June 01, 2026
The computation of the numerical index of classical Banach spaces is one of the original problems in the theory. In this paper, we compute the numerical index of two-dimensional real \(\ell_p\)-spaces for all \(p\ge1\). More precisely, we prove that \[n(\ell_p^2)=v(J), \qquad J= \begin{pmatrix} 0&1\\ -1&0 \end{pmatrix},\] confirming the conjectured formula in the two-dimensional real case.
The numerical index is a classical isometric invariant of Banach spaces, introduced to quantify the relation between the norm and the numerical radius of bounded linear operators. Let \(X\) be a Banach space over the real field, let \(X^*\) denote its dual space, and let \(\mathcal{L}(X)\) denote the space of all bounded linear operators from \(X\) into itself. For \(T\in\mathcal{L}(X)\), the numerical radius of \(T\) is defined by \[v(T) := \sup\bigl\{ |x^*(Tx)|: x\in X,\;x^*\in X^*,\;\|x\|=\|x^*\|=x^*(x)=1 \bigr\}.\] The numerical radius is a seminorm on \(\mathcal{L}(X)\) and satisfies \(v(T)\le \|T\|\) for every \(T\in\mathcal{L}(X)\). The numerical index of \(X\) is the number \[n(X) := \inf\bigl\{ v(T):T\in\mathcal{L}(X),\;\|T\|=1 \bigr\}.\]
The concept was introduced in the seminal work of Duncan, McGregor, Pryce and White [1], after a question of Lumer, and was developed systematically in the monographs of Bonsall and Duncan [2], [3]. One of the original problems in the theory was to compute the numerical index of classical \(L_p\)-spaces. In the real case, this problem remains open for \(1<p<\infty\), \(p\ne2\). In the two-dimensional case, it was conjectured that the numerical index of \(\ell_p^2\) is attained by the rotation by \(90^\circ\), which corresponds to the operator \[J= \begin{pmatrix} 0&1\\ -1&0 \end{pmatrix}.\] In this paper, we prove the conjecture.
Theorem 1. For every \(p\ge1\), \[n(\ell_p^2)=v(J).\]
The cases \(p=1\) and \(p=2\) are classical. Moreover, as recalled in [4], the range \(1<p<2\) reduces by duality to the range \(p>2\). Thus, in the proof, we always work under the assumption \(p>2\).
Several general facts about the numerical index are classical. In the real case, Hilbert spaces of dimension at least two have numerical index zero, while \(L_1(\mu)\)-spaces and \(C(K)\)-spaces have numerical index one. For \(1<p<\infty\), \(p\ne2\), the situation is much more subtle. The exact value of \(n(L_p(\mu))\) is not known in general, and even finite-dimensional cases have required separate arguments.
For \(1<p<\infty\), set \(M_p:=v(J)\) on \(\ell_p^2\). For the two-dimensional problem, Martín and Merí [5] proved general estimates for \(n(\ell_p^2)\). If \(q=p/(p-1)\) is the conjugate exponent, then \[\max\{2^{-1/p},2^{-1/q}\}M_p \le n(\ell_p^2)\le M_p,\] and \(M_p=M_q\) by duality. Since the upper estimate is realized by the rotation \(J\), this led to the conjecture that equality should hold for every \(1<p<\infty\).
The conjecture was subsequently verified in several ranges of \(p\). Merí and Quero [6] studied numerical indices for absolute and symmetric norms on the plane and proved, as a consequence, that \[n(\ell_p^2)=M_p \qquad \text{for } \frac{3}{2}\le p\le3.\] Monika and Zheng [7] refined these methods and proved the equality in the larger interval \[1+\alpha_0\le p\le \alpha_1,\] where \(\alpha_0\approx0.4547\) and \(\alpha_1\) is determined by \[\frac{1}{1+\alpha_0}+\frac{1}{\alpha_1}=1.\] More recently, Merí and Quero [4] used Riesz–Thorin interpolation to extend the equality for \[\frac{6}{5}\le p\le\frac{3}{2} \qquad\text{and}\qquad 3\le p\le6.\] Theorem 1 completes the computation of the numerical index of real two-dimensional \(\ell_p\)-spaces.
We briefly describe the organization of the paper. In Section 2, we introduce the preliminary results, notation, and constants used in the proof. Section 3 contains the main auxiliary lemmas and explains how they imply Theorem 1. The proof of Theorem 1 is then given in Section 4. The remaining sections are devoted to the proofs of the auxiliary lemmas stated in Section 3.
We shall use the following formula for the numerical radius of an operator on \(\ell_p^2\).
Lemma 1 (Lemma 1 in [4]). Let \(1<p<\infty\) and \(T= \begin{pmatrix} a&b\\ c&d \end{pmatrix}\) be an operator on \(\ell_p^2\). Then \[v(T) = \max\left\{ \max_{0\le t\le1} \frac{\left\lvert a+dt^p\right\rvert+\left\lvert bt+ct^{p-1}\right\rvert}{1+t^p}, \, \max_{0\le t\le1} \frac{\left\lvert d+at^p\right\rvert+\left\lvert ct+bt^{p-1}\right\rvert}{1+t^p} \right\}.\]
We shall also use the following elementary invariance property (see [4]). If \(U\colon X\to X\) is a surjective linear isometry, then for every \(T\in\mathcal{L}(X)\), \[\label{eq:invariance} \|U^{-1}TU\|=\|T\|, \qquad v(U^{-1}TU)=v(T).\tag{1}\] Define \[L(t):=\frac{t^{p-1}}{1+t^p},\qquad M(t):=\frac{t}{1+t^p},\qquad N(t):=\frac{1-t^p}{1+t^p},\] and \[R(t):=M(t)-L(t)=\frac{t-t^{p-1}}{1+t^p}, \qquad 0\le t\le1.\] For \(p>2\), a standard calculus argument shows that the function \(R\) has a unique maximizer in \((0,1)\), which we denote by \(t_0\). We set \[L_0:=L(t_0),\qquad M_0:=M(t_0),\qquad N_0:=N(t_0),\qquad R_0:=R(t_0)=v(J).\] By standard symmetry reductions, which are carried out in the proof of Theorem 1, it is enough to study operators of the form \[\label{eq:32T} T= \begin{pmatrix} a & 1\\ -(1-c) & -d \end{pmatrix}, \qquad a,d\ge0, \quad c\in[0,1].\tag{2}\] For convenience, we introduce the parameters \[\alpha:=\frac{a+d}{2}, \qquad \beta:=\frac{d-a}{2}, \qquad \eta:=\frac{p-2}{p}.\] We also define \[\label{eq:Hacd} H(a,c,d):= \max\left\{ cL_0+\left\lvert \beta-\alpha N_0\right\rvert, \, \left\lvert \beta+\alpha N_0\right\rvert-cM_0 \right\}.\tag{3}\]
The proof of Theorem 1 is based on two estimates. Our goal is to show that \[\frac{v(T)}{\left\lVert T\right\rVert}\ge v(J)\] for every matrix \(T\) of the form 2 , where \(J\) is the rotation by \(90^\circ\). The argument separates into a lower bound for the numerical radius and an upper bound for the operator norm.
The first estimate gives a quantitative lower bound for \(v(T)\) by evaluating the numerical radius at the point where \(J\) attains its numerical radius.
Lemma 2. Let \(a,d\ge0\), \(c\in[0,1]\), and \(T\) defined as in (2 ). Then \[v(T)\ge v(J)+H(a,c,d).\]
Since \(H(a,c,d)\geq 0\), Lemma 2 shows that, among matrices of the form 2 , the numerical radius is minimized by the rotation \(J\). Moreover, \(H(a,c,d)\) provides a quantitative estimate of how the numerical radius increases as the parameters \(a,d,c\) move away from the rotational case. A key feature of \(H\), which follows directly from its definition in 3 , is the homogeneity \[\label{eq:32homogeneity32of32H} H(\lambda a,\lambda c,\lambda d) = \lambda H(a,c,d), \qquad \lambda\ge0,\tag{4}\] which will play an important role in the proof of the main theorem.
The second estimate provides an upper bound for the operator norm of \(T\).
Lemma 3. Let \(a,d\ge0\), \(c\in[0,1]\), and \(T\) defined as in (2 ). Then, \[\left\lVert T\right\rVert\le \frac{v(J)+H(a,c,d)}{v(J)}.\]
Combining Lemmas 2 and 3 yields \[\frac{v(T)}{\left\lVert T\right\rVert} \ge v(J)\] for every matrix \(T\) of the form 2 . Since standard symmetry and normalization arguments reduce the problem to this class of matrices, Theorem 1 follows.
In order to prove Lemma 3, we will show that it suffices to analyze the case \(c=1\), which corresponds to an upper triangular matrix. Indeed, we can express \(T\) as the convex combination \[T=(1-c)J+cS,\] where \[S= \begin{pmatrix} a/c & 1\\ 0 & -d/c \end{pmatrix}.\] Then, the convexity of the operator norm together with the homogeneity of \(H(a,c,d)\) given in (4 ) can be used to reduce the problem to estimating the norm of \(S\).
Lemma 4. Let \(A,B\ge0\), and consider \[S=\begin{pmatrix}A&1\\0&-B\end{pmatrix}.\] Then \[\left\lVert S\right\rVert\le \frac{v(J)+H(A,1,B)}{v(J)}.\]
Theorem 1 follows from Lemmas 2 and 3 once we show that it is enough to study operators of the form 2 .
Proof of Theorem 1. We first reduce to the case \(p>2\). The cases \(p=1\) and \(p=2\) are classical: \(n(\ell_1^2)=v(J)=1\), while \(n(\ell_2^2)=v(J)=0\). Now let \(1<p<\infty\), \(p\ne2\), and let \(q=p/(p-1)\). Since \((\ell_p^2)^*=\ell_q^2\), the identity \(v(T^*)=v(T)\) gives \(n(\ell_p^2)=n(\ell_q^2)\). Moreover \(v_{\ell_p^2}(J)=v_{\ell_q^2}(J)\) [4]. Thus the range \(1<p<2\) follows from the range \(q>2\), and it remains to prove the result for \(p>2\). By the reduction used by Merí and Quero in the proof of [4], it is enough to consider operators of the form \[T= \begin{pmatrix} a&b\\ -c&-d \end{pmatrix}, \qquad a,b,c,d\ge0.\] Next, we note that we may assume \(b\geq c\). Indeed, consider the isometry \[S= \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}.\] By the invariance property (1 ), replacing \(T\) by \(-S^{-1}TS\) does not change the quotient \(v(T)/\|T\|\). A direct computation gives \[-S^{-1} T S = \begin{pmatrix} d&c\\ -b&-a \end{pmatrix}.\] Thus this replacement preserves the class of matrices under consideration and interchanges the roles of \(b\) and \(c\). Next, by a density argument, it suffices to consider the case \(b>0\). Indeed, if \(b=0\), then the assumption \(b\ge c\) forces \(c=0\). Since we only need to consider nonzero operators, we may also assume that \(a\) and \(d\) are not both zero. For \(b_n>0\) with \(b_n \to 0\), set \[T_n = \begin{pmatrix} a&b_n\\ 0&-d \end{pmatrix}.\] Since \(v(T_n) \to v(T)\), \(\|T_n\|\to \|T\|\), and \(\|T\|>0\), if the desired inequality holds for \(T_n\), then we deduce that \[\frac{v(T)}{\|T\|} = \lim_{n \to \infty} \frac{v(T_n)}{\|T_n\|} \geq v(J).\] Assume now that \(b>0\). Since the quotient \(v(T)/\left\lVert T\right\rVert\) is invariant under multiplication by a positive scalar, we divide by \(b\). We obtain \[\frac{1}{b} T = \begin{pmatrix} a/b&1\\ -c/b&-d/b \end{pmatrix}.\] Since \(0\le c/b\le1\), this matrix is of the normalized form 2 . The result now follows for every \(p>2\) by combining Lemma 2 and Lemma 3. ◻
We first prove the lower bound for the numerical radius.
Proof of Lemma 2. By Lemma 1, the numerical radius of an operator \(T\) defined as in (2 ) is given by \[v(T)=\max\left\{ \max_{0\le t\le1} \frac{\left\lvert a-dt^p\right\rvert+t-(1-c)t^{p-1}}{1+t^p}, \; \max_{0\le t\le1} \frac{\left\lvert d-at^p\right\rvert+\left\lvert -(1-c)t+t^{p-1}\right\rvert}{1+t^p} \right\},\] where we have used \(p>2\) to remove the absolute value. We evaluate both maxima at \(t=t_0\) and reformulate in terms of \(L_0\), \(M_0\), \(N_0\), and \(R_0\). For the first term, using \[a=\alpha-\beta, \qquad d=\alpha+\beta,\] we obtain \[a-dt_0^p = (\alpha-\beta)-(\alpha+\beta)t_0^p = \alpha(1-t_0^p)-\beta(1+t_0^p).\] Hence \[\frac{\left\lvert a-dt_0^p\right\rvert}{1+t_0^p} = \left\lvert \alpha\frac{1-t_0^p}{1+t_0^p}-\beta \right\rvert = \left\lvert \alpha N_0-\beta\right\rvert.\] Moreover, \[\frac{t_0-(1-c)t_0^{p-1}}{1+t_0^p} = \frac{t_0-t_0^{p-1}}{1+t_0^p} + c\frac{t_0^{p-1}}{1+t_0^p} = R_0+cL_0.\] Therefore, \[\frac{\left\lvert a-dt_0^p\right\rvert+t_0-(1-c)t_0^{p-1}}{1+t_0^p} = R_0+cL_0+\left\lvert \alpha N_0-\beta\right\rvert.\] For the second term, \[d-at_0^p = (\alpha+\beta)-(\alpha-\beta)t_0^p = \alpha(1-t_0^p)+\beta(1+t_0^p),\] and hence \[\frac{\left\lvert d-at_0^p\right\rvert}{1+t_0^p} = \left\lvert \alpha\frac{1-t_0^p}{1+t_0^p}+\beta \right\rvert = \left\lvert \alpha N_0+\beta\right\rvert.\] Moreover, \[\begin{align} \frac{\left\lvert (1-c)t_0-t_0^{p-1}\right\rvert}{1+t_0^p} \geq \frac{(1-c)t_0-t_0^{p-1}}{1+t_0^p} = \frac{t_0-t_0^{p-1}}{1+t_0^p} - c\frac{t_0}{1+t_0^p} = R_0-cM_0. \end{align}\] Therefore, \[\frac{\left\lvert d-at_0^p\right\rvert+\left\lvert -(1-c)t_0+t_0^{p-1}\right\rvert}{1+t_0^p} \ge R_0-cM_0+\left\lvert \alpha N_0+\beta\right\rvert.\] Taking the maximum of the two lower bounds gives the result. ◻
We next prove the upper bound for the normalized sign class, assuming the upper triangular estimate of Lemma 4.
Proof of Lemma 3. Assume first that \(0<c\le1\). Write \[T=(1-c)J+cS, \qquad S= \begin{pmatrix} a/c & 1\\ 0 & -d/c \end{pmatrix}.\] By convexity of the operator norm, \[\left\lVert T\right\rVert \le (1-c)\left\lVert J\right\rVert+c\left\lVert S\right\rVert = (1-c)+c\left\lVert S\right\rVert.\] Applying Lemma 4 to \(S\) and the homogeneity property (4 ), we deduce that \[\left\lVert S\right\rVert \le \frac{v(J)+H(a/c,1,d/c)}{v(J)} = 1+ \frac{H(a,c,d)}{c v(J)}.\] Consequently, \[\begin{align} \left\lVert T\right\rVert \leq (1-c)+ c\left(1 + \frac{H(a,c,d)}{c v(J)} \right) = 1+\frac{H(a,c,d)}{v(J)}. \end{align}\] If \(c=0\), we argue by approximation. For \(0<c_n\le1\) with \(c_n\downarrow0\), set \[T_n := \begin{pmatrix} a & 1\\ -(1-c_n) & -d \end{pmatrix}.\] By the case already proved, \[\left\lVert T_n\right\rVert \le \frac{v(J)+H(a,c_n,d)}{v(J)}.\] Since \(\|T_n\|\to \|T\|\) and \(H(a,c_n,d)\to H(a,0,d)\), letting \(n\to\infty\) gives \[\left\lVert T\right\rVert \le \frac{v(J)+H(a,0,d)}{v(J)}.\qedhere\] ◻
We first record a simple norm criterion.
Lemma 5. Let \(A,B\ge0\) and \(K\ge1\). If \[A\le K-1 \qquad\text{and}\qquad B^p\le K^p-K^{p-1},\] then \[\left\|\begin{pmatrix}A&1\\0&-B\end{pmatrix}\right\|\le K.\]
Proof. For \(A,B\ge0\), the norm of this operator may be computed on vectors with nonnegative coordinates. Thus, a parametrization of the positive quadrant of the unit sphere of \(\ell_p^2\) by \[\frac{(1,s)}{(1+s^p)^{1/p}}, \qquad s\ge0,\] shows that \[\label{eq:norm-of-S} \left\|\begin{pmatrix}A&1\\0&-B\end{pmatrix}\right\|^p =\sup_{s\ge0}\frac{(A+s)^p+B^ps^p}{1+s^p}.\tag{5}\] Hence it is enough to show \[(A+s)^p+B^ps^p\le K^p(1+s^p) \qquad\text{for all }s\ge0.\] Since \(A\le K-1\), we have \[(A+s)^p\le (K-1+s)^p.\] Now write \[\frac{K-1+s}{K} = \frac{K-1}{K}\cdot 1+\frac{1}{K}\cdot s.\] By convexity of \(t\mapsto t^p\), \[\left(\frac{K-1+s}{K}\right)^p \le \frac{K-1}{K}+\frac{s^p}{K}.\] Multiplying by \(K^p\), we get \[(K-1+s)^p \le K^{p-1}(K-1+s^p).\] Using the assumption on \(B\) we obtain \[\begin{align} (A+s)^p+B^ps^p& \leq K^{p-1}(K-1+s^p)+K^{p-1}(K-1)s^p = K^{p-1}(K-1)+K^ps^p \\ &\leq K^p(1+s^p).\qedhere \end{align}\] ◻
We also need the following identities and inequalities, which we prove in Section 7.
Lemma 6. The constants \(L_0,M_0,N_0,R_0\) and \(\eta\) satisfy:
\(M_0-L_0=R_0\).
\(M_0+L_0=\dfrac{R_0N_0}{\eta}\).
\(L_0=\frac{R_0}{2}\left(\frac{N_0}{\eta}-1\right)\), and \(M_0=\frac{R_0}{2}\left(\frac{N_0}{\eta}+1\right).\)
\(\dfrac{L_0}{M_0}=t_0^{p-2}\).
\(R_0\le N_0\).
\(N_0^2\ge \eta\).
\(\dfrac{R_0(1+N_0)}{\eta+N_0}\le t_0\).
Proof of Lemma 4. Fix \(n\ge0\) and suppose that \(A,B\ge0\) satisfy \(H(A,1,B)\le n.\) Recall that \(v(J)=R_0\). We will show that \[\label{eq:levels} \left\| \begin{pmatrix} A&1\\ 0&-B \end{pmatrix} \right\| \le \frac{R_0+n}{R_0}.\tag{6}\] Let \[\alpha=\frac{A+B}{2}, \qquad \beta=\frac{B-A}{2},\] and define \[\Phi:=\beta-\alpha N_0, \qquad \Psi:=\beta+\alpha N_0.\] The constraint \(H(A,1,B)\le n\) is equivalent to \[\label{eq:constraint} \left\lvert \Phi\right\rvert\le n-L_0, \qquad \left\lvert \Psi\right\rvert\le n+M_0.\tag{7}\] In particular, if the region is nonempty, then \(n\ge L_0\). Solving for \(\alpha\) and \(\beta\), we get \[\alpha=\frac{\Psi-\Phi}{2N_0}, \qquad \beta=\frac{\Phi+\Psi}{2}.\] Using \(A=\alpha-\beta\) and \(B=\alpha+\beta\), we can write \(A\) and \(B\) in terms of \(\Phi\) and \(\Psi\). Namely, \[A(\Phi,\Psi)= \frac{1}{2}\left[ \Psi\left(\frac{1}{N_0}-1\right) - \Phi\left(\frac{1}{N_0}+1\right) \right],\] and \[B(\Phi,\Psi)= \frac{1}{2}\left[ \Psi\left(\frac{1}{N_0}+1\right) - \Phi\left(\frac{1}{N_0}-1\right) \right].\] Since \(0<N_0<1\), both \((N_0^{-1} +1)\) and \((N_0^{-1} - 1)\) are positive. Hence, the formulas above show that \(A(\Phi,\Psi)\) and \(B(\Phi,\Psi)\) are increasing functions of \(\Psi\) and decreasing functions of \(\Phi\). Therefore for every feasible \((\Phi, \Psi)\) satisfying 7 , the corresponding variables \(A(\Phi,\Psi)\) and \(B(\Phi,\Psi)\) are dominated by \(A_*:=A(\Phi_*,\Psi_*)\) and \(B_*:=B(\Phi_*,\Psi_*)\), where \[\Phi_*:=-(n-L_0),\qquad \Psi_*:=n+M_0.\] Furthermore, by 5 , the norm of the triangular matrix is increasing in both \(A\) and \(B\). Therefore it is enough to prove the desired inequality (6 ) for the corner \((A_*,B_*)\). A simple computation gives \[A_* = \frac{n}{N_0} + \frac{M_0(1-N_0)-L_0(1+N_0)}{2N_0}, \qquad B_* = \frac{n}{N_0} + \frac{M_0(1+N_0)-L_0(1-N_0)}{2N_0}.\] Using \(M_0-L_0=R_0\), we can write this corner as \[A_*=\widetilde{A}+\frac{n-L_0}{N_0}, \qquad B_*=\widetilde{B}+\frac{n-L_0}{N_0},\] where \[\widetilde{A}:=\frac{(M_0+L_0)(1-N_0)}{2N_0}, \qquad \widetilde{B}:=\frac{(M_0+L_0)(1+N_0)}{2N_0}.\] Hence, by the triangle inequality and \(R_0\le N_0\), \[\begin{align} \left\| \begin{pmatrix} A_*&1\\ 0&-B_* \end{pmatrix} \right\| &\le \left\| \begin{pmatrix} \widetilde{A}&1\\ 0&-\widetilde{B} \end{pmatrix} \right\| + \frac{n-L_0}{N_0} \left\| \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} \right\| \leq \left\| \begin{pmatrix} \widetilde{A}&1\\ 0&-\widetilde{B} \end{pmatrix} \right\| + \frac{n-L_0}{R_0}. \end{align}\] We claim that \[\label{eq:special-base} \left\| \begin{pmatrix} \widetilde{A}&1\\ 0&-\widetilde{B} \end{pmatrix} \right\| \le 1+\frac{L_0}{R_0}.\tag{8}\] Note that the result follows from this claim since for any \(A\), \(B\geq 0\) with \(H(A,1,B)\leq n\) we have \[\left\| \begin{pmatrix} A&1\\ 0&-B \end{pmatrix} \right\| \leq \left\| \begin{pmatrix} A_*&1\\ 0&-B_* \end{pmatrix} \right\| \le 1+\frac{L_0}{R_0}+\frac{n-L_0}{R_0} = \frac{R_0+n}{R_0}.\] Thus, it remains to prove 8 . Observe that by Lemma 6, we have
\[K: = 1+\frac{L_0}{R_0} \overset{\mathrm{(i)}}{=} \frac{M_0}{R_0} \overset{\mathrm{(iii)}}{=} \frac{1}{2}\left(\frac{N_0}{\eta}+1\right).\] By Lemma 5, it suffices to show \[\widetilde{A}\le K-1 \quad and \quad \widetilde{B}^p\le K^p-K^{p-1}.\] For the first inequality, again using Lemma 6, \[\widetilde{A} = \frac{(M_0+L_0)(1-N_0)}{2N_0} \;\overset{\mathrm{(ii)}}{=}\; \frac{R_0(1-N_0)}{2\eta} \;\overset{\mathrm{(v)}}{\le}\; \frac{N_0(1-N_0)}{2\eta} \;\overset{\mathrm{(vi)}}{\le}\; \frac{N_0-\eta}{2\eta} \;\overset{\mathrm{(iii)}}{=}\; \frac{L_0}{R_0} \; = \; K-1.\] For the second inequality, first observe that \[\frac{\widetilde{B}}{K} = \frac{(M_0+L_0)(1+N_0)}{2N_0K} \;\overset{\mathrm{(ii)}}{=}\; \frac{R_0(1+N_0)}{\eta+N_0} \;\overset{\mathrm{(vii)}}{\le}\; t_0.\] Therefore, \[\left(\frac{\widetilde{B}}{K}\right)^p \le t_0^p \le t_0^{p-2} \;\overset{\mathrm{(iv)}}{=}\; \frac{L_0}{M_0} \;\overset{\mathrm{(i)}}{=}\; 1-\frac{1}{K}.\] Multiplying by \(K^p\), we obtain \[\widetilde{B}^p\le K^p-K^{p-1}.\] Thus Lemma 5 gives 8 , and the proof is complete. ◻
In order to prove Lemma 6, we introduce the change of variables \[t^p=e^{-2z}, \qquad z\ge0.\] Using \(\eta=(p-2)/p\), we compute \[M(t)=\frac{e^{\eta z}}{2\cosh z}, \qquad L(t)=\frac{e^{-\eta z}}{2\cosh z}, \qquad N(t)=\tanh z, \qquad R(t)=M(t)-L(t)=\frac{\sinh(\eta z)}{\cosh z}.\] Let \(z_0\) be defined by \(t_0^p=e^{-2z_0}\), where \(t_0\) is the maximizer of \(R(t)\).
Proof of Lemma 6. First, we record a useful identity for the maximizer \(z_0\). Differentiating \(R(t)\) gives that \(z_0\) is characterized by \[\eta\cosh(\eta z_0)\cosh z_0 = \sinh(\eta z_0)\sinh z_0.\] Equivalently, \[\label{eq:critical} \tanh(\eta z_0)\tanh z_0=\eta.\tag{9}\] Note that (i) is immediate from the definition of \(L_0\), \(M_0\), and \(R_0\). Next, since \[N_0=\tanh z_0, \qquad M_0+L_0=\frac{\cosh(\eta z_0)}{\cosh z_0}, \qquad R_0=\frac{\sinh(\eta z_0)}{\cosh z_0},\] the identity (9 ) gives \[\eta(M_0+L_0)=N_0R_0.\] Dividing by \(\eta\) proves (ii). Moreover, combining the above identity with \(M_0-L_0=R_0\) gives (iii). Furthermore, (iv) follows directly from \[\frac{L(t)}{M(t)}=t^{p-2}.\] Next, observe that for any \(0\leq t\leq 1\) we have \[N(t)-R(t)=\frac{1-t^p-t+t^{p-1}}{1+t^p} =\frac{(1-t)(1+t^{p-1})}{1+t^p}\geq 0.\] Taking \(t=t_0\) proves (v). The inequality (vi) follows from the identity (9 ) and the fact that the function \(\tanh\) is increasing. Indeed, since \(\eta <1\) notice that \[\eta = \tanh(\eta z_0)\tanh z_0 \leq \tanh^2 z_0 = N_0^2.\] It remains to prove (vii). Using the hyperbolic expressions above, we get \[\frac{R_0(1+N_0)}{\eta+N_0} = \frac{\frac{\sinh(\eta z_0)}{\cosh z_0}(1+\tanh z_0)}{\eta+\tanh z_0}.\] The critical identity 9 gives \[\eta+\tanh z_0 = \tanh z_0\bigl(1+\tanh(\eta z_0)\bigr).\] Hence \[\frac{R_0(1+N_0)}{\eta+N_0} = \frac{\sinh(\eta z_0)}{\cosh z_0} \frac{1+\tanh z_0}{\tanh z_0(1+\tanh(\eta z_0))}.\] Using \[1+\tanh z=\frac{e^z}{\cosh z},\] we obtain \[\frac{1+\tanh z_0}{1+\tanh(\eta z_0)} = \frac{e^{z_0}\cosh(\eta z_0)}{e^{\eta z_0}\cosh z_0} = e^{(1-\eta)z_0}\frac{\cosh(\eta z_0)}{\cosh z_0}.\] Thus \[\frac{R_0(1+N_0)}{\eta+N_0} = e^{(1-\eta)z_0} \frac{\sinh(\eta z_0)\cosh(\eta z_0)}{\sinh z_0\cosh z_0} = e^{(1-\eta)z_0} \frac{\sinh(2\eta z_0)}{\sinh(2z_0)}.\] Since \(0<\eta<1\), \[\frac{\sinh(2\eta z_0)}{\sinh(2z_0)} = e^{-2(1-\eta)z_0} \frac{1-e^{-4\eta z_0}}{1-e^{-4z_0}} \le e^{-2(1-\eta)z_0}.\] Consequently, \[\frac{R_0(1+N_0)}{\eta+N_0} \le e^{-(1-\eta)z_0}.\] Finally, since \(1-\eta=2/p\) and \(t_0^p=e^{-2z_0}\), we have \[e^{-(1-\eta)z_0}=e^{-2z_0/p}=t_0.\] This proves (vii), and the proof is complete. ◻
The author thanks Alicia Quero for helpful discussions and comments related to this work.
Department of Mathematics, Michigan State University. Email: chiclan1@msu.edu.↩︎