In $\mathbb{R}^S$, where $S < \infty$, if there is a linear functional such that $F(x) = 0$, $\forall x \in M$ and $F(y) > 0$, $\forall k \in K$ where $K$ and $M$ are linear subspaces, then can I assert that there is a strictly positive vector $\phi$ such that:
$$ F(z) = <x,\phi>, \forall z \in \mathbb{R}^S $$
I suppose this is an application of Riesz Representation Theorem, I am unsure about the positivity part. Thanks.
Let $B = \{e_{1}, \ldots, e_{S} \}$ an orthonormal base for $\mathbb{R}^{S}$ and let $ \{e_{r_{1}}, \ldots, e_{r_{n}} \} \subset B$ an base ortonormal for $M$, respectively. Let's define $F_{K}: K \longrightarrow \mathbb{R}$ as $x \longmapsto F_{K}(x) = F(x)$ then $F_{K}(x) > 0$ for all $x\in K -\{0\}$.
On the other hand, $K$ is an vectorial space with inner product (the canonical inner product in $\mathbb{R}^{S}$) then for the Riesz representation theorem there is a vector $w\in K$ such that $F_{K}(x) = <x, w>$. Note that $<w, w > = F_{K}(w) >0$, i. e. $w$ strictly positive.
Now, if $y \in (\mathbb{R}^{S} - K)$ then $F(y) = <y, w> = 0$, in particular, $0 = F(y) = <y, w>$ for all $y\in M$.