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Dec 17, 2021 at 11:51 history edited Mr.xue CC BY-SA 4.0
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Dec 12, 2021 at 1:53 history edited Mr.xue CC BY-SA 4.0
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Dec 12, 2021 at 1:22 comment added Mr.xue @username I think that I first need to prove the inequality, then using standard Osgood lemma to prove Lemma 1.2. Is it right?
Dec 11, 2021 at 19:13 comment added username You did not read Lemma 1.2 correctly. Read it again (suppose further..).
Dec 10, 2021 at 10:46 comment added Mr.xue @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.
Dec 10, 2021 at 9:54 comment added username 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?
Dec 10, 2021 at 8:19 history edited YCor CC BY-SA 4.0
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Dec 10, 2021 at 8:10 history asked Mr.xue CC BY-SA 4.0