# Upper bound for an integral of positive functions

Let $\gamma \colon \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$ be a function that is continuous and increasing with $\gamma (0) = 0$. Also, let $w \colon \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$ be an integrable function, that is to say, $\int_{0}^{\infty} w(\tau) \mathrm{d} \tau \leq r$ for some $r > 0$. The question is "does there exist some continuous function $\eta \colon \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$ with $\eta(0)=0$ such that the following holds $$\int_{0}^{\infty} \gamma(w(\tau)) \mathrm{d} \tau \leq \eta(r)$$ for any $\gamma(.),w(.)$ and $r$ defined above?"

## 1 Answer

Certainly not. Let $\gamma(t)=\sqrt{t}$, $w(t)=\frac1{t^2+1}$. Then $\gamma$ is continuous and increasing with $\gamma(0)=0$. We also have $$\int_0^\infty w(\tau)\,d\tau=\frac\pi2.$$ But $$\int_0^\infty\gamma(w(\tau))\,d\tau=\int_0^\infty\frac1{\sqrt{\tau^2+1}}\,d\tau=\infty.$$

• Thanks martin for your prompt response. – Navid Noroozi Jun 30 '13 at 12:31
• You are welcome. A way that often helps me think about this kind of questions is to phrase them in terms of series instead of integrals; in most cases the answer will be the same, and they are easier for my intuition. – Martin Argerami Jun 30 '13 at 19:15