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The Kruskal-Katona theorem states that if $\mathcal{A}\subseteq [n{]}^{(r)}$ has the unique binomial representationthen its lower shadow obeysAn initial segment of colexicographic order has exactly this shadow.

Here is the UV-compression proof of the Kruskal-Katona theorem. For disjoint equal-size sets $U,V$, the UV-compression $C_{U,V}$ replaces the pattern $V$ by $U$ when the image is not already in the family. If the family is not an initial colexicographic segment, choose a changing pair with $\max U<\max V$ and $|U|$ minimal. Minimality ensures that for every $x\in U$ an appropriate smaller compression $C_{U\setminus \{x\},V\setminus \{y\}}$ already fixes the family. The Shadow lemma for UV-compressions then givesMeanwhile the binary weight ${\sum }_{A\in \mathcal{A}}{\sum }_{i\in A}{2}^i$ strictly decreases. Iterating must terminate, and a terminal family is an initial colexicographic segment. Computing that segment's shadow from its binomial representation proves the theorem.

We next classify pairs that decrease the shadow for every uniform family. The answer, including the identity case, isFor $|U|=0$ the operation is the identity. For $|U|=1$, the hypotheses of the Shadow lemma for UV-compressions reduce to stability under the empty compression and therefore hold for every family.

To see failure for larger pairs, relabel freely. If $|U|=|V|=2$, write $U=\{u_1,u_2\}$ and $V=\{v_1,v_2\}$. The familyhas a three-point lower shadow, whereas its compression replaces $v_1{v}_2$ by $u_1{u}_2$ and has a four-point lower shadow. If the common size is $m\ge 3$, takeThe old two shadows overlap once and have size $2m-1$. Compression replaces $V$ by $U$; the two resulting $m$-sets intersect in only $u_1$, so their lower shadows are disjoint and have size $2m$.

For the two specified pairs, the answer is no in both cases, even after assuming the family is left-compressed.

For $(U,V)=(345,126)$, takeThis family is left-compressed. Its compression replaces $126$ by $345$. The old shadow isof size nine, while the new shadow replaces the last two pairs by $34,35,45$ and has size ten.

For $(U,V)=(145,236)$, take the left-compressed familyIts shadow consists of $12,13,23$ and the nine pairs having one element in $\{1,2,3\}$ and one in $\{4,5,6\}$, so it has size twelve. Compression replaces $236$ by $145$; all twelve old shadow pairs remain and $45$ is added. The new shadow therefore has size thirteen.

Let $c(A)=[n]\setminus A$ and reverse the order of the ground set by $j\mapsto n+1-j$. The resulting bijection sends lexicographic order on $[n{]}^{(r)}$ to colexicographic order on $[n{]}^{(n-r)}$. It also sends the iterated upper shadow of an $r$-uniform family at level $r+1$ to the lower shadow of the corresponding $(n-r)$-uniform family, up to the same harmless reversal of the ground set.

The Kruskal-Katona theorem says that this lower shadow is smallest for an initial colex segment. Undoing the complement and reversal therefore says that the upper shadow of a family of fixed size in $[n{]}^{(r)}$ is smallest for the initial lex segment.

Every positive integer $m$ has a unique binomial representationDefineThe Kruskal-Katona theorem says that every $r$-uniform family $\mathcal{A}$ of size $m$ hasand the initial segment of length $m$ in colexicographic order attains equality.

We prove the inequality by simultaneous mathematical induction on the ground-set size and on $r$. We use the elementary binomial-shadow arithmetic lemmaTo verify the lemma, greedily remove the largest term $\left(\genfrac{}{}{0ex}{}{q}{r}\right)$ from the binomial representation of $a+b$. If $a$ reaches that term first, apply the induction hypothesis to the remainders; otherwise transfer the excess to $b$. In the crossing case Pascal's identitygives exactly the two terms on the right. This induction also proves uniqueness of the greedy representation.

Choose the largest ground-set element $x$ and splitThe members of the lower shadow that avoid $x$ form $\partial {\mathcal{A}}_0\cup {\mathcal{A}}_1$, while those containing $x$ are the sets $B\cup \{x\}$ with $B\in \partial {\mathcal{A}}_1$. Hence, writing $a=|{\mathcal{A}}_0|$ and $b=|{\mathcal{A}}_1|$,This completes the induction.

Finally, the initial colex segment first contains all $r$-subsets of $[a_r]$, followed recursively by sets containing $a_r+1$ whose remaining elements form the appropriate initial segment at level $r-1$. Its shadow has the same recursive decomposition, and therefore has exactly $f_r(m)$ members. This proves sharpness.