I want to show that for a proper subspace of linear functionals $W^* \subset V^*$ on a vector space V, we can find $v \in V \setminus \{0\} \text{ with } w^*(v) = 0 \; \forall w^* \in W^*$.
This is fairly simple in the finite dimensional case, but I was wondering about this in general.
I have come up with the following proof for the finite dimensional case:
Pick a basis of linear functionals $\{ \phi_1, ..., \phi_m \}$ for $W^*$. Extend to a basis $\{\phi_1, ..., \phi_m, \psi_1, ... , \psi_m \} $ for $V^*$. Then $\{ \phi_1^*, ..., \phi_m^*, \psi_1^*, ... , \psi_m^* \}$ is a basis for $(V^*)^*$. We have natural isomorphism $V \simeq (V^*)^*$.
So we have that $\phi_i^* = \tilde{\phi}_{v_i}$ and $\psi_j^* = \tilde{\psi}_{w_j}$. Where $\tilde{\phi}_{v_i} \in (V^*)^*$ is map $\tilde{\phi}_{v_i}(\lambda) = \lambda(v_i)$ and the vectors $\{ v_1, ..., v_n, w_1, ..., w_m \}$ form a basis. Then $\forall i \in \{1, ..., n\}$ we have that $$ \phi_i(w_1) = \tilde{\psi}_{w_1}(\phi_i) = \psi_1^*(\phi_j) = 0$$ By construction of $\psi_1^*$.
I see that this fails right at the beginning when we pick a basis (but can probably work around this just by picking a $\psi_1 \in V^* \setminus W^*$). Then my main worry is that we no longer have surjectivity in the evaluation map $V \simeq (V^*)^*$. Can this be worked around??
This is false in every infinite-dimensional vector space.
Say $V$ is an infinite-dimensional vector space over $F$. Say $B=(e_j)$ is a basis for $V$. It's easy to see that $V^*$ is naturally isomorphic ("naturally" given $B$) to the space of all functions from $B$ to $F$: If $f:B\to F$ define $L_f\in V^*$ by $$L_f(\sum a_j e_j)=\sum a_j f(e_j);$$then $L_f\in V^*$ and it's easy to see that the map $f\mapsto L_f$ is an isomorphism.
Let $W^*$ be the space of all $L_f$ such that $f(e_j)=0$ except for finite many $e_j$. Then $W^*$ is a proper subspace of $V^*$, but if $v\in V$ and $L_fv=0$ for every $L_f\in W^*$ then $v=0$.