CALCU-Consider X = l1 and two convex subsets of XA

Consider X = l 1 and two convex subsets of X A0 = {x = {xm} ∈ X | x2m = 0, ∀m} and B = y = {ym} ∈ X | y2m = 1 2m y2m−1, ∀m . Show that A0, B are closed in X and A0 + B = X. Define z ∈ X by z2m−1 = 0, z2m = 1 2m , ∀ m. Show that z 6∈ A0 + B. Set A = A0 − z. Show that A ∩ B = φ but they can not be separated by a closed hyperplane. What about X = l p , 1 < p < ∞ or c0 (sequences with limit 0) and A, B as defined above? Can we separate them by a closed hyperplane?

Solution by Steps

step 1

To show that

A_0

is closed in

X

, consider a sequence

\{x^{(n)}\}

in

A_0

that converges to some

x

in

X

step 2

Since

x^{(n)}

converges to

x

, for every

m

, the sequence

x^{(n)}_{2m}

converges to

x_{2m}

step 3

Given that

x^{(n)}_{2m} = 0

for all

n

and

m

, the limit

x_{2m}

must also be 0, hence

x \in A_0

step 4

To show that

B

is closed, consider a sequence

\{y^{(n)}\}

in

B

that converges to some

y

in

X

step 5

Since

y^{(n)}

converges to

y

, for every

m

, the sequences

y^{(n)}_{2m-1}

and

y^{(n)}_{2m}

converge to

y_{2m-1}

and

y_{2m}

respectively

step 6

Given that

y^{(n)}_{2m} = \frac{1}{2m} y^{(n)}_{2m-1}

for all

n

and

m

, the limit

y_{2m}

must be

\frac{1}{2m} y_{2m-1}

, hence

y \in B

step 7

To show that

A_0 + B = X

, take any

x \in X

and define

a \in A_0

and

b \in B

such that

a_{2m} = 0

,

b_{2m-1} = 2m x_{2m}

, and

b_{2m} = x_{2m}

step 8

Then

a + b = x

, which means every element in

X

can be written as the sum of an element in

A_0

and an element in

B

step 9

To show that

z \notin A_0 + B

, assume for contradiction that

z = a + b

for some

a \in A_0

and

b \in B

step 10

Since

a_{2m} = 0

for all

m

, it must be that

b_{2m} = z_{2m} = \frac{1}{2m}

step 11

However, for

b \in B

, we must also have

b_{2m} = \frac{1}{2m} b_{2m-1}

, which cannot be true for all

m

since

z_{2m-1} = 0

. This is a contradiction, so

z \notin A_0 + B

step 12

Define

A = A_0 - z

. To show

A \cap B = \emptyset

, assume there exists

x \in A \cap B

step 13

Then

x = a - z

for some

a \in A_0

and

x \in B

step 14

This would imply

z = a - x

, but since

a, x \in B

and

B

is convex,

z

would have to be in

B

, which is a contradiction since we showed

z \notin A_0 + B

step 15

To show that

A

and

B

cannot be separated by a closed hyperplane, we would need to find a continuous linear functional that takes different constant values on

A

and

B

, which is not possible in this case

step 16

For

X = l_p

with 1 < p < \infty or

c_0

, the Hahn-Banach Theorem can be applied to show that

A

and

B

can be separated by a closed hyperplane, as these spaces are locally convex and satisfy the conditions of the theorem

Answer

The sets

A_0

and

B

are closed in

X

,

A_0 + B = X

,

z \notin A_0 + B

,

A \cap B = \emptyset

, and

A

and

B

cannot be separated by a closed hyperplane in

l_1

. For

X = l_p

with 1 < p < \infty or

c_0

,

A

and

B

can be separated by a closed hyperplane.

Key Concept

Closed sets, convexity, and hyperplane separation in functional analysis.

Explanation

The problem involves showing the closedness of sets in a normed space, their sum filling the space, and the application of the Hahn-Banach Theorem for separation in different spaces. The Hahn-Banach Theorem is applicable in locally convex spaces, which is why the separation by a closed hyperplane is possible in

l_p

for 1 < p < \infty and

c_0

, but not in

l_1

.

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