Inequality of $h$-vectors of shellable simplicial complexes I've been studying the article of Bjorner entitled "Homology and shellability of matroid complexes". At a certain point he states an exercise that says:

Let $\Delta$ be a shellable simplicial complex and $h_i$ be the coefficients of the $h$-vector of $\Delta$. Prove that, for each $i\geq 1$, $h_i\geq 2$ implies $h_{i-1}\geq 2$.

I was able to prove that if $h_i\geq 1$ then $h_{i-1}\geq 1$ using a standard result that says that for each facet $F$ (different of the first) and its restriction $\mathscr{R}(F)$ induced by the shelling order, it holds that $\mathscr{R}(F) = \mathscr{R}(G)\cup \{x\}$ for another facet $G$.
This result seems to be well-known, but I'm not being able to prove it. I've put in MSE and included a bounty there but nobody answered so far.
 A: Does Björner state the fundamental result that if $\Delta$ is a shellable simplicial complex, then there is a multicomplex $\Gamma$ (a collection of multisets such that if $F\in\Gamma$ and $G\subset F$, then $G\in\Gamma$) such that $h_i$ is the number of $i$-element multisets in $\Gamma$? (See Chapter 12 of my book Algebraic Combinatorics, second ed., for a treatment of this topic.) Suppose that a multicomplex $\Gamma$ contains only one $(i-1)$-element multiset $G$. If $G$ has more than one distinct element, then there is no $i$-element multiset $F$ whose only $(i-1)$-element multisubset is $G$. If $G$ consists of $i-1$ copies of an element $x$, then the only choice for $F$ is the multiset consisting of $i$ copies of $x$.
A: To provide another answer (or rather a hint, as it is an exercise): You can indeed, as Richard suggested, use the characterization theorem using multicomplexes. More elementary, but a little less insightful, you can use the following idea: the contribution to $h_i$ appears when along a shelling, the minimal new face is of cardinality face. Consider the complementary face, of cardinality $d-i$ (if the complex is of dimension $d-1$). This is the minimal "old" face. Note that if the minimal old faces do not coincide, then by your argument, you immediately get two generators of $h_i-1$ in the respective stars of old faces.
But if they do have an intersection, you can pass to the link of the intersection face. Hence, you end up in the case where $i=$ dimension of the complex $+1$.
Now, if the minimal new faces (the entire cardinality $i$-face, as we know now) of both generators intersect, you can once again pass to a link, unless $i=2$. You end up in one of two cases: Either the two generators are disjoint, or $i=2$ and you are in a graph. Either case is a somewhat easy exercise.
