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Given $X$ a linear normed space over $K$ . $I$ be arbitrary indexing set , $\{f_\alpha: \alpha \in I\}\subset X$ and a family $\{c_\alpha: \alpha\in I\} \subset K$, I want to know that there exists exactly one bounded linear functional $f'\in X'$ with

  • $f'(f)=c_\alpha\quad \mbox{ for all}\quad \alpha \in I$

  • $||f||\le M\quad \mbox{ for some}\quad M\geq 0$

exists when for every finite subfamily $J \subset I$ and every choice of the member $\{\beta_\alpha :\alpha \in J\} \subset K$ the following inequality holds: $$\left|\sum_{\alpha \in J}\beta_\alpha c_\alpha\right|\le M\left\|\sum_{\alpha\in J} \beta_\alpha f_\alpha\right\|_X$$ I have been trying to solve this problem , to be frank i don't understand the question fully , I need a big deal of help to solve and understand this problem . Thank you very much .

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Are the $f_\alpha$ linearly independent? And $J$ assumed finite? – Davide Giraudo Nov 29 '12 at 21:27
@DavideGiraudo : As per question i found , it says nothing about the dependency of $f_\alpha$ , but yes $J$ is finite . – Theorem Nov 29 '12 at 21:30
up vote 1 down vote accepted

I assume that $\{f_\alpha:\alpha\in I\}$ are linearly independent. Consider $X_0=\mathrm{span}\{f_\alpha:\alpha\in I\}$ and define $$ f'':X_0\to K: \sum\limits_{\alpha\in I}\beta_\alpha f_\alpha\mapsto \sum\limits_{\alpha\in I}\beta_\alpha c_\alpha f_\alpha $$ Take arbitrary $x\in X_0$, then $x=\sum\limits_{\alpha\in J}\beta_\alpha f_\alpha$ for some finite $J\subset I$ and $\{\beta_\alpha:\alpha\in J\}\subset K$. In this case $$ |f''(x)|= \left|f''\left(\sum\limits_{\alpha\in J}\beta_\alpha f_\alpha\right)\right|= \left|\sum\limits_{\alpha\in J}\beta_\alpha c_\alpha\right|\leq M\left\Vert \sum\limits_{\alpha\in J}\beta_\alpha f_\alpha\right\Vert=M\Vert x\Vert $$ Since $x\in X_0$ is arbitrary we conclude that $f''\in X_0^*$ and $\Vert f''\Vert\leq M$. Then by Hahn-Banach theorem you can extend $f''$ to the $f'\in X^*$ with $\Vert f'\Vert\leq M$ and $f'|_{X_0}=f''$. In this case $f'(f_\alpha)=f''(f_\alpha)=c_\alpha$, so we have built the desired functional $f'$.

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Can you give me more explanation , and comments about the steps that u have written . – Theorem Nov 29 '12 at 21:38
@Theorem see my edits – Norbert Nov 29 '12 at 21:53
Thank you, i will let you know if i have some confusion. Is the extension always unique , normally when i see hahn banach extension theorem , its not stated ? – Theorem Nov 29 '12 at 22:05
It is not unique in general. Uniqueness is equivalent to non existence of segment in the unit sphere of dual space. – Norbert Nov 29 '12 at 22:10
Is it possible to prove the existence of the unique linear functional as above without using the last given inequality? – Theorem Nov 30 '12 at 8:38

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