Let $[a, b]$ be a nonempty interval., $o \in C^1([a, b])$ be such that $o>0$ and $o'<0$ and assume we found some $v \in L^\infty(\mathbb{R})$ such that \begin{equation}\tag{1} \int_a^b \varphi' v o ~\mathrm{d}x \leq 0 \quad \forall \varphi \in C_0^\infty([a, b], [0, \infty))=:U, \quad \frac{o(b)}{o(a)} \leq v \leq 1 \text{ a.e. } \end{equation} and $$\tag{2} \int^b_a \varphi'vo~\mathrm{d}x < 0 $$ for at least one $\varphi \in U$. My question is whether we can find $\zeta \in L^\infty(\mathbb{R})$ with $\int_a^b \zeta ~\mathrm{d}x >0$ such that $v+\delta \zeta$ still fulfils $(1)$ for $\delta$ arbitrarily small. I also don't know whether $(1)$ and $(2)$ can actually be satisfied at once.

I thought about $\zeta$ adding a very small constant where $v<1$ (on sets with measure greater than $0$) which should lead to the integral inequality not being fulfiled. If one could exclude the case of $v=1$ on a set with measure greater than zero, then we could take $v = \zeta$. I also tried to choose $\zeta$ such that $\zeta o$ is constant which would render the integral condition zero. But then we would very likely violate the box constraints. Of course, $\zeta$ can also be negative, but I would not know how to construct such example.

Thank you in advance!

Let $[a, b]$ be a nonempty interval, $o \in C^1([a, b])$ be such that $o>0$ and $o'<0$ and assume we found some $v \in L^\infty(\mathbb{R})$ such that \begin{equation}\tag{1}\label{1} \int_a^b \varphi' v o ~\mathrm{d}x \leq 0 \quad \forall \varphi \in C_0^\infty([a, b], [0, \infty))=:U, \quad \frac{o(b)}{o(a)} \leq v \leq 1 \text{ a.e. } \end{equation} and $$\tag{2}\label{2} \int^b_a \varphi'vo~\mathrm{d}x < 0 $$ for at least one $\varphi \in U$. My question is whether we can find $\zeta \in L^\infty(\mathbb{R})$ with $\int_a^b \zeta ~\mathrm{d}x >0$ such that $v+\delta \zeta$ still fulfils $(1)$ for $\delta$ arbitrarily small. I also don't know whether \eqref{1} and \eqref{2} can actually be satisfied at once.

I thought about $\zeta$ adding a very small constant where $v<1$ (on sets with measure greater than $0$) which should lead to the integral inequality not being fulfiled. If one could exclude the case of $v=1$ on a set with measure greater than zero, then we could take $v = \zeta$. I also tried to choose $\zeta$ such that $\zeta o$ is constant which would render the integral condition zero. But then we would very likely violate the box constraints. Of course, $\zeta$ can also be negative, but I would not know how to construct such example.