Proof of the consecutive unique-difference bound

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Let $B\subset\mathbb Z$ have $m\ge2$ elements and suppose that every integer $1\le d\le L$ has exactly one representation $d=b'-b$ with $b,b'\in B$. If $L=0$, the bound is immediate. Otherwise put $u=\lfloor(L+1)/2\rfloor\ge1$, so $2u-1\le L\le2u$.

For every integer $j$ define

$$F(j)=|B\cap[j,j+u-1]|, \qquad r(d)=|\{(b,b')\in B^2:b'-b=d\}|.$$

All sums below have finite support. Counting the intervals that contain two specified points gives

$$\sum_j F(j)^2 =mu+2\sum_{d=1}^{u-1}(u-d)r(d) =u(m+u-1). \tag{1}$$

For a point $b$ in $[j,j+u-1]$ and a point $b'$ in the adjacent interval $[j+u,j+2u-1]$, the positive difference $d=b'-b$ lies between $1$ and $2u-1$. There are $u-|u-d|$ choices of $j$ for this pair. Thus

$$\sum_j F(j)F(j+u) =\sum_{d=1}^{2u-1}(u-|u-d|)r(d)=u^2. \tag{2}$$

The only representation counts used in (1) and (2) are $r(d)=1$ for $1\le d\le2u-1$. Translation of the index $j$ then gives

$$\sum_j\bigl(F(j+u)-F(j)\bigr)^2=2u(m-1). \tag{3}$$

For each $a\in\{0,\ldots,u-1\}$ let $f_a(n)=F(a+nu)$ for $n\in\mathbb Z$. These values count $B$ in a partition of the integers into disjoint intervals, so $f_a\ge0$ and $\sum_n f_a(n)=m$. Write

$$T_a=\sum_n f_a(n)^2,\qquad E_a=\sum_n(f_a(n+1)-f_a(n))^2.$$

The bijection $j=a+nu$ and (1), (3) yield

$$\frac1u\sum_a T_a=m+u-1,\qquad \frac1u\sum_a E_a=2(m-1). \tag{4}$$

Fix a positive integer $k$ and average $k$ translates of each sequence:

$$g_a(n)=\frac1k\sum_{i=0}^{k-1}f_a(n+i).$$

Nonnegativity and the mass identity give $0\le g_a(n)\le m/k$ and $\sum_n g_a(n)=m$. Therefore $\sum_n g_a(n)^2\le m^2/k$. Telescoping consecutive translates and applying the triangle inequality in $\ell^2(\mathbb Z)$ give

$$\|f_a-g_a\|_2 \le\frac1k\sum_{i=0}^{k-1}\|f_a-f_a(\,\cdot+i)\|_2 \le\frac{k-1}{2}\sqrt{E_a}.$$

Now use the triangle inequality for arrays indexed by $(a,n)$, with squared norm $u^{-1}\sum_{a,n}|\cdot|^2$. Equation (4) gives, for every positive integer $k$,

$$\sqrt{m+u-1} \le\frac{m}{\sqrt k}+(k-1)\sqrt{\frac{m-1}{2}}. \tag{5}$$

Take $k=\lceil m^{1/3}\rceil$. Since $k\ge m^{1/3}$ and $k-1\le m^{1/3}$, the right side is at most $(1+1/\sqrt2)m^{5/6}$. Squaring and using $L\le2u$ proves

$$L\le2u \le2(1+1/\sqrt2)^2m^{5/3}-2m+2 =(3+2\sqrt2)m^{5/3}-2m+2.$$

This proves the claim. No step bounds the support diameter or uses uniqueness for any difference larger than $L$. $\square$