Subdigraphs of prescribed size and out‐degree

In 2006, Noga Alon raised the following open problem: Does there exist an absolute constant c>0 $c\gt 0$ such that every 2n $2n$ ‐vertex digraph with minimum out‐degree at least s $s$ contains an n $n$ ‐vertex subdigraph with minimum out‐degree at least s2−c $\frac{s}{2}-c$ ? In this note, we answer this natural question in the negative, by showing that for arbitrarily large values of n $n$ there exists a 2n $2n$ ‐vertex tournament with minimum out‐degree s=n−1 $s=n-1$ , in which every n $n$ ‐vertex subdigraph contains a vertex of out‐degree at most s2−12+o(1)log3(s) $\frac{s}{2}-\left(\frac{1}{2}+o(1)\right){\mathrm{log}}_{3}(s)$ .


| INTRODUCTION
The following open problem was raised by Alon in 2006 in the article [2] on digraph splitting.
Problem 1 (cf.[2, problem 4.1]).Does there exist an absolute constant c > 0 such that the following holds for all pairs of natural numbers n s ( , )? Every n 2 -vertex digraph D of minimum out-degree at least s contains an n-vertex subdigraph of minimum out-degree at least c − s 2 .
Using an elegant probabilistic argument, Alon proved in [2] that a weaker bound holds, as follows: Every n 2 -vertex digraph of minimum out-degree at least s contains an n-vertex subdigraph of minimum out-degree at least O s s − ( log ) . Furthermore, in [3] he proved that a generalization of Problem 1 to asymmetric sizes of the subdigraphs does not hold.
In this short note, we resolve Problem 1 in the negative, by proving the following result via an explicit construction.
Theorem 1.For an infinite sequence of numbers n ∈ , there exists a n 2 -vertex tournament whose minimum out-degree is s n = − 1, while the minimum out-degree of every n-vertex subdigraph is at most .
Following the notation from [2], for every s ∈ denote by d s ( ) the largest integer such that every n 2 -vertex digraph of minimum out-degree at least s contains an n-vertex subdigraph of minimum out-degree at least d s ( ).Alon's result in [2] and Theorem 1 yield It remains an interesting open problem to close the gap between the lower and upper bounds.As the digraphs constructed in this paper to prove that d s s s ( ) < 2 − Ω(log ) ∕ are tournaments, and since tournaments are a widely studied and natural class of digraphs, it may also be of interest to study the quantity d s ( ) , defined as the largest integer d such that every n 2 -vertex tournament of minimum out-degree at least s contains an n-vertex subtournament of minimum out-degree at least d.Interestingly, in this setting the s log -factor in Alon's lower bound on d s ( ) can be removed using discrepancy-theoretic methods: It follows directly from theorem 7.1 in the paper [1] by Alon, Bang-Jensen, and Bessy that there exists an absolute constant c such that every n 2 -vertex tournament of minimum out-degree s contains an n-vertex subtournament of minimum out-degree at least c s − s 2 . In consequence, the currently bestknown bounds in the case of tournaments are T Given a discrepancy-theoretic intuition, it seems reasonable to suspect that maybe both d s ( ) and d s ( ) by roughly s Θ( ).However, the author has no tangible evidence in this direction.
Terminology.All digraphs considered in this paper have no loops or parallel arcs.For a digraph D we denote by V D ( ) its vertex set and by of vertices we denote by D X [ ] the subdigraph of D induced by X , that is, consisting of the vertex set X and all arcs of D going between vertices of X .By δ D ( ) , we denote the minimum out-degree of D, which is the smallest out-degree occurring in D if D is nonempty, and defined as 0 when D is the empty digraph.
Theorem 1 is an immediate consequence of the following result.
Theorem 2. For every integer k 0 ≥ there exists a tournament T k on 3 k vertices in which every vertex has out-degree 3 − 1 2 k , and such that the following holds: For every set X V T ( ) ≤ .
To deduce Theorem 1 from Theorem 2, pick k 0 ≥ arbitrarily, let , and let D k be the n 2 -vertex tournament obtained from T k by deleting an (arbitrarily selected) vertex.Since deleting a vertex can lower the out-degree by at most 1, we have that

≤
. Thus, the statement of Theorem 1 follows from Theorem 2.
Let us now go about proving Theorem 2. The tournaments T ( ) ≥ have the following simple recursive definition: • T 0 is the one-vertex-tournament.
• For every integer k 0 ≥ , the tournament T k+1 is obtained from the disjoint union of three isomorphic copies of T k on vertex sets A B C , , k k k by adding all possible arcs from A k to B k , from B k to C k and from C k to A k .
It is immediate from this definition that for every k 0 ≥ the digraph T k is a tournament on 3 k vertices and that every vertex has the same in-and out-degree, hence, the out-degree of each vertex equals 3 − 1 2 k .We now prove the statement of Theorem 2 by induction on k.
Proof of Theorem 2. For k = 0, every induced subdigraph of T 1 has minimum out-degree 0 by definition, which is equal to , and thus the induction basis holds.For the induction step, assume that for some integer k 0 , and let us show the corresponding statement for T k+1 .

So let
. Then possibly after relabeling, we may assume w.l.o.g. that X = B ∅ and X A ≠ ∅.Pick a vertex x X A ∈ .Since all its out-neighbors within X are also contained in A k , and since the copy of T k induced on Next, suppose that x x x , , > 0 . We then have, by construction of Among the three numbers x x , A B and x C , at least two are of size at least 3 + 1 2 k , or at least two are of size at most 3 − 1 2 k .Without loss of generality (possibly after relabeling), we may assume that x A and x B have this property.We now proceed according to the two possible cases.
. We want to estimate as follows:

≤
To estimate the minimum out-degree of Since adding a single vertex can increase the minimum outdegree by at most 1, we conclude that ( ) Putting things together, we obtain: as desired, and this concludes the proof in the first case.

≤
. We can now estimate the minimum out-degree in X via thus proving the desired bound on the minimum degree also in the second case.This concludes the proof of the theorem.□

Case 2
we can apply the inductive assumption to the setX A in the tournament T A [ ] k k +1, which is isomorphic to T k .We conclude that δ T X (