For the -tournament Moon [ 184 ] proved the following extension of Landau's theorem.
for , with equality for .
Later Kemnitz and Dulff [ 144 ] reproved this theorem.
The proof of Kemnitz and Dulff is based on the following lemma, which is an extension of a lemma due to Thomassen [ 270 ].
Lemma 25.5 (Thomassen [ 270 ]) Let be a vertex of maximum score in an -tournament . If is a vertex of different from , then there is a directed path from to of length at most 2.
Proof. ([ 144 ]) Let be all vertices of such that . If then for the length of path . Otherwise if there exists a vertex , such that then . If for all then there are arcs which implies , a contradiction to the assumption that has maximum score.
To prove the sufficiency of (25.7) we assume that the sequence is a counterexample to the theorem with minimum and smallest with that choice of . Suppose first that there exists an integer , such that
Because the minimality of , the sequence is the score sequence of some -tournament .
Consider the sequence defined by , . Because of by assumption it follows that
which implies . Since is nondecreasing also is a nondecreasing sequence of nonnegative integers.
For each with it holds that
with equality for since by assumption
Therefore the sequence fulfils condition (25.8), by the minimality of , is the score sequence of some -tournament . By forming the disjoint union of and we obtain a -tournament with score sequence in contradiction to the assumption that is counterexample.
Now we consider the case when the inequality in condition (25.8) is strict for each , . This implies in particular .
The sequence is a nondecreasing sequence of nonnegative integers which fulfils condition (25.8). Because of the minimality of , is the score sequence of some -tournament . Let denote a vertex of with score and a vertex of with score . Since has maximum score in there is a directed path from to of length at most 2 according to Lemma 25.5. By reversing the arcs of the path we obtain an -tournament with score sequence . This contradiction completes the proof.