# Thread: Annihilators of Banach Spaces

1. ## Annihilators of Banach Spaces

Hi All,

Let X be a normed vector space with Banach dual space X'. Let $\displaystyle f_{1}, \ ... \ f_{n}$ be in X' such that $\displaystyle Ker(f) \supset \bigcap_{i} Ker (f_{i})$. Show that

$\displaystyle T: (f_{1}(x),...,f_{n}(x))\rightarrow f(x)$

is a well-defined linear map on a subspace of $\displaystyle \mathbb{F}^{n}$. Extending T to all of $\displaystyle \mathbb{F}^{n}$, show that there are $\displaystyle \alpha_{1},...,\alpha_{n} \in \mathbb{F}$ such that $\displaystyle f=\sum_{j=1}^{n} \alpha_{j}f_{j}$

Conclude that, for any finite-dimensional subspace M of X', $\displaystyle ( M_{o} ) ^{o} =M$

Now, by definition, for $\displaystyle M \subset X, N \subset X'$

$\displaystyle M^{o}= \{f \in X': \ f(x)=0 \ \forall x \in M\}$
$\displaystyle N_{o}=\{ x \in X; \ f(x)=0 \ \forall f \in N \}$

Now, I can show that T is well-defined and linear on the subspace $\displaystyle (ranf_{1}) x ... x (ranf_{n})=:K$.
By the Hahn-Banach theorem, I can extend this to a function $\displaystyle G \in $$\mathbb{F}^{n}$$'$ such that $\displaystyle G|_{K}=T$ and $\displaystyle \|G\|_{ ( \mathbb{F}^{n} )'} = \|T\|_{K'}$. But where do I go from here?

2. The key thing that you need to notice is that every linear functional $\displaystyle T$ on $\displaystyle \mathbb{F}^{n}$ is of the form $\displaystyle Ty = \alpha_1y_1 + \alpha_2y_2 + \ldots + \alpha_ny_n\;\,(y = (y_1,y_2,\ldots,y_n)\in\mathbb{F}^{n})$, for some scalars $\displaystyle \alpha_1,\, \alpha_2,\,\ldots, \alpha_n$. All you have to do then is to take $\displaystyle (y_1,y_2,\ldots,y_n) = (f_1(x),f_2(x),\ldots,f_n(x))$.

3. Thanks a lot for this!
Could you please check I've got the last part, ($\displaystyle ( M_{o} ) ^{o} =M$), right?

So, want to show $\displaystyle M \subset ( M_{o} ) ^{o}\ , \ so \ take \ f \in M \leq X' \ \Rightarrow f \in X'$ Want to show $\displaystyle f \in ( M_{o} ) ^{o}$,

i.e. that (by defn of $\displaystyle ( M_{o} ) ^{o}$ ) $\displaystyle f \in X' \ and \ f(x)=0 \ \forall x \in M_{o}$
i.e. (by defn of $\displaystyle M_{o}$ ) $\displaystyle f(x)=0 \ \forall x \in X : (f(x)=0 \forall f \in M )$
which is true, as we started with $\displaystyle f \in M$

Similarly for the other direction. What I don't get - where did we use the $\displaystyle Ker(f) \supset \bigcap_{i} Ker (f_{i})$ bit? And where in this last part did we use any of the previous bits of the question?

4. Originally Posted by Mimi89
What I don't get - where did we use the $\displaystyle Ker(f) \supset \bigcap_{i} Ker (f_{i})$ bit?
The place where you needed $\displaystyle \ker(f) \supset \bigcap_{i} \ker (f_{i})$ was right at the beginning of the question, in order to show that $\displaystyle T$ is well defined. If the definition $\displaystyle T: (f_{1}(x),...,f_{n}(x))\rightarrow f(x)$ is to make sense, it must be true that $\displaystyle f(x)$ depends only on $\displaystyle f_{1}(x),...,f_{n}(x)$. So you need to be sure that if $\displaystyle (f_{1}(x),...,f_{n}(x)) = (f_{1}(y),...,f_{n}(y))$ then $\displaystyle f(x) = f(y)$. But that is equivalent to saying that if $\displaystyle f_i(x-y) = 0$ for i=1,2,...,n, then $\displaystyle f(x-y)=0$. That in turn is equivalent to the condition $\displaystyle \ker(f) \supset \bigcap_{i} \ker (f_{i})$.

5. Originally Posted by Mimi89
Conclude that, for any finite-dimensional subspace M of X', $\displaystyle ( M_{o} ) ^{o} =M$.
For the last part of the question, you are given a finite-dimensional subspace $\displaystyle M\subset X'$. Let $\displaystyle f_1,\ldots,f_n$ be a basis for $\displaystyle M$. Then

$\displaystyle M_o = \{ x \in X: f(x)=0\ \forall f \in M \} = \{x\in X:f_i(x)=0\ (1\leqslant i\leqslant n)\} = \bigcap_{i} \ker (f_{i}),$

and $\displaystyle (M_o)^o = \{f\in X':f(M_o)=0\} = \{f\in X':\ker(f)\supset \bigcap_{i} \ker (f_{i})\}.$

But it follows from the earlier part of the question that that last condition implies that $\displaystyle f$ is a linear combination of $\displaystyle f_1,\ldots,f_n$, which says that $\displaystyle f\in M$. Therefore $\displaystyle (M_o)^o\subset M$. You have shown how to prove the reverse inclusion $\displaystyle M\subset (M_o)^o$, and so $\displaystyle (M_o)^o = M$.