Proof. Since \(\mu_0(p)\geq 0\), \(\left|\sum_{j=1}^mz_j\right|^2\leq \left(\sum_{j=1}^m \sqrt{t_j}\right)^2\) for \(t_j\equiv |z_j|^2\). Thus, it suffices to prove the following bound on the probabilities \(\{t_j\}\): \[\label{eq:acute-ineq-t} -\sum_{j=1}^m t_j \log_2 t_j -\mu_0(p)\left(\sum_{j=1}^m \sqrt{t_j}\right)^2\geq -\mu_1(p)\tag{5}\] Theorem 2 of [2] does exactly this. Considering the minimizers of the left hand side of 5 , [2] use Lagrange multipliers to show that every extrema must have at most two coordinate values, reducing 5 to a soluble single variable inequality in \(q\) with one additional discrete multiplicity parameter. ◻

Proof. Suppose \(z \in \R^m\). Since \(\|z\|_2=1\), \(\{t_j\}_{j\in[m]}\) is a probability distribution for \(t_j\equiv |z_j|^2\) as before. Now, consider its \(\alpha\)-Rényi entropy: \[H_\alpha(t)=\frac{1}{1-\alpha}\log_2 \sum_{j=1}^mt_j^\alpha\] For \(\alpha \in (0,1)\), Theorem 4 gives that \(\sum_{j=1}^{m}t_j^\alpha=\|x\|_{2\alpha}^{2\alpha}\geq M(m,\alpha)\). Then, \[H_\alpha(t)\geq \frac{1}{1-\alpha}\log_2 M(m,\alpha)\xrightarrow{\alpha\to1^-} -\sum_{j=1}^m t_j \log_2 t_j \geq \lim_{\alpha \to 1^-}\frac{\log_2 M(m,\alpha)}{1-\alpha}\] By [3], \(d(\alpha)\in (6,7)\) for \(\alpha\) sufficiently close to \(1\). Thus, for \(m\leq 6\), \(M(m,\alpha)= 2^{1-\alpha}\) which proves this branch of Theorem 3. For \(m\geq 7\), by L’Hôpital’s rule, \[\begin{align} \lim_{\alpha \to 1^-}\frac{\log_2 M(m,\alpha)}{1-\alpha}&=-\frac{d}{d\alpha}\log_2 (m^{-\alpha}((m-1)^\alpha+(m-1)^{1-\alpha}))\bigg|_{\alpha=1}\\ &=\log_2 m-\frac{(m-1)^\alpha-(m-1)^{1-\alpha}}{(m-1)^\alpha+(m-1)^{1-\alpha}}\log_2(m-1)\bigg|_{=1}\\ &=\log_2 m -\frac{m-2}{m}\log_2(m-1) \end{align}\] as desired, proving Theorem 3 for \(z\in \R^m\). We now argue that this gives the claim for \(z\in \C^m\). Let \(z=x+iy\) for \(x,y\in \R^m\), and let \(C_m\) denote the right hand side of ?? . Letting \(\lambda\equiv \|x\|_2^2\in (0,1)\), we can write: \[|z_j|^2=\lambda \frac{x_j^2}{\lambda}+(1-\lambda)\frac{y_j^2}{1-\lambda}\] where \(\{x_j^2/\lambda\}_{j\in [m]}\) and \(\{y_j^2/(1-\lambda)\}_{j\in[m]}\) are probability distributions. Then, by the concavity of Shannon entropy, \[-\sum_{j=1}^m|z_j|^2 \log_2 |z_j|^2 \geq -\lambda \sum_{j=1}^m \frac{x_j^2}{\lambda^2}\log_2 \frac{x_j^2}{\lambda^2}-(1-\lambda) \sum_{j=1}^m \frac{y_j^2}{\lambda^2}\log_2 \frac{y_j^2}{\lambda^2}\geq \lambda C_m+(1-\lambda) C_m=C_m\] where we use the fact that \(\frac{x}{\sqrt \lambda},\frac{y}{\sqrt{1-\lambda}}\in \R^m\), have unit \(\ell^2\) norm, and have zero sum, meaning Theorem 3 for real variables applies. Of course, if \(\lambda \in \{0,1\}\), this is exactly the real case. This argument yields the proof for complex variables and is a simple specialization of a broader real reduction trick given in Lemma 1 of [2], which we state as Lemma 1 in our later proof of the obtuse case. ◻

Proof. This follows from direct computation and standard calculus. ◻

Proof. Let the zero coordinate be \(x_k=0\). Then, \(\langle x,e_k\rangle=0\) meaning the unit vector \(e_k\) is tangent to \(\S^{m-1}\) and by first-order optimality, \[0=\nabla_{e_k}f_{{\tilde \nu_0}}(x)=-\cancel{2x_k (\ln x_k^2 +1)}+2{\tilde \nu_0} S(x)\] where we use the convention \(0\ln 0\equiv 0\). Then, Lemma 2 gives: \[0={\tilde \nu_0} S(x)=x_j(\ln x_j^2 + 1+\lambda)\] meaning every \(x_j\in \{0,\pm s\}\) for some \(s>0\). This along with the fact that \(S(x)=0\) implies that there are the same number of positive and negative coordinates, say \(r\) of each. Then, \(1=\|x\|_2^2=2rs^2\), meaning: \[f_{{\tilde \nu_0}}(x)=-2r\cdot s^2\ln s^2 +{\tilde \nu_0} \cdot 0^2 =\ln(2r)\] Since we seek minimizers, we have that \(r=1\), forcing \(s=2^{-1/2}\) and \(m\geq 3\) as well. ◻

Proof. By Lemma 2, every \(x_j\) is a non-zero root of \(g(t)\equiv t\ln t^2 + (1+\lambda )t +b=0\) where \(b\equiv -{\tilde \nu_0}S(x)\). Taking derivatives, \(g'(t)=\ln t^2 +\lambda+3\) and \(g''(t)=2/t\), meaning \(g\) is strictly convex on \((0,\infty)\), strictly concave on \((-\infty,0)\), with \(g'\) having at most one root on each domain, meaning \(g\) has at most two roots on each domain. We now refine this using the limiting behavior \(\lim_{t\to \pm \infty}g(t)=\pm \infty\) and \(\lim_{t\to 0^\pm}=b\).

Suppose that \(b\leq 0\) and that there are two positive roots \(0<s<t\). By Rolle’s theorem, there is a \(c\in (s,t)\) with \(g'(c)=0\), and since \(g''>0\) on \((0,\infty)\), \(g'<0\) on \((0,c)\), meaning \(g(s)<0\) since \(b\leq 0\). This contradicts the fact that \(s\) is a root, meaning there is at most one positive root if \(b\leq 0\). By a similar argument, if \(b\geq 0\), there is at most one negative root. ◻

Proof. We first show that \(x\) has at least one positive coordinate. Suppose every \(x_j<0\). Then, since \(S(x)<0\), by Lemma 4, there is at most one negative root, meaning \(x_j=t=-m^{-1/2}\) since \(\|x\|_2=1\) as well. Since \(m\geq 2\), there exists a \(y\in \R^m\) such that \(\langle x,y\rangle =0\) and \(S(y)=0\). Then, by Lemma 2, \(\ln t^2 + 1+\lambda ={\tilde \nu_0}S(x)/t={\tilde \nu_0}m\) meaning: \[\frac{1}{2} y^\top \nabla^2_x \mathcal{L} _{{\tilde \nu_0}}(x;\lambda)y=-(\ln t^2+3+\lambda)\|y\|_2^2 =-({\tilde \nu_0}m+2)\|y\|_2^2<0\] which contradicts the Hessian condition in Lemma 2. Thus, \(x\) has at least one positive coordinate, and by negation, this proves it must have at least one negative coordinate as well.

WLOG, by negation, let \(S(x)\geq 0\). We now show that no positive value can occur in two or more coordinates of such \(x\). For example, suppose \(x_i=x_k=s>0\) for indices \(i\neq k\). Then, defining \(y\equiv e_i-e_k\), \(\langle x,y\rangle=0\) and \(S(y)=0\). Then, by Lemma 2, \(\ln s^2 +1+\lambda ={\tilde \nu_0} S(x)/s\), meaning: \[\frac{1}{2} y^\top \nabla^2_x \mathcal{L} _{{\tilde \nu_0}}(x;\lambda)y=-2(\ln s^2+3+\lambda)=-2\left(\frac{{\tilde \nu_0} S(x)}{s}+2\right)<0\] which contradicts the Hessian condition in Lemma 2.

Together with Lemma 4, these facts prove the claim. ◻

Proof. It suffices to show that if \(x\) has three distinct values, \(v\) in Lemma 5 has multiplicity \(1\). Define \(r\equiv s_1/s_0\) and \(c\equiv {\tilde \nu_0} S(x)/s_0\). Then, Lemma 2 gives \(\ln s_0^2 +1 +\lambda =-c\) and \(\ln s_1^2 +1+\lambda =-c/r\). Subtracting and re-arranging, \[\label{eq:obtuse-c40r41} c=\frac{2r\ln r }{r-1}\tag{7}\] Now, suppose \(s_1\) has multiplicity at least two, meaning there exist distinct indices \(i,k\in [m]\) such that \(x_i=x_k=-s_1\). Then, let \(y=e_i-e_k\) which satisfies \(\langle x,y\rangle=0\) and \(S(y)=0\). Then, \[\frac{1}{2} y^\top \nabla^2_x \mathcal{L} _{{\tilde \nu_0}}(x;\lambda)y=-2(\ln s_1^2 + 3+\lambda)=2(c/r-2)=4\left(\frac{\ln r }{r-1}-1\right)<0\] since \(\ln r < r-1\) for \(r>1\). This contradicts the Hessian condition in Lemma 2, meaning \(v\) can not have multiplicity at least two. ◻

Proof. We do this by first parameterizing this family as \(x=s_0\cdot (-1,\dots,-1,-r,s)\) where \(r\equiv s_1/s_0>1\) as before, \(s\equiv s_2/s_0>0\), and \(k\equiv m-2\geq 1\).

Since \(\|x\|_2=1\), we have that \(s_0=(k+r^2+s^2)^{-1/2}\), and \(S(x)=s_0(s-k-r)\) by summation. Taking \(c\equiv {\tilde \nu_0} S(x)/s_0\) as in the proof of Lemma 6, 7 gives: \[\label{eq:obtuse-c040r44s41} {\tilde \nu_0}=\frac{c}{s-k-r}=\frac{2r\ln r }{(r-1)(s-k-r)}\tag{8}\] Now, 7 for \(r>1\) gives \(c>0\), meaning \(s>k+r\) since \(c,{\tilde \nu_0},s_0>0\) implies \(S(x)>0\). Applying Lemma 2, we have \(\ln s_0^2 + 1+\lambda=-c\) and \(\ln s_2^2+1+\lambda=c/s\). Subtracting, \[\label{eq:obtuse-s40r41} \frac{s\ln s}{1+s}=\frac{c}{2} =\frac{r\ln r }{r-1}\tag{9}\] where we equate with 7 . Thus, our family is defined by a single parameter \(r>1\) where \(s\) is given by 9 , \({\tilde \nu_0}\) is forced by 8 , and \(s>k+r\) is necessary so that \(S(x),{\tilde \nu_0}>0\).

At this point, we seek to construct a \(y\in \R^m\) such that \(\langle x,y\rangle=0\) with \(y^\top \nabla^2_x \mathcal{L}_{{\tilde \nu_0}}(x;\lambda)y<0\) and \(\nabla \mathcal{L}_{{\tilde \nu_0}}(x;\lambda)=0\), the existence of which would exclude all minimizers from the three value family via Lemma 2. In particular, we show that there exists such a \(y\) that is uniform on each coordinate group. Let \(y=(\alpha,\dots,\alpha,\beta,\gamma)\in \R^m\) such that: \[\label{eq:obtuse-gamma-elim} 0=\langle x,y\rangle =-ks_0\alpha-rs_0\beta+ss_0\gamma \Rightarrow \gamma=\frac{k\alpha+r\beta}{s}\tag{10}\] Recall that \(\nabla \mathcal{L}_{{\tilde \nu_0}}(x;\lambda)=0\) gives: \[\label{eq:obtuse-three-gradient-zero} \ln s_0^2 + 1+\lambda=-c,\quad \ln s_1^2 +1 +\lambda =-c/r,\quad \ln s_2^2+1+\lambda=c/s\tag{11}\] Noting that the diagonal term in the Hessian expression in ?? is the left hand side of 11 up to an additive constant of \(+2\), \[\label{eq:obtuse-hessian-M} \begin{align} y^\top \nabla_x^2 \mathcal{L}_{{\tilde \nu_0}}(x;\lambda)y&=-2(k\alpha^2 (2-c)+\beta^2 (2-c/r)+\gamma^2 (2+c/s))+2{\tilde \nu_0}(k\alpha+\beta +\gamma)^2\\ &=M_{11}\alpha^2+2M_{12}\alpha\beta +M_{22}\beta^2 \end{align}\tag{12}\] where substituting 10 and grouping terms gives: \[\begin{align} M_{11} &\equiv k h_0+\frac{k^2}{s^2}h_2+2{\tilde \nu_0} k^2\left(1+1/s\right)^2\\ M_{12} &\equiv \frac{kr}{s^2}h_2+2{\tilde \nu_0} k\left(1+1/s\right)\left(1+r/s\right)\\ M_{22} &\equiv h_1+\frac{r^2}{s^2}h_2+2{\tilde \nu_0}\left(1+r/s\right)^2 \end{align}\] with \(h_0\equiv -2(2-c)\), \(h_1\equiv -2(2-c/r)\), and \(h_2\equiv -2(2+c/s)\). Since 12 is exactly a quadratic form of the matrix \(M\), it suffices to show that \(\det M<0\). Substituting 8 , expanding, and factoring, we arrive at the following expression which one can verify by direct expansion: \[\det M =M_{11}M_{22}-M_{12}^2 =-\frac{4k(k+r^2+s^2)}{rs^3(s-k-r)}P_{k}(r,s,c)\] where \[\begin{align} P_k(r,s,c)&\equiv-c^2 k-c^2 r^2-c^2 s^2+2ckr-2cks-2cr^2s+2cr^2+2crs^2+2cs^2+4krs+4r^2 s-4rs^2\\ &=P_1(r,s,c)+(k-1)(2r-c)(c+2s) \end{align}\] Since \(s>k+r\), \(r>1\), and \(k\geq 1\), \(\det M <0\) is equivalent to \(P_k(r,s,c)>0\). By 7 , \(r>1\) implies \(\ln r<r-1\) meaning \(c>0\) and \(2r-c=2r(1-\ln (r)/(r-1))>0\), so \((k-1)(2r-c)(c+2s)>0\). Thus, it suffices to show \(P_1(r,s,c)>0\).

We now re-parameterize to derive a scalar inequality involving parameters restricted to bounded domains. Specifically, let \(\sigma=1/r\in (0,1)\) and \(t=s/r>1\), though one can show \(t\in (1,t_0)\) for some unique \(t_0\). Then, 9 becomes: \[\label{eq:obtuse-sigma-t} \frac{rt(\ln r+\ln t)}{1+rt}=\frac{r\ln r }{r-1}\Rightarrow t(r-1)\ln t =(1+rt-t(r-1))\ln r \Rightarrow \frac{t\ln t }{1+t}=\frac{\ln r }{r-1}=-\frac{\sigma \ln \sigma}{1-\sigma }\equiv \kappa\tag{13}\] From 7 , we also have that \(c=2\kappa/\sigma\). Substituting this parameterization into \(P_1(r,s,c)\), \[\label{eq:obtuse-K} \begin{align} P_1(r,s,c)&=-\frac{4}{\sigma^4 }K(\sigma,t)\\ K(\sigma,t)&\equiv \kappa^2(\sigma^2+t^2+1)+\kappa(\sigma^2(t-1)-\sigma(t^2+1)-t(t-1))+\sigma t(t-\sigma-1) \end{align}\tag{14}\] Thus, it suffices to show that \(K(\sigma,t)<0\) on the curve 13 .

We now use yet another re-parameterization to reduce this to a simple algebraic inequality. Let \(\alpha\equiv \kappa/t\) and \(\beta \equiv \kappa/\sigma\). We first claim that \(0<\alpha<\kappa<1<\beta\) and \(\alpha e^\alpha=\kappa e^{-\kappa}=\beta e^{-\beta}\). Manipulating, \[\kappa=\frac{t\ln t}{1+t}\Rightarrow \alpha=\frac{\ln t }{1+t}\Rightarrow \ln (\kappa/\alpha)=\alpha(1+\kappa/\alpha)\Rightarrow \alpha e^\alpha=\kappa e^{-\kappa}\] Similarly, \[\kappa=-\frac{\sigma\ln \sigma}{1-\sigma}\Rightarrow \beta=-\frac{\ln \sigma}{1-\sigma}\Rightarrow -\ln(\kappa/\beta)=\beta(1-\kappa/\beta)\Rightarrow \kappa e^{-\kappa}=\beta e^{-\beta}\] Since \(\sigma \in (0,1)\), \(\beta =-\frac{\ln \sigma}{1-\sigma}>1\), and \(\alpha < \kappa\) since \(t>1\), which concludes our initial claim. Substituting this parameterization into 14 , one can verify by direct expansion that: \[K(\sigma,t)=K(\kappa/\beta,\kappa/\alpha)=\frac{\kappa^2}{\alpha^2\beta^2} \alpha(1+\alpha)\kappa(1-\kappa)\beta(\beta-1)\left(\frac{1}{\kappa(1-\kappa)}-\frac{1}{\alpha(1+\alpha)}-\frac{1}{\beta(\beta-1)}\right)\] Noting that the entire pre-factor is positive, it suffices to show that: \[\frac{1}{\kappa(1-\kappa)}<\frac{1}{\alpha(1+\alpha)}+\frac{1}{\beta(\beta-1)}\] We prove in Lemma 20 that this is indeed a neat algebraic fact for any \(0<\alpha<\kappa<1<\beta\) with \(\alpha e^\alpha=\kappa e^{-\kappa}=\beta e^{-\beta}\). ◻

Proof. From Lemma 6, \(x=(-\sqrt {q'},\cdots,-\sqrt {q'},\sqrt {p'})\) for some \(p',q'>0\). Then, \(\|x\|_2=1\) implies \(q'=(1-p')/(m-1)\), and \(S(x)\geq 0\) implies \(\sqrt{p'}\geq (m-1)\sqrt{q'}\), meaning \(p'\geq (m-1)/m\). Letting \(r'\equiv \sqrt{p'/q'}\geq m-1\), we have: \[\label{eq:obtuse-pqr} p'=\frac{r'^2}{r'^2+m-1},\quad q'=\frac{1}{r'^2+m-1}\tag{15}\] meaning: \[\begin{align} -\sum_{j=1}^m x_j^2 \ln x_j^2 &=-p'\ln p' -(m-1)q'\ln q' =\ln (r'^2+m-1)-\frac{2r'^2\ln r'}{r'^2+m-1}\equiv H_m(r')\\ S(x)^2&=(\sqrt {p'}-(m-1)\sqrt {q'})^2=\frac{(r'-m+1)^2}{r'^2+m-1}\equiv R_m(r') \end{align}\] We now argue that \(H_m\) as a function of \(R_m\) is convex. Differentiating, one can verify that: \[H_m'(r')=-\frac{4(m-1)r' \ln r' }{(m-1+r'^2)^2},\quad R_m'(r')=\frac{2(m-1)(r'-m+1)(r'+1)}{(m-1+r'^2)^2}>0\] Then, one can also verify that: \[\begin{align} \label{eq:obtuse-two-HmRm-slope} \frac{dH_m}{dR_m}&=-\frac{2r' \ln r' }{(r'-m+1)(r'+1)}\\ \frac{d}{dr'}\left(\frac{dH_m}{dR_m}\right)&=\frac{2N_m(r')}{(r'-m+1)^2(r'+1)^2} \end{align}\tag{16}\] where \(N_m(r')\equiv (m-1+r'^2)\ln r' -(r'-m+1)(r'+1)\). We now show that \(r' \geq m-1\) implies this numerator is non-negative. \[\begin{align} N_m(m-1)&=m(m-1)\ln (m-1) \geq 0\\ N_m'(m-1)&=2r' \ln r' +\frac{m-1+r'^2}{r' }-(r'+1)-(r'-m+1)\bigg|_{r'=m-1}=2(m-1)\ln (m-1)\geq 0\\ N_m''(r')&=2\ln r' +1-\frac{m-1}{r'^2}\geq 0\quad \end{align}\] Thus, \(N_m'(r')\geq 0\) for \(r' \geq m-1\), and \(N_m(r')\geq 0\) for \(r' \geq m-1\) as well. Since \(R_m'(r')>0\) and \(\frac{d}{dr'}(dH_m/dR_m)\geq 0\), it follows that \(dH_m/dR_m\) is increasing as a function of \(R_m\), and is thus convex in \(R_m\).

Applying convexity at the point \(r'=r\), by the slope in 16 , \[\label{eq:obtuse-two-convex-at-r} H_m(r')\geq H_m(r)+\frac{dH_m}{d R_m}\bigg|_{r'=r}(R_m(r')-R_m(r))=H_m(r)-\tilde \nu_1(p)(R_m(r')-R_m(r))\tag{17}\] where one can verify the last equality by substituting \(r'=r\equiv \sqrt{p/q}\) into 16 . Re-arranging 17 , it follows that: \[\label{eq:obtuse-two-convex-bound} \begin{align} f_{{\tilde \nu_0}(p)}(x)&=-\sum_{j=1}^mx_j^2\ln x_j^2 +{\tilde \nu_0}(p)S(x)^2=H_m(r')+{\tilde \nu_0}(p)R_m(r')\\ &\geq H_m(r)+{\tilde \nu_0}(p)R_m(r)\\ &=-p\ln p -(m-1)q\ln q +\frac{\sqrt{pq}(\ln p-\ln q)}{(\sqrt p-(m-1)\sqrt q)(\sqrt p+\sqrt q)}(\sqrt p-(m-1)\sqrt q)^2\\ &={\tilde \nu_1}(p) \end{align}\tag{18}\] where one can verify the last equality. It is immediate that equality exclusively occurs for \(r'=r\). ◻

Proof. From Lemmas 3, 6, 7, and 8, we have that the two relevant competing local minima are \((2^{-1/2},-2^{-1/2},0,\cdots,0)\), which attains \(f_{{\tilde \nu_0}(p)}(x)=\ln 2\), and \((-\sqrt q,\cdots,-\sqrt q,\sqrt p)\), which gives \(f_{{\tilde \nu_0}(p)}(x)= {\tilde \nu_1}(p)\) for any \(p> (m-1)/m\). From 17 and 18 , \[\label{eq:thm-obtuse-59-c1} \begin{align} {\tilde \nu_1}(r)&=f_{{\tilde \nu_0}(r)}(x)=H_m(r)+{\tilde \nu_0}(r) R_m(r)=H_m(r)-\frac{dH_m}{dR_m}(r)R_m(r)\\ {\tilde \nu_1}'(r)&={\tilde \nu_0}'(r)R_m(r) \end{align}\tag{19}\] However, since the proof of Lemma 8 implies that \(R_m(r)>0\) and \(\frac{d}{dr}(\frac{dH_m}{dR_m}(r))> 0\) for \(p>(m-1)/m\), we have that \({\tilde \nu_1}\) is decreasing in \(r\), meaning it is decreasing in \(p\), since the analogous relation of 15 for \(p\) and \(r\) implies that \(p\) increases with \(r\). Now, from Equation (66) of [2], we have that \({\tilde \nu_1}(p(m))=\ln 2\). This immediately gives the claim about ?? for \(m\leq 6\). For \(m\geq 7\), taking \(p\to^+(m-1)/m\) or equivalently \(r\to^+m-1\), from 19 , \[\lim_{r\to (m-1)^+}{\tilde \nu_1}(r)\leq H_m(m-1)=\ln m -\frac{m-2}{m}\ln(m-1)\equiv \ln 2-g(m)\] Differentiating, \[g'(m)=\frac{2(m-1)\ln(m-1)-m}{m^2(m-1)}>0\] since the numerator is clearly positive and increasing for \(m\geq 7\). Since \(g(7)>0\) as well by computation, \(g(m)>0\). Since \({\tilde \nu_1}(r)\) is decreasing, it follows that \({\tilde \nu_1}(p)<\ln 2\) for all \(p\in ((m-1)/m,1]\) with \(m\geq 7\), proving the claim about ?? for \(m\geq 7\).

Simplifying ?? for \(p=p(m)\) with \(m\leq 6\), we precisely obtain ?? and its claimed minimizers. ◻

Proof. Separating the coordinates by sign, WLOG, we can write \[x=(u_1,\dots,u_n,-v_1,\dots,v_m,0,\dots,0)\] where \(n+m\leq d\) and \(u_i,v_j\geq 0\) for \(i\in [n],j\in [m]\). For convenience, let \(s\equiv S(u)=S(v)\). By Fact [fact:xp-subadd], \(f(x)=x^{2\alpha}\) is sub-additive. Then, \[\begin{align} \|x\|_{2\alpha}^{2\alpha}=\sum_{i\in [n]}u_i^{2\alpha}+\sum_{i\in [m]}v_i^{2\alpha}\geq S(u)^{2\alpha}+S(v)^{2\alpha}=2s^{2\alpha} \end{align}\] From the \(\|x\|_2=1\) constraint, \[\begin{align} 1&=\|x\|_2^2=\sum_{i\in [n]}u_i^2 +\sum_{i\in [m]}v_i^2\leq S(u)^2+S(v)^2=2s^2 \end{align}\] Then, \(\|x\|_{2\alpha}^{2\alpha}\geq 2s^{2\alpha}\geq 2(1/2)^{\alpha}=2^{1-\alpha}\). ◻

Proof. When \(d=2\), \(d^{-\alpha}((d-1)^\alpha+(d-1)^{1-\alpha})=2^{1-\alpha}\), meaning \(M(2,\alpha)=2^{1-\alpha}\). Then, \(S(x)=0\) gives \(x_2=-x_1\), meaning \(\|x\|_2^2=1\) implies \(x_1=2^{-1/2}\). Thus, \(\|x\|_{2\alpha}^{2\alpha}=2x_1^{2\alpha}=2^{1-\alpha}\). ◻

Proof. This follows by direct computation and standard calculus. ◻

Proof. Let \(k\in [d]\) such that \(x_k=0\). Then, from Lemma 11, \(\p_k \c L(x;\lambda_1,\lambda_2)=0\) gives \(\lambda_1=0\). Then, \[0=\p_j \c L(x;\lambda_1,\lambda_2)=2x_j(\alpha |x_j|^{2(\alpha-1)}-\lambda_2)\] meaning \(x_j\in\{0,\pm s\}\) for some \(s>0\). Since \(\|x\|_2^2=1\) and \(S(x)=0\), there are the same number of positive and negative coordinates, say \(r\geq 1\) of each. Then, \(1=\|x\|_2^2=2rs^2\), meaning \[\|x\|_{2\alpha}^{2\alpha}=2r s^{2\alpha}=(2r)^{1-\alpha}\] Then, \(r=1\) minimizes this value for \(\alpha \in (0,1)\) and maximizes it for \(\alpha >1\). Since \(1=2rs^2\), \(s= 2^{-1/2}\), proving the claim. ◻

Proof. Since the form in Lemma 12 lies within the claimed family, it suffices to prove the claim for \(x\) with no zero coordinate. From Lemma 11, there exists a \(\lambda_1,\lambda_2\in \R\) such that each \(x_j\) is a root of the function \[f(t;\lambda_1,\lambda_2)\equiv 2\alpha t |t|^{2(\alpha-1)}-2\lambda_2 t-\lambda_1\] We now claim that \(f(t)\) has at most three non-zero roots, which is equivalent to showing that \(g(t)=\pm \lambda_1\) has at most three total solutions where \(g(t)=2\alpha t^{2\alpha-1}-2\lambda_2 t\) for \(t>0\). Taking a derivative, \[g'(t)=2\alpha (2\alpha-1)t^{2\alpha-2}-2\lambda_2\] If \(\lambda_2\leq 0\), then \(g'(t)>0\) for \(t>0\), meaning \(g(t)=|\lambda_1|\) and \(g(t)=-|\lambda_1|\) each have at most one solution. Supposing \(\lambda_2>0\), \(g'(t)\) has at most one root for \(t>0\). Thus, if \(\alpha \in (1/2,1)\), \(g\) increases from \(0\) to a maximum and then tends to \(-\infty\), in which case \(g(t)=|\lambda_1|\) has at most two solutions while \(g(t)=-|\lambda_1|\) has at most one. If \(\alpha >1\), \(g\) decreases from \(0\) to a minimum and then tends to \(+\infty\), in which case \(g(t)=|\lambda_1|\) has at most one solution while \(g(t)=-|\lambda_1|\) has at most two.

Thus, there are at most three total solutions in each case, meaning \(f(t;\lambda_1,\lambda_2)\) has at most three non-zero roots and thus \(x\) has at most three distinct coordinate values. The claimed family follows by taking \(-x\) instead of \(x\) if \(x\) has two distinct negative values. ◻

Proof. This follows from the fact that the proof of the Equal Variables Theorem in Chapter 5 of [12] is proved by way of contradiction, showing that any \(x\) not of the claimed form can be replaced with an \(x\) of the form that achieves a strictly better value of \(S_n\). ◻

Proof. From Lemma 12, it suffices to consider \(x\) with no zero coordinate. Then, from Lemma 13, we have that \[x=(-s_0,\dots,-s_0,s_1,\dots,s_1,s_2,\dots,s_2)\] where \(0< s_1\leq s_2\), \(s_0>0\), and \(s_i\) is repeated \(k_i\) times, for \(i\in \{0,1,2\}\). We seek to apply Theorem 5 on the positive coordinates to show that \(k_2=1\), and we begin by defining the positive block and negative block to be the following vectors: \[\begin{align} x^+&=(s_1,\dots,s_1,s_2,\dots,s_2)\in \R^{d-k_0}\\ x^-&=(s_0,\dots,s_0)\in \R^{k_0} \end{align}\] Since \(S(x)=0\) and no coordinates are zero, \(k_0\geq 1\), so fix the negative block by choosing \(s_0\in (0,1]\) and \(k_0\in [d]\). Then, the zero sum and \(L^2\) constraint give the following feasible set over the positive block. \[\begin{align} {\c S}(s_0,k_0)&\equiv \left\{x^+\in \R^{d-k_0}:x^+_i> 0,\quad \sum_{i=1}^{d-k_0}x^+_i=k_0s_0,\quad \sum_{i=1}^{d-k_0}(x^+_i)^2=1-k_0s_0^2\right\} \end{align}\] Then, \[\begin{align} \inf_{x\in {\c S}}\|x\|_{2\alpha}^{2\alpha}&=\inf_{\substack{ s_0\in(0,1],\, k_0\in [d]\\ {\c S}(s_0,k_0)\neq\emptyset }}\left(k_0s_0^{2\alpha}+\inf_{x^+\in {\c S}(s_0,k_0)}\|x^+\|^{2\alpha}_{2\alpha}\right)\quad (1/2<\alpha<1)\\ \sup_{x\in {\c S}}\|x\|_{2\alpha}^{2\alpha}&=\sup_{\substack{ s_0\in(0,1],\, k_0\in [d]\\ {\c S}(s_0,k_0)\neq\emptyset }}\left(k_0s_0^{2\alpha}+\sup_{x^+\in {\c S}(s_0,k_0)}\|x^+\|^{2\alpha}_{2\alpha}\right)\quad (\alpha > 1) \end{align}\]

First, consider \(\alpha \in (1/2,1)\) and the infimum over the positive block for any choice of \(s_0,k_0\) such that \({\c S}(s_0,k_0)\neq \emptyset\). Suppose \(d-k_0\geq 3\). We seek to apply Theorem 5 for \(f(x)=-x^{2\alpha}\), \(k=2\), and any \(a\in {\c S}(s_0,k_0)\) in order to maximize \(S_{d-k_0}=-\|x^+\|_{2\alpha}^{2\alpha}\). We verify that \(g''(x)=f^{(3)}(x)=-2\alpha(2\alpha-1)(2\alpha-2)x^{2\alpha-3}>0\) for \(\alpha \in (1/2,1)\), meaning \(g\) is strictly convex. \(f\) is clearly continuous at \(0\) and differentiable on \((0,\infty)\). The conditions on \(a\) are satisfied by construction of the feasible set \({\c S}(s_0,k_0)\), and the entries of \(x^+\) are positive and non-decreasing since \(s_2\geq s_1\). Then, by the second statement of Theorem 5, the maximum of \(S_{d-k_0}\), or equivalently \(\inf_{x^+\in {\c S}(s_0,k_0)}\|x^+\|_{2\alpha}^{2\alpha}\), is attained only by \(x^+\) of the form: \[x^+=(x^+_1,\dots,x_1^+,x_{d-k_0}^+)\] where \(0< x_{1}^+\leq x_{d-k_0}^+\). If \(d-k_0=2\), by permutation, \(x^+=(x_1^+,x_{2}^+)\) where \(x_{2}^+\geq x_1^+> 0\), and if \(d-k_0=1\), \(x^+=(x_{d-k_0}^+)\). Thus, regardless of \(d-k_0\), \(x^+_{i}=x_j^+\) for any \(i,j\in [1,d-k_0-1]\), and \(x^{+}_i\leq x^+_{d-k_0}\) for \(i \in [1,d-k_0]\).

Next, consider \(\alpha > 1\) and the supremum over the positive block for any choice of \(s_0,k_0\) such that \({\c S}(s_0,k_0)\neq \emptyset\). Suppose \(d-k_0\geq 3\). We again seek to apply Theorem 5, this time for \(f(x)=x^{2\alpha}\), \(k=2\), and any \(a\in {\c S}(s_0,k_0)\) in order to maximize \(S_{d-k_0}=\|x^+\|_{2\alpha}^{2\alpha}\). We verify that \(g''(x)=2\alpha(2\alpha-1)(2\alpha-2)x^{2\alpha-3}>0\) for \(\alpha > 1\), meaning \(g\) is strictly convex, and the rest of the conditions follow in the same manner as the \(\alpha \in (1/2,1)\) case. Then, by the second statement of Theorem 5, the maximum of \(S_d\), or equivalently \(\sup_{x^+\in {\c S}(s_0,k_0)}\|x^+\|_{2\alpha}^{2\alpha}\), is attained only by \(x^+\) of the form: \[x^+=(x^+_1,\dots,x_1^+,x_{d-k_0}^+)\] where \(0< x_{1}^+\leq x_{d-k_0}^+\). Note that this is the same form as the \(\alpha \in (1/2,1)\) case, and by the same argument as before, the \(d-k_0\in \{1,2\}\) cases are subsumed by this.

Then, identifying \(s_1\equiv x_1^+\) and \(s_2\equiv x_{d-k_0}^+\), we see that \(k_2\leq 1\). Since this form holds on the positive block for any valid choice of \((s_0,k_0)\), the claim follows, where we use the fact that \(s_1=s_2\) is permitted to compress the \(k_2=0\) case. ◻

Proof. Since the form in Lemma 12 achieves \(\|x\|_{2\alpha}^{2\alpha}=2^{1-\alpha}\), it suffices to consider \(x\) with no zero coordinate. From Lemma 14, \[\begin{align} \|x\|_{2\alpha}^{2\alpha}\geq \inf \{k_0s_0^{2\alpha}+(d-k_0-1)s_1^{2\alpha}+s_2^{2\alpha}:k_0s_0=(d-k_0-1)s_1+s_2,\, k_0s_0^2+(d-k_0-1)s_1^2+s_2^2=1\} \end{align}\] where \(0< s_1\leq s_2\), \(0< s_0\), and \(2\leq k_0\leq d-1\). By Lemma [fact:lp-monotonicity] for \(p=2\alpha<2=q\), \[\begin{align} \left((d-k_0-1)s_1^{2\alpha}+s_2^{2\alpha}\right)^{1/2\alpha}\geq \left((d-k_0-1)s_1^2+s_2^2\right)^{1/2}=(1-k_0s_0^2)^{1/2} \end{align}\] Then, we can lower bound the objective as \[k_0s_0^{2\alpha}+(d-k_0-1)s_1^{2\alpha}+s_2^{2\alpha}\geq k_0s_0^{2\alpha}+(1-k_0s_0^2)^{\alpha}\] This motivates the substitution \(u=k_0s_0^2> 0\) such that the right hand side is \(k_0^{1-\alpha}u^{\alpha}+(1-u)^{\alpha}\). We now determine the feasible set of \(u\). From the two constraints, \[\begin{align} 1-u=1-k_0s_0^2=(d-k_0-1)s_1^2+s_2^2\leq ((d-k_0-1)s_1+s_2)^2=(k_0s_0)^2=k_0u\Rightarrow u\geq \frac{1}{k_0+1} \end{align}\] Further, by Cauchy-Schwarz, \[\begin{align} k_0u&=((d-k_0-1)s_1+s_2)^2\leq (d-k_0)((d-k_0-1)s_1^2+s_2^2)=(d-k_0)(1-u)\Rightarrow u\leq \frac{d-k_0}{d} \end{align}\] Thus, \[\begin{align} \|x\|_{2\alpha}^{2\alpha}&\geq \inf \{k_0s_0^{2\alpha}+(d-k_0-1)s_1^{2\alpha}+s_2^{2\alpha}:k_0s_0=(d-k_0-1)s_1+s_2,\, k_0s_0^2+(d-k_0-1)s_1^2+s_2^2=1\}\\ &\geq \inf \left\{f(u;k_0):\frac{1}{k_0+1}\leq u\leq \frac{d-k_0}{d},\,2\leq k_0\leq d-1\right\} \end{align}\] where \(f(u;k_0)=k_0^{1-\alpha}u^\alpha+(1-u)^{\alpha}\). Now, \[\begin{align} f''(u;k_0)&= k_0^{1-\alpha}\alpha(\alpha-1)u^{\alpha-2}+\alpha(\alpha-1)(1-u)^{\alpha-2}<0 \end{align}\] on \(u\in (0,1)\), since \(\alpha < 1\). Thus, \(f(u;k_0)\) is strictly concave on \((0,1)\), meaning \[\begin{align} \|x\|_{2\alpha}^{2\alpha}&\geq \inf \left\{f(u;k_0):\frac{1}{k_0+1}\leq u\leq \frac{d-k_0}{d},\,2\leq k_0\leq d-1\right\}\\&=\min_{2\leq k_0\leq d-1}\min \left \{f\left(\frac{1}{k_0+1};k_0\right),f\left(\frac{d-k_0}{d};k_0\right)\right\} \end{align}\] We now directly calculate the minimum of each branch. For \(t\in [2,d-1]\), \[\begin{align} f_l(t)&\equiv f\left(\frac{1}{t+1};t\right)=\frac{t^{\alpha}+t^{1-\alpha}}{(t+1)^{\alpha}}\\ f_r(t)&\equiv f\left(\frac{d-t}{d};t\right)=d^{-\alpha}(t^{1-\alpha}(d-t)^\alpha+t^\alpha) \end{align}\] Since \(d\geq 3\) and \(k_0\geq 2\), by Lemmas 23 and 22, we are done. ◻

Proof. As in Lemma 15, it suffices to consider \(x\) with no zero coordinate, meaning we are considering the form \((-s_0,s_1,\dots,s_1,s_2)\) such that \(0< s_1\leq s_2\) and \(s_0 > 0\). Writing \(t\equiv s_2/s_1\geq 1\), the constraints become: \[\begin{align} 0&=-s_0+(d-2)s_1+s_2\Rightarrow s_0=(d-2+t)s_1\\ 1&=s_0^2 +(d-2)s_1^2+s_2^2=(d-2+t)^2s_1^2+(d-2)s_1^2+t^2s_1^2\Rightarrow s_1^2=\frac{1}{(d-2+t)^2 +d-2+t^2 } \end{align}\] Then, the objective becomes \[\begin{align} s_0^{2\alpha}+(d-2)s_1^{2\alpha}+s_2^{2\alpha}&=((d-2+t)^{2\alpha}+(d-2)+t^{2\alpha})s_1^{2\alpha}=\frac{(d-2+t)^{2\alpha}+(d-2)+t^{2\alpha}}{((d-2+t)^{2} +d-2+t^2 )^\alpha}\equiv f(t) \end{align}\] Thus, \(\|x\|_{2\alpha}^{2\alpha}\geq \min_{t\geq 1}f(t)\). By Lemma 24, this minimum is exactly the claim lower bound. ◻

Proof. We follow an analogous argument to Lemma 15 and consider \(x\) with no zero coordinate. By Lemma [fact:lp-monotonicity] for \(p=2<2\alpha=q\), \[\begin{align} ((d-k_0-1)s_1^{2\alpha}+s_2^{2\alpha})^{1/(2\alpha)}\leq ((d-k_0-1)s_1^2+s_2^2)^{1/2}=(1-k_0s_0^2)^{1/2} \end{align}\] Then, by the same substitution \(u=k_0s_0^2\) and an analogous argument, \[\begin{align} \|x\|_{2\alpha}^{2\alpha}\leq \max_{2\leq k_0\leq d-1}\max \left\{f\left(\frac{1}{k_0+1};k_0\right),f\left(\frac{d-k_0}{d};k_0\right)\right\} \end{align}\] where \(f(u;k_0)=k_0^{1-\alpha}u^\alpha+(1-u)^\alpha\) as before. Then, by Lemma 23 and Lemma 22, which is valid for \(d\geq 4\) and \(k_0\geq 2\), we have that these maximums give exactly the claimed upper bound. ◻

Proof. By an analogous argument to Lemma 16, we can consider \(x\) with no zero coordinate and substitute \(t=s_2/s_1\geq 1\), yielding: \[\|x\|_{2\alpha}^{2\alpha}\leq \max_{t\geq 1}\frac{(d-2+t)^{2\alpha}+(d-2)+t^{2\alpha}}{((d-2+t)^2+d-2+t^2)^\alpha}\] By Lemma 24, this maximum is exactly the claimed upper bound. ◻

Proof. For any \(\alpha \in (1/2,1)\cup(1,\infty)\), by Lemma 14, any global extremum with no zero coordinate has the form: \[x=(-s_0,\dots,-s_0,s_1,\cdots,s_1,s_2)\] where \(s_0\geq 0\) and \(0\leq s_1\leq s_2\), where \(s_i\) occurs \(k_i\) times for \(i\in \{0,1\}\).

For \(d\geq 4\), by Lemma 17 and 18, we are done for \(\alpha > 1\), and by Lemmas 15 and 16, we are done for \(\alpha \in (1/2, 1)\) as well. Now, for \(d=3\), since \(k_2=1\), \(k_0\in \{1,2\}\). If \(k_0=1\), then Lemmas 16 and 18 together finish the lower and upper bounds. If \(k_0=2\), then \(x=(-s_0,-s_0,s_2)\), where \(s_2=2s_0\) from the zero sum constraint. Then, \[\begin{align} 1&=\|x\|^2_2=6s_0^2 \\ \|x\|_{2\alpha}^{2\alpha}&=(2+2^{2\alpha})s_0^{2\alpha}=(2+2^{2\alpha})6^{-\alpha}=3^{-\alpha}((3-1)^\alpha+(3-1)^{1-\alpha}) \end{align}\] which satisfies the claimed bound as well as the equality case for \(d=3\) and \(k_0=2\).

We now consider the equality case more fully. Since Lemma 12 handles any global extremum with a zero coordinate, consider a global extremum, \(x\), with no zero coordinate that saturates the bound. For \(d\geq 4\) and \(k_0\geq 2\), the \(\ell^p\)-monotonicity from Lemma [fact:lp-monotonicity] used on the positive block in Lemmas 15 and 17 must then be an equality: \[(d-k_0-1)s_1^{2\alpha}+s_2^{2\alpha}= ((d-k_0-1)s_1^2+s_2^2)^\alpha\] which only occurs if there is at most one non-zero value in the positive block. Since \(x\) has no zero coordinate, we must have the form \(x=(-s_0,\dots,-s_0,s_2)\). Since \(\|x\|_2=1\) and \(S(x)=0\) imply \(s_2=(d-1)s_0\) and \(d(d-1)s_0^2=1\), the claimed non-zero value equality family in Theorem 4 follows. For \(d\geq 3\) and \(k_0=1\), Lemma 24 used in both Lemmas 16 and 18 states that equality occurs only for \(t=1\) or \(t\to\infty\) where \(t=s_2/s_1\), unless \((d,\alpha)=(3,2)\) in which case all \(t\geq 1\) saturate the bound. Discarding \(t\to \infty\) as it corresponds to \(s_1\to 0\) though \(x\) has no zero coordinate, \(t=1\) gives \(x=(-s_0,\dots,-s_0,s_2)\) and thus the non-zero desired equality family. The exception \((d,\alpha)=(3,2)\) is derivative of the fact that every \(x\in \R^3\) with \(S(x)=0\) and \(\|x\|_2=1\) has \(\|x\|_4^4=1/2\). We can see this with the following manipulation: \[\begin{align} x_1x_2+x_2x_3+x_1x_3&=\frac{S(x)^2-\|x\|_2^2}{2}=-1/2 \\ x_1^2x_2^2+x_2^2x_3^2+x_1^2x_3^2&=(x_1x_2+x_2x_3+x_1x_3)^2-2x_1x_2x_3S(x)=1/4\\ \|x\|_4^4 &=\|x\|_2^4 -2(x_1^2x_2^2+x_2^2x_3^2+x_1^2x_3^2)=1/2 \end{align}\] This proves every case of Theorem 4. ◻

Proof. We proceed by first upper bounding \(\alpha\) and \(\beta\) in terms of \(x\) and then proving the resulting single variable inequality.

We begin with \(\beta\), observing that \(xe^{-x}=\beta e^{-\beta}\) is equivalent to saying that the logarithmic mean of \(x\) and \(\beta\) is \(1\). Recall that \(L^3\geq G^2A\) where \(L, G, A\) denote the logarithmic, geometric, and arithmetic mean, with equality attained only when the two inputs are equal [22]. Then, by the quadratic formula, \[\label{lemma:obtuse-alg-beta-ub} x\beta \frac{x+\beta}{2}<1\Rightarrow x\beta^2 +x^2\beta -2<0 \Rightarrow \beta < \frac{-x+\sqrt{x^2+8/x}}{2}\equiv B(x)\tag{20}\] We now continue to \(\alpha\), and claim that: \[\label{lemma:obtuse-alg-alpha-ub} \alpha < \frac{x}{1+2x+x^3/2}\equiv A(x)\tag{21}\] Since \(t\mapsto te^t\) is strictly increasing, it suffices to show that \(A(x)e^{A(x)}>xe^{-x}\), or equivalently: \[G(x)\equiv x+\frac{x}{1+2x+x^3/2}-\ln (1+2x+x^3/2)>0\] Differentiating, \[G'(x)=1+\frac{1+2x+x^3/2-x(2+3x^2/2)}{(1+2x+x^3/2)^2}-\frac{2+3x^2/2}{1+2x+x^3/2}=\frac{x^2(x-1)(x^3-2x^2+6x-10)}{(x^3+4x+2)^2}\] where one can verify the last equality by direct expansion. Now, consider the polynomial \(p(x)\equiv x^3-2x^2+6x-10\), whose derivative is \(p'(x)=3x^2-4x+6>3-4+6>0\) on \((0,1)\). Since \(p(1)=-5\), \(p(x)\) is negative on \((0,1)\), meaning \(G'(x)>0\) on \((0,1)\). Since \(G(0)=0\) as well, \(G(x)>0\) on \((0,1)\), proving 21 . Applying the upper bounds in 20 and 21 , one can verify by direct expansion that: \[\label{eq:obtuse-alg-ratx} \begin{align} \frac{1}{\alpha(1+\alpha)}+\frac{1}{\beta(\beta-1)}-\frac{1}{x(1-x)}&>\frac{1}{A(x)(1+A(x))}+\frac{1}{B(x)(B(x)-1)}-\frac{1}{x(1-x)}\\ &=\frac{x^2(Q(x)-x^{3/2}\sqrt{x^3+8}\,P(x))}{x(x-1)(x^{3/2}-\sqrt{x^3+8})(x^{3/2}+2\sqrt x-\sqrt{x^3+8})(x^3+6x+2)} \end{align}\tag{22}\] where \[\label{eq:obtuse-alg-PQ} \begin{align} P(x)&\equiv x^5+7x^3+4x^2+10x+14\\ Q(x)&\equiv x^8+7x^6+8x^5+10x^4+42x^3+8x^2+40x-8 \end{align}\tag{23}\] Analyzing the denominator of 22 , the first and last terms are positive on \((0,1)\) while the second and third terms are negative. We also claim that the fourth is negative: \[x^{3/2}+2\sqrt x<\sqrt{x^3+8}\Leftrightarrow x^{3}+4x^2+4x<x^3+8\Leftrightarrow x(x+1)<2\] which is clearly true on \((0,1)\). Thus, the denominator of 22 is negative, meaning it suffices to show that the numerator is negative, or equivalently: \[\label{eq:obtuse-alg-PQ-ineq} Q(x)<x^{3/2}\sqrt{x^3+8}\, P(x)\tag{24}\] Now, since \(P(x)>0\) from 23 , if \(Q(x)<0\), we are done since the numerator is immediately negative. Thus, suppose \(Q(x)\geq 0\), meaning proving 24 is equivalent to proving its square: \[\label{eq:obtuse-alg-PQ-ineq2} 0<x^3(x^3+8)P(x)^2-Q(x)^2=16(1-x)^2T(x)\tag{25}\] where \(T(x)=x^6+2x^5+20x^3-24x^2+32x-4\), which can be verified by direct expansion. From 23 , \(Q'(x)>0\) on \((0,1)\) and \(Q(1/7)=-11502504/5764801<0\), meaning \(x>1/7\). Then, \[T'(x)=6x^5+10x^4+60x^2-48x+32=6x^5+10x^4+60(x-2/5)^2+112/5>0\] meaning \(T(x)>0\) on \((0,1)\), since \(T(1/7)=16479/117649>0\) as well. This proves 25 , and thus the lemma. ◻

Proof. We desire to show that for \(x,y\geq 0\), \((x+y)^p\leq x^p+y^p\). Taking \(t=y/x\geq 0\), WLOG, it suffices to prove that \((1+t)^p\leq 1+t^p\). Let \(f(t)=1+t^p-(1+t)^p\). Then, \[\begin{align} f'(t)=p(t^{p-1} -(1+t)^{p-1})\geq 0 \end{align}\] since \(t\mapsto t^{p-1}\) is a decreasing function for \(p\in (0,1]\). Since \(f(0)=0\), \(f(t)\geq 0\) for all \(t\geq 0\), and we are done. ◻

Proof. By Hölder’s inequality, since \(q/p>1\) and \(q/(q-p)>1\) are valid conjugate exponents, \[\begin{align} \|x\|_p^p=\sum_{i=1}^{n}|x_i|^p\cdot 1\leq \left(\sum_{i=1}^{n}|x_i|^{p\cdot q/p}\right)^{p/q}\left(\sum_{i=1}^{n}1^{q/(q-p)}\right)^{(q-p)/q}=\|x\|_q^{p}n^{1-p/q} \end{align}\] which proves the upper bound after taking \((\cdot)^{1/p}\) on both sides. For the lower bound, let \(y=x/\|x\|_p\) such that \(\|y\|_p=1\), meaning \(|y_i|\leq 1\) for each \(i\in [n]\). Then, since \(p<q\), \[\begin{align} \|y\|_q^q&=\sum_{i=1}^{n}|y_i|^q\leq \sum_{i=1}^{n}|y_i|^p=\|y\|_p^p=1 \end{align}\] Thus, \(\|x\|_q=\|x\|_p\|y\|_q\leq \|x\|_p\), proving the lower bound. ◻

Proof. Let \(L(\alpha)=2^{1-\alpha}\) and \(R(\alpha,d)=d^{-\alpha}((d-1)^\alpha+(d-1)^{1-\alpha})\). Consider \(d=2\). \[\begin{align} R(\alpha,2)&=2^{-\alpha}((2-1)^\alpha+(2-1)^{1-\alpha})=2^{1-\alpha}=L(\alpha) \end{align}\] meaning \(d=2\) is a root of \(D(d)\). We proceed by substituting \(t\equiv (d-1)^{-1}\in (0,1]\). Then, \[\begin{align} R(\alpha,d)&=(1+t^{-1})^{-\alpha}(t^{-\alpha}+t^{\alpha-1})=(1+t)^{-\alpha}(1+t^{2\alpha-1})\equiv f(t) \end{align}\] Differentiating, \[\begin{align} f'(t)&=-\alpha(1+t)^{-\alpha-1}(1+t^{2\alpha-1})+(2\alpha-1)t^{2\alpha-2}(1+t)^{-\alpha} =(1+t)^{-\alpha-1}g(t) \end{align}\] where \[\begin{align} g(t)&\equiv (\alpha-1)t^{2\alpha-1}+(2\alpha-1)t^{2\alpha-2}-\alpha\\ g'(t)&=(\alpha-1)(2\alpha-1)t^{2\alpha-2}+(2\alpha-1)(2\alpha-2)t^{2\alpha-3}=(2\alpha-1)(\alpha-1)\underbrace{(t+2)t^{2\alpha-3}}_{>0} \end{align}\] We now break into cases for \(\alpha\).

Consider \(\alpha \in (1/2,1)\). Then, \(g'(t)<0\) on \((0,1]\). Further, \(g(1)=2\alpha-2<0\) and \[\begin{align} \lim_{t\to 0^+}g(t)&=\lim_{t\to0^+}\frac{2\alpha-1}{t^{2(1-\alpha)}}=\infty \end{align}\] which all together mean that \(g(t)\) has a unique zero at some \(t^*\in (0,1)\), at which it changes from positive to negative. Since \({\rm sign}(f'(t))={\rm sign}(g(t))\), it follows that \(f(t)\) has a unique maximum at \(t=t^*\). Now, evaluating the boundary, for \(\alpha \in (1/2,1)\), \[\begin{align} f(1)&=2^{1-\alpha}>1,\quad \lim_{t\to 0^+}f(t)=1 \end{align}\] Together with the fact that \(f'(t)\) has a unique zero in \((0,1)\) which yields a maximum of \(f(t)\), it follows that there exists a unique \(t'\in (0,1)\) such that \(f(t')=2^{1-\alpha}\), with \(t'\leq t^*\). Further, \(f'(t)>0\) on \((0,t^*)\) implies that \(f(t)<f(t')=2^{1-\alpha}\) for \(t<t'\). Finally, since \(f(t')=f(1)=2^{1-\alpha}\) and \(f'\) has a unique zero on \((t',1)\) that is a local maxima of \(f(t)\), it follows that \(f(t)\geq 2^{1-\alpha}\) for \(t \in (t',1)\). Thus, for \(t\in (0,1]\), \[\min\{2^{1-\alpha},f(t)\}=\begin{cases} f(t)& t < t'\\ 2^{1-\alpha} & t \geq t' \end{cases}\Rightarrow \min\{2^{1-\alpha},R(\alpha,d)\}=\begin{cases} R(\alpha,d) & d > d(\alpha)\\ 2^{1-\alpha} & d \leq d(\alpha) \end{cases}\] where \(d(\alpha)=1+1/t'\) is the unique root of \(D(d)\) in \((2,\infty)\). Since \(d=2\) is the only other relevant root, \(d(\alpha)\) is the largest root. This shows the claim for \(\alpha \in (1/2,1)\).

Next, consider \(\alpha>1\). Then, \(g'(t)>0\) on \((0,1]\), \(g(1)=2\alpha-2>0\), and \[\begin{align} \lim_{t\to 0^+}g(t)&=-\alpha<0 \end{align}\] which all together mean that \(g(t)\) has a unique zero at some \(t^*\in (0,1)\), at which it changes from negative to positive. This corresponds to a unique minimum of \(f(t)\) at \(t=t^*\). Recall that \(\lim_{t\to 0^+}f(t)=1\) and \(f(1)=2^{1-\alpha}<1\) for \(\alpha > 1\). Together with the fact that \(f'(t)\) has a unique zero on \((0,1)\) at which \(f(t)\) is a local minimum, it follows that there exists a unique \(t'\in (0,1)\) such that \(f(t')=2^{1-\alpha}\), with \(t'\leq t^*\). Since \(f'(t)<0\) on \((0,t^*)\), we have that \(f(t)>f(t')=2^{1-\alpha}\) for \(t<t'\). Finally, since \(f(t')=f(1)=2^{1-\alpha}\) and \(f'(t)\) has a unique zero on \((t',1)\) that is a local minima of \(f(t)\), it follows that \(f(t)\leq 2^{1-\alpha}\) for \(t\in (t',1)\). Thus, for \(t\in (0,1]\), \[\max\{2^{1-\alpha},f(t)\}=\begin{cases} f(t)& t < t'\\ 2^{1-\alpha} & t \geq t' \end{cases}\Rightarrow \max\{2^{1-\alpha},R(\alpha,d)\}=\begin{cases} R(\alpha,d) & d > d(\alpha)\\ 2^{1-\alpha} & d \leq d(\alpha) \end{cases}\] which proves the claim for \(\alpha > 1\). Thus, we are done. ◻

Proof. The first statement is a standard fact from calculus. The second statement follows by applying Rolle’s theorem on each interval \([r_i,r_{i+1}]\) for each \(i\in [n-1]\). ◻

Proof. Taking derivatives, \[\begin{align} f'(t)&=d^{-\alpha}((1-\alpha)t^{-\alpha}(d-t)^\alpha -\alpha t^{1-\alpha}(d-t)^{\alpha-1} +\alpha t^{\alpha-1} )\\ f''(t)&=d^{-\alpha}((1-\alpha)(-\alpha t^{-\alpha-1}(d-t)^\alpha-\alpha t^{-\alpha }(d-t)^{\alpha-1})\\ &-\alpha(1-\alpha)t^{-\alpha}(d-t)^{\alpha-1}+\alpha(\alpha-1)t^{1-\alpha}(d-t)^{\alpha-2}+\alpha(\alpha-1) t^{\alpha-2})\\ &=-\alpha(1-\alpha)d^{-\alpha}\underbrace{(t^{-\alpha-1}(d-t)^\alpha +t^{-\alpha}(d-t)^{\alpha-1}+t^{-\alpha}(d-t)^{\alpha-1}+t^{1-\alpha}(d-t)^{\alpha-2}+t^{\alpha-2})}_{>0} \end{align}\] on \(t\in [2,d-1]\). Thus, \(f''(t) < 0\) for \(\alpha \in (1/2,1)\) and \(f''(t)>0\) for \(\alpha >1\). Thus, \[\begin{align} \min_{t\in [2,d-1]}f(t)&=\min\{f(2),f(d-1)\}\quad (\alpha \in (1/2,1))\\ \max_{t\in [2,d-1]}f(t)&=\max \{f(2),f(d-1)\}\quad (\alpha > 1) \end{align}\] Now, notice that \(f(d-1)=d^{-\alpha}((d-1)^{1-\alpha}+(d-1)^\alpha)\). Thus, it suffices to show that \(f(2)\geq 2^{1-\alpha}\) for \(\alpha \in (1/2,1)\) and \(f(2)\leq 2^{1-\alpha}\) for \(\alpha > 1\).

We begin with the lower bound for \(\alpha \in (1/2,1)\). Manipulating, \[\begin{align} 2^{1-\alpha}&\leq f(2)=d^{-\alpha}(2^{1-\alpha}(d-2)^{\alpha}+2^{\alpha}) \\ \Leftrightarrow d^\alpha -(d-2)^\alpha &\leq 2^{2\alpha -1} \end{align}\] Now, consider \(h(d)=d^\alpha-(d-2)^\alpha\). Then, \[\begin{align} h'(d)=\alpha(d^{\alpha-1} -(d-2)^{\alpha-1})<0 \end{align}\] since \(t\mapsto t^{\alpha-1}\) is decreasing for \(\alpha < 1\). This means that \(\max_{d\geq 4}h(d)=h(4)=4^\alpha-2^\alpha\). Thus, it suffices to prove that \(4^\alpha-2^\alpha \leq 2^{2\alpha-1}\), or equivalently \(g(\alpha)=4^\alpha/2-2^\alpha \leq 0\). Now, for \(\alpha \in (1/2,1)\), \[\begin{align} g'(\alpha)&=4^\alpha\ln 2 -2^\alpha\ln2=(4^\alpha-2^\alpha)\ln 2 > 0 \end{align}\] Thus, \(\max_{\alpha \in (1/2,1)}g(\alpha)=g(1)=4/2-2=0\), meaning \(g(\alpha)\leq 0\) for all \(\alpha \in (1/2,1)\), which proves the lower bound case.

Now, consider the upper bound claim that \(f(2)\leq 2^{1-\alpha}\) for \(\alpha > 1\). By an analogous argument to before, it suffices to show that \(h(d)\geq 2^{2\alpha-1}\). Now, \(h'(d)=\alpha (d^{\alpha-1}-(d-2)^{\alpha-1})>0\) since \(t\to t^{\alpha-1}\) is increasing for \(\alpha > 1\). Then, \(\min_{d\geq 4}h(d)=h(4)\). Thus, by a similar manipulation as before, it suffices to show that \(g(\alpha)\geq 0\). Now, recall that \(g'(\alpha)=(4^\alpha-2^\alpha )\ln 2> 0\) for \(\alpha > 1\) as well. Thus, \(\min_{\alpha>1}g(\alpha)=g(1)=0\), proving the upper bound case. ◻

Proof. Evaluating the boundary, \[\begin{align} f(1)& =2^{1-\alpha}, \quad f(d-1) =d^{-\alpha}((d-1)^\alpha+(d-1)^{1-\alpha}) \end{align}\] which are exactly the left and right terms in the claimed equalities. Thus, it suffices to show that \(f(t)\) does not have a local minimum on \((1,d-1)\) for \(\alpha \in (1/2,1)\) and does not have a local maximum on \((1,d-1)\) for \(\alpha >1\). Differentiating, \[\begin{align} f'(t)&=-\alpha (t+1)^{-\alpha-1}(t^\alpha+t^{1-\alpha})+(t+1)^{-\alpha}(\alpha t^{\alpha-1}+(1-\alpha)t^{-\alpha}) =\underbrace{t^{-\alpha}(t+1)^{-\alpha-1}}_{>0}g(t) \end{align}\] where \(g(t)=\alpha t^{2\alpha-1}-(2\alpha-1)t+(1-\alpha)\). We now consider the sign of \(g(t)\), which is also the sign of \(f'(t)\). \[\begin{align} g(1)& =2(1-\alpha)\\ g'(t)&=\alpha(2\alpha-1)t^{2\alpha-2}-(2\alpha-1)= (2\alpha-1)(\alpha t^{2\alpha-2}-1) \end{align}\] For \(\alpha \in (1/2,1)\), \(g(1)> 0\) and \(g'(t)<0\) for \(t\geq 1\), since \(\alpha t^{2\alpha-2}\leq \alpha <1\). Then, \(g\) is positive then negative on \(t\geq 1\), as is \(f'\), meaning \(f\) does not have a local minimum. For \(\alpha > 1\), \(g(1)<0\) and \(g'(t)>0\), since \(\alpha t^{2\alpha -2}\geq \alpha > 1\). Then, \(g\) is negative then positive on \(t\geq 1\), as is \(f'\), meaning \(f\) does not have a local maximum. Thus, we are done. ◻

Proof. Evaluating the boundary, \[\begin{align} f(1)&=\frac{(d-1)^{2\alpha}+d-1}{((d-1)^2+d-1)^\alpha}=\frac{(d-1)((d-1)^{2\alpha-1}+1)}{d^{\alpha}(d-1)^{\alpha}}=d^{-\alpha}((d-1)^{\alpha}+(d-1)^{1-\alpha})\\ \lim_{t\to \infty}f(t)&=\lim_{t\to \infty}\frac{t^{2\alpha}+t^{2\alpha}}{(t^2 + t^2 )^\alpha}=2^{1-\alpha} \end{align}\] which are exactly the left and right terms in the claimed equalities. Thus, it suffices to show that \(f(t)\) does not have a local minimum on \((1,\infty)\) for \(\alpha \in (1/2,1)\) and also does not have a local maximum on \((1,\infty)\) for \(\alpha > 1\). As we will see shortly, we are able to do this with the single exception of \((d,\alpha)=(3,2)\), for which \(f(t)\) is constant. Taking derivatives, \[\begin{align} f'(t)&=\frac{((d-2+t)^{2}+d-2+t^2)^{\alpha}2\alpha((d-2+t)^{2\alpha-1}+t^{2\alpha-1})}{((d-2+t)^{2}+d-2+t^2)^{2\alpha}}\\ &\quad -\frac{((d-2+t)^{2\alpha}+(d-2)+t^{2\alpha})\alpha ((d-2+t)^{2}+d-2+t^2)^{\alpha-1}(2 (d-2+t)+2t)}{((d-2+t)^{2}+d-2+t^2)^{2\alpha}}\\ &=2\alpha ((d-2+t)^{2}+d-2+t^2)^{-\alpha-1}g(t) \end{align}\] where \[\label{eq:m1-min-g-h} \begin{align} g(t)&\equiv ((d-2+t)^{2}+(d-2)+t^2)((d-2+t)^{2\alpha-1}+t^{2\alpha-1})\\&-((d-2+t)^{2\alpha}+(d-2)+t^{2\alpha}) ( (d-2+t)+t) \\ &=(d-2)h(t;d-2,2\alpha-1) \end{align}\tag{26}\] where \(h(t;n,\beta)=(1-t)(n+t)^{\beta}+(n+t+1)t^{\beta}-(n+2t)\) with \(n\geq 1\), \(t\geq 1\), and \(\beta \in (0,1)\) or \(\beta>1\), depending on if \(\alpha \in (1/2,1)\) or \(\alpha>1\) respectively. The manipulation in 26 can be verified by direct expansion.

Since \(\alpha>0\) and \(d\geq 3\), we clearly see that \({\rm sign}(f')={\rm sign}(h)\). We then proceed to analyze the sign of \(h\) on \(t\in [1,\infty)\). We first make the change of variables \(x=1/t\in (0,1]\) such that \(h(t)=t^{\beta+1}j(1/t)\) where \[\begin{align} j(x)\equiv 1+(n+1)x-(1-x)(1+nx)^\beta -(2+nx)x^\beta \end{align}\] and \(j(0)=j(1)=0\). We desire to show that \(j(x)\) has at most one zero on \(x\in (0,1)\). Differentiating, \[\begin{align} j'(x)&=n+1+(1+nx)^\beta -\beta n(1-x)(1+nx)^{\beta-1} -nx^\beta -\beta(2+nx)x^{\beta-1} \\ j''(x)&=\beta n(1+nx)^{\beta-1}+\beta n (1+nx)^{\beta-1} -\beta(\beta-1)n^2 (1-x)(1+nx)^{\beta-2} -n\beta x^{\beta-1}-n\beta x^{\beta-1}\\ &\quad -\beta(\beta-1)(2+nx)x^{\beta-2}\\ &=\beta x^{\beta-1}k(1/x) \end{align}\] where \[\begin{align} k(t)&\equiv 2n(n+t)^{\beta-1}+(\beta-1)n^2 (1-t)(n+t)^{\beta-2}-(\beta-1)(n+2t)-2n\\ &=n(n+t)^{\beta-2}A(t;n,\beta)-B(t;n,\beta) \end{align}\] with \(A(t;n,\beta)=(2-n(\beta-1))t+n(\beta+1)\) and \(B(t;n,\beta)=2(\beta-1)t+n(\beta+1)\), all of which one can verify by direct expansion. We first show that \(j''(x)\) has at most one zero on \(x\in (0,1)\) by showing that \(k(t)\) has at most one zero on \(t\in (1,\infty)\). Now, consider \(k(t)=0\), or equivalently, \[\begin{align} 1=\ell(t)\equiv \frac{B(t;n,\beta)}{A(t;n,\beta)n(n+t)^{\beta-2}} \end{align}\] where we divide the two terms in \(k(t)\). Then, \[\begin{align} \p_t \ln \ell(t)&=\frac{2(\beta-1)}{2(\beta-1)t+n(\beta+1)} \\&\quad -\frac{n(\beta-2)(n+t)^{\beta-3}((2-n(\beta-1))t+n(\beta+1))+n(n+t)^{\beta-2}(2-n(\beta-1))}{n(n+t)^{\beta-2}((2-n(\beta-1))t+n(\beta+1))}\\ &=\frac{(\beta-1)m(t)}{(n+t)A(t;n,\beta)B(t;n,\beta)} \end{align}\] where \[m(t)=2(\beta-2)(n(\beta-1)-2)\cdot t^2 +n(\beta+1)((n-2)(\beta-1)+2)\cdot t+n^2(\beta+1)(n+2-\beta)\] which one can verify by direct expansion.

We now split into cases for \(\beta\). Consider \(\beta\in (0,1)\). We claim that \(k(t)\) has at most one zero on \((1,\infty)\). Since \(t>1\), \(n\geq 1\), and \(\beta \in (0,1)\), \[\begin{align} A(t;n,\beta)=\underbrace{(2-n(\beta-1))}_{\geq 0}t+n(\beta+1)>0 \end{align}\] Then, if \(B(t;n,\beta)\leq 0\), \(k(t)>0\), meaning there are no zeros. Thus, instead consider \[t\in \mathcal{D}_B=\{t:B(t;n,\beta)>0\}\] Now, notice that the coefficient of \(t^2\) in \(m(t)\), \(2(\beta-2)(n(\beta-1)-2)\), is positive for \(\beta\in (0,1)\). Further, one can check that the discriminant of the quadratic \(m(t)\) factors: \[\begin{align} \Delta&\equiv (n(\beta+1)((n-2)(\beta-1)+2))^2-4\cdot 2(\beta-2)(n(\beta-1)-2)\cdot n^2(\beta+1)(n+2-\beta)\\ &=n^2(\beta+1)(\beta-3)(\beta(n+2)-5n-4)(\beta(n+2)-n-4) \end{align}\] Since \(n\geq 1\) and \(\beta \in (0,1)\), the first two factors are both positive while the rest are negative, meaning \(\Delta < 0\). Thus, \(m(t)>0\) everywhere. Then, for \(t\in \mathcal{D}_B\), \(A,B,m(t)>0\) and \(\beta-1<0\), meaning \(\ell(t)>0\) and \(\p_t\ln \ell(t)<0\). Since \(\p_t \ln \ell(t)=\ell'(t)/\ell(t)\), it follows that \(\ell'(t)<0\), meaning there is at most \(1\) value of \(t\) where \(\ell(t)=1\), or equivalently \(k(t)=0\).

Now, since \(k(t)\) has at most one zero on \((1,\infty)\), \(j''(x)\) has at most one zero on \((0,1)\). Now, suppose \(j(x)\) had two distinct zeros \(r_1,r_2\in (0,1)\) with \(r_1<r_2\). Then, since \(j(0)=j(1)=0\), by Rolle’s theorem, \(j'(x)\) must have zeros at some \(r_1'\in (0,r_1),r_2'\in (r_1,r_2),r_3'\in (r_2,1)\). Applying Rolle’s theorem again, \(j''(x)\) must have a zero at some \(r''_1\in (r_1',r_2')\) and \(r''_2\in (r_2',r_3')\), which contradicts the fact that \(j''(x)\) has at most one zero in \((0,1)\). Thus, \(j(x)\) has at most one zero on \((0,1)\), meaning \(h(t;n,\beta)\) has at most one zero on \((1,\infty)\). Then, since \(\beta <1\), \[\begin{align} \lim_{t\to \infty}h(t;n,\beta)=\lim_{t\to\infty}((1-t)(n+t)^\beta+(n+t+1)t^\beta -(n+2t))=\lim_{t\to \infty}(-2t+O(t^\beta))=-\infty \end{align}\] meaning \(h\) is either always negative or positive then negative, as are \(g(t)\) and \(f'(t)\). Thus, \(f(t)\) is either always decreasing or increasing and then decreasing, meaning it does not have a local minimum for \(t\in (1,\infty)\). This finishes the \(\alpha\in (1/2,1)\), or equivalently \(\beta \in (0,1)\), case.

Now, consider \(\beta > 1\), with \(n\geq 2\) for now. We first desire to show that \(k(t)\) has at most one zero on \((1,\infty)\).

Since \(t>1, n\geq 2,\) and \(\beta >1\), \[\begin{align} B(t;n,\beta)=2(\beta-1)t+n(\beta+1)>0 \end{align}\] Suppose \(A\leq 0\). Then, \(k(t)=n(n+t)^{\beta-2}A-B<0\), meaning \(k(t)\) has no zero in this regime. Thus, instead consider \[t \in \mathcal{D}_A=\{t:A(t;n,\beta)>0\}\] on which we show that \(m(t)>0\), by splitting into cases.

Suppose \(\beta\in (1,1+2/n]\subset (1,2]\) such that \(n(\beta-1)\leq 2\). Then, since \(n\geq 2\) as well, \[\begin{align} m(t)=2\underbrace{(\beta-2)}_{\leq 0}\underbrace{(n(\beta-1)-2)}_{\leq 0}\cdot t^2 +n(\beta+1)\underbrace{((n-2)(\beta-1)+2)}_{>0}\cdot t+n^2(\beta+1)\underbrace{(n+2-\beta)}_{>0}>0 \end{align}\] Now, suppose \(n(\beta-1)> 2\), or equivalently \(\beta >1+2/n\), and further suppose that \(\beta \geq 2\). Then, \[\begin{align} \mathcal{D}_A=\{t:A(t;n,\beta)=(2-n(\beta-1))t+n(\beta+1)>0\}=(1,t_A) \end{align}\] where \(t_A=\frac{n(\beta+1)}{n(\beta-1)-2}\). Then, \[\begin{align} m'(t)&=\underbrace{4(\beta-2)(n(\beta-1)-2)}_{\geq 0}+\underbrace{n(\beta+1)((n-2)(\beta-1)+2)}_{>0}>0\\ m(1)&=2(\beta-2)(n(\beta-1)-2)+n(\beta+1)((n-2)(\beta-1)+2)+n^2(\beta+1)(n+2-\beta)\\ &=(n+1)(\beta(n^2-4)+n^2+8)\\ &>0 \end{align}\] where one can check that \(m(1)\) factors as such. Then, \(m(1)>0\) and \(m'(t)>0\) for \(t\in(1,t_A)\) imply that \(m(t)>0\) for \(t\in (1,t_A)\).

Finally, suppose \(n(\beta-1)>2\) and \(\beta \in (1,2)\), meaning \(n\geq 3\) and \(\mathcal{D}_A=(1,t_A)\) as before. Then, \[\begin{align} m''(t)&=4(\beta-2)(n(\beta-1)-2)<0\\ m(1)&=(n+1)(\beta(n^2-4)+n^2+8)>0\\ m(t_A)&=\frac{2(\beta-2)n^2(\beta+1)^2}{n(\beta-1)-2}+\frac{n^2(\beta+1)^2((n-2)(\beta-1)+2)}{n(\beta-1)-2}+n^2(\beta+1)(n+2-\beta)\\ &=\frac{n^2(\beta+1)(n+1)}{n(\beta-1)-2}(2(\beta-1)+\underbrace{n(\beta-1)-2}_{>0})\\ &>0 \end{align}\] which one can verify by direct computation. Then, since \(m''(t)<0\) on \((1,t_A)\), \(m(t)\geq \min\{m(1),m(t_A)\}>0\) for any \(t \in \mathcal{D}_A=(1,t_A)\). Thus, in all cases, \(m(t)>0\). Then, for \(t\in \mathcal{D_A}\), \(A,B,m(t),(\beta-1)>0\), meaning \[\p_t \ln \ell(t)=\frac{(\beta-1)m(t)}{((n+t)AB)}>0,\quad \ell(t)=\frac{B}{An(n+t)^{\beta-2}}>0\Rightarrow \ell'(t)=\ell(t)\p_t\ln \ell(t)>0\] which means that \(\ell(t)=1\Leftrightarrow k(t)=0\) has at most one solution for \(t\in \mathcal{D}_A\). Thus, \(k(t)\) has at most one zero on \((1,\infty)\). By the same argument via Rolle’s theorem as before, \(h(t)\) has at most one zero on \((1,\infty)\). We next claim that \(h(t)\) can not be positive then negative.

Suppose \(n\geq 3\). For \(t=1\), \[\begin{align} h(1;n,\beta)&=(1-t)(n+t)^\beta +(n+t+1)t^\beta-(n+2t)\bigg|_{t=1}=0\\ h'(1;n,\beta)&=-(n+t)^\beta +\beta(1-t)(n+t)^{\beta-1}+t^\beta + \beta (n+t+1) t^{\beta-1}-2\bigg|_{t=1}\\ &=\beta(n+2)-(n+1)^\beta-1\\ h'(1;n,1)&=0\\ \p_\beta h'(1;n,1)&=n+2-\ln(n+1)(n+1)^\beta \bigg|_{\beta=1}\\&=n(1-\ln(n+1))+(2-\ln(n+1))\\ &\leq (1-\ln 4)n+(2-\ln 4)\\ &<0\\ \p_\beta^2 h'(1;n,\beta)&=-\ln^2(n+1)(n+1)^\beta \\&< 0 \end{align}\] Then, since \(\p_\beta^2 h'(1;n,\beta)<0\), \(\p_\beta h'(1;n,1)<0\), and \(h'(1;n,1)=0\), It follows that \(h'(1;n,\beta)<0\) for all \(\beta > 1\). Then, since \(h(1;n,\beta)=0\), there exists \(t'\) such that \(h(t;n,\beta)<0\) for all \(t\in (1,t')\). Thus, \(h(t)\) starts negative and thus can not start positive then negative.

Next, suppose \(n=2\). Suppose further that \(\beta \geq 2\). Then, \[\begin{align} h'(1;2,\beta)&=\beta(n+2)-(n+1)^\beta -1 \bigg|_{n=2}=4\beta-3^\beta -1<0 \end{align}\] since \(3^\beta > 4\beta\) for \(\beta \geq 2\). Then, since \(h(1;n,\beta)=0\), as argued before, \(h(t;2,\beta)\) starts negative and thus can not start positive and then negative.

Finally, suppose \(n=2\) and \(\beta \in (1,2)\). Then, \[\begin{align} \lim_{t\to \infty}h(t;2,\beta)&=\lim_{t\to \infty}((1-t)(2+t)^\beta +(t+3)t^\beta -(2+2t))\\ &=\lim_{t\to \infty}(t^\beta ((1-t)(1+2/t)^\beta +t+3))-O(t))\\ &=\lim_{t\to \infty}(t^\beta ((1-t)(1+2\beta/t+O(t^{-2})) +t+3))-O(t))\\ &=\lim_{t\to \infty}(t^\beta (4-2\beta )+O(t))\\ &=\infty \end{align}\] meaning \(h(t;2,\beta)\) ends positive, and thus can not be positive and then negative.

Thus, \(h(t)\) has at most one zero on \((1,\infty)\) and can not be positive and then negative. Thus, \(h(t)\) is either always positive or negative and then positive, as is \(g(t),f'(t)\), meaning \(f(t)\) does not have a local maximum on \((1,\infty)\). This finishes the \(\beta > 1\) case for \(n\geq 2\).

The final case is \(\beta > 1\), with \(n=1\). Then, one can verify by direct computation that: \[\begin{align} k(t)&=2n(n+t)^{\beta -1}+(\beta-1)n^2 (1-t)(n+t)^{\beta-2}-(\beta-1)(n+2t)-2n\bigg|_{n=1}\\ &=(1+t)^{\beta-2}((3-\beta)t+\beta+1)-(2(\beta-1)t+\beta+1)\\ k'(t)&=(\beta-2)(1+t)^{\beta-3}((3-\beta)t+\beta+1)+(3-\beta)(1+t)^{\beta-2}-2(\beta-1)\\ &=(\beta-1)(p(t)-2) \end{align}\] where \(p(t)\equiv (1+t)^{\beta-3}((3-\beta)t+\beta-1)\). Then, \[\begin{align} p'(t)&=(\beta-3)(1+t)^{\beta-4}((3-\beta)t+\beta-1)+(3-\beta)(1+t)^{\beta-3}\\ &=-(\beta-3)(\beta-2)(t-1)(1+t)^{\beta-4} \end{align}\] which one can also verify by direct computation. We now show that \(h(t)\) is either strictly positive or strictly negative on \((1,\infty)\) via casework on \(\beta\).

Suppose \(\beta \in (1,2]\). Then, \(p'(t)\leq 0\), since \(t> 1\), meaning \(k''(t)=(\beta-1)p'(t)\leq 0\). Then, \[\begin{align} p(1)&=2^{\beta-3}((3-\beta)+\beta-1)=2^{\beta -2 }<2\Rightarrow k'(1)=(\beta-1)(p(1)-2)<0 \end{align}\] Thus, \(k'(t)<0\) for \(t\geq 1\). Then, \[\begin{align} k(1;\beta )&=2^{\beta-2}((3-\beta)+\beta+1)-(2(\beta-1)+\beta+1)=2^\beta-3\beta+1\\ k(1;1)&=0\\ \p_\beta k(1;\beta)&=\ln(2)2^{\beta }-3 \end{align}\] Now, \(\p_\beta k(1;1)=0\) has solution \(\beta^*\equiv \log_2(3/\ln(2))\approx 2.11>2\), so for \(\beta \in (1,2]\), \(\p_\beta k(1;1)<0\), meaning \(k(1;\beta)<0\) for all \(\beta \in (1,2]\). Then, since \(k'(t)<0\) as well, \(k(t;\beta)<0\), meaning \(j''(x)<0\). Since \(j(0)=j(1)=0\) from before, this implies \(j(x)>0\) for \(x\in (0,1)\), meaning \(h(t)>0\) for \(t\in (1,\infty)\), as is \(g(t),f'(t)\), meaning \(f(t)\) can not have a local maximum on \((1,\infty)\).

Next, suppose \(\beta\in (2,3)\). Then, \(p'(t)>0\) for \(t>1\), \(p(1)=2^{\beta-2}<2\) still, and \[\lim_{t\to \infty}p(t)=\lim_{t\to \infty}(1+t)^{\beta-3}((3-\beta)t+\beta-1)=\infty\] Then, \(k'(t)=(\beta-1)(p(t)-2)\) has \(k''(t)>0\), \(k'(1)<0\), and \(\lim_{t\to \infty}k'(t)>0\). This means that \(k'(t)\) changes sign exactly once, from negative to positive. Now, consider \(k(1;\beta)=2^\beta-3\beta +1\), which we looked at in the previous case. Recall that \(\beta^*=\log_2(3/\ln(2))\) is the only extrema. \[\begin{align} k(1;3)&=k(1;1)=0\\ k(1,\beta^*)&=3/\ln(2)-3\log_2(3/\ln(2))+1\approx -1.01<0 \end{align}\] Now, \(\p_\beta k(1;\beta)=\ln(2)2^\beta-3\) must be negative for \(\beta < \beta^*\) and positive for \(\beta > \beta^*\). Then, since \(k(1,\beta^*)<0\), since \(k(1;3)=k(1;1)=0\), \(k(1;\beta)<0\) for all \(\beta \in (1,3)\). Then, since \(k(1)<0\) and \(k(t)\) decreases then increases, \(k(t)\) has at most one zero on \((1,\infty)\). Then, by the Rolle’s theorem argument from before, \(h(t)\) has at most one zero on \((1,\infty)\). Then, \[\begin{align} h(1;1,\beta)&=(1-t)(1+t)^\beta +(t+2)t^\beta -(1+2t)\bigg|_{t=1}=0\\ h'(1;1,\beta)&=-(1+t)^\beta +\beta(1-t)(1+t)^{\beta-1}+t^\beta +\beta (t+2)t^{\beta-1}-2\bigg|_{t=1}\\ &=-2^\beta +3\beta-1\\ &>0 \end{align}\] where we use the fact that we previously showed that \(k(1;\beta)=2^\beta -3\beta + 1<0\) for \(\beta > 1\). Further, \[\begin{align} \lim_{t\to \infty}h(t;1,\beta)&=\lim_{t\to \infty}((1-t)t^\beta(1+\beta/t+O(t^{-2}))+t^{\beta+1}+2t^{\beta}-1-2t)\\ &=\lim_{t\to \infty}(t^{\beta}(-t+1-\beta +O(t^{-1}))+t^{\beta+1}+2t^{\beta}-1-2t)\\ &=\lim_{t\to \infty}((3-\beta)t^{\beta}+O(t^{\beta-1}))\\ &=\infty \end{align}\] Since \(h(1)=0\), \(\lim_{t\to \infty}h(t)=\infty\), \(h'(1)>0\), and \(h(t)\) has at most one zero on \((1,\infty)\), it follows that \(h(t)\) is always positive on \((1,\infty)\).

Next, suppose \(\beta=3\). Then, \[\begin{align} h(t;1,3)&=(1-t)(1+t)^3 +(t+2)t^3 -(1+2t)\\ &=1+3t+3t^2+t^3 -t-3t^2-3t^3 -t^4 +t^4 +2t^3 -1-2t\\ &=0 \end{align}\] meaning \(g(t)=f'(t)=0\), which implies that \(f(t)\) can not have a local maximum on \((1,\infty)\).

Finally, suppose \(\beta>3\), and again we analyze the zeros of \(k(t)\). Then, \[\begin{align} k(t)=0\Leftrightarrow (1+t)^{\beta-2}((3-\beta)t+\beta+1)=2(\beta-1)t+\beta+1 \end{align}\] Observing that the right hand side and \((1+t)^{\beta-2}\) are both positive, \((3-\beta)t+\beta+1>0\) as well for the root, or equivalently \(t\in (1,t')\) where \(t'=(\beta+1)/(\beta-3)\). Now, for \(\beta > 3\) and \(t>1\), \(p'(t)<0\), and recall \(p(1)=2^{\beta-2}>2\). Further, \[\begin{align} p(t')&=\left(1+\frac{\beta+1}{\beta-3}\right)^{\beta-3}\left((3-\beta)\frac{\beta+1}{\beta-3}+\beta-1\right)=-2 \left(1+\frac{\beta+1}{\beta-3}\right)^{\beta-3}<0 \end{align}\] Thus, regarding \(k'(t)=(\beta-1)(p(t)-2)\), \(k'(1)>0\), \(k'(t')<0\), and \(k''(t)<0\) for \(t\in (1,\infty)\). Thus, \(k'(t)\) on \((1,t')\) changes sign exactly once, from positive to negative. Now, consider \(k(1;\beta)=2^{\beta}-3\beta +1\). As seen before, \(\p_\beta k(1;\beta)>0\) for \(\beta > \beta^*\approx 2.11\) and \(k(1;3)=0\). Thus, \(k(1;\beta)>0\). Then, since \(k(1)>0\) and \(k(t)\) is increasing then decreasing on \((1,t')\), \(k(t)\) has at most one zero in \((1,t')\), and \((1,\infty)\) for that matter, as does \(h(t)\) by the Rolle’s theorem argument. Then, recall that \(h(1;1,\beta)=0\), and \[h'(1;1,\beta)=-2^{\beta}+3\beta -1=-k(1;\beta)<0\] From the previous case, we also have that \(\lim_{t\to \infty}h(t;1,\beta)=\lim_{t\to \infty}((3-\beta)t^{\beta}+O(t^{\beta-1}))=-\infty\). Thus, if \(h(1)=0\), \(h'(1)<0\), \(\lim_{t\to \infty}h(t)=-\infty\), and \(h(t)\) has at most one zero on \((1,\infty)\), it follows that \(h(t)<0\) on \((1,\infty)\).

To summarize, we have that for \(\beta > 1\) and \(n=1\), \(h(t)\) is either strictly positive or strictly negative on \((1,\infty)\) as long as \(\beta \neq 3\). Then, this is also the case for \(g(t)\) and \(f'(t)\), meaning \(f(t)\) is monotone and thus does not admit local maxima in \((1,\infty)\). The only exception is \(\beta =3\), for which \(f'(t)=0\). This corresponds to the exceptional equality case for \((d,\alpha)=(3,2)\). ◻