Let A be an n× n real matrix with n>2. For matrices A and B, define langle A,B rangle=E_x langle Ax,Bx rangle=frac1noperatornametr(A^mathsf TB), where x is uniformly distributed on S^n-1, and define the dual Hardy norm by |A|_H^1_=suplangle A,B rangle:|B|_H^1≤ 1. Let P_2 be a rank-two projection, and set c_n=fraclangle P_2,P_2 rangle|P_2|_H^1|P_2|_H^4. Near-isometric duality conjecture. For every n× n matrix A with n>2, c_n|A|_H^4≤|A|_H^1_≤|A|_H^4. This conjectures that the near-isometric duality between the H^1 and H^4 Hardy norms observed for 2×2 matrices persists in higher dimensions, with the sharp multiplicative loss occurring in the lower bound and determined by a rank-two projection. The parser provides no evidence that the conjecture has been resolved.

Near-isometric duality conjecture for Hardy norms of matrices

Less than 1 year old · traced to 2026

Let $A$ be an $n\times n$ real matrix with $n>2$ . For matrices $A$ and $B$ , define

\langle A,B\rangle=\mathbb{E}_x\langle Ax,Bx\rangle=\frac{1}{n}\operatorname{tr}(A^{\mathsf T}B),

where $x$ is uniformly distributed on $S^{n-1}$ , and define the dual Hardy norm by

\|A\|_{H^1_*}=\sup\{\langle A,B\rangle:\|B\|_{H^1}\leq 1\}.

Let $P_2$ be a rank-two projection, and set

c_n=\frac{\langle P_2,P_2\rangle}{\|P_2\|_{H^1}\|P_2\|_{H^4}}.

Near-isometric duality conjecture. For every $n\times n$ matrix $A$ with $n>2$ ,

c_n\|A\|_{H^4}\leq\|A\|_{H^1_*}\leq\|A\|_{H^4}.

This conjectures that the near-isometric duality between the $H^1$ and $H^4$ Hardy norms observed for $2\times2$ matrices persists in higher dimensions, with the sharp multiplicative loss occurring in the lower bound and determined by a rank-two projection. The parser provides no evidence that the conjecture has been resolved.

References

Primary source

Leonid V. Kovalev, “Hardy-type norms of matrices”, arXiv:2607.17373 (2026).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed progress

A posted, unverified calculation claims to disprove the proposed sharp lower bound in seven dimensions, while the separate upper bound remains open.

Kovalev’s 2026 paper formulates the near-isometric duality conjecture for matrices in dimensions $n>2$ , with the lower constant determined by a rank-two projection. It records numerical evidence but no proof of the conjecture.

Known results

In dimension $2$ , Brevig, Ortega-Cerdà, and Seip proved $\|A\|_{H^4}\leq\|A\|_{H^1_*}\leq C_2\|A\|_{H^4}$ , with optimal $C_2=\frac{\pi}{2\sqrt[4]{6}}$ .
For $n>2$ , Kovalev reports numerical evidence but states that even $\|A\|_{H^1_*}\leq\|A\|_{H^4}$ remains open.

August 2026 posted attempt

An unverified calculation claims that a positive-definite $7\times7$ example has $\|A\|_{H^1_*}/\|A\|_{H^4}<c_7$ , disproving the conjectured lower constant and yielding an open family of alleged counterexamples. It explicitly leaves the separate upper inequality open.

Current status (as of August 2026): The two-dimensional analogue is established; a dimension-seven counterexample to the proposed lower constant is claimed but unverified, and the higher-dimensional upper bound remains open.

Sources

CounterexampleThis solution needs a summarySee full solutionHide full solution

Counterexample to the conjectured sharp lower bound. The separate upper bound remains open.

Use the conventions of Conjecture 6.1 in Leonid V. Kovalev, Hardy-type norms of matrices, arXiv:2607.17373. In dimension $n=7$ , write

\Phi(C)=\mathbb E_{X\sim\operatorname{Unif}(S^6)}|CX|, \qquad \langle A,C\rangle=\frac17\operatorname{tr}(A^{\mathsf T}C).

1. A positive definite matrix and its exact dual norm

Set

\begin{aligned} B&=\operatorname{diag}(15,1,1,1,1,1,1),\\ A&=\nabla_F\Phi(B)=\operatorname{diag}(a,b,b,b,b,b,b), \end{aligned}

where $\nabla_F$ denotes the ordinary, unnormalized Frobenius gradient. Convexity and positive homogeneity imply

\operatorname{tr}(A^{\mathsf T}C)\leq\Phi(C) \qquad\text{for every }C\in\mathbb R^{7\times7},

with equality at $C=B$ . Consequently, the source-normalized dual norm is

\boxed{\|A\|_{H^1_*}=\frac17.}

2. Exact evaluation of the gradient

Let

Y=\sqrt{14}\log\!\left(15+4\sqrt{14}\right).

Direct evaluation of the spherical gradient gives

\begin{aligned} a&=\frac{458152875}{1438646272} -\frac{90922725}{80564191232}Y,\\[2mm] b&=-\frac{55440375}{2877292544} +\frac{905195225}{161128382464}Y. \end{aligned}

These identities require no numerical quadrature. Since $X_1^2$ has the $\operatorname{Beta}(1/2,3)$ distribution, define

J_j=\int_0^1\frac{u^{2j}}{\sqrt{1+224u^2}}\,du.

Elementary integration yields

\begin{aligned} J_0&=\frac{Y}{56},\\ J_j&=\frac{15-(2j-1)J_{j-1}}{448j} \qquad(j\geq1),\\ a&=\frac{225}{8}(J_1-2J_2+J_3),\\ b&=\frac5{16}(J_0-3J_1+3J_2-J_3). \end{aligned}

3. Strict failure of the conjectured lower constant

The spherical fourth-moment formula gives

\|A\|_{H^4}^{4}=\frac{R}{63}, \qquad R=(a^2+6b^2)^2+2(a^4+6b^4).

The conjectured rank-two-projection constant in dimension seven is

c_7=\frac{32\,14^{1/4}\sqrt3}{35\pi}.

Therefore the proposed inequality $c_7\|A\|_{H^4}\leq\|A\|_{H^1_*}$ is equivalent to

R\leq\frac{625\pi^4}{2097152}.

In fact, exact rational interval arithmetic certifies

\begin{aligned} R&>0.0290375624309815956692515439,\\ \frac{625\pi^4}{2097152} &<0.0290301713448770157207370796,\\ R-\frac{625\pi^4}{2097152} &>\frac7{1000000}>0. \end{aligned}

The certification uses integer-square-root bounds for $\sqrt{14}$ , the positive-remainder expansion

\log z =2\sum_{j=0}^{N-1}\frac{u^{2j+1}}{2j+1}+\varepsilon_N, \qquad u=\frac{z-1}{z+1}, \qquad 0<\varepsilon_N<\frac{2u^{2N+1}}{(2N+1)(1-u^2)},

and Machin's identity

\pi=16\arctan\!\left(\frac15\right) -4\arctan\!\left(\frac1{239}\right).

Equivalently,

\frac{\|A\|_{H^1_*}}{\|A\|_{H^4}} =0.974983033437249971\ldots <0.975045085220861666\ldots =c_7.

Both $a$ and $b$ are positive, so $A$ is positive definite. Continuity gives a full-dimensional open family of counterexamples. Only the proposed lower constant is disproved; the separate upper inequality remains open.