We address a supersaturation problem in the context of forbidden subposets. A family $\mathcal{ F}$ of sets is said to contain the poset $P$ if there is an injection $i:P \rightarrow \mathcal{ F}$ such that $p \le_P q$ implies $i(p) \subset i (q)$. The poset on four elements $a,b,c,d$ with $a,b \le c,d$ is called butterfly. The maximum size of a family $\mathcal{ F} \subseteq 2^{ [n]}$ that does not contain a butterfly is $\Sigma(n,2)=\binom{ n}{ \lfloor n/2 \rfloor}+\binom{ n}{ \lfloor n/2 \rfloor+1}$ as proved by De Bonis, Katona, and Swanepoel. We prove that if $\mathcal{ F} \subseteq 2^{ [n]}$ contains $\Sigma(n,2)+E$ sets, then it has to contain at least $(1-o(1))E(\lceil n/2 \rceil +1)\binom{ \lceil n/2\rceil}{ 2}$ copies of the butterfly provided $E\le 2^{ n^{ 1-\varepsilon}}$ for some positive $\varepsilon$. We show by a construction that this is asymptotically tight and for small values of $E$ we show that the minimum number of butterflies contained in $\mathcal{ F}$ is exactly $E(\lceil n/2 \rceil +1)\binom{ \lceil n/2\rceil}{ 2}$.
Categories: math.CO