The real three-dimensional Hlawka interval $[\log 3/\log(3/2),\,4]$

For a real exponent $p$ and vectors $x,y,z \in \mathbb{R}^3$ the coordinate norm is

and the triple deficit and pair-deficit sum are

For $t \in [1/2,2]$ put $A_p(t) = (t^p+2)^{1/p}$ and $B_p(t) = \bigl(2|1-t|^p+2^p\bigr)^{1/p}$, and let

be the cyclic constant, i.e. the comparison ratio forced by the cyclic family $(-t,1,1),(1,-t,1),(1,1,-t)$.

The theorem asserts the three-dimensional real Hlawka bound for exponents strictly above the Hlawka threshold $p_0 = \log 3/\log(3/2) \approx 2.70951$ and up to $4$:

No normalization, equal-norm or nonvanishing hypothesis is imposed, and the statement is made in dimension three only; the passage to arbitrary finite dimension is a separate theorem.

This is the part of the low-exponent range where the problem is no longer classical. Below $p_0$ the constant equals $1$ and the assertion reduces to Hlawka's inequality; it is known that Hlawka's inequality fails in $\ell_p$ for every $p > p_0$, so throughout the present interval the constant genuinely exceeds one, rising from $K_{p_0} = 1$ to $K_4 \approx 1.46711$. The left endpoint is included so that the interval joins continuously to the classical range: at $p = p_0$ the statement is still Hlawka's inequality, since $K_{p_0} = 1$.

The statement is sharp rather than merely sufficient. Because $2A_p(t) - B_p(t)$ is strictly positive on the parameter range, the pair-deficit sum of every cyclic triple is positive, so each cyclic triple yields a legitimate lower bound $\sup_t R_p(t) = K_p$ for any admissible constant. The target constant is thus the best possible, and a proof of the statement cannot proceed by any argument that leaves slack in the constant.

Structurally the interval is the genuinely new portion of the compact range below the established cutoff: uniform bounds are known for complex exponents $p \ge 6$ and hence for real exponents $p \ge 6$, and the three-dimensional real bound is known for $p \ge 6$ from the windowed confinement analysis and for $p \ge 8$ from the diagonal construction. What remains below $6$ is the three-dimensional interval $[2,6]$, and the present statement is its upper part $[p_0,4]$, the part above the classical Hlawka threshold.