Prove: $(L_0)^0 = L$

Question

Let $W \subseteq V$ and $L \subseteq V^{*}$.
Let $W^0 = \{ f\in V^{*} \mid f(w)=0 ~~ \forall w \in W \}$ be the annihilator of $W$.
Let $L_0 = \{ v\in V \mid f(v)=0 ~~ \forall f \in L \}$.

1. How can I show that $(L_0)^0=L$ ?
2. What is the name of $L_0$ ? (Like $W^0$ is the annihilator of $W$)

Thank you.

Answer 1

Lemma: Suppose $\Lambda_1,...,\Lambda_n$ and $\Lambda$ are linear functionls on a vector space $X$. Let $N=\bigcap_{i=1}^n \ker \Lambda_i$ and $\Lambda(N)=\{0\}$. Then there are scalras $c_1,...,c_n$ such that $\Lambda=\sum_{i=1}^nc_i\Lambda_i$.

Proof: Let $\Phi$ be our scalar field. Define $\pi: X\to \Phi^n$ by $\pi(x)=(\Lambda_1x,...,\Lambda_nx)$. Note that $\pi(x)=\pi(x')$ implies $\Lambda x=\Lambda x'$, so we have well-defined linear functional $f:\pi(X)\to\Phi$ given by $f\big(\pi(x)\big)=\Lambda x$. Extend the linear functional $f$ to a linear functional $F$ on $\Phi^n$: there are scalars $c_1,..., c_n\in \Phi$ with $F(u_1,...,u_n)= c_1u_1+...c_nu_n.$ Thus, $$\Lambda x= f\big(\pi(x)\big)=F(\Lambda_1 x,...,\Lambda_n x)=\sum_{i=1}^n c_i \Lambda_i x.$$

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To prove $(i)$ consider $L_0=\bigcap_{f\in L}f^{-1}(0)=\bigcap_{f\in \text{span}(L)}f^{-1}(0)$. Note that if we assume $V$ is finite-dimensional, then $V^*$ will be finite-dimensional, so $\text{span}(L)$ also. Now, considering a basis $\{\Lambda_1,...,\Lambda_n\}$ of $\text{span}(L)$ we have $$g\in (L_0)^0\iff g(L_0)=0$$$$\iff g\bigg(\bigcap_{f\in L}f^{-1}(0)\bigg)=0\iff g\bigg(\bigcap_{f\in \text{span}(L)}f^{-1}(0)\bigg)=0$$$$\iff g\bigg(\bigcap_{i=1}^n\Lambda_i^{-1}(0)\bigg)=0\iff g=\sum_{i=1}^n c_i \Lambda_i\text{ for some scalars }c_i$$$$\iff g\in \text{span}(L).$$ So, if I assume from the beginning that $L$ is a vector space then $\text{span}(L)=L$, and we are done.

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$(ii)$ Note that $L_0$ is the intersections of some hyperspaces of $V$. In a vector space of dimension $n$, a subspace of dimension $n-1$ is called a hyperspace.

Answer 2

If $f \in L$, then $f(v) = 0$ for every $v \in L_0$ (by definition of $L_0$), that is, $f$ annihilates every element of $L_0$; and this means precisely that $f \in (L_0)^0$. Thus $L \subseteq (L_0)^0$.

To see that it is not necessarily true the equality $L = (L_0)^0$, take $V$ to be the real vector space of all continuous functions from $[0,1]$ to $\mathbb R$, and for each $n \in \mathbb N$ (the zero is included) consider $\phi_n \in V^*$ such that $$(\forall f \in V) \quad \phi_n(f) = \int_0^1 x^nf(x)dx.$$

Now, in virtue of the Weierstrass's Theorem we have that if $L := \{\phi_n : n \in \mathbb N\}$, then $L_0 = \{0\}$ and so $(L_0)^0 = \{0\}^0 = V^* \supsetneq L$.

However, the equality $L = (L_0)^0$ is true when $V$ is finite-dimensional, as @Mathlover points in his answer.