Consider a $d$-regular, $k$-uniform hypergraph: the elements $S$ of its set $E$ of edges are subsets of $V$ of size $k$, and each vertex $v\in V$ is in $d$ edges. We can then define its adjacency tensor $\mathscr{A} = (a_{i_1\dotsb i_k})$ by $a_{i_1\dotsb i_k}=1$ for $\{i_1,\dotsc,i_k\}\in E$ and $a_{i_1\dotsb i_k}=1$ otherwise.
There is a standard definition of the spectral radius $\rho(\mathscr{A})$ of a tensor $\mathscr{A}$ (see, e.g., https://www.degruyter.com/document/doi/10.1515/math-2020-0143/html). My question is whether one can define an operator norm of $\mathscr{A}$ so that (a) it makes intuitive sense, (b) it is bounded by $\rho(\mathscr{A})$.
Let me ask this question in a precise form. Is it the case that, for any $f_1,\dotsc,f_k:V\to \mathbb{C}$, $$|\sum_{S\in E} \prod_{v\in S} f_i(v)| \leq \frac{\rho(\mathscr{A})}{(k-1)!} |V| \prod_{i=1}^k |f_i|_2,$$
where $|\cdot|_2$ is the $\ell^2$ norm on functions $V\to \mathbb{C}$ with respect to the uniform probability measure on $V$? (Rationale for the factor $(k-1)!$: the example of constant functions gives us $\rho(\mathscr{A})\geq d (k-1)!$, and would also make the inequality tight as stated, if in fact $\rho(\mathscr{A})=d (k-1)!$.) If not, is there a sensible way to modify this inequality so that it holds -- tightly, if possible?
Assuming that the inequality above (modified or not) holds in some overly simple way: what if we replace $\mathscr{A}$ by its restrictions to functions $f_1,\dotsc,f_k:V\to \mathbb{C}$ orthogonal to constant functions, and define $\rho(\mathscr{A})$ accordingly? Is the inequality above still correct (for functions orthogonal to constant functions)?
