I'm working on this exercise (not homework) and I would gladly welcome some hints for how to solve it!
Exercise: Let $\{x_1,\dots,x_n\}$ be a set of linearly independent elements of a normed vector space $X$. Let $c_1,\dots,c_n \in \mathbb{C}$. Show that there exists $f\in X^\ast$ such that $f(x_i)=c_i$.
My idea:
I consider $M = span\{x_1,…,x_n\}$, which is a subspace of $X$. Any $x\in M$ can be written $x=\sum_1^n \lambda_k x_k$, for some $\lambda_1,…,\lambda_n \in \mathbb{C}$. Define $f:M \rightarrow \mathbb{C}$ by $f(x_i)=c_i$ for $i=1,…,n$. Then $$f(x) = \sum_1^n \lambda_k f(x_k) = \sum_1^n \lambda_k c_k.$$
If I can find a semi-norm $p:X \rightarrow \mathbb{R}$ such that $|f(x)| \leq p(x)$ for any $x \in M$, then by Hahn-Banach Theorem we would be done.
Thanks in advance!
Best Answer
You still need a twist in your argument. You can define a norm $p$ on $M$ by $$ p(\sum_{k=1}^n\lambda_kx_k)=\sum_{k=1}^n|\lambda_k|. $$ Now, as $M$ is finite-dimensional, all norms on it are equivalent. This in particular tells us that there exists a constant $c$ such that $p(x)\leq c\|x\|$ for all $x\in M$ (since $\|\cdot\|$ is another norm on $M$). So you have $$ |f(x)|\leq\,c\,\max\{|c_1|,\ldots,|c_n|\}\,\|x\|,\ \ x\in M, $$ and now you can apply Hahn-Banach.
Or a slightly more direct approach would be to notice that since $M$ is finite dimensional, every functional is continuous, and thus $f$ is necessarily bounded in $M$.