Show that $x_0 \in \overline{\langle M \rangle } $ if and only if $f(x_0) = 0, \; \forall f \in X^* : f|_M = 0$.

Question

Exercise :

Let $X$ be a normed space and $M \subset X$. Show that an element $x_0 \in X$ belongs to the set $\overline{\langle M \rangle}$ if and only if $f(x_0) = 0$ for all $f \in X^*$ such that $f|_M = 0$.

Attempt :

First of all, clarifying that $\overline{\langle M \rangle}$ denotes the closure of the linear hull of $m \in M$, thus it is the set :

$$\overline{\langle M \rangle} = \overline{\{\lambda \cdot m | m \in M , \lambda \in \mathbb R\}}$$

I have the intuition that what we want to prove essentialy is that $\langle M \rangle$ is dense over $X$, but I'm not sure if that's correct.

Other than that, I can't seem to find any other way around it, so I would really appreciate any tips or elaborations.

Answer 1

$\newcommand{\span}{\operatorname{span}}$No, the exercise is not about proving that $\span M$ is dense. The set $M$ could be a singleton for instance. A sketch of a proof could be as follows:

"$\Rightarrow$": Let $x\in \overline{\span M}$, then there is a sequence $x_n\in \span M$ such that $x_n\to x$. Moreover if $f|_M=0$ for some $f\in X^*$ then what can you say about $f(x)$? Use the continuity of $f$.

"$\Leftarrow$": Assume that $f(x)=0$ for all $f\in X^*$ such that $f|_M=0$. Assume that $x\notin \overline{\span M}$ then there is some ball $B(x,\delta)$ such that $B(x,\delta)\cap\overline{\span M}=\emptyset$. In particular for every $z\in \span M$ we have $\|x-z\|\geq \delta$.

We will make use of this. First notice that each $z\in \span(M\cup x)$ can be written as follows \begin{align} z=\lambda x+\sum_{i=1}^n \mu_i m_i \end{align} for some $\lambda$, $\mu_i\in\mathbb K$ and $m_i\in M$. We define the operator $g:\span(M\cup x)\to\mathbb K$ as follows: \begin{align} g(z)=\lambda \end{align} It is easy to see that it is linear. Moreover $g|_M=0$ and $g(x)=1$ by definition. It is also bounded, since (assuming $\lambda\neq 0$) \begin{align} \frac{|g(z)|}{\|z\|}=\frac{|\lambda|}{\|\lambda x + \sum_{i}\mu_i m_i \|}=\frac{1}{\|x+\sum_{i}\frac{\mu_i}{\lambda} m_i\|}\leq \frac 1 \delta \end{align} This bound trivially holds if $\lambda=0$. Hahn Banach now says that there is an extension of $g$, call it $f\in X^*$ satisfying $f|_{\span(M\cup \{x\})}=g$. This $f$ satisfies $f(x)=1$ while $f|_M=0$ a contradiction.