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?



Continuous dual basis in a normed space
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.


