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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. )

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

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  • $\begingroup$ But why is the original $f$ bounded? $\endgroup$ – Keshav Srinivasan Feb 8 '19 at 19:00
  • $\begingroup$ It's a linear functional on a finite dimensional normed space. $\endgroup$ – Rigel Feb 8 '19 at 19:24
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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 $X$.

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  • $\begingroup$ Your f is not a functional. $\endgroup$ – Infinite Nov 26 '17 at 10:53

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