proof of equivalence of definitions of valuation

proof of equivalence of definitions of valuation

We will start with a lemma:

Lemma

Let

|

⋅

|

be a

valuation

according to the definition of the parent entry with constant

C

≤

2

. Then

|

∑

i

=

1

n

x

i

|

≤

2

⁢

n

⁢

max

i

=

1

n

⁡

|

x

i

|

.

Proof:

We will start with the case where

n

is a

power of two

:

n

=

2

r

We shall prove that

|

∑

i

=

1

2

r

x

i

|

≤

2

r

⁢

max

i

=

1

n

⁡

|

x

i

|

by induction. The assertion is certainly true when

n

=

2

by hypothesis. Assume that it also holds when

n

=

2

r

-

1

. Then we have

|

∑

i

=

1

n

x

i

|

=

|

∑

i

=

1

2

r

-

1

x

i

+

∑

i

=

2

r

-

1

+

1

2

r

x

i

|

≤

2

⋅

2

r

-

1

⁢

max

i

=

1

2

r

-

1

⁡

|

x

i

|

+

2

⋅

2

r

-

1

⁢

max

i

=

2

r

-

1

+

1

2

r

⁡

|

x

i

|

≤

2

⋅

2

r

⁢

max

i

=

1

2

r

⁡

|

x

i

|

To deal with the case where

n

is not a power of two, we shall pad the sum with zeros. That is to say, we shall define

x

i

=

0

when

i

>

n

. Let

r

be the greatest

integer

such that

2

r

≤

n

. Then

2

⁢

n

>

2

r

and we have

|

∑

i

=

1

n

x

i

|

=

|

∑

i

=

1

2

r

x

i

|

≤

2

r

⁢

max

i

=

1

2

r

⁡

|

x

i

|

≤

2

⁢

n

⁢

max

i

=

1

n

⁡

|

x

i

|

Q.E.D.

Corollary

Let

|

⋅

|

be a valuation according to the definition of the parent entry with constant

C

≤

2

. Then, if

n

is a

positive integer

,

|

n

|

<

2

⁢

n

.

Proof:

Write

n

=

∑

i

=

1

n

1

. Then, by the lemma,

|

n

|

=

|

∑

i

=

1

n

1

|

<

2

⁢

n

⁢

|

1

|

However, since

|

⋅

|

is a valuation

|

1

|

≠

0

since

1

≠

0

and

|

1

|

=

|

1

|

⋅

|

1

|

so

|

1

|

=

1

, hence

|

n

|

≤

2

⁢

n

.

Q.E.D.

Having established this lemma, we will now use it to prove the main theorem:

Theorem

If

|

⋅

|

is a valuation according to the definition of the parent entry with constant

C

≤

2

, then

|

⋅

|

satisfies the identity

|

x

+

y

|

≤

|

x

|

+

|

y

|

.

Proof:

Let

n

be a positive integer. Then we have

|

x

+

y

|

n

=

|

(

x

+

y

)

n

|

=

|

∑

m

=

0

n

(

n

m

)

⁢

x

m

⁢

y

n

-

m

|

Using the lemma, we can bound this:

|

x

+

y

|

n

≤

2

⁢

(

n

+

1

)

⁢

max

m

=

0

n

⁡

|

(

n

m

)

⁢

x

m

⁢

y

n

-

m

|

Using the corollary to the lemma, we can bound the

binomial coefficient

to obtain

|

x

+

y

|

n

≤

4

⁢

(

n

+

1

)

⁢

max

m

=

0

n

⁡

(

n

m

)

⁢

|

x

|

m

⁢

|

y

|

n

-

m

Using the obvious

inequality

max

m

=

0

n

⁡

(

n

m

)

⁢

|

x

|

m

⁢

|

y

|

n

-

m

≤

∑

m

=

0

n

(

n

m

)

⁢

|

x

|

m

⁢

|

y

|

n

-

m

,

we obtain

|

x

+

y

|

n

≤

∑

m

=

0

n

4

⁢

(

n

+

1

)

⁢

(

n

m

)

⁢

|

x

|

m

⁢

|

y

|

n

-

m

=

4

⁢

(

n

+

1

)

⁢

(

|

x

|

+

|

y

|

)

n

.

If either

x

=

0

or

y

=

0

, then the theorem to be proven is trivial. If not, then

|

x

|

+

|

y

|

≠

0

and we can divide by

(

|

x

|

+

|

y

|

)

n

to obtain

(

|

x

+

y

|

|

x

|

+

|

y

|

)

n

≤

4

⁢

(

n

+

1

)

By the theorem on the

growth of exponential function

, it follows that this inequality could not hold for all

n

unless

|

x

+

y

|

|

x

|

+

|

y

|

≤

1

,

in other word, unless

|

x

+

y

|

≤

|

x

|

+

|

y

|

.

Q.E.D.

Title

proof of equivalence of definitions of valuation

Canonical name

ProofOfEquivalenceOfDefinitionsOfValuation

Date of creation

2013-03-22 14:56:00

Last modified on

2013-03-22 14:56:00

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

12

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 11R99

Classification

msc 12J20

Classification

msc 13A18

Classification

msc 13F30

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