弱赋范图是边传递的
Weakly norming graphs are edge-transitive
摘要：
设$\mathcal{H}$是$[0,1]^2$上的有界可测对称函数类.对于函数$h \in \mathcal{H}$和图$G$，定义了\[ t_G(h) \; = \; \int \cdots \int \prod_{\{v_i,v_j\} \in E(G)} h(x_i,x_j) \: dx_1 \cdots dx_n \: . \]，并证明了若要使$t_G(|h|)^{1/|E(G)|}$成为$\mathcal{H}$上的范数，则图$G$必须是边传递的.
英文摘要：
Let $\mathcal{H}$ be the class of bounded measurable symmetric functions on $[0,1]^2$. For a function $h \in \mathcal{H}$ and a graph $G$ with vertex set $\{v_1,\ldots,v_n\}$ and edge set $E(G)$, define \[ t_G(h) \; = \; \int \cdots \int \prod_{\{v_i,v_j\} \in E(G)} h(x_i,x_j) \: dx_1 \cdots dx_n \: . \] Answering a question raised by Conlon and Lee, we prove that in order for $t_G(|h|)^{1/|E(G)|}$ to be a norm on $\mathcal{H}$, the graph $G$ must be edge-transitive.
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