Axler Linear Algebra Done Right 3.f.26

Question

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?

Answer 1

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 $f=0$.

Thus, if

• $V$ 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 $$ \{f \in V : \forall \varphi \in \Gamma, \ \varphi(f)=0\}^0 = V’. $$