On the normalized Shannon capacity of a union

TL;DR

This paper investigates the normalized Shannon capacity of graph unions and proves that it cannot be significantly greater than 1 when both component graphs have capacities close to zero, answering a question posed by Alon.

Contribution

The paper proves that the normalized Shannon capacity of a union of graphs cannot exceed 1 minus a small epsilon if each graph's capacity is less than epsilon, resolving Alon's question.

Findings

Normalized Shannon capacity of graph unions is bounded by 1 minus epsilon.

Graphs with small individual capacities cannot have a union with capacity close to 1.

Theoretical limitation on the capacity increase through graph union.

Abstract

Let $G_1 \times G_2$ denote the strong product of graphs $G_1$ and $G_2$, i.e. the graph on $V(G_1) \times V(G_2)$ in which $(u_1,u_2)$ and $(v_1,v_2)$ are adjacent if for each $i=1,2$ we have $u_i=v_i$ or $u_iv_i \in E(G_i)$. The Shannon capacity of $G$ is $c(G) = \lim_{n\to \infty} \alpha (G^n)^{1/n}$, where $G^n$ denotes the $n$-fold strong power of $G$, and $\alpha (H)$ denotes the independence number of a graph $H$. The normalized Shannon capacity of $G$ is $C(G) = \frac {\log c(G)}{\log |V(G)|}$. Alon asked whether for every $\epsilon > 0$ there are graphs $G$ and $G'$ satisfying $C(G), C(G') < \epsilon$ but with $C(G + G') > 1 - \epsilon $. We show that the answer is no.