Hahn Banach Theorem implying existence of a nonzero linear functional taking 0 in a linear subspace

Question

I am reading this paper. In the proof of theorem 1, it is stated

By the Hahn-Banach theorem, there is a bounded linear functional on $C(I_n)$, call it $L$, with the property that $L\ne 0$ but $L(R) = L(S) = 0$.

$C(I_n)$ is space of continuous functions on $[0,1]^n$. $S$ is a linear subspace in it. $R$ is the closure of $S$.

Can you explain to me why this statement is true?

Answer 1

Let $M=\{f+af_0:a\in \mathbb R\}$ where $f_0$ is any fixed element not in $R$. Define $T(f+af_0)=a$. If we show that this is continuous on the space spanned by $R \cup \{f_0\}$ we can use Hahn Banach Theorem to get a continuous linear functional which is $0$ on $R$ and has the value $1$ at $f_0$. I will let you verify that $T$ is well defined. Suppose $f_n+a_nf_0 \to g$. If $(a_n)$ is unbounded it has a subsequence ${a_{n'}}$ converging to $\pm \infty$. Dividing by this we get $\frac {f_n'} {a_{n'}} +f_0=0$ which shows that $-f_0$ is the limit of sequence from $R$ which is a contardiction. Hence $(a_n)$ is bounded and it has subsequence converging to some $a$. we then get $f_n+a_nf_0 \to g=f+af_0$ for some $f \in R$ and $a=T(g)=\lim a_{n'} =\lim T(f_{n'}+a_{n'}f_0)$. By arguing with subsequences we see that $T$ is continuous.

Answer 2

This is true if $R \neq C(I_n)$. We shall assume that $C(I_n)$ is a real space.

Choose $x_0 \in C(I_n) \setminus R$. $C(I_n)$ is locally convex, $\{x_0\}, R \subseteq C(I_n)$ are disjoint, nonempty and convex sets. Furthermore, $\{x_0\}$ is compact and $R$ is closed, so by the geometric Hahn-Banach theorem we can find a bounded functional $L$ on $C(I_n)$ and constants $\gamma_1, \gamma_2 \in \mathbb{R}$ such that

$$ \forall y \in R : L(x_0) \leq \gamma_1 < \gamma_2 \leq L(y). $$

Note that $L \mid_R \equiv 0$, since otherwise we could (by linearity of $L$ and $R$) find a $y_0 \in R$ with $L(y_0) < \gamma_2$, contradicting the inequality above.

Since $L(x_0) < 0$ we have $L \neq 0$, but $L \mid_R \equiv 0$, and since $S \subseteq R$ we also get $L \mid_S \equiv 0$.

This statement holds in general as long as $R \neq X$ is a linear and closed subspace of a real or complex, locally convex topological vector space $X$.