The least uniform complex coordinate Hlawka constant for p ≥ 85

For every real number $p \ge 85$ ($p$ need not be an integer, and $p = 85$ itself is included), the statement asserts that the real number $c(p)$ defined below is the least element of the set $S_p$. This set consists of all real numbers $C$ such that, for every $n \in \mathbb{N} = \lbrace 0, 1, 2, \dots \rbrace$ and all vectors $x, y, z \in \mathbb{C}^n$,

$$\Delta_3(x,y,z) \le C\big[\Delta_2(x,y) + \Delta_2(x,z) + \Delta_2(y,z)\big],$$

where

$$\Delta_2(u,v) = \lVert u\rVert_p + \lVert v\rVert_p - \lVert u+v\rVert_p, \qquad \Delta_3(x,y,z) = \lVert x\rVert_p + \lVert y\rVert_p + \lVert z\rVert_p - \lVert x+y+z\rVert_p .$$

Vectors are added coordinatewise. For $x = (x_1, \dots, x_n)$,

$$\lVert x\rVert_p = \Big(\sum_{i=1}^{n} |x_i|^p\Big)^{1/p},$$

where $|x_i|$ is the complex modulus and the powers are real powers. Because $p \ge 85 > 0$, we have $0^p = 0$, so this is the ordinary $\ell^p$-norm on $\mathbb{C}^n$. For $n = 1$ it equals $|x_1|$, and on $\mathbb{C}^0$ (a single point) it equals $0$. Rearranged, the inequality is equivalent to

$$C\big(\lVert x+y\rVert_p + \lVert x+z\rVert_p + \lVert y+z\rVert_p\big) \le (2C-1)\big(\lVert x\rVert_p + \lVert y\rVert_p + \lVert z\rVert_p\big) + \lVert x+y+z\rVert_p .$$

Being the least element amounts to two claims:

1. Membership: the inequality holds with $C = c(p)$ for every $n$ and all $x, y, z \in \mathbb{C}^n$.
2. Minimality: every real $C$ for which the inequality holds for every $n$ and all $x, y, z \in \mathbb{C}^n$ satisfies $c(p) \le C$. Equivalently, for each real $C < c(p)$ there exist some $n$ (which may depend on $C$) and some $x, y, z \in \mathbb{C}^n$ for which the inequality fails. This claim concerns constants that are valid in all dimensions at once. Nothing is asserted about the best constant in any single fixed dimension.

The constant. For real $t$, define

$$A_p(t) = (t^p + 2)^{1/p}, \qquad B_p(t) = \big(2|1-t|^p + 2^p\big)^{1/p}, \qquad R_p(t) = \frac{3A_p(t) - 3^{1/p}|2-t|}{6A_p(t) - 3B_p(t)},$$ $$c(p) = \sup \lbrace R_p(t) : \tfrac12 \le t \le 2 \rbrace .$$

The supremum is taken over the closed interval, both endpoints included. On that interval $t > 0$ and $|2-t| = 2-t$, and $c(p)$ depends only on $p$.

In the formal system, a quotient with zero denominator equals $0$, and the supremum of an empty or unbounded-above set of reals also equals $0$. Neither convention applies here. For $\tfrac12 \le t \le 2$ we have $|1-t| \le 1$, so

$$B_p(t)^p = 2|1-t|^p + 2^p \le 2 + 2^p < 2^p(t^p+2) = (2A_p(t))^p .$$

Hence the denominator $3\big(2A_p(t) - B_p(t)\big)$ is strictly positive and $R_p$ is continuous on the interval. So $c(p) = \max_{1/2 \le t \le 2} R_p(t)$, and this maximum is attained.

Scope and edge cases.

• The only hypothesis is $p \ge 85$. There are no other parameters or assumptions, and nothing is asserted for $p < 85$.
• $n = 0$ is included. There every norm is $0$, so the inequality reads $0 \le 0$, which holds for every $C$ and imposes no constraint.
• One and the same $C$ must work for every $n$ simultaneously.
• $x, y, z$ are arbitrary complex vectors, including zero, equal or parallel vectors.
• In both claims, $C$ ranges over all real numbers with no sign restriction.