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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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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
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1 Answer

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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