In the book of Nadir Jeevanjee „An Introduction to Tensors and Group Theory for Physicists“ it is stated as an exercise that:
2.17 Given a basis $\{e_i\}_{i=1,...,n}$ , under what circumstances do we have $e^i = \tilde e_i$ for all $i$,
where $e^i$ is the dual vectors to $e_i$ and the metric dual is defined as $\tilde e_i \equiv (e_i|\cdot) $ such that $\tilde e_i(w)=(e_i|w)$.
I know that this is the case for Hilbert space because in QM we always use bra's as dual vectors and for taking inner products. However I could neither generalize it, nor could I prove that this is the case for Hilbert space.
The dual basis $(e^i \mid 1 \le i \le n)$ to $(e_i)$ is defined by the property $$ e^i(e_j) = \delta_{ij} \qquad 1 \le i, j \le n $$ So if we want $e^i = \tilde e_i$, we must have for every $j$: $$ (e_i|e_j) = \tilde e_i(e_j) = e^ie_j = \delta_{ij} $$ So, the basis $(e_i)$ must be orthonormal.
Note, that the metric dual $\tilde e_i$ is only defined if we have a scalar product, hence the question makes sense only in (pre-)Hilbert spaces