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?


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.

  • $\begingroup$ Thnx! - Although, I was expecting $u^*_i(u_j)=\delta_{ij}$, for all $i,j$. $\endgroup$ – smyrlis Jan 6 '14 at 4:21
  • $\begingroup$ Yes, my construction achieves $u_i^*(u_j) = \delta_{ij}$ for all $i, j$. Each $u_{k+1}$ is chosen to lie in the intersection of the kernels of all $u_i^*$ for $i \leq k$, so we get $u_i^*(u_j) = 0$ for $i < j$; beyond that, we only have to worry about $u_{k+1}^*(u_j)$ for $j \leq k+1$, and the construction explicitly handles that. $\endgroup$ – Nik Weaver Jan 6 '14 at 4:53

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