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Let $\gamma$ be a positive, nondecreasing, continuous, function defined on $[0,\infty]$. Suppose that $\gamma(x+y)\le C(\gamma(x)+\gamma(y))$. In addition, suppose $$ \int_{2}^{\infty}\frac{dr}{\gamma(r)}=\infty.$$ suppose further that $$ f(t)\le f(0)+\int_0^t\gamma(f(r))dr+\gamma \Big(\int_0^tf(r)dr\Big). $$ How to show that $$ \gamma\Big(\int_0^t f(r)dr\Big)\le C(t+1)\int_0^t \gamma(f(r))dr $$

Actually, this question come from Lemma 1.2 in this paper.

My effort: Following the hint in this paper, I want to show that $\gamma$ cannot grow faster than $t^2$. However, I can't rule out the possibility that $$\limsup_{t\to +\infty}\frac{\gamma(t)}{t^2}=\infty.$$

I have no idea how to proceed it. Thanks for any help.

Update: If $C\le 1$, I can give a proof. \begin{align*} \gamma(\int_0^t f(r)dr)&=\gamma(\sum_{k=1}^n\int_{\frac{(k-1)t}{n}}^{\frac{kt}{n}}f(r)dr)\\ &\le \sum_{k=1}^{n}C^k\gamma(f(\xi_k))\frac{t}{n},\xi_k\in (\frac{k-1}{n}t,\frac{k}{n}t) \end{align*} If $C\le 1$, we can send $n$ to $\infty$ and get the proof.

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  • $\begingroup$ Here is a.n idea. Suppose $n=2^k\Leftrightarrow k = \ln n/\ln2$. Then $$ \gamma(n)\leq 2C\gamma(2^{k-1})\leq \cdots \leq 2^k C^k = n C^{\ln n/\ln2} =n^{1+\ln C/\ln2}. $$ It doesn't show $t^2$, but it shows no larger than $t^{1+\frac{\ln C}{\ln 2}}$. Is that enough? $\endgroup$
    – username
    Commented Dec 10, 2021 at 9:54
  • $\begingroup$ @username Thanks for your reply. I think your idea is helpful to solve this question. However, I also don't know how to complete the rest part of the proof. $\endgroup$
    – Mr.xue
    Commented Dec 10, 2021 at 10:46
  • $\begingroup$ You did not read Lemma 1.2 correctly. Read it again (suppose further..). $\endgroup$
    – username
    Commented Dec 11, 2021 at 19:13
  • $\begingroup$ @username I think that I first need to prove the inequality, then using standard Osgood lemma to prove Lemma 1.2. Is it right? $\endgroup$
    – Mr.xue
    Commented Dec 12, 2021 at 1:22

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