Continuous dual basis in a normed space Let $X$ be a normed space (not necessarily complete) and $\{w_n\}_{n\in\mathbb N}\subset X$ be
a linearly independent set. Let $\{x^*_n\}_{n\in\mathbb N}$ be a set of linear functionals on $X$ with the property that $x_i^*(w_j)=\delta_{ij}$. These linear functional are not necessarily bounded.
My question is the following: Is there any linear transformation (an analogue to Gramm-Schmidt procedure) that creates a new linearly independent set $\{u_n\}_{n\in\mathbb N}$, such that
$$
\mathrm{span}\,\{u_n\}_{n\in\mathbb N}=\mathrm{span}\,\{w_n\}_{n\in\mathbb N} \subset X,
$$ 
and for which the existence of a bounded dual set $\{u^*_n\}_{n\in\mathbb N}\subset X^*$, i.e., $u_i^*(u_j)=\delta_{ij}$, is guaranteed?
 A: Sure. We can ensure that ${\rm span}(u_1, \ldots, u_k) = {\rm span}(w_1, \ldots, w_k)$ and $\|u_k^*\| = 1$ for each $k$. The construction goes by induction on $k$. For $k = 1$ let $u_1 = \frac{1}{\|w_1\|}w_1$ and use Hahn-Banach to find $u_1^* \in X^*$ such that $\|u_1^*\| = u_1^*(u_1) = 1$. Inductively suppose $u_1, \ldots, u_k$ and $u_1^*, \ldots, u_k^*$ are chosen so that ${\rm span}(u_1, \ldots, u_k) = {\rm span}(w_1, \ldots, w_k)$ and so that $\|u_i^*\| = 1$ and $u_i^*(u_j) = \delta_{ij}$ for $1 \leq i,j \leq k$. Then we can find $u_{k+1} \in \bigcap_{i=1}^k {\rm ker}\, u_i^*$ such that ${\rm span}(u_1, \ldots, u_{k+1}) = {\rm span}(w_1, \ldots, w_{k+1})$; an explicit formula is
$$u_{k+1} = w_{k+1} - \sum_{i=1}^k u_i^*(w_{k+1})u_i.$$
Define a linear functional $\tilde{u}_{k+1}^*$ on this span such that $\tilde{u}_{k+1}(u_i) = \delta_{i,k+1}$ for $1 \leq i \leq k+1$; by scaling $u_{k+1}$ we can ensure that $\|\tilde{u}_{k+1}^*\| = 1$. Then use Hahn-Banach to extend $\tilde{u}_{k+1}^*$ to $X$ without increasing its norm.
