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A k-Sperner family meets every maximal chain in a Boolean lattice in at most $k$ members. The random-chain calculation in part (a) therefore gives the generalized LYM inequality $L(\mathcal{A})\le k$. Define the proportion of rank $i$ that is present byThen ${\sum }_i{x}_i=L(\mathcal{A})\le k$ and $w(\mathcal{A})={\sum }_{i}u(i)x_i$. This reduces the weighted theorem for k-Sperner families to allocating at most $k$ units of mass among ranks, each of capacity one.

To handle ties explicitly, let $t>0$ be the $k$th largest value among the $u(i)$, let $H=\{i:u(i)>t\}$, and put $h=|H|<k$. ThenThe last expression is exactly the sum of the $k$ largest rank weights, with multiplicity. Writing those values in descending order as $u_{(1)},\dots ,u_{(n+1)}$, the answer is

Choose $k$ distinct ranks whose $u(i)$ values are the $k$ largest, resolving any tie arbitrarily, and take the union of their full levels. Along a strict chain in a partially ordered set, cardinality strictly increases, so this set family is a k-Sperner family. Its weight is the sum of the selected rank weights because a full level of rank $i$ has $\left(\genfrac{}{}{0ex}{}{n}{i}\right)$ members. Thus the bound in the weighted theorem for k-Sperner families is always attained: