For $i=1,2$, let $\phi_i:R_+\to R_+$ be a continuous function such that $\phi_i(0)=0$ and define

$$\gamma_i(l) := \int_0^l\frac{dm}{\phi_i(m)}.$$

Assume that $(\phi_1,\phi_2)$ satisfy the following conditions:


  1. There exists some $l^*\in R_+$ such that \begin{equation} \phi_2(l) - \phi_1(l) \begin{cases} > 0 \text{ for } l\in (0,l^*), \\ < 0 \text{ for } l\in (l^*,+\infty). \end{cases} \end{equation}
  2. $\phi_i\in\mathcal{C}^1(R_+^{\ast}\backslash\{l^{\ast}\})\cap\mathcal{C}^0(R_+)$ is strictly increasing on $R_+$ for $i=1,2$.

  3. $\gamma_i(0+)=0$ and $\gamma_i(+\infty)=+\infty$ for $i=1,2$.

For $i=1,2$ define $H_i\in\mathcal{C}^0(R_+)$ by

\begin{eqnarray} H_2'\big((\phi_2(l))\big)&=&e^{\gamma_2(l)}\int_{l}^{+\infty}e^{-\gamma_2(m)}F''(m)dm+F'(l),~ \forall l\in R_+, \\ H_1'\big((\phi_1(l))\big)&=&\begin{cases} e^{\gamma_1(l)}\int_{l^{\ast}}^{+\infty}e^{-\gamma_1(m)}F''(m)dm-e^{\gamma_1(l)}\int_{l^{\ast}}^{+\infty}e^{\gamma_2(l^{\ast})-\gamma_1(l^{\ast})-\gamma_2(m)}F''(m)dm,~ \forall l\in [0,l^{\ast}), \\ e^{\gamma_1(l)}\int_{l}^{+\infty}e^{-\gamma_1(m)}F''(m)dm+F'(l)-H_2'\big((\phi_1(l))\big),~ \forall l\in [l^{\ast},+\infty), \end{cases} \end{eqnarray}

where $F\in \mathcal{C}^2(R_+)$ is convex such that $F(0)=H_1(0)=H_2(0)=0$. We may show easily that $H_2\in\mathcal{C}^2(R_+^{\ast})\cap\mathcal{C}^0(R_+)$ and $H_1\in\mathcal{C}^2(R_+^{\ast}\backslash\{l^{\ast}\})\cap\mathcal{C}^1(R_+^{\ast})\cap\mathcal{C}^0(R_+)$. Moreover, we may show that $H_2$ and $H_1+H_2$ are also convex. Now my question is whether it can be proved that the inequality

$$H_2'(\phi_1)(s-\phi_2)+H_2(\phi_2)+H_1(s)-\big(H_1'(\phi_1)+H_2'(\phi_1)\big)(s-\phi_1)-\big(H_1(\phi_1)+H_2(\phi_1)\big)\ge 0~~~~~ (\star)$$

holds for all $l\in [l^{\ast},+\infty)$ and $0\le s<\phi_2(l)$. Here $\phi_i=\phi_i(l)$.

In fact, the objective is to prove that $(\star)$ holds for $0\le s <\phi_1(l)$. It is easy to prove that $(\star)$ holds for the two cases:

  1. $l\in [0,l^{\ast})$ and $0\le s <\phi_1(l)$

  2. $l\in [l^{\ast}, +\infty)$ and $\phi_2(l)\le s <\phi_1(l)$

But it is not trivial for $l\in [l^{\ast},+\infty)$ and $0\le s<\phi_2(l)$. I'm not even sure whether the proposition is right or not. If someone has an idea to prove or to find some counterexample, please let me know. Thx so much!


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