How much "room" in inequality $\displaystyle \int_a^b \varphi' ov ~\mathrm{d}x \leq 0$

Question

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.

Answer 1

Let's upgrade Iosif's comment to an answer.

Let $\chi$ be a smooth bump function supported in $[-\epsilon,\epsilon]$. For any $\varphi\in U$ with support within $[a+\epsilon,a-\epsilon]$, then $\chi*\varphi$ is also in $U$. So $$ \int (\chi*\varphi)(x) v(x) o(x) ~dx = 0 \implies \int_{a+\epsilon}^{b-\epsilon} \int_{-\epsilon}^{\epsilon} \varphi'(x) \chi(y) v(x+y) o(x+y) ~dy ~dx = 0$$

The output of the $y$ integral is a smooth function on $[a+\epsilon,b-\epsilon]$, and hence Iosif's argument shows that it must be constant.

Taking a limit as $\epsilon\to 0$ using $\chi$ an approximation to identity, we find that $vo$ must be almost everywhere constant.

Answer 2

$\newcommand{\al}{\alpha}\newcommand{\be}{\beta}\newcommand{\vpi}{\varphi}\newcommand{\thh}{\theta}\newcommand{\I}{\mathscr I}\newcommand{\J}{\mathscr J}$The conjunction of your conditions, \begin{equation}\tag{1}\label{A1} \int_a^b \varphi' uv ~\mathrm{d}x \leq 0 \quad \forall \varphi \in C_0^\infty([a, b], [0, \infty))=:U, \quad \frac{u(b)}{u(a)} \leq v \leq 1 \text{ a.e. } \end{equation} and \begin{equation} \tag{2}\label{A2} \int^b_a \psi'uv < 0 \end{equation} for at least one $\psi \in C_0^\infty([a,b],[0,\infty))$ (where $u:=o$), cannot hold. Therefore, assuming that \eqref{A1} and \eqref{A2} hold, any conclusion will hold.

Indeed, without loss of generality $[a,b]=[0,1]$. Take any disjoint open subintervals $I$ and $J$ of the interval $[0,1]$ of equal lengths such that $I$ is to the left of $J$. Let $\thh:=1_I-1_J$. Then there is a sequence $(\vpi_k)$ in $U=C_0^\infty([0,1],[0,\infty))$ such that $\vpi'_k\to\thh$ in $L^1[0,1]$ (as $k\to\infty$). Since the function $uv$ is bounded, from \eqref{A1} we get \begin{equation*} 0\ge\int_0^1\vpi'_k uv\to\int_0^1\thh uv=\int_I uv-\int_J uv, \end{equation*} so that \begin{equation*} \int_I uv\le\int_J uv. \tag{10}\label{10} \end{equation*}

For natural $n$ and $j\in[n]:=\{1,\dotsc,n\}$, let \begin{equation*} I_{n,j}:=((j-1)/n,j/n),\quad \I_{n,j}:=\int_{I_{n,j}}uv,\quad \J_{n,j}:=\int_{I_{n,j}}\psi'uv. \end{equation*} By the conditions $u(0)>u(1)>0$ and $u\in C^1[0,1]$, the second condition in \eqref{A1}, and in view of \eqref{10}, for all $j\in[n]$ \begin{equation} \frac{u(1)+O(1/n)}n\ge \I_{n,n}\ge\I_{n,j}\ge\I_{n,1}\ge(u(0)-O(1/n))\frac{u(1)}{u(0)}\frac1n =(1-O(1/n)) \frac{u(1)}n \end{equation} and hence \begin{equation} \I_{n,j}=\frac{C+O(1/n)}n, \end{equation} where $C:=u(1)$ and the constants in $O(\cdot)$ depend only on the function $u$. Also, for all $j\in[n]$ we have $\psi'=\psi'(j/n)+O(1/n)$ on the intervals $I_{n,j}$ and hence \begin{equation} \J_{n,j}=(\psi'(j/n)+O(1/n))\I_{n,j}=(\psi'(j/n)+O(1/n))\frac{C+O(1/n)}n =C\psi'(j/n)\frac1n+O(1/n^2), \end{equation} where the constants in $O(\cdot)$ depend only on the functions $u$ and $\psi$. So, \begin{equation} \int_0^1\psi'uv=\sum_{j\in[n]}\J_{n,j}\to C\int_0^1\psi'=0 \end{equation} as $n\to\infty$.

Thus, $\int_0^1\psi'ub=0$, and your condition \eqref{A2} cannot hold. $\quad\Box$