Discrete Geometry and Extremal Graph Theory - IMSA

We begin by proving the following.
Theorem 1. A graph G on n vertices without a 4-cycle contains at most cn 3/2 edges for
some constant c > 0.
Proof of Theorem 1. Let d 1 , d 2 , . . . , d n be the degrees of the vertices v 1 , v 2 , . . . , v n of G.
Let E be the number of edges of G, so 2E = ∑ n
i=1 d i. For a pair of vertices v 1 and v 2 , note
that there can be at most one other vertex v 3 connected to both v 1 and v 2 . Otherwise, for
if v 4 was also joined to v 1 and v 2 , we would have a 4-cycle with vertices v 1 , v 3 , v 2 , v 4 . Thus,
more generally, for a vertex v i connected to d i other vertices, we count every pair of these d i
vertices. If we repeat this count for all the vertices v 1 , v 2 , . . . , v n , note that our total count
must less than the total number of pairs of vertices, ( n
2)
. This is because every pair of some
d i vertices connected to v i can be uniquely associated with v i . In other words, we have the
following inequality.
( )
d1
+
2
(
d2
) ( ) (
dn n
+ · · · ≤ .
2 2 2)
We claim that ( d 1
) (
2 +
d2
) (
2 + · · ·
dn
)
2 ≥
2E 2
− E, so it will follow that
n
( )
2E 2 n
n − E ≤ n(n − 1)
=
2 2
⇒ 4E 2 − 2En ≤ n 2 (n − 1).
Thus, ( )
2E − n 2
2 ≤ n 2 (n − 1) + n2 ≤ 4 n3 , so we have 2E ≤ n 3/2 + n . Finally, we conclude
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that E ∼ < n 3/2 (and actually, c ≈ 1 will suffice for sufficiently large n).
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Now, we give two proofs of the inequality ( d 1
) (
2 +
d2
) (
2 + · · · +
dn
)
2 ≥
2E 2
− E. n
• First Proof. By the Quadratic-Arithmetic Mean Inequality (or Cauchy-Schwartz), we
have
( )
d1
+
2
( )
d2
+ · · · +
2
( )
dn
= 1 2 2 (d2 1 + d 2 2 + · · · + d 2 n) − 1 2 (d 1 + d 2 + · · · + d n )
≥ 1
2n (2E)2 − E.
• Second Proof. As a function, f(x) = ( )
x
2 =
x(x−1)
is convex, so by Jensen’s inequality,
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we have ( ) ( ) ( ) 2E
)
d1 d2
dn
+ + · · · + ≥ n(
n
= 2E2
2 2
2 2 n − E.
This concludes our proof.
The “Θ-graph” is the union of three internally disjoint (simple) paths that have the same
two end vertices, as shown below.
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