A uniform twelve-point counterexample to norm monotonicity

Conjecture 3.6 of Liu and Slade, arXiv:2605.15545v1, is false. Take the step distribution uniform on

$$\mathcal U=\{u\in\mathbb Z^2:u_1^2+u_2^2=25\},$$

and take $x=(4,3)$. We will prove that $z\mapsto |x|_z$ is neither nondecreasing nor nonincreasing on $(0,1)$.

The source's Example 3.7 treats a nonuniform distribution; the equal-weight support below satisfies the uniformity assumption of the conjecture.

The idea is that the two endpoint limits of this norm are both $5$, but its fourth-moment expansion forces it strictly below $5$ near $z=1$.

1. The support satisfies the hypotheses

The set $\mathcal U$ consists of the twelve points

$$\begin{gathered} (\pm5,0),\quad(0,\pm5),\\ (\pm4,\pm3),\quad(\pm3,\pm4), \end{gathered}$$

with independent signs. Give each point probability $1/12$. This distribution is invariant under coordinate sign changes and coordinate interchange, and it has finite support. Its support generates $\mathbb Z^2$: indeed,

$$2(4,3)+2(4,-3)-3(5,0)=(1,0),$$

and interchanging coordinates gives $(0,1)$.

Every support point is a strict vertex of its convex hull. For fixed $u\in\mathcal U$, the Cauchy–Schwarz inequality gives $u\cdot v\le25$ for every $v\in\mathcal U$, with equality only when $v=u$. Thus the linear functional $v\mapsto u\cdot v$ exposes $u$ uniquely. In particular, every support point lies on the required boundary.

2. The defining variational formula

Here $e_1=(1,0)$, and $|\cdot|$ without a subscript denotes the Euclidean norm.

Let $U$ have this distribution and write

$$M(\mu)=\mathbb E e^{\mu\cdot U}.$$ $$K_z=\{\mu\in\mathbb R^2:M(\mu)\le z^{-1}\}.$$

For the random-walk norm in the conjecture, the mass $m_z>0$ and the norm are characterized by

$$M(m_z e_1)=z^{-1}.$$ $$|x|_z=\frac{1}{m_z}\max_{\mu\in K_z}\mu\cdot x.$$

The axis atoms make $K_z$ compact, and

$$\begin{aligned} M(me_1)={}&\frac{1+\cosh(5m)}6\\ &+\frac{\cosh(4m)+\cosh(3m)}3. \end{aligned}$$

Consequently $m\mapsto M(me_1)$ is strictly increasing from $1$ to infinity on $[0,\infty)$. We may therefore parameterize $z$ by

$$z(m)=M(me_1)^{-1},\qquad m>0,$$

so that $m_{z(m)}=m$ and $z(m)\uparrow1$ as $m\downarrow0$.

3. An analytic expansion of the actual maximizer

Set $n=x/5=(4/5,3/5)$, so $|n|=1$. Introduce the moments

$$\begin{aligned} s&=\mathbb E U_1^2=\frac{25}{2},\\ A&=\mathbb E U_1^4=\frac{433}{2},\\ B&=\mathbb E U_1^2U_2^2=96. \end{aligned}$$

Let $I$ denote the two-by-two identity matrix. Coordinate symmetry gives $\mathbb E UU^{\mathsf T}=sI$. Define

$$\begin{aligned} Q(v)&=\mathbb E(v\cdot U)^4\\ &=A(v_1^4+v_2^4)+6Bv_1^2v_2^2. \end{aligned}$$

For $t>0$, put

$$G(t,v)=\frac{M(\sqrt t\,v)-M(\sqrt t\,e_1)}{t}.$$

Since the support is finite and centrally symmetric, the power series of $M(mv)$ contains only even powers of $m$. It follows that $G$ extends real-analytically to a neighborhood of $(0,n)$, with

$$\begin{aligned} G(t,v)={}&\frac{s}{2}(|v|^2-1)\\ &+\frac{t}{24}\bigl(Q(v)-A\bigr)+O(t^2). \end{aligned}$$

Here the remainder is analytic locally in both $t$ and $v$. Consider the equations

$$G(t,v)=0,\qquad \nabla_vG(t,v)=\lambda n.$$

At $t=0$ they hold with $v=n$ and $\lambda=s$. Their Jacobian with respect to $(v_1,v_2,\lambda)$ at that point is

$$\begin{pmatrix} sn_1&sn_2&0\\ s&0&-n_1\\ 0&s&-n_2 \end{pmatrix},$$

whose determinant is $s^2(n_1^2+n_2^2)=625/4$. The analytic implicit-function theorem therefore supplies analytic solutions $v(t)$ and $\lambda(t)$ near $0$, with $v(0)=n$ and $\lambda(0)=s>0$.

These solutions give the global maximizer, not merely a stationary point. For $t>0$,

$$\nabla_v^2G(t,v)=\nabla^2M(\sqrt t\,v)$$

is positive definite, since the support spans $\mathbb R^2$. If $G(t,w)\le0$, convexity and $G(t,v(t))=0$ give

$$\lambda(t)n\cdot\bigl(w-v(t)\bigr) \le G(t,w)\le0.$$

For all sufficiently small positive $t$, we have $\lambda(t)>0$, and hence $n\cdot w\le n\cdot v(t)$. The feasible set $G(t,w)\le0$ is precisely $K_{z(\sqrt t)}/\sqrt t$. Thus

$$|x|_{z(\sqrt t)}=5n\cdot v(t).$$

Differentiating $G(t,v(t))=0$ at $t=0$ yields

$$s\,n\cdot v'(0)+\frac{Q(n)-A}{24}=0.$$

Since $n_1^2+n_2^2=1$, the required fourth-moment difference is

$$\begin{aligned} A-Q(n)&=2(A-3B)n_1^2n_2^2\\ &=-\frac{20592}{625}. \end{aligned}$$

It follows that

$$\begin{aligned} n\cdot v'(0)&=\frac{A-Q(n)}{24s}\\ &=-\frac{1716}{15625}. \end{aligned}$$

The analyticity established above now gives

$$|x|_{z(m)}=5-\frac{1716}{3125}m^2+O(m^4).$$

The remainder divided by $m^2$ tends to zero. Therefore $|x|_{z(m)}<5$ for every sufficiently small positive $m$, and also

$$\lim_{z\uparrow1}|x|_z=5.$$

4. The other endpoint and the failure of monotonicity

Here the limit as $z\downarrow0$ can be proved directly. Put $L=\log(1/z)$ and

$$h_z=\max_{\mu\in K_z}\mu\cdot x.$$

Because $u_1\le5$ for every support point and $(5,0)$ has probability $1/12$,

$$\frac{e^{5m_z}}{12}\le M(m_ze_1)\le e^{5m_z}.$$

Consequently

$$\frac L5\le m_z\le\frac{L+\log12}{5}.$$

The atom at $x$ similarly gives $h_z\le L+\log12$. For the opposite bound, take $\mu=Lx/25$. Since $u\cdot x\le25$ for all $u\in\mathcal U$, every summand in $M(\mu)$ is at most $e^L$, so $\mu\in K_z$ and $\mu\cdot x=L$. Hence $h_z\ge L$. Combining these inequalities gives

$$\begin{gathered} \frac{5L}{L+\log12}\le |x|_z,\\ |x|_z\le\frac{5(L+\log12)}L. \end{gathered}$$

Thus $\lim_{z\downarrow0}|x|_z=5$ as well.

A monotone function on $(0,1)$ whose two endpoint limits both equal $5$ must be identically $5$. Our expansion proves that this norm is strictly smaller than $5$ near $z=1$. More explicitly, choose such a point $z_2$; the two endpoint limits supply $z_1<z_2<z_3$ with

$$|x|_{z_1}>|x|_{z_2}<|x|_{z_3}.$$

This disproves the conjectured monotonicity while satisfying all its uniform-support and boundary hypotheses.