Comments on $g(x)\xrightarrow{x\to\infty}\infty$ Implies $g'(x)\leq g^{1+\varepsilon}(x)$
Parent
$g(x)\xrightarrow{x\to\infty}\infty$ Implies $g'(x)\leq g^{1+\varepsilon}(x)$
Recently in my ordinary differential equations class we were given the following problem:
Suppose $g:(0,\infty)\to\mathbb{R}$ is an increasing function of class $C^{1}$ such that $g(x)\xrightarrow{x\to\infty}\infty$. Show that for every $\varepsilon>0$ the inequality $g^{\prime}(x)\leq g^{1+\varepsilon}(x)$ upholds, outside a set of finite length.
I thought about using Grönwall's inequality , but i did not got any useful result.
Post
Consider $g^{-\epsilon}$. Then its first derivative is $D_x \; g^{-\epsilon} = \epsilon g^{-1-\epsilon} g'$. Then $g^{-\epsilon} > 0$ and tends to 0, and ${D_x \; g^{-\epsilon}} < 0 $.
If $D_x \; g^{-\epsilon} < -\epsilon$ on a set of infinite measure, then $\int^{x}_0 D_x , g^{-\epsilon} \quad dx $ (which differs from $g^{-\epsilon}$ by a constant) would diverge to negative infinity. So $D_x \; g^{-\epsilon} \ge -\epsilon$ outside a set of finite measure, from which your desired inequality follows immediately.
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