Lattice approach
This is an approach to proving (weak) FUNC in the lattice formulation by distinguishing various regimes of lattices. Let [math]\displaystyle{ \mathcal{A} }[/math] be a finite lattice containing [math]\displaystyle{ m }[/math] join-irreducibles, and write [math]\displaystyle{ n=|\mathcal{A}| }[/math].
Throwing in a new atom below every join-irreducible creates a new lattice in which abundances differ by less than a factor of 2. So for the purposes of weak FUNC, it should be possible to assume wlog that [math]\displaystyle{ \mathcal{A} }[/math] is atomistic.
Case 1: Short maximal chains
If [math]\displaystyle{ \mathcal{A} }[/math] has a maximal chain of length [math]\displaystyle{ l }[/math], then a Knill-type argument will establish the existence of a join-irreducible element with relative abundance of at most [math]\displaystyle{ 1-\frac{1}{l-1} }[/math].
Case 2: Equal maximal chains
The opposite extreme is that all maximal chains have equal length, i.e. that [math]\displaystyle{ \mathcal{A} }[/math] satisfies the Jordan-Dedekind chain condition. In this case, [math]\displaystyle{ \mathcal{A} }[/math] is graded. We distinguish two subcases depending on the height [math]\displaystyle{ h }[/math] and the width [math]\displaystyle{ w }[/math]; since [math]\displaystyle{ (h-1)(w-1)\geq n-1 }[/math], they can't both be too small. Using the JD chain condition, do arguments along the lines of Dilworth's theorem or Mirsky's theorem give more information?
Case 2a: large height
The height of [math]\displaystyle{ \mathcal{A} }[/math] can be at most [math]\displaystyle{ h=m+1 }[/math]. If this value is reached, then [math]\displaystyle{ \mathcal{A} }[/math] has a prime element by the dual of Theorem 15 in this paper, and we are done since every prime element is rare.
Case 2b: large width
In this case, can try to choose a large maximal antichain and a join-irreducible element [math]\displaystyle{ x }[/math] such that [math]\displaystyle{ \mathcal{A}_x }[/math] contains as few elements of the antichain as possible.
