proof of Mantel’s theorem

proof of Mantel’s theorem

Let

G

be a triangle-free graph. We may assume that

G

has at least three vertices and at least one edge; otherwise, there is nothing to prove. Consider the set

P

of all functions

c

:

V

⁢

(

G

)

→

ℝ

+

such that

∑

v

∈

V

⁢

(

G

)

c

⁢

(

v

)

=

1

. Define the total weight

W

⁢

(

c

)

of such a function by

W

⁢

(

c

)

=

∑

u

⁢

v

∈

E

⁢

(

G

)

c

⁢

(

u

)

⋅

c

⁢

(

v

)

.

By declaring that

c

≤

c

*

if and only if

W

⁢

(

c

)

≤

W

⁢

(

c

*

)

we make

P

into a poset.

Consider the function

c

0

∈

P

which takes the constant value

1

|

V

⁢

(

G

)

|

on each vertex. The total weight of this function is

W

⁢

(

c

0

)

=

∑

u

⁢

v

∈

E

⁢

(

G

)

1

|

V

⁢

(

G

)

|

⋅

1

|

V

⁢

(

G

)

|

=

|

E

⁢

(

G

)

|

|

V

⁢

(

G

)

|

2

,

which is

positive

because

G

has an edge. So if

c

≥

c

0

in

P

, then

c

has support on an

induced subgraph

of

G

with at least one edge.

We claim that a

maximal element

of

P

above

c

0

is supported on a copy of

K

2

inside

G

. To see this, suppose

c

≥

c

0

in

P

. If

c

has support on a

subgraph

larger than

K

2

, then there are nonadjacent vertices

u

and

v

such that

c

⁢

(

u

)

and

c

⁢

(

v

)

are both positive. Without loss of generality, suppose that

∑

u

⁢

w

∈

E

⁢

(

G

)

c

⁢

(

w

)

≥

∑

v

⁢

w

∈

E

⁢

(

G

)

c

⁢

(

w

)

.

(*)

Now we push the function off

v

. To do this, define a function

c

*

:

V

⁢

(

G

)

→

ℝ

+

by

c

*

⁢

(

w

)

=

{

c

⁢

(

u

)

+

c

⁢

(

v

)

w

=

u

0

w

=

v

c

⁢

(

w

)

otherwise.

Observe that

∑

w

∈

V

⁢

(

G

)

c

*

⁢

(

w

)

=

1

, so

c

*

is still in the poset

P

.
Furthermore, by inequality (*) and the definition of

c

*

,

W

⁢

(

c

*

)

=

∑

u

⁢

w

∈

E

⁢

(

G

)

c

*

⁢

(

u

)

⋅

c

*

⁢

(

w

)

+

∑

v

⁢

w

∈

E

⁢

(

G

)

c

*

⁢

(

v

)

⋅

c

*

⁢

(

w

)

+

∑

w

⁢

z

∈

E

⁢

(

G

)

c

*

⁢

(

w

)

⋅

c

*

⁢

(

z

)

=

∑

u

⁢

w

∈

E

⁢

(

G

)

[

c

⁢

(

u

)

+

c

⁢

(

v

)

]

⋅

c

⁢

(

w

)

+

0

+

∑

w

⁢

z

∈

E

⁢

(

G

)

c

⁢

(

w

)

⋅

c

⁢

(

z

)

=

∑

u

⁢

w

∈

E

⁢

(

G

)

c

⁢

(

u

)

⋅

c

⁢

(

w

)

+

∑

v

⁢

w

c

⁢

(

v

)

⋅

c

⁢

(

w

)

+

∑

w

⁢

z

∈

E

⁢

(

G

)

c

⁢

(

w

)

⋅

c

⁢

(

z

)

=

W

⁢

(

c

)

.

Thus

c

*

≥

c

in

G

and is supported on one less vertex than

c

is. So let

c

be a maximal element of

P

above

c

0

. We have just seen that

c

must be supported on

adjacent vertices

u

and

v

. The weight

W

⁢

(

c

)

is just

c

⁢

(

u

)

⋅

c

⁢

(

v

)

; since

c

⁢

(

u

)

+

c

⁢

(

v

)

=

1

and

c

has maximal weight, it must be that

c

⁢

(

u

)

=

c

⁢

(

v

)

=

1

2

. Hence

1

4

=

W

⁢

(

c

)

≥

W

⁢

(

c

0

)

=

|

E

⁢

(

G

)

|

|

V

⁢

(

G

)

|

2

,

which gives us the desired inequality:

|

E

⁢

(

G

)

|

≤

|

V

⁢

(

G

)

|

2

4

.

Title

proof of Mantel’s theorem

Canonical name

ProofOfMantelsTheorem

Date of creation

2013-03-22 13:03:04

Last modified on

2013-03-22 13:03:04

Owner

mps (409)

Last modified by

mps (409)

Numerical id

6

Author

mps (409)

Entry type

Proof

Classification

msc 05C75

Classification

msc 05C69

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