \leqslant \sup _{x\in \mathcal {A}_{N,\alpha }} |x-m_*(N)| + |m_*(N) - m_*| \leqslant N^{-\frac{1}{4} + \alpha } + O\left( N^{-\frac{1}{4}}\right) \nonumber \\&\quad = O\left( N^{-\frac{1}{4} + \alpha }\right) , \end{aligned}$$

by Lemma B.10. Following (B.22), we have

$$\begin{aligned} \sup _{x\in \mathcal {A}_{N,\alpha }} |H_N'(x)| = O\left( N^{-\frac{3}{4} + 3\alpha }\right) . \end{aligned}$$

The rest of the proof is exactly same as that of Lemma B.7. $\square$

The following lemma provides estimates of the first four derivatives of the function H at the maximizer $m_*(N)$ for a perturbation of a p-special point. This key result is used in the proof of Lemma 3.4.

Lemma B.10

Let $(\beta ,h)$ be a p-special point and $h_N := h+\bar{h}N^{-\frac{3}{4}}$ for some $\bar{h} \in \mathbb {R}$. If $m_*$ and $m_*(N)$ denote the unique global maximizers of $H := H_{\beta ,h,p}$ and $H_N := H_{\beta ,h_N,p}$ respectively, then we have the following:

$$\begin{aligned}&N^\frac{1}{4}(m_*(N) - m_*)&= -\left( \frac{6 \bar{h}}{H^{(4)}(m_*)}\right) ^\frac{1}{3} + O\left( N^{-\frac{1}{4}}\right) , \end{aligned}$$

(B.25)

$$\begin{aligned}&N^\frac{1}{2}H''(m_*(N))&= \frac{1}{2}\left( 6 \bar{h}\right) ^\frac{2}{3}\left( H^{(4)}(m_*)\right) ^{\frac{1}{3}} + O\left( N^{-\frac{1}{4}}\right) , \end{aligned}$$

(B.26)

$$\begin{aligned}&N^\frac{1}{4}H^{(3)}(m_*(N))&= -\left( 6 \bar{h}\right) ^\frac{1}{3}\left( H^{(4)}(m_*)\right) ^\frac{2}{3} + O\left( N^{-\frac{1}{4}}\right) , \end{aligned}$$

(B.27)

$$\begin{aligned}&H^{(4)}(m_*(N))&= H^{(4)}(m_*) + O\left( N^{-\frac{1}{4}}\right) . \end{aligned}$$

(B.28)

Proof

Let us start by noting that

$$\begin{aligned} H'(m_*(N)) = H_N'(m_*(N)) - \bar{h} N^{-\frac{3}{4}} = - \bar{h} N^{-\frac{3}{4}}. \end{aligned}$$

On the other hand, by a Taylor expansion of $H'$ around $m_*$ and using the fact $H'(m_*) = H''(m_*) = H^{(3)}(m_*) = 0$ (see Lemma B.2), we have

$$\begin{aligned} H'(m_*(N)) = \tfrac{1}{6}(m_*(N)-m_*)^3 H^{(4)}(\zeta _N), \end{aligned}$$

where $\zeta _N$ lies between $m_*(N)$ and $m_*$. Hence,

$$\begin{aligned} N^\frac{3}{4} (m_*(N)-m_*)^3 = -\frac{6\bar{h}}{H^{(4)}(\zeta _N)}. \end{aligned}$$

Now, it follows from the proof of Lemma B.4, part (1), that $m_*(N) \rightarrow m_*$, and hence, $\zeta _N \rightarrow m_*$. This implies that

$$\begin{aligned} \lim _{N \rightarrow \infty } N^{\tfrac{1}{4}} (m_*(N) - m_*) = -\left( \frac{6 \bar{h}}{H^{(4)}(m_*)}\right) ^\frac{1}{3} . \end{aligned}$$

(B.29)

By a 5-term Taylor expansion of $H'(m_*(N))$ around $m_*$, one obtains

$$\begin{aligned} \tfrac{1}{6}(m_*(N)-m_*)^3 H^{(4)}(m_*) + \tfrac{1}{24}(m_*(N)-m_*)^4 H^{(5)}(\zeta _N') = -\bar{h} N^{-\frac{3}{4}}. \end{aligned}$$

(B.30)

for some sequence $\zeta _N'$ lying between $m_*(N)$ and $m_*$. From (B.30) and (B.29), we have

$$\begin{aligned} N^\frac{3}{4} (m_*(N) - m_*)^3&= -\frac{6 \bar{h}}{H^{(4)}(m_*)} - \frac{N^\frac{3}{4}(m_*(N) - m_*)^4 H^{(5)}(\zeta _N')}{4H^{(4)}(m_*)}\nonumber \\&= -\frac{6\bar{h}}{H^{(4)}(m_*)} + O\left( N^{-\frac{1}{4}}\right) . \end{aligned}$$

(B.31)

(B.25) now follows from (B.31), and (B.26), (B.27), (B.28) follow by substituting (B.25) into the following expansions

$$\begin{aligned}&H''(m_*(N)) = \tfrac{1}{2}\left( m_*(N) - m_*\right) ^2 H^{(4)}(m_*) + O\left( (m_*(N)-m_*)^3\right) , \\&H^{(3)}(m_*(N)) = \left( m_*(N) - m_*\right) H^{(4)}(m_*) + O\left( (m_*(N)-m_*)^2\right) , \end{aligned}$$

and $H^{(4)}(m_*(N)) = H^{(4)}(m_*) + O(m_*(N) - m_*)$. $\square$

The final lemma shows that if a function has non-vanishing curvature at a unique point of maxima, then for every sufficiently small open interval I around that point of maxima, it attains its maximum on $I^c$ at either of the endpoints of I. This fact is used in the proofs of Lemmas 3.1 and 3.7 .

Lemma B.11

Let $A \subseteq [-1,1]$ be a closed interval. Suppose that $f: A \mapsto \mathbb {R}$ is continuous on A and twice continuously differentiable on $\mathrm {int}(A)$. Suppose that there exists $x_*\in \mathrm {int}(A)$ such that $f(x_*) > f(x)$ for all $x \in A{\setminus } \{x_*\}$, and $f''(x_*) < 0$. Then, there exists $\eta > 0$ such that for all $0<\varepsilon \leqslant \eta$, f attains maximum on the set $A{\setminus } (x_*-\varepsilon ,x_*+\varepsilon )$ at either $x_*-\varepsilon$ or $x_*+\varepsilon$.

Proof

Since $f''$ is continuous on $\mathrm {int}(A)$ and negative at $x_*$, there exists $\delta > 0$ such that $f''(x)<0$ for all $x \in (x_*-\delta ,x_*+\delta )$. Hence, $f'$ is strictly decreasing on $(x_*-\delta ,x_*+\delta )$. Since $f'(x_*) = 0$, we have $f'(x) > 0$ for all $x \in (x_*-\delta ,x_*)$ and $f'(x) < 0$ for all $x \in (x_*,x_*+\delta )$. Hence, f is strictly increasing on $(x_*-\delta ,x_*]$ and strictly decreasing on $[x_*,x_*+\delta )$.

Suppose now, towards a contradiction, that the lemma is not true. Then, there is a sequence $\varepsilon _n \rightarrow 0$ such that neither $x_*-{\varepsilon }_n$ nor $x_*+{\varepsilon }_n$ is a point of maximum of f on $A{\setminus } (x_*-{\varepsilon }_n,x_*+{\varepsilon }_n)$. Let $x_n \in A{\setminus } [x_*-{\varepsilon }_n,x_*+{\varepsilon }_n]$ be such that $f(x_n) = \sup _{x \in A{\setminus } (x_*-{\varepsilon }_n,x_*+{\varepsilon }_n)} f(x)$, which exists by the continuity of f and compactness of the set $A{\setminus } (x_*-{\varepsilon }_n,x_*+{\varepsilon }_n)$. Since $f(x_*-{\varepsilon }_n) \leqslant f(x_n) \leqslant f(x_*)$ for all n, and f is continuous, it follows that $f(x_n) \rightarrow f(x_*)$. If $x_{n_k}$ is a convergent subsequence of $x_n$ converging to some $y \in A$, then by continuity of f, we have $f(y) = f(x_*)$. This implies that $y = x_*$. Therefore, there exists k such that $x_{n_k} \in (x_*-\delta ,x_*+\delta ){\setminus } \{x_*\}$ and ${\varepsilon }_{n_k} < \delta$. For this k, we have $f(x_{n_k})< \max \{f(x_*-{\varepsilon }_{n_k}), f(x_*+{\varepsilon }_{n_k})\}$. This contradicts the fact that $x_{n_k}$ maximizes f on the set $A{\setminus } (x_*-{\varepsilon }_{n_k},x_*+{\varepsilon }_{n_k})$, completing the proof of Lemma B.11. $\square$