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!

  • 1
    $\begingroup$ $f$ is continuous on $M$, so $f$ is bounded. Does boundedness ring a bell for a good candidate of $p$? $\endgroup$
    – user27126
    Dec 17 '12 at 10:53
  • $\begingroup$ @Sanchez what boundedness are you referring for? $\endgroup$
    – Juniven
    Dec 17 '12 at 10:58
  • $\begingroup$ @jun, $f$ is bounded on $M$. $\endgroup$
    – user27126
    Dec 17 '12 at 11:01
  • $\begingroup$ Sanchez: That makes sense! Since $|f(x)| < \infty$ there must exist a constant $C$ such that $|f(x)|< C\|x\| =: p(x)$ where $\|\cdot \|$ is the norm on $X$. $\endgroup$ Dec 17 '12 at 11:12
  • $\begingroup$ @DoubleTrouble, there you go :) It may be a good idea for you to write it up as an answer too. $\endgroup$
    – user27126
    Dec 17 '12 at 11:17

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 $M$.

  • $\begingroup$ I was actually thinking along that path (all norm are equivalent on M) at first, but was confused by something. However, you made it very clear. Thanks! Your second point fits my argument better, and is the missing puzzle bit! Thank you. $\endgroup$ Dec 17 '12 at 17:44

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!

  • $\begingroup$ A problem I mentioned still exists, namely that when you say $|f(x)| < \infty$, it's not clear what $x$ you are talking about. It's also unclear why $|f(x)| \leq C \| x \|$ follows. So, the latter follows from the continuity of $f$ on $M$, which is important here. $\endgroup$
    – user27126
    Dec 17 '12 at 11:41
  • $\begingroup$ Okay, I was thinking that $|f(x)|< \infty$ for every $x\in M$ implied that there exists a constant $C$ s.t. $|f(x)| \leq C\|x \|$ for every $x\in M$. But I think I now understand why that is incorrect, e.g. you could have a sequence $\{y_n \}\subset M$ with $|f(y_n)| = n$. But then I don't understand how I should argue to see that $f$ is bounded/continuous. $\endgroup$ Dec 17 '12 at 12:29
  • $\begingroup$ I get it now, from Martin answer above! :) $\endgroup$ Dec 17 '12 at 17:44

Set $L_k=span_{i \neq k}\{x_i\}$ for each $k=1,\ldots, n$ and observe that each $L_k$ is a closed subspace of $X$. By the Hahn-Banach Theorem there exist functionals $f_k \in X^*$ such that $f_k(L_k)=0, \ \ f_k(x_k)=1$ for each $k=1,\ldots ,n$. Now the functional $f=\sum c_if_i$ satisfies the desired property.


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