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

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. )

• i guess that with X' you mean the dual space right? – JayTuma Nov 26 '17 at 10:09
• yes, $X'$ is dual of $X$ (X'=set of all bounded linear functionals on X) – Infinite Nov 26 '17 at 10:29

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 $X$.
• But why is the original $f$ bounded? – Keshav Srinivasan Feb 8 '19 at 19:00
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 $X$.