Formulating the Kruskal-Katona for upper shadows instead of lower shadows For $  \mathscr{F}\subset[n]^{(r)}$ we define:
the lower shadow of $ \mathscr{F}$ as $\delta\mathscr{F}$ := {B $\in$ $[n]^{(r-1)} : B \subset A$ for some $A \in \mathscr{F}$}
the upper shadow of $ \mathscr{F}$ as $\delta ^{+}\mathscr{F}$ := {B $\in$ $[n]^{(r+1)} : A \subset B$ for some $A \in \mathscr{F}$}
I have the Kruskal-Katona theorem: Let  $\mathscr{A} \subset [n]^{(k)} $ and let $\mathscr{B}$ be the family consisting of the first $|\mathscr{A}|$ elements of $[n]^{(k)} $ in colex order. Then $|\delta \mathscr{A}| \ge |\delta \mathscr{B}|$
And I want to formulate the corresponding result for the upper shadows. I know it's something to do with taking complements within each layer, and I think the results is meant to be:
 Let  $\mathscr{A} \subset [n]^{(k)} $ and let $\mathscr{B}$ be the family consisting of the first $|\mathscr{A}|$ elements of $[n]^{(k)} $ in lex order. Then $|\delta ^+ \mathscr{A}| \ge |\delta ^+ \mathscr{B}|$, but I can't see any way to get there. 
 A: The Kruskal-Katona theorem for upper shadows can be stated as follows:
Theorem:
Let  $\mathcal{F}\subseteq [n]^{(r)}$ and let $\mathcal{A}$ be the family consisting of the final $|\mathcal{F}|$ elements of $[n]^{(r)}$ in colex order. Then $|\partial^+\mathcal{F}|\geq|\partial^+\mathcal{A}|$


Proof:
For a family $\mathcal{A}\subseteq [n]^{(r)}$, let $\mathcal{A'}$ be defined as $\mathcal{A'}:=\{[n]\setminus A: A\in \mathcal{A}\}\in[n]^{(n-r)}$
The intuition for why this is the formulation for upper shadows might go along the lines of:

*

*Taking a shadow of $\mathcal{A'}$ is like taking an upper shadow of $\mathcal{A}$

*Minimising the upper shadow of $\mathcal{A}$ is therefore like minimising the lower shadow of $\mathcal{A'}$

*By Kruskal-Katona for lower shadows, we want to choose $\mathcal{A'}$ to be an initial segment of the colex.

*If $\mathcal{A'}$ is an initial segment of colex of $[n]^{(n-r)}$, then $\mathcal{A}$ is a final segment of colex of $[n]^{(r)}$
Thus the proof is reduced to formalising the intuition above and showing that $\partial(\mathcal{A'})=\partial^+(\mathcal{A})$, and that $A<_{colex}B\iff[n]\setminus B <_{colex} [n]\setminus A$, where $X<_{colex}Y$ means X precedes Y in colex order.
