# Existence of only one bounded linear functional.

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

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 $f'$.