Upper shadow of a union-closed family

Question

Is the following claim true?

Let $\mathcal{F} \subseteq 2^{[n]}$ be a union-closed family. That is, if $A,B \in \mathcal{F}$, then $A \cup B \in \mathcal{F}$. Then

$$|\partial^+\mathcal{F}\setminus \mathcal{F}|\leq 2^{n-1},$$

where $\partial^+ \mathcal{F}$ is the upper shadow of $\mathcal{F}$.

Edit: To be explicit about what is the upper shadow of $\mathcal{F}$:

$\partial^+ \mathcal{F}=\{A\cup \{x\}|A \in \mathcal{F}, x\in [n]\setminus A\}$

Edit: Ok, so the above statement turned out to be true. What about the following generalization?

Similar to the above definition of the upper shadow, we can also define, for any $k$, the $k$th upper-shadow of a family $\mathcal{F}\subseteq 2^{[n]}$ as follows:

$\partial_k^+\mathcal{F}:=\{A \cup K|A \in \mathcal{F}\wedge K\subseteq [n]\setminus A \wedge |K|=k\}$.

Denote $\partial_{\leq k}^+\mathcal{F}:=\bigcup_{i=1}^k\partial_i^+\mathcal{F}$.

Now, Let $\mathcal{F}\subseteq 2^{[n]}$ be a union-closed family. Is it true that for any $k \in [n]$, $|\partial_{\leq k}^+\mathcal{F}\setminus \mathcal{F} |\leq 2^n-2^{n-k}$? We know it's true for $k=1$, but I wonder if it's true in general. If true, this would be tight for any $k$, since one can take the family $\mathcal{F}=2^{[n-k]}$ as a subfamily of $2^{[n]}$.

Also, it would be interesting to find bounds on $|\partial_k^+\mathcal{F}\setminus \mathcal{F}|$.I can prove that for any $k$ this quantity is at most $2^{n-1}$, and I know that for $k=1$ this is tight, but for larger $k$ perhaps the bound is lower. For instance, for $k=2$ the best I can do is $\frac{3}{8}2^n$, by taking $\mathcal{F}=2^{[n-3]}$.

Answer 1

I ended up proving that this is indeed the case. Theorem $1.3$ in this paper:

https://arxiv.org/abs/1708.01434