Proof about dual space

Question

Let $X$ a vectorial space y let $\Gamma \subset X^{\ast}$. We will say that $\Gamma$ is total in $X$ if $f(x)=0$, $\forall f \in \Gamma$ implies that $x=0$. I have to prove that if $\Gamma$ is total in $X$ and $x_1, x_2, \cdots, x_n$ are linearly independent in $X$, then exist $f_1, f_2, \cdots, f_n$ in $\Gamma$ such that $f_i(x_j)=\delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta function, that is $$f_i(x_j)=\delta_{ij}=\begin{cases}1 \qquad i=j \\ 0 \qquad i \ne j \end{cases}$$

My attempt. I have to define $f_i: X \to \mathbb{K}$, where $\mathbb{K}=\mathbb{C}$ or $\mathbb{R}$. The issue is that I have no guarantee that the space is finite dimensional. Since if it were of finite dimension the problem would be easier and i'm stuck here. I would really appreciate some help.

Answer 1

Consider the map $$T:f\in\Gamma \mapsto (f(x_1), \ldots,f(x_n))\in \mathbb K^n.$$ I then claim that $T$ is onto. Arguing by contradiction, suppose not, so there is some nonzero linear functional vanishing on the range if $T$. In other words, there is $(a_1,\ldots,a_n)$, not all zeros, such that, for every $f$ in $\Gamma$, $$ 0 = \sum a_if(x_i) = f\left(\sum a_ix_i\right). $$ By hypothesis, $\sum a_ix_i=0$, contradicting the linear independence of the $x_i$. This shows that $T$ is onto, from where the conclusion follows easily.

Answer 2

As stated, the assertion is false: You have to assume in addition that $\Gamma$ is a linear subspace. Otherwise, $X=\mathbb R$, $\Gamma=\{id\}$, $n=1$, $x_1=1/2$ is a counterexample.

Under the additional hypothesis that $\Gamma$ is a linear subspace, the assertion can be shown by induction by $n$.

For the induction step, let $x_1,\dotsc,x_n$ be given. For $i=1,\dotsc,n$ we have to show that there is $f_i$ with $f_i(x_j)=\delta_{ij}$ $(j=1,\dotsc,n)$. By renumbering, it suffices to consider the case $i=n$

By induction hypothesis, there do exist $g_1,\dotsc,g_{n-1}$ such that $g_i(x_j)=\delta_{ij}$ $(i,j=1,\dotsc,n)$. Since the $x_i$ are linearly independent, $x:=x_n-\sum_{i=1}^{n-1}g_i(x_n)x_i\ne0$. So there is some $f\in\Gamma$ such that $f(x)=1$.

Now $f_n=f-\sum_{i=1}^{n-1}f(x_i)g_i$ has the required property.