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If X is a normed linear space ,$x_{1},x_{2},…,x_{n}\in X$ are linear independent,$a_{1},a_{2},…,a_{n}\in F$ are arbitrary,then there exist $f\in X^{\ast } $ such that $f\left( x_{k}\right)= a_{k}$,$k=1,2,…,n$.

I have tried to find a $M\in F$ such that for aritrary $t_{1},t_{2},…,t_{n}\in F$,$\left| \sum _{k=1}^{n} t_{j}a_{j}\right|\leq M\left\|\sum _{k=1}^{n} t_{j}x_{j}\right\|$ but it seemed failed.

Is it necessary to show that $\overline {f}$ is bounded?$\overline {f}\in span\ \left\{ x_{1},x_{2},…,x_{n}\right\}$ $\overline {f}\left( x_{k}\right)= a_{k}$

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Since you are still new here, I want to give you some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. Additionally, many would consider your post rude because it is a command ("Show..."), not a request for help, so please consider rewriting it. – Zev Chonoles Dec 25 '12 at 14:31
The requirement "such that ..." defines a functional $g \colon Y \to F$ on the subspace $Y$ spanned by $x_1,\dots,x_k$ such that $g(x_k) = a_k$. Can you show that $g$ is bounded? Then apply Hahn-Banach and get $f$. – Martin Dec 25 '12 at 14:31
up vote 2 down vote accepted


$$M={\rm span} \{x_1,\dots,x_n\}.$$

Note that

$$\bar{f}:M\rightarrow\mathbb{R}$$ defined by giving its values on the basis $\{x_1,\dots,x_n\}$ of $M$: $$\bar{f}(x_k)=a_k,$$ belongs to $M^{*}$ (it is trivial since $M$ is finite dimensional).

Now, thanks to the Hahn-Banach theorem, you can extend $\bar{f}$ to the continuous linear functional $$f:X\rightarrow\mathbb{R}.$$

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Is it necessary to show that f is bounded?I have just learned that theorem. – Jebei Dec 25 '12 at 14:43
You do not need to do that since the boundedness of $f$ is the point of the Hahn-Banach theorem. – Godot Dec 25 '12 at 14:47
What about $\overline {f}$ IS it must bounded? – Jebei Dec 25 '12 at 14:50
Yes $\bar{f}$ have to be bounded. In short: you are extending a bounded linear functional acting on a subspace of $X$ to a bounded linear functional acting on a whole space $X$. – Godot Dec 25 '12 at 14:56

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