# functional analysis – \$ conv (A) = { sum_ {i = 1} ^ n lambda _i x_i: sum_ {i = 1} ^ n lambda_i = 1, lambda_i ge 0, x_i in A, i = 1, cdots, n } \$

Leave $$X$$ be a normed vector space and $$A$$ be a subset of $$X$$. $$conv (A)$$
It is called the intersection of all convex subsets of $$X$$ that contain
$$A$$

a) Show that conv (A) is a convex set

b) show that

$$conv (A) = { sum_ {i = 1} ^ n lambda _i x_i: sum_ {i = 1} ^ n lambda_i = 1, lambda_i ge 0, x_i in A, i = 1, cdots, n }$$

c) yes $$A$$ it is compact then it is $$conv (A)$$ compact?

d) Show that if $$A subseteq mathbb {R} ^ n$$ it is compact then $$conv (A)$$ is
compact

a) Take two elements $$a$$ Y $$b$$ at the intersection of all convex subsets of $$X$$ that contain $$A$$. Now take $$lambda a + (1- lambda) b$$. It is contained in each of the subsets of $$X$$ that contain $$A$$, therefore it is contained in the intersection. Q.E.D.

second)

$$leftarrow$$:

Suppose by induction that $$sum_ {i = 1} ^ n lambda_i x_i$$ It belongs to conv (A). We must prove that $$sum_ {i = 1} ^ {k + 1} lambda_i x_i$$, with $$sum_ {i = 1} ^ {k + 1} = 1$$ It also belongs.

$$sum_ {i = 1} ^ {k + 1} lambda_i x_i = sum_ {i = 1} ^ {k} lambda_i x_i + lambda_ {k + 1} x_ {k + 1} = sum_ {i = 1} ^ {k} lambda_i x_i + (1- sum_ {i = 1} ^ n lambda_i) x_ {k + 1}$$

Now choose $$delta$$ such that $$delta sum_ {i = 1} ^ {k} lambda_i = 1$$, so

$$frac { delta} { delta} left ( sum_ {i = 1} ^ {k} lambda_i x_i + (1- sum_ {i = 1} ^ n lambda_i) x_ {k + 1 } right = = left ( frac {1} { delta} sum_ {i = 1} ^ {k} ( delta lambda_i) x_i + frac {1} { delta} ( delta-1) x_ {k + 1} right) = \ frac {1} { delta} x + left (1- frac {1} { delta} right) x_ {k + 1}$$

which is a collection of elements of $$A$$ that adds up to $$1$$

$$rightarrow$$:

I can only say that $$x in conv (A)$$, so $$x = 1x + 0 cdot all$$ Therefore, it is a combination that adds to $$1$$ of elements of $$A$$?

by do), give me a clue: show that conv (A union B) is the image through a continuous function of the compact $${( alpha, beta; alpha, beta ge 0, alpha + beta = 1) } times A times B$$. I do not understand how to show this.

re) Some clue?