# Axler Linear Algebra Done Right 3.f.26

Here's the problem: Suppose $$V$$ is finite-dimensional and $$\Gamma$$ is a subspace of $$V'$$. Show that $$\Gamma=\{v \in V:\varphi(v)=0 \forall \varphi \in \Gamma\}^0.$$

Note that $$V'$$ denotes the dual space of $$V$$. I've found the proof, but I'm wondering why this statement falls apart for when $$V$$ is infinite dimensional. Could one give a counterexample?

Is a well-known consequence of the Stone-Weierstrass theorem that if $$f \colon [0,1] \to \Bbb R$$ is a continuous function such that $$\forall n \in \Bbb N \quad \int_0^1 x^nf(x)dx = 0,$$ then $$f=0$$.
• $$V$$ is the real vector space of continuous functions from $$[0,1]$$ to $$\Bbb R$$;
• for each $$n \in \Bbb N$$ we define $$\varphi_n \in V’$$ by $$\forall f \in V \quad \varphi_n(f) = \int_0^1 x^nf(x)dx\,;$$
• and $$\Gamma$$ is the span of $$\{\varphi_n : n \in \Bbb N\}$$;
then $$\{f \in V : \forall \varphi \in \Gamma, \ \varphi(f)=0\} = \{0\}.$$ Hence $$\{f \in V : \forall \varphi \in \Gamma, \ \varphi(f)=0\}^0 = V’.$$