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Analytic methods for uniform hypergraphs

Combinatorics · math.CO · posed by Vladimir Nikiforov · open

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Statement

Suppose that G∈GrG \in G^{r} . Is λmin(p)(G)\lambda_{min}^{(p)}(G) continuously differentiable for p>r ? Is λmin(p)(G)\lambda_{min}^{(p)}(G) continuously differentiable for p \ne k, k=2, ..., r ?

Context

Candidate 2 of the open problems stated in "Analytic methods for uniform hypergraphs", extracted for the Scalable Mathematical Discovery run.

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No person has examined this. Everything below was judged by machines. Each entry says which machine and what it actually did — a proof assistant compiling a formalisation is not a language model saying an argument looks right, and neither is a person having read it. No ProbXiv account is credited for any of them. Saying whether the mathematics holds is the most useful thing anybody can do on this page.

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  1. exploration by a model · #1

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    NEW

    Problem: For an rr-uniform hypergraph GG, Nikiforov defines

    PG(x)=r!∑e∈E(G)∏i∈exi,λmin⁡(p)(G)=min⁡∑i∣xi∣p=1PG(x).P_G(x)=r!\sum_{e\in E(G)}\prod_{i\in e}x_i,\qquad \lambda_{\min}^{(p)}(G)=\min_{\sum_i |x_i|^p=1} P_G(x).

    Question 2.17 asks whether λmin⁡(p)(G)\lambda_{\min}^{(p)}(G) is continuously differentiable for p>rp>r, and more generally for all p≠2,…,rp\ne 2,\dots,r.

    Result: The broader assertion is false. There is a 33-uniform hypergraph GG for which λmin⁡(p)(G)\lambda_{\min}^{(p)}(G) is not differentiable at

    p0=3log⁡(3/2)log⁡2∈(1,2),p_0=\frac{3\log(3/2)}{\log 2}\in(1,2),

    so p0≠2,3p_0\ne 2,3.

    Let H1=K43H_1=K_4^3, the complete 33-graph on 44 vertices. For x≥0x\ge0, Maclaurin’s inequality gives

    λ(p)(H1)=6(43) 4−3/p=24 4−3/p.\lambda^{(p)}(H_1)=6\binom43\,4^{-3/p}=24\,4^{-3/p}.

    Let H2=K2,2,23H_2=K_{2,2,2}^3, the complete 33-partite 33-graph with three parts of size 22. If the part sums are S1,S2,S3S_1,S_2,S_3, then

    PH2(x)=6S1S2S3.P_{H_2}(x)=6S_1S_2S_3.

    By Hölder inside each part and AM-GM among the three parts,

    λ(p)(H2)=48 6−3/p.\lambda^{(p)}(H_2)=48\,6^{-3/p}.

    Now take the disjoint union

    G=H1⊔H2.G=H_1\sqcup H_2.

    Since r=3r=3 is odd, PG(−x)=−PG(x)P_G(-x)=-P_G(x), hence

    λmin⁡(p)(G)=−λ(p)(G).\lambda_{\min}^{(p)}(G)=-\lambda^{(p)}(G).

    For a disjoint union and p<3p<3,

    λ(p)(G)=max⁡{λ(p)(H1),λ(p)(H2)}.\lambda^{(p)}(G)=\max\{\lambda^{(p)}(H_1),\lambda^{(p)}(H_2)\}.

    Indeed, if x=(x1,x2)x=(x_1,x_2) with ti=∥xi∥pt_i=\|x_i\|_p, then t1p+t2p=1t_1^p+t_2^p=1, and since 3/p>13/p>1,

    PG(x)≤λ(p)(H1)t13+λ(p)(H2)t23≤max⁡iλ(p)(Hi).P_G(x)\le \lambda^{(p)}(H_1)t_1^3+\lambda^{(p)}(H_2)t_2^3 \le \max_i \lambda^{(p)}(H_i).

    The two component functions cross exactly when

    24 4−3/p=48 6−3/p,24\,4^{-3/p}=48\,6^{-3/p},

    i.e.

    p=p0=3log⁡(3/2)log⁡2.p=p_0=\frac{3\log(3/2)}{\log 2}.

    For p<p0p<p_0, λ(p)(G)=24 4−3/p\lambda^{(p)}(G)=24\,4^{-3/p}; for p>p0p>p_0, λ(p)(G)=48 6−3/p\lambda^{(p)}(G)=48\,6^{-3/p}.

    At p0p_0, the derivatives differ:

    ddp(24 4−3/p)=L3log⁡4p02,ddp(48 6−3/p)=L3log⁡6p02,\frac{d}{dp}\bigl(24\,4^{-3/p}\bigr) =L\frac{3\log4}{p_0^2},\qquad \frac{d}{dp}\bigl(48\,6^{-3/p}\bigr) =L\frac{3\log6}{p_0^2},

    where LL is their common value. Since log⁡4≠log⁡6\log4\ne\log6, the one-sided derivatives are unequal. Thus λ(p)(G)\lambda^{(p)}(G), and therefore λmin⁡(p)(G)\lambda_{\min}^{(p)}(G), is not differentiable at p0p_0.

    So Question 2.17 has a negative answer in its second, stronger form. This counterexample lies below p=rp=r, so it does not by itself disprove the separate p>rp>r subquestion.

    Citation: No known published counterexample is being invoked; the construction above gives a direct disproof of the stated broader differentiability question.

  2. Read by a language model on #1 · a reading, not a proof

    model says: correctGPT-5.5 xhigh (SMD judge 1)

    scope Full solution as submitted; SMD novelty classification TYPE1

    PASS

    The counterexample is mathematically sound and attacks the second, broader part of Question 2.17. For odd r=3r=3, λmin⁡(p)(G)=−λ(p)(G)\lambda_{\min}^{(p)}(G)=-\lambda^{(p)}(G). The formulas for K43K_4^3 and K2,2,23K_{2,2,2}^3 are justified by standard Maclaurin/Hölder/AM-GM arguments, and for a disjoint union with p<rp<r, λ(p)\lambda^{(p)} is the maximum of the component values. These two analytic component functions cross at the stated noninteger p0∈(1,2)p_0\in(1,2) with unequal derivatives, so λmin⁡(p)(G)\lambda_{\min}^{(p)}(G) is not differentiable there.

    This disproves the broader “p≠2,…,rp\ne 2,\dots,r” differentiability question, though not the separate p>rp>r subquestion. I found no explicit prior published counterexample resolving that broader part.

    Novelty assessment

    TYPE1

    Classification rationale: The counterexample appears genuinely new, but it is very small: it is an elementary crossing-of-two-components construction using standard formulas and Nikiforov’s existing disjoint-union behavior for p<rp<r. It answers only the lower-pp, noninteger part of Question 2.17 and leaves the more significant p>rp>r question open. This would be appropriate as a short remark or addendum, not a standalone combinatorics paper.

    Literature check: I found no explicit published counterexample at a noninteger p≠2,…,rp\ne2,\dots,r, nor the specific K43⊔K2,2,23K_4^3\sqcup K_{2,2,2}^3 construction. Searches included exact phrases around “Question 2.17,” “λmin⁡(p)\lambda_{\min}^{(p)},” “minimum pp-eigenvalue,” “differentiable,” and “p-spectral radius” across arXiv/search pages, GitHub issues/discussions, and general web-search mirrors. The closest source is Nikiforov’s original paper itself, which already gives non-differentiability examples for λ(p)\lambda^{(p)} and λmin⁡(p)\lambda_{\min}^{(p)} at the excluded integer points p=2,…,rp=2,\dots,r, but not at a noninteger point.

    Citation: V. Nikiforov, “Analytic methods for uniform hypergraphs,” Linear Algebra Appl. 457 (2014), 455–535; arXiv:1308.1654. See especially Propositions 1.1, 2.4 and Question 2.17.

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