The following question is motivated by the recent breakthrough [result](https://arxiv.org/abs/2211.09055) by Justin Gilmer on the union-closed sets (aka Frankl) conjecture.


Let $\mathcal{F}\subseteq\mathcal{P}(\mathbb{N})$ be a finite, union-closed family, i.e. $A,B\in\mathcal{F}\Rightarrow A\cup B\in\mathcal{F}$. Let $A\_{\mathcal{F}}$ and $B\_{\mathcal{F}}$ be two independent random variables, uniformly distributed over the elements of $\mathcal{F}$. Denoting by $H$ the entropy, we clearly have
$$H(A\_{\mathcal{F}}\cup B\_{\mathcal{F}})\leq H(A\_{\mathcal{F}})=\ln|\mathcal{F}|,$$
since the entropy is maximized by the uniform distribution. I am wondering whether a sharper bound of the form $H(A\_{\mathcal{F}}\cup B\_{\mathcal{F}})\leq \lambda H(A\_{\mathcal{F}})$, for some $\lambda<1$, still continue to hold. The intuition is that the distribution $A\_{\mathcal{F}}\cup B\_{\mathcal{F}}$ should "deviate" enough from the uniform bound in order to get a non-trivial upper bound on its entropy. Denoting by $\mathrm{UC}$ the collection of all finite, union-closed families $\mathcal{F}\subseteq\mathcal{P}(\mathbb{N})$ (with $|\mathcal{F}|>1$), we then consider the following quantity:


$$\lambda:=\sup\_{\mathcal{F}\in UC}\frac{H(A\_{\mathcal{F}}\cup B\_{\mathcal{F}})}{H(A\_{\mathcal{F}})}=\sup\_{\mathcal{F}\in UC}\frac{H(A\_{\mathcal{F}}\cup B\_{\mathcal{F}})}{\ln|\mathcal{F}|}.$$


We have $\lambda\leq 1$, and by considering the union-closed families $\mathcal{P}[n]\setminus\{\emptyset\}$, we get the lower bound $\lambda\geq 0.82$.


**Is $\lambda=1$ or $\lambda<1$? In the latter case, it is possible to provide an explicit, non-trivial upper bound?**



Entropy upper bound for the union of uniform distributions over union-closed families
The best upper bound is $\lambda=1$. Here is a simple family of examples:
Let $\mathcal F\_n$ be $\Big\{\{1,\ldots,i\}\colon 1\le i\le n\}\Big\}$. That is $\mathcal F\_n$ is the collection of all initial segments of $\{1,\ldots,n\}$, which is clearly union-closed. Clearly $|\mathcal F\_n|=n$, so that $H(A)=\log n$. Also if $A:=[1,i]$ and $B:=[1,j]$ are random elements of $\mathcal F\_n$, then $[1,i]\cup[1,j]=[1,\max(i,j)]$ so that $\mathcal P(A\cup B)=\{1,\ldots,k\}=(2k-1)/n^2$. We now estimate $H(A\cup B)$. This is given by
$$
H(A\cup B)=-\sum\_{j=1}^n \frac{2j-1}{n^2}\log \frac{2j-1}{n^2}.
$$
We will approximate this by a Riemann sum to show that for large $n$,
$H(A\cup B)\approx H(A)-\log 2+\frac 12$.
To see this,
we have
$$
H(A\cup B)=\sum\_{j=1}^n\frac{2j-1}{n^2}\log n-\frac 1n
\sum\_{j=1}^n \frac{2j-1}n\log\frac{2j-1}n.
$$
Hence
$$
H(A\cup B)\approx \log n - \int\_{0}^1 2x\log(2x)\,dx,
$$
giving $H(A\cup B)\approx\log n-\log 2+\frac 12$ as claimed.


