The Tang-Shu Zhang Schatten Norm Formula Fails Below the Hilbertian Exponent

Zijian Zeng, Houde Liu, and Kuru Ratnavelu

Schatten norms, matrix absolute values, sharp constants, rank-one matrices, counterexamples, and a phase transition at \(p=2\)

Abstract. For \(m\ge 2\), let \(c_p(m)\) be the all-dimensional best constant in \[ \left\|\sum_{k=1}^{m}A_k\right\|_p \le c_p(m)\left\|\sum_{k=1}^{m}|A_k|\right\|_p. \] Quanyu Tang and Shu Zhang conjectured an explicit formula for every finite \(p>1\). We prove that their formula fails for every \(1<p<2\). For each such \(p\), an explicit pair of real \(2\times2\) rank-one matrices gives a strictly larger ratio. The proof is analytic and reduces to one concavity estimate. At \(p=3/2\) we additionally give a fixed algebraic pair whose comparison is certified by seven strict rational inequalities. A transverse perturbation analysis identifies \(p=2\) as the exact local phase transition: below two the conjectured equiangular family is a non-smooth saddle, whereas at and above two it is locally maximizing in that direction. On the positive side, we prove the conjectured sharp bound for every family of rank-at-most-one summands when \(2\le p<\infty\), and classify all equality cases. We also prove the corresponding endpoint statement for \(p=\infty\). Finally, for arbitrary complex matrices we establish the conjectured sharp constant in the case \(m=2,p=4\).

1. Introduction

For \(A\in M_n(\mathbb C)\), write \[ |A|=(A^*A)^{1/2}, \] and let \(\|A\|_p=(\operatorname{Tr}|A|^p)^{1/p}\) be the Schatten \(p\)-norm for \(1\le p<\infty\); \(\|\cdot\|_\infty\) denotes the operator norm. For fixed \(m,n\ge1\), define \[ c^{\mathrm{abs}}_p(m,n) =\sup_{(A_1,\ldots,A_m)\ne0} \frac{\left\|\sum_{k=1}^{m}A_k\right\|_p} {\left\|\sum_{k=1}^{m}|A_k|\right\|_p}, \qquad A_k\in M_n(\mathbb C). \tag{1} \] We use the dimension-free notation \[ c_p(m)=\sup_{n\ge1}c^{\mathrm{abs}}_p(m,n). \tag{2} \] The all-zero family is excluded; the denominator otherwise cannot vanish.

Tang and Shu Zhang proved \[ c_1(m)=1,\qquad c_2(m)=\sqrt{\frac{1+\sqrt m}{2}},\qquad c_\infty(m)=\sqrt m, \] and proposed a formula for the remaining exponents. For finite \(p>1\), consider \[ g_{p,m}(x):=x^p-2x-(m-1)=0. \tag{3} \] This equation has exactly one root \(x_{p,m}>1\). Their conjectured value is \[ C^{\mathrm{TZ}}_{p,m} =\frac{\sqrt{x_{p,m}(x_{p,m}+m-1)}} {(x_{p,m}^{p}+m-1)^{1/p}}. \tag{4} \] The lower bound \(c_p(m)\ge C^{\mathrm{TZ}}_{p,m}\) is attained by a rank-one equiangular family.

The dimension parameter matters at fixed size. Writing \(d=\min\{m,n\}\), Teng Zhang subsequently obtained \[ c^{\mathrm{abs}}_1(m,n)=1,\qquad c^{\mathrm{abs}}_2(m,n)=\sqrt{\frac{1+\sqrt d}{2}},\qquad c^{\mathrm{abs}}_\infty(m,n)=\sqrt d, \] together with \(c^{\mathrm{abs}}_p(m,n)\le d^{1/2-1/(2p)}\) for \(1\le p\le\infty\). Bourin and Lee also highlighted the question for Schatten exponents other than two. Neither result asserts formula (4) for general \(p\).

Theorem 1.1 (Failure throughout the sub-Hilbertian range). For every \(1<p<2\), \[ c^{\mathrm{abs}}_p(2,2)>C^{\mathrm{TZ}}_{p,2}. \] More explicitly, let \(x=x_{p,2}>1\) solve \(x^p-2x-1=0\), and put \[ s=\frac{x-1}{x+1}, \qquad \delta=\frac12x^{-p/(2-p)}. \] With \[ e=\begin{pmatrix}1\\0\end{pmatrix},\qquad u=\begin{pmatrix}1-\delta\\\sqrt{2\delta-\delta^2}\end{pmatrix}, \qquad v=\begin{pmatrix}s\\\sqrt{1-s^2}\end{pmatrix}, \] the two real rank-one matrices \(A_1=ee^T\) and \(A_2=uv^T\) satisfy \[ \frac{\|A_1+A_2\|_p}{\||A_1|+|A_2|\|_p} >C^{\mathrm{TZ}}_{p,2}. \] In particular, \([4/3,5/3]\) is an explicit closed rational interval of failure. More generally, every closed subinterval of \((1,2)\) has this property.

Theorem 1.2 (Exact counterexample). For \(m=n=2\) and \(p=3/2\), there are real rank-one matrices \(A_1,A_2\) such that \[ \frac{\|A_1+A_2\|_{3/2}}{\||A_1|+|A_2|\|_{3/2}} >\frac{207}{200}>C^{\mathrm{TZ}}_{3/2,2}. \] Consequently, Conjecture 3.1 of Quanyu Tang and Shu Zhang is false.

Theorem 1.3 (Sharp rank-one bound). Let \(m\ge2\), \(2\le p<\infty\), and let \(A_1,\ldots,A_m\in M_n(\mathbb C)\) have rank at most one and not all vanish. Then \[ \left\|\sum_{k=1}^{m}A_k\right\|_p \le C^{\mathrm{TZ}}_{p,m} \left\|\sum_{k=1}^{m}|A_k|\right\|_p. \tag{5} \] The constant is sharp in the dimension-free rank-one problem and is attained whenever \(n\ge m\). Equality holds precisely, up to common input and output unitaries, a common positive scale, and harmless phase changes in rank-one factorizations, when \[ A_k=r u v_k^*,\qquad \langle v_j,v_k\rangle= \begin{cases} 1,&j=k,\\ s_{p,m},&j\ne k, \end{cases} \] where \(r>0\), \(u\) is a unit vector, and \[ s_{p,m}=\frac{x_{p,m}-1}{x_{p,m}+m-1}. \] In particular, equality requires \(n\ge m\).

Theorem 1.4 (The full case \(m=2,p=4\)). Let \(A,B\in M_n(\mathbb C)\), and let \(x>1\) solve \[ x^4-2x-1=0. \] Then \[ \|A+B\|_4^4 \le \frac{x^2(x+1)}{2}\,\||A|+|B|\|_4^4. \tag{6} \] The constant is sharp in the dimension-free problem, is attained for every \(n\ge2\), and equals \((C^{\mathrm{TZ}}_{4,2})^4\).

2. Failure throughout \(1<p<2\)

Proof of Theorem 1.1. Fix \(1<p<2\), write \(q=p/2\in(1/2,1)\), and use the parameters in Theorem 1.1. Since \(x>1\), one has \(0<s<1\) and \(0<\delta<1/2\), so the displayed vectors \(e,u,v\) are real unit vectors.

Let \(U=(e,u)\) and \(V=(e,v)\). Their Gram matrices \[ L=U^TU=\begin{pmatrix}1&1-\delta\\1-\delta&1\end{pmatrix}, \qquad G=V^TV=\begin{pmatrix}1&s\\s&1\end{pmatrix} \] commute. Consequently, the squared singular values of \(A_1+A_2=UV^T\) are \[ (2-\delta)(1+s),\qquad \delta(1-s), \] whereas the eigenvalues of \(|A_1|+|A_2|=VV^T\) are \(1+s\) and \(1-s\). Hence, if \(R_p(\delta)\) denotes the attained ratio, then \[ R_p(\delta)^p =\frac{[(2-\delta)(1+s)]^q+[\delta(1-s)]^q} {(1+s)^p+(1-s)^p}. \tag{7} \]

At \(\delta=0\), this is the Tang-Shu Zhang value. Indeed, \[ 1+s=\frac{2x}{x+1},\qquad 1-s=\frac{2}{x+1},\qquad x^p+1=2(x+1), \] and direct substitution gives \[ R_p(0)^p=\frac12x^q(x+1)^{q-1} =(C^{\mathrm{TZ}}_{p,2})^p. \tag{8} \]

It remains to compare the numerators in (7). Their difference is \[ \Delta_p(\delta) =(1+s)^q\big((2-\delta)^q-2^q\big)+(1-s)^q\delta^q. \] For \(0<\delta\le1\), the mean value theorem and \(0<q<1\) give \[ 2^q-(2-\delta)^q\le q\delta\le\delta. \] Since \((1-s)/(1+s)=x^{-1}\), it follows that \[ \Delta_p(\delta) \ge(1+s)^q\delta\big(x^{-q}\delta^{q-1}-1\big). \tag{9} \] The choice \(\delta=\frac12x^{-q/(1-q)}\) satisfies \[ x^{-q}\delta^{q-1}=2^{1-q}>1. \] Thus \(\Delta_p(\delta)>0\), and (7)-(8) yield \(R_p(\delta)>C^{\mathrm{TZ}}_{p,2}\). □

Proposition 2.1 (Local phase transition at \(p=2\)). Fix \(p>1\), put \(q=p/2\), and let \(s=(x_{p,2}-1)/(x_{p,2}+1)\). For \[ u_\theta=(\cos\theta,\sin\theta)^T, \qquad v=(s,\sqrt{1-s^2})^T, \] set \(A_1=ee^T\), \(A_2(\theta)=u_\theta v^T\), and \(N_p(\theta)=\|A_1+A_2(\theta)\|_p^p\). Then \[ N_p(\theta)-N_p(0) =2^{-q}(1-s)^q|\theta|^p -q2^{q-2}(1+s)^q\theta^2 +o(|\theta|^p+\theta^2). \tag{10} \] Consequently, the equiangular Tang-Shu Zhang candidate is transversely locally minimizing when \(1<p<2\), but transversely locally maximizing when \(p\ge2\). Below two it is a non-smooth saddle. The critical exponent is exactly \(p=2\).

Proof. The squared singular values are \((1+\cos\theta)(1+s)\) and \((1-\cos\theta)(1-s)\). Therefore \[ N_p(\theta) =(1+s)^q(1+\cos\theta)^q+(1-s)^q(1-\cos\theta)^q. \] The expansions \[ (1+\cos\theta)^q =2^q-q2^{q-2}\theta^2+O(\theta^4) \] and \[ (1-\cos\theta)^q =2^{-q}|\theta|^{2q}+O(|\theta|^{2q+2}) \] give (10). If \(p<2\), the positive \(|\theta|^p\) term dominates the quadratic term; if \(p>2\), the negative quadratic term dominates. For \(p=2\), \[ N_2(\theta)-N_2(0)=-2s(1-\cos\theta)<0 \] for nonzero sufficiently small \(\theta\). Moreover, when \(1<p<2\), \[ N_p''(\theta) \sim p(p-1)2^{-q}(1-s)^q|\theta|^{p-2}\longrightarrow+\infty. \] Finally, in the common-left equiangular direction the ratio reduces to \[ F_{p,2}(y)=\frac{\sqrt{y(y+1)}}{(y^p+1)^{1/p}}. \] Its logarithmic derivative has the sign of \(2y+1-y^p\), which changes from positive to negative at \(y=x_{p,2}\). Hence the same candidate is locally maximizing in the equiangular direction, completing the saddle and phase-transition assertions. □

3. An exact \(2\times2\) counterexample

Set

\[ e=\begin{pmatrix}1\\0\end{pmatrix},\qquad u=\begin{pmatrix}39/40\\\sqrt{79}/40\end{pmatrix},\qquad v=\begin{pmatrix}5/8\\\sqrt{39}/8\end{pmatrix}. \]

These are real unit vectors. Define

\[ A_1=ee^T=\begin{pmatrix}1&0\\0&0\end{pmatrix}, \qquad A_2=uv^T= \begin{pmatrix} 39/64&39\sqrt{39}/320\\ \sqrt{79}/64&\sqrt{3081}/320 \end{pmatrix}. \tag{11} \]

Both matrices have rank one and unique nonzero singular value equal to one. Consequently, \(|A_1|=ee^T\) and \(|A_2|=vv^T\).

Proof of Theorem 1.2. Let \(U=(e,u)\) and \(V=(e,v)\). Then \(A_1+A_2=UV^T\), and \[ L=U^TU=\begin{pmatrix}1&39/40\\39/40&1\end{pmatrix}, \qquad G=V^TV=\begin{pmatrix}1&5/8\\5/8&1\end{pmatrix}. \] The squared singular values of \(UV^T\) are \[ \frac{1027}{320},\qquad \frac{3}{320}, \tag{12} \] whereas the eigenvalues of \(|A_1|+|A_2|=VV^T\) are \[ \frac{13}{8},\qquad\frac{3}{8}. \tag{13} \] Thus, if \[ R=\frac{\|A_1+A_2\|_{3/2}}{\||A_1|+|A_2|\|_{3/2}}, \] then \[ R^{3/2} =\frac{(1027/320)^{3/4}+(3/320)^{3/4}} {(13/8)^{3/2}+(3/8)^{3/2}}. \tag{14} \]

We first prove \(R>207/200\) using rational arithmetic only. Positivity allows us to raise each proposed enclosure to the fourth or second power. The exact differences are \[ \left(\frac{1027}{320}\right)^3 -\left(\frac{11989}{5000}\right)^4 =\frac{29144230879351}{40000000000000000}>0, \tag{15} \] \[ \left(\frac{3}{320}\right)^3 -\left(\frac{301}{10000}\right)^4 =\frac{124819571}{40000000000000000}>0, \tag{16} \] \[ \left(\frac{4143}{2000}\right)^2 -\left(\frac{13}{8}\right)^3 =\frac{773}{8000000}>0, \tag{17} \] and \[ \left(\frac{2297}{10000}\right)^2 -\left(\frac{3}{8}\right)^3 =\frac{5543}{200000000}>0. \tag{18} \] Put \[ A_0=\frac{11989}{5000}+\frac{301}{10000}=\frac{24279}{10000}, \] \[ B_0=\frac{4143}{2000}+\frac{2297}{10000}=\frac{5753}{2500}, \qquad q=\frac{207}{200}. \] A final exact comparison gives \[ A_0^2-q^3B_0^2 =\frac{1172956601313}{50000000000000}>0. \tag{19} \] Equations (15)-(19) imply \(R>q\).

It remains to put the conjectured constant below the same rational separator. Let \(x>1\) solve \(x^{3/2}-2x-1=0\), and write \(x=t^2\). The unique positive root of \[ h(t)=t^3-2t^2-1=0 \tag{20} \] lies above \(4/3\), where \(h\) is strictly increasing. Moreover, \[ h\left(\frac{1103}{500}\right) =\frac{310727}{125000000}>0, \tag{21} \] so \(t<T:=1103/500\). Writing \(C=C^{\mathrm{TZ}}_{3/2,2}\) and using \(t^3=2t^2+1\), we obtain \[ C^6=\frac{t^6}{16(t^2+1)}=:H(t). \] The function \(H\) is strictly increasing for \(t>0\), since \[ H'(t)=\frac{t^5(2t^2+3)}{8(t^2+1)^2}>0. \] Finally, \[ q^6-H(T) =\frac{26731151399029597}{18772595200000000000}>0. \tag{22} \] It follows that \(C<q<R\). □

Remark 3.1. Numerically, \[ R=1.0364136587048904\ldots, \qquad C^{\mathrm{TZ}}_{3/2,2}=1.0346539518514341\ldots. \] These decimals play no role in the proof.

4. The sharp rank-one problem for \(p\ge2\)

Proof of Theorem 1.3. Write \[ A_k=r_ku_kv_k^*,\qquad r_k\ge0, \] where \(u_k,v_k\) are unit vectors whenever \(r_k>0\). Form the column matrices \[ U=(\sqrt{r_1}u_1,\ldots,\sqrt{r_m}u_m), \qquad V=(\sqrt{r_1}v_1,\ldots,\sqrt{r_m}v_m), \] and their Gram matrices \[ L=U^*U,\qquad G=V^*V. \] Then \[ \sum_kA_k=UV^*,\qquad \sum_k|A_k|=VV^*, \tag{23} \] and \[ \operatorname{diag}L=\operatorname{diag}G=(r_1,\ldots,r_m), \qquad \operatorname{Tr}L=\operatorname{Tr}G=:R>0. \tag{24} \]

Let \(K=L^{1/2}GL^{1/2}\). The nonzero eigenvalues of \(K\) are the squared singular values of \(UV^*\), while the nonzero eigenvalues of \(G\) are the eigenvalues of \(VV^*\). Therefore \[ \left\|\sum_kA_k\right\|_p^p=\operatorname{Tr}K^{p/2}, \qquad \left\|\sum_k|A_k|\right\|_p^p=\operatorname{Tr}G^p. \tag{25} \] Since \(p/2\ge1\), \[ \operatorname{Tr}K^{p/2}\le(\operatorname{Tr}K)^{p/2}. \tag{26} \] If \(\lambda=\lambda_{\max}(G)\), then \[ \operatorname{Tr}K=\operatorname{Tr}(LG) \le\lambda\operatorname{Tr}L=R\lambda. \tag{27} \]

List the \(m\) eigenvalues of \(G\), including zeros, as \(\lambda=\lambda_1\ge\lambda_2\ge\cdots\ge\lambda_m\ge0\). Convexity gives \[ \operatorname{Tr}G^p \ge\lambda^p+\frac{(R-\lambda)^p}{(m-1)^{p-1}}. \tag{28} \] If \(\lambda=R\), the ratio is at most one. Otherwise put \[ y=\frac{(m-1)\lambda}{R-\lambda}. \] Since \(\lambda\ge R/m\), one has \(y\ge1\). Combining (25)-(28) yields \[ \frac{\left\|\sum_kA_k\right\|_p} {\left\|\sum_k|A_k|\right\|_p} \le F_{p,m}(y) :=\frac{\sqrt{y(y+m-1)}}{(y^p+m-1)^{1/p}}. \tag{29} \] Direct differentiation gives \[ \frac{d}{dy}\log F_{p,m}(y) =\frac{(m-1)(2y+m-1-y^p)} {2y(y+m-1)(y^p+m-1)}. \] For \(p\ge2\), the function \(y^p-2y-(m-1)\) has exactly one zero \(x_{p,m}>1\). Thus \(F_{p,m}\) has a unique maximum at \(x_{p,m}\), proving (5).

To track equality, put \(\beta=(R-\lambda)/(m-1)\). Equality in the scalar maximization and in (28) forces \[ \lambda=x_{p,m}\beta, \qquad \operatorname{spec}(G)=\{\lambda,\beta,\ldots,\beta\}. \] The top eigenspace is one-dimensional. Equality in (27) forces the range of \(L\) into this eigenspace. If \(w\) is its unit eigenvector, then \[ L=Rww^*, \qquad G=\beta I+(\lambda-\beta)ww^*. \tag{30} \] The common diagonal condition gives \(|w_k|^2=1/m\) and \(r_k=R/m\). After simultaneous phase changes, take \(w=m^{-1/2}(1,\ldots,1)^T\). Equation (30) then says that the \(u_k\) coincide and \[ \langle v_j,v_k\rangle =\frac{\lambda-\beta}{R} =\frac{x_{p,m}-1}{x_{p,m}+m-1} \qquad(j\ne k). \] Conversely, this family takes equality at every step. Its Gram matrix is positive definite, so equality requires \(n\ge m\). □

Corollary 4.1 (Rank-one endpoint). If \(A_1,\ldots,A_m\) have rank at most one, then \[ \left\|\sum_kA_k\right\|_\infty \le\sqrt m\left\|\sum_k|A_k|\right\|_\infty. \] The constant is sharp in the dimension-free sense. Equality is possible only when \(n\ge m\), and then holds precisely for equal nonzero singular values, a common one-dimensional range, and pairwise orthogonal right vectors, modulo the same unitary and phase symmetries as above.

Proof. Use the notation above and set \(\lambda=\lambda_{\max}(G)\). Then \[ \|UV^*\|_\infty \le\|U\|_\infty\|V^*\|_\infty =\sqrt{\|L\|_\infty\lambda} \le\sqrt{R\lambda}. \] Since \(R=\operatorname{Tr}G\le m\lambda\), while \(\|VV^*\|_\infty=\lambda\), the inequality follows. Equality requires \(L\) to have rank one and \(G=\lambda I_m\), giving the stated configuration. □

5. The full \(m=2,p=4\) problem

Proof of Theorem 1.4. Put \(H=|A|\) and \(K=|B|\). Extend the partial isometries in the polar decompositions to unitaries, and write \(A=UH\), \(B=VK\). With \(W=U^*V\), unitary invariance reduces the numerator to \[ S=H+WK. \] Set \[ a=\operatorname{Tr}H^4, \qquad b=\operatorname{Tr}K^4, \qquad d=\operatorname{Tr}H^2K^2. \] The Hilbert-Schmidt triangle inequality applied to \[ SS^*=H^2+HKW^*+WKH+WK^2W^* \] gives \[ \|S\|_4^4\le(\sqrt a+\sqrt b+2\sqrt d)^2. \tag{31} \]

Introduce \[ e=\operatorname{Tr}H^3K, \qquad f=\operatorname{Tr}HK^3, \qquad g=\operatorname{Tr}HKHK. \] Cyclically collecting all words in the noncommutative expansion gives \[ \operatorname{Tr}(H+K)^4 =a+b+4(e+f)+4d+2g. \tag{32} \] Two lower bounds are needed. First, \[ e+f-2d =\operatorname{Tr}(H-K)H(H-K)K =\|H^{1/2}(H-K)K^{1/2}\|_2^2\ge0. \tag{33} \] Second, the three-factor Schatten Holder inequality gives \[ \sqrt d=\|HK\|_2 \le\|H^{1/2}\|_8\|H^{1/2}K^{1/2}\|_4\|K^{1/2}\|_8 =a^{1/8}g^{1/4}b^{1/8}. \] Thus, when \(ab>0\), \[ g\ge\frac{d^2}{\sqrt{ab}}. \tag{34} \] If \(ab=0\), one of \(H,K\) vanishes and the theorem is immediate. Combining (32)-(34), \[ \operatorname{Tr}(H+K)^4 \ge a+b+12d+\frac{2d^2}{\sqrt{ab}}. \tag{35} \]

Let \(A_0=\sqrt a\), \(B_0=\sqrt b\), and define \[ z=\frac{A_0+B_0}{\sqrt{A_0B_0}}\ge2, \qquad s=\sqrt{\frac{d}{A_0B_0}}\in[0,1]. \] The upper bound on \(s\) is the Hilbert-Schmidt Cauchy-Schwarz inequality \(d\le\sqrt{ab}\). From (31) and (35), \[ \frac{\|A+B\|_4^4}{\||A|+|B|\|_4^4} \le\Phi(z,s) :=\frac{(z+2s)^2}{z^2-2+12s^2+2s^4}. \tag{36} \]

If \(s\ge1/2\), then \[ 2(z^2-2+12s^2+2s^4)-(z+2s)^2 =(z-2s)^2+16s^2+4s^4-4\ge0, \] so \(\Phi(z,s)\le2\). Suppose \(0\le s\le1/2\). The sign of \(\partial\Phi/\partial z\) is the sign of \[ s^4+6s^2-zs-1. \] Since \(z\ge2\), \[ s^4+6s^2-zs-1 \le s^4+6s^2-2s-1\le2s-1\le0. \] Consequently, \[ \Phi(z,s)\le\Phi(2,s) =f(s):=\frac{2(1+s)^2}{1+6s^2+s^4}. \tag{37} \] The sign of \(f'(s)\) is the sign of \[ 1-6s-2s^3-s^4. \] Hence \(f\) has a unique maximizer \(s_0\in(0,1/2)\), characterized by \[ s_0^4+2s_0^3+6s_0-1=0. \tag{38} \] The first branch cannot dominate because \(f(1/6)>2\).

Set \(x=(1+s_0)/(1-s_0)\). Then \[ (x^4-2x-1)(1-s_0)^4 =2(s_0^4+2s_0^3+6s_0-1)=0, \] and \[ f(s_0) =\frac{x^2(x+1)^2}{x^4+1} =\frac{x^2(x+1)}{2}, \tag{39} \] where the last equality uses \(x^4+1=2(x+1)\). Equations (36)-(39) prove (6).

For sharpness, choose unit vectors \(r_1,r_2\) with \(\langle r_1,r_2\rangle=s_0\), choose a unit vector \(u\), and set \[ A=ur_1^*,\qquad B=ur_2^*. \] Then \[ \|A+B\|_4^4=4(1+s_0)^2 \] and \[ \||A|+|B|\|_4^4 =(1+s_0)^4+(1-s_0)^4 =2(1+6s_0^2+s_0^4). \] Their ratio is \(f(s_0)\), proving sharpness. □

Remark 5.1. Orthogonal direct sums of the two-dimensional extremal block give higher-dimensional equality examples. The proof above does not attempt a complete classification of all equality cases for Theorem 1.4.

Reproducibility and disclosure

The proof for the full interval \(1<p<2\) is analytic; a standard-library high-precision script checks the closed formulas at representative rational exponents. The fixed \(p=3/2\) certificate consists of the seven positive rational differences in (15)-(22) and uses exact arithmetic only. Both scripts accompany the manuscript. Low-dimensional numerical searches were used for exploration and adversarial testing only.

OpenAI Codex assisted with proof exploration, counterexample search, adversarial checking, and manuscript preparation. The submitting author is responsible for the correctness of every statement and for compliance with the target journal's authorship and disclosure policies.

References

1. Jean-Christophe Bourin and Eun-Young Lee, Triangle inequalities for the operator symmetric modulus, Proceedings of the American Mathematical Society, Early View, July 2026.
2. Quanyu Tang and Shu Zhang, Generalizing Lee's conjecture on the sum of absolute values of matrices, Linear Algebra and its Applications 731 (2026), 196-204.
3. Teng Zhang, Operator symmetric moduli and sharp triangle inequalities, Journal of the London Mathematical Society 114(2) (2026), e70672.

Author information

Zijian Zeng
Institute of Computer Science and Digital Innovation, UCSI University,
Kuala Lumpur 56000, Malaysia
1002266693@ucsiuniversity.edu.my

Houde Liu
Tsinghua Shenzhen International Graduate School, Tsinghua University,
Shenzhen 518055, China
liu.hd@sz.tsinghua.edu.cn

Kuru Ratnavelu
Institute of Computer Science and Digital Innovation, UCSI University,
Kuala Lumpur 56000, Malaysia
Institute of Mathematical Sciences, University of Malaya,
Kuala Lumpur 50603, Malaysia
Kurunathan@ucsiuniversity.edu.my