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Exercise 2.6.13

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1.

We can check that

f + g

and

cf

are elements in

S^{0}

f

and

g

are elements in

S^{0}

since

(f + g) (x) = f (x) + g (x) = 0

and

(cf) (x) = cf (x) = 0

. And the zero function is an element in

S^{0}

2.

Let

{v_{1}, v_{2}, \dots, v_{k}}

be the basis of

W

. Since

x \notin W

we know that

{v_{1}, v_{2}, \dots, v_{k + 1} = x}

is an independent set and hence we can extend it to a basis

{v_{1}, v_{2}, \dots, v_{n}}

for

V

. So we can define a linear transformation

T

such that

f (v_{i}) = δ_{i (k + 1)}

. And thus

f

is the desired functional.

3.

Let

W

be the subspace span

(S)

. We first prove that

W^{0} = S^{0}

. Since every function who is zero at

W

must be a function who is zero at

S

. we know

W^{0} \subset S^{0}

. On the other hand, if a linear function has the property that

f (x) = 0

for all

x \in S

, we can deduce that

f (y) = 0

for all

y \in W =

span

(S)

. Hence we know that

W^{0} \supset S^{0}

and

W^{0} = S^{0}

. Since

{(W^{0})}^{0} = {(S^{0})}^{0}

and span

(ψ (S)) = ψ (W)

by the fact

ψ

is an isomorphism, we can just prove that

{(W^{0})}^{0} = ψ (W)

Next, by Theorem 2.26 we may assume every element in ${(W^{0})}^{0} \subset V^{∗∗}$ has the form $\hat{x}$ for some $x$ . Let $\hat{x}$ is an element in ${(W^{0})}^{0}$ . We have that $\hat{x} (f) = f (x) = 0$ if $f \in W^{0}$ . Now if $x$ is not an element in $W$ , by the previous exercise there exist some functional $f \in W^{0}$ such that $f (x) \neq 0$ . But this is a contradiction. So we know that $\hat{x}$ is an element in $ψ (W)$ and ${(W^{0})}^{0} \subset ψ (W)$ .

For the converse, we may assume that $\hat{x}$ is an element in $ψ (W)$ . Thus for all $f \in W^{0}$ we have that $\hat{x} (f) = f (x) = 0$ since $x$ is an element in $W$ . So we know that ${(W^{0})}^{0} \supset ψ (W)$ and get the desired conclusion.

4.

It’s natural that if

W_{1} = W_{2}

then we have

W_{1}^{0} = W_{2}^{0}

. For the converse, if

W_{1}^{0} = W_{2}^{0}

then we have

ψ (W_{1}) = {(W_{1}^{0})}^{0} = {(W_{2}^{0})}^{0} = ψ (W_{2})

and hence

W_{1} = ψ^{- 1} ψ (W_{1}) = ψ^{- 1} ψ (W_{2}) = W_{2}

by the fact that $ψ$ is an isomorphism.

5.

f

is an element in

{(W_{1} + W_{2})}^{0}

, we have that

f (w_{1} + w_{2}) = 0

for all

w_{1} \in W_{1}

and

w_{2} \in W_{2}

. So we know that

f (w_{1} + 0) = 0

and

f (0 + w_{2}) = 0

for all proper

w_{1}

and

w_{2}

. This means

f

is an element in

W_{1}^{0} \cap W_{2}^{0}

. For the converse, if

f

is an element in

W_{1}^{0} \cap W_{2}^{0}

, we have that

f (w_{1} + w_{2}) = f (w_{1}) + f (w_{2}) = 0

for all

w_{1} \in W_{1}

and

w_{2} \in W_{2}

. Hence we have that

f

is an element in

{(W_{1} + W_{2})}^{0}

jephianlin

2011-06-27 00:00

Exercise 2.6.13

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