[sub]Proof[/sub]

Let H be a supporting hyperplane on A, we know that so it is also compact and convex.
Then there is a point x in the intersection of the hyperplane and A such that it is an extreme point of H intersection A, then x is also an extreme point of A.

[sub]Krein-Milman Theorem[/sub]
Let E be a convex and compact set on R^n, then the convex hull of the set of extreme points of E is the set E itself.

[sub]Proof:[/sub]
Its clear that E is a subset of the convex hull of the set of all extreme points of E since the extreme points belong to E.
Now we have to see that .
Lets use induction over k the dimension of E
If k =0 then E is a single point
If K = 1 then E is a closed line segment.
Lets use as induction hypothesis that the theorem is true for k-1 with k<n.
Let x be the relative interiorof E. Then by our lemma there is a supporting hyperplane of E that contains x.
Then is compact and convex with , so by our induction hypothesis we have that x is a convex combination of the extreme points of . Since the extreme points of are extreme in E we have that :

If x is in the relative interior of E then there is a line segment that contains x that also intersects the boundary of E. Then E and the line segment can be expressed as a convex combination of extreme points of E then

Then:
.

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This is by far my favorite theorem, Ive been tryin to use it along with Radon’s theorem or directly with Caratheodory’s but I havent found anything good enough.
:)

Proof
Let y in the boundary of S, then we define the set Sy as the points on every chord that has as a boundary the point y, and that the distance to y its less than 2 thirds of the lenght of the chord.

Let y,z,w be on the boundary of S and lets define Sw,Sz and Sy as before.

Using the triange with vertices wzy we build the medians and define x as the centroid of the triangle, now we’ll show that .
This is because
but
then
then
, using the same argument we can show that with w’,z’ and y’ the boundary of the median with vertex w,y,z and a the midpoint of yz.
Now we have that
Then using Helly’s theorem
Let bb’ be a chord that includes x0, then
Then,
This proofs our lemma.

Theorem ( Blaschke’s ) :
Every convex, compact set S with width 1 contains a circle of radius 1/3.

Proof:
By the last lemma there exists on S.
Lets show that .
Let y in the boundary of and let ww’ be a chord that includes x0, then, since the width of the set is 1, all chords have lenght greather or equal than 1.
Now , then . QED.
———-
I hope you comment, Ill try to upload about other topics, maybe some algebra or number theory.
‘Till next time :)

Proof:
Let
Since we know that where z1,…,zn are the roots of the polynomial .

We can also write
Now:

If z is a root of the derivative, then:
This means that:

Then :

Which means:

Then:

But with this we know that z is a convex combination of all the roots, this is because :

Then by this we know that z is in the convex hull of

QED.

Example:
Proof that the diameter of the set of roots of the polynomial :
cannot be less than

Proof:
We know that :
Now we just use that:
and
Now its clear that:
.
Since the roots of the derivative are inside the convex hull generated by the roots of the original polynomial, then the diameter between them cannot be less than the diameter between the set of the roots of the second derivative of the polynomial.

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The example was one of the problems on my exam. Good luck I liked this theorem and I didnt forget about it :).

Proposition:

is multiplicative and

Proof:

Let gcd(m,n)=1

If or
Then

If , , then

If n or m = 1 proof is clear.

Now, let , since is multiplicative then F(n) is multiplicative.

It is clear that

Also:

.

( Möbius function is 0 on )

Now, if we use prime factorization on n, if n>1:

we can do this because F(n) is multiplicative.

——–
I <3 Möbius function

The convex hull for a set A ( Im using real A a subset of ) is the minimal convex set that contains A.
We’ll now define:

Lemma:

Let with A a convex set and then

Proof:

We’ll use induction over k.
If k=1 then x*1 is in A ( we choose x that way ).
Now we’ll asume that it works for every k.
Now, let where and

We wanna see that x is in A.

Since then at least there is one , lets say its .

Let
and
.

Then we know by our induction hypothesis that y is in A .
Now, since A is a convex set and then .

This shows that x is in A.

Definition:

A point its a convex combination of points if there is such that

Theorem:

Let then c(A) ( The convex hull of the set A ) is the set of all convex combinations of all points in A.

Proof:

Let B the set of all convex combinations of points in A.
We now have to show that B=c(A) (convex hull of A )

1)

Let x be an element of B, then with and , since A is in c(A) then , then because c(A) is convex.
This shows that .

2)

Note that if B is a convex set then A is a subset of B,this is because if x is in A then there is a such that px will be on B. Then c(A) is a subset of B

We have to show that B is convex.

Let x and y be in B
Then and where and and .
Let with such that
Then , it is clear now that
This shows that , which means B is convex.

Then B=c(A).

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‘Till next time :)

Old site was about giving application to pure mathematics, but I decided I’ll just use this for notes and results and demonstrations I find interesting.

Ill try to update as much as I can.

Angel.