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?