Weakly norming graphs are edge-transitive

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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