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Statement

Setting (p. 24). An edge-weighted graph is a triple $G=(V,E,W)$ with $E\subset\{(v_1,v_2):v_1,v_2\in V,\ v_1\ne v_2\}$ and real weights $W=\{w(e)\}_{e\in E}$. For a coloring $\theta:E\to\{-1,1\}$,

$$\operatorname{disc}(G,\theta)=\max_{V'\subset V}\Bigl|\sum_{e=(v_1,v_2)\in E,\ v_i\in V'}\theta(e)w(e)\Bigr|,$$

and $\operatorname{disc}(G)=\min_\theta\operatorname{disc}(G,\theta)$.

Theorem 7 (p. 26). Let $G=(V,E,W)$ be an arbitrary edge-weighted graph. Then, with universal constants,

$$\operatorname{disc}(G)\asymp\mathsf E_\theta\max_{V'\subset V}\Bigl|\sum_{e=(v_1,v_2)\in E,\ v_i\in V'}\theta(e)w(e)\Bigr|\asymp\sum_{v\in V}\Bigl(\sum_{e\in E:\,v\in e}w(e)^2\Bigr)^{1/2},$$

the expectation being over all colorings $\theta:E\to\{\pm1\}$.

Edges are unordered. The definition writes an edge as a pair $(v_1,v_2)$, but the theorem is derived from Theorem 6 by viewing a graph on $n$ vertices as $K_n$ with zero weights on the missing edges (p. 26), where each unordered pair carries one weight and one sign. The theorem is read with one edge per unordered pair. It cannot hold if $E$ may contain both $(u,v)$ and $(v,u)$ as separate edges (an observation of this page): give both weight $1$ and opposite signs, and every signed sum vanishes, while the right side is positive.

Unit weights (computed here). On $K_n$ with every weight $1$, each vertex meets $n-1$ edges and the right side is $n(n-1)^{1/2}$, of order $n^{3/2}$ for $n\ge2$.

Source. Sergey V. Astashkin and Konstantin V. Lykov, Random unconditional convergence of Rademacher chaos in $L_\infty$ and sharp estimates for discrepancy of weighted graphs and hypergraphs, arXiv:2412.20107v1 [math.PR], 28 December 2024; Section 6 (pp. 24--27), the definitions on p. 24 and Theorem 7 on p. 26. The edition read is identified on the source card.

Read depth. Claims checked: the definitions, the statement and the derivation from Theorem 6 were read clause by clause on the page images. Nothing here is independently reviewed.

Proof pointer

P. 26. Theorem 6 holds with constants independent of $n$ and the weights, and its right side is equivalent to the vertex sum on the right here (the paper says one can readily check this; the Theorem 6 page records the constants $1/2$ and $1$). A graph on $n$ vertices is $K_n$ with zero weights on its non-edges; zero weights change neither side, so Theorem 6 applies. The paper notes that this also covers the bipartite case of Theorem 5.

Dependencies

Theorem 6 (p. 26).

Bears on

• Problem 1028: with unit weights on $K_n$, read with one sign per unordered pair as above, the theorem gives $\operatorname{disc}(K_n)$ of order $n(n-1)^{1/2}$, that is $n^{3/2}$, for every $n\ge2$ with universal constants that are not made explicit. This is the order of the problem's unordered-edge reading, which the paper attributes to Erdős and Spencer (p. 25); the theorem gives no leading constant or exact value, and the separate-signs ordered-pair reading falls outside it, as the observation above shows.