{$x_1, x_2, ...,x_n$} is linearly independent in normed space $X$. Show for any scalars $a_1, ..., a_n$ there is $f$ in dual $X'$ with $f(x_i) = a_i$

Question

Let $X$ be a normed space, $n\in N$ and {$x_1, x_2, ...,x_n$} be a linearly independent set in $X$. Prove that for any scalars $\alpha_1, \alpha_2, ..., \alpha_n$ there exists $f$ in the dual space $X'$ such that $f(x_i) = \alpha_i$; $ i = 1, 2, ..., n$.

(I tried it by using the theorem: For any non-zero $x_0$ in a normed space $X$ we have a linear functional $f$ such that $||f||=1 $ and $f(x_0)=||x_0||$. But I could not succeed. )

Answer 1

I think you can define $f$ on the finite dimensional subspace $$ M := \text{span}\{x_1, \ldots, x_n\} $$ by $$ f\left(\sum_{j=1}^n \lambda_j x_j\right) = \sum_{j=1}^n \lambda_j \alpha_j. $$ (The definition is well-posed since every $x\in M$ has a unique decomposition of the form $\sum\lambda_j x_j$.)

Then you can use the Hahn-Banach theorem to extend $f$ to the whole $X$.

Answer 2

Let $V$ be the smallest linear subspace of $X$ such that $\{\alpha_1^{-1} x_1, ..., \alpha_n^{-1} x_n \}\subset X.$ For $x=\sum_j t_j \alpha^{-1}_j x_j $ we define $$f(x)=\sum_j t_j.$$ Then we extend $f$ to whole $X$.