Preserving inequality through inner product

Let $f, g, h \in H$ be functions mapping from $X$ to $\mathbb{R}$. Define an inner product $\langle \cdot, \cdot \rangle_H:H \times H \to F$. Under what conditions (if any) does $$0<f(x) < g(x), \quad 0<h(x), \quad\forall x \in X$$ imply $$\langle f, h \rangle_H < \langle g, h\rangle_H$$ ?

Assume for all $u,v\in H$ such that $u>0,v>0$ we have $$\langle u,v\rangle_H > 0$$
Then $g-f > 0$, $g-f \in H$ and $$\langle g,h \rangle_H - \langle f,h \rangle_H = \langle g-f,h \rangle_H > 0$$
and we can prove $$\langle g,h\rangle_H > \langle f,h\rangle_H$$
Assume that there exist an $u>0$ and a $v>0$ such that $\langle u,v\rangle_H \leq 0$, then take $h=u,f=u/2,g=v$ then $h,f,g\in H$, $0<f<g$ and $$\langle g-f ,h\rangle_H = \langle u/2 ,v\rangle_H = \frac{1}{2}\langle u ,v\rangle_H \leq 0$$ Therefore the condition is nessesary and sufficient
• Great answer. I think in your last line you didn't mean to put the $>$ symbols there. – Student Jun 1 '18 at 7:29