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This question might be a duplicate of Zeta function's asymptotic $\zeta(\sigma+it) = \mathcal{O}(|t|^{1-\sigma+\epsilon})$, but I cannot understand the answer there, and I would like to address my concern more clearly.

The question is of the following proposition in Stein and Shakarchi's Complex Analysis.

Suppose $s = \sigma + it$ with $\sigma, t\in\mathbb{R}$. Then for each $\sigma_0$, $0\leq \sigma_0\leq 1$, and every $\epsilon > 0$, there exists a constant $c_\epsilon$ so that $|\zeta(s)|\leq c_\epsilon |t|^{1 - \sigma_0 + \epsilon}$, if $\sigma_0 \leq \sigma$ and $|t| \geq 1$.

The key conclusion in the proof is that when $\sigma = \mathrm{Re}(s) \geq \sigma_0$, we have $$|\zeta(s)| \leq \frac{1}{|s - 1|} + 2|s|^{1-\sigma_0 + \epsilon} \sum_{n = 1}^\infty \frac{1}{n^{1 + \epsilon}}.$$ I wonder how this implies the conclusion to be proven.

The first term can be bounded trivially, so the second term is $C_\epsilon|s|^{1 - \sigma_0 + \epsilon}$ since the series converges. However, I suspect that $|s|^{1 - \sigma_0 + \epsilon} \leq A |t|^{1 - \sigma_0+\epsilon}$ would not hold for all $\sigma \geq \sigma_0$. Just take some fixed $t > 1$ and let $\sigma$ get to infinity. The first term goes to infinity while the second term is fixed. From the statement of the proposition, $c_\epsilon$ seems to be independent of $\sigma$.

Bo Liu
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  • As you say, you need to show that $|s|^{1 - \sigma_0 + \epsilon} \leq A |t|^{1 - \sigma_0+\epsilon}$. Taking $\sigma\to\infty$ isn't a good idea, because that increases $|s|$ without bound while leaving $|t|$ fixed. Note however that you have the hypothesis $\sigma\ge\sigma_0$; and the proposition is trivial for $\sigma\ge2$ by the uniform convergence of $\sum_{n=1}^\infty n^{-s}$. So you may assume that $\sigma_0\le\sigma\le2$, and then the inequality follows from $\sigma^2+t^2 \le Bt^2$ under the additional assumption $|t|\ge1$. – Greg Martin May 09 '21 at 23:49
  • @GregMartin Sorry just found I had a typo in the last paragraph I wrote and I corrected it. I still wonder how it is trivial for $\sigma \geq 2$. Note that $|s| = \sqrt{\sigma^2 + t^2}$, when $\sigma$ goes to infinity, does not $|s|$ exceed far beyond $|t|$? Thank you. – Bo Liu May 09 '21 at 23:51
  • @GregMartin Maybe I am having some difficulty understanding the proposition actually. I wonder if the stated $c_\epsilon$ only depends on $\sigma_0$ and $\epsilon$. If it does, it surely should hold for all $\sigma$ and $t$ in the described region. Then for fixed $t$ could I increase $\sigma$ and the inequality should still hold? Otherwise, if we cannot choose $\sigma$ arbitrarily, $c_\epsilon$ should be dependent on the choice of $\sigma$. Please correct me if I am understanding it incorrectly and I appreciate it much. – Bo Liu May 09 '21 at 23:59
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    I think I understand a little bit now... It seems that I should go from the definition of $\zeta(s)$ when $\sigma$ is larger than something... Then $|\zeta(s)|$ can be estimated by value irrelevant of $t$. – Bo Liu May 10 '21 at 01:03

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