A characterization of the annihilator of a subspace Let $V$ be a vector space and let $W\subset V$ be a subspace. Let $W^0=\{f \in V^*: f[W]=\{0\}\}$. It's easy to see that $$W=\bigcap_{f \in W^{0}}\ker f.$$
Now suppose $Y$ is a subspace of $V^\ast$ such that the following holds:
$$W=\bigcap_{f \in Y}\ker f.$$
Is it true that $Y=W^0$? It's easy to show that $Y\subset W^0$. If not, does the additional hypothesis of $\dim V<\infty$ makes it work?
 A: The result is false if $\dim Y = \infty$ (which would necessarily imply that $\dim V = \infty$ as well). Here is an example:
Let $V = L^1(\mathbb{R})$ and $W = \{0\}$. Then clearly $W^{0} = V^*$. Now let $Y = C_C^{\infty}(\mathbb{R})$ be the set of $C^{\infty}$ functions with compact support. This is a subspace of $V^*$, where a function $f\in C_C^{\infty}(\mathbb{R})$ is associated with the functional $g\mapsto\int\limits_{\mathbb{R}}{fg}$. We have the following fact:

If $g\in L^1(\mathbb{R})$ and $\int\limits_{\mathbb{R}}{fg} = 0$ for all $f\in C_C^{\infty}(\mathbb{R})$, then $g = 0$.

This fact implies that $\bigcap\limits_{f\in Y}{\ker f} = \{0\} = W$. But $Y\ne V^*$, since $V^*$ is $L^{\infty}(\mathbb{R})$ under the same identification made above if we interpret $V^*$ as the set of continuous linear functionals on $V$, and $V^*$ is some even bigger space if we include noncontinuous functionals as well.
A: I can give you a proof in finite dimension. Take $\{f_1,\cdots, f_n\}$ a basis (actually a spanning set is enough) for $Y$. Then $$\bigcap_{i=1}^n\ker f_i = \bigcap_{f \in Y}\ker f = W.$$The $\subseteq$ follows because the $f_i$ span $Y$ and $\supseteq$ is trivial. Let's see that if $f$ annihilates $W$, then $f$ is spanned by the $f_i$. Consider the linear map $T\colon V \to \Bbb K^n$ given by $$T(x) = (f_1(x),\cdots,f_n(x)).$$Then $\ker T = W$. Then $f \in W^0 = (\ker T)^0$, but we know that $$(\ker T)^0 = {\rm Im}(T^\top),$$right? Meaning $f = T^\top(g)$ for some $g \in (\Bbb K^n)^\ast$. Now write $g = \sum_{i=1}^n\lambda_i \pi_i$ for some constants $\lambda_i \in \Bbb K$ and the projections $\pi_i$. Finally: $$f(v) = T^\top(g)(v) = g(f_1(v),\cdots,f_n(v)) = \sum_{i=1}^n\lambda_i \pi_i(f_1(v),\cdots,f_n(v)) = \sum_{i=1}^n\lambda_if_i(v).$$
I think the result is false in infinite dimension but I don't know a counter-example right now. I essentially solved exercise $6$ here, which I'm sure you will find interesting to read.

I just realized that only $Y$ need to have finite dimension, not $V$.
