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The Hahn-Banach theorem (as defined on Kreyszig) is as follows:

Let $V$ be a real vector space and $\phi$ be a sublinear functional on $V$. Let $f$ be a linear functional on a subspace $W$ of $V$ which satisfies $$ f(x) \leq \phi(x) \text{, } \forall x \in W $$

Then $f$ has a linear extension $\tilde{f}$ from $W$ to $V$ such that $\tilde{f}(x) = f(x)$, $\forall x \in W$ and $$ \tilde{f}(x) \leq \phi(x) \text{, } \forall x \in V $$

Indeed, $\tilde{f}$ is a linear functional on $V$.

Last night, while reviewing this paper, I went through the proof of Theorem 1 and while I understood how the conclusion was reached, there was one assumption which made use of the Hahn-Banach theorem that I did not quite understand:

"By Hahn-Banach theorem, there exists a bounded linear functional $L$, on $C(I_n)$, having the property that $L \neq 0$ but $L(\overline{S}) = L(S) = 0$"

Note that $C(I_n)$ denotes the space of continuous functions defined on the $n$-dimensional unit hypercube $I_n = [0,1]^n$. Moreover, S denotes a linear subspace and $\overline{S}$ denotes its closure which is again a (closed) linear subspace.

I do not understand exactly how the property $L \neq 0$ but $L(\overline{S}) = L(S) = 0$ came about simply by the Hahn-Banach theorem. Am I missing something?

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Since $\overline{S}$ is a closed linear subspace and since $C(I_n)$ is a normed linear space and hence in particular a locally convex topological vector space so you can use the following Theorem which you can find in Rudin's Functional Analysis [Page 59, Theorem 3.5]

$\textbf{Theorem}$: Let $Y$ be a closed subspace of a locally convex topological vector space $X$, and $x_0 \in X$. If $x_0 \notin Y$, then there exists $f \in X^*$ such that $f(x_0) \neq 0$ but $f|_Y \equiv 0$.

$\textit{Note:}$ I am assuming that $\overline{S} \neq C(I_n)$. Though it is a consequence of the Hahn-Banach Theorem but it requires a lot of effort to get this far, which you can see when you refer the book.

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You need to assume $\overline{S} \ne C(I_n)$, otherwise the statement is obviously false. Consider the subspace $W$ spanned by $\overline{S}$ and some element $y \notin \overline{S}$, and the linear functional $f$ defined on $W$ by $f(a y + s) = a$ for $a \in \mathbb R$, $s \in \overline{S}$. Let $\delta = dist(y, \overline{S}) > 0$. Then $\|a y + s\| \ge |a| \delta$, i.e. $|f(x)| \le \|x\|/\delta$. Use Hahn-Banach with $\phi(x) = \|x\|/\delta$.

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