Exercise: Application of Hahn-Banach Theorem

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!

Solution 1:

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 $M$.

Solution 2:

With some guidance from Sanchez, I managed to figure it out! We have

$$|f(x)| = |\sum_1^n \lambda_k c_k | < \infty,$$

Hence, for any $x \in M$ we have $|f(x)| < \infty$. Since $f$ is bounded there exists a constant $C$ such that $|f(x)| \leq C\|x\|$ for all $x \in M$, where $\| \cdot \|$ is the norm on $X$. If we define $p(x) = C\|x \|$, then it will clearly be a semi-norm on $X$ and we can use the Hahn-Banach Theorem to get the desired result!