F is a linear functional in V' a linear vector space which operates on $\phi\in V$. Show that there is a one-to one correspondence between F and $f\in V $ such that
$F(\phi)=(f,\phi)$
where $V$ is also a linear vector space. $(,)$ represents inner product. Any hint will be helpful.
Attempt: I don't even know if this is right but...
If $F(\phi)=c$ , $c\in \mathbb{C}$ then let $(f_1, \phi)=c$ and $(f_2, \phi)=c$ $\rightarrow$ $(f_1-f_2, \phi)=0$
But this only means orthogonality.
EDIT: so, from hint
let $\{e_i\}$ be a linearly independent basis for vectorspace $V$
$F(\Phi) = F (\sum_n \phi_n e_n) =\sum_n \phi_n F(e_n) $
$(f,\Phi) = \sum_n f_n^* \phi_n$
Therefore if $F(\phi)=c$ , $c\in \mathbb{C}$ and $(f,\Phi)=c$ then
$\sum_n (f_n^* -F(e_n)) \phi_n =0$
implies (?) $f_n^* =F(e_n)$ $\forall n$?
EDIT Ok. So the above implies that if some $(e_i,\Phi)$ vanishes then we cannot have a unique $f$. So is this solved by specifying at the beginning that $\{e_i\}$ is a basis where $\phi_i=(e_i,\Phi)$ is non vanishing for all i?