A rough estimate for $ F(\mu):=\Bigg[\frac{4}{\pi}+\frac{\sqrt{2}}{\pi}\bigg(8+\pi\log K+\frac{\pi}{2}\log(1+\mu)\bigg)^{1/2}\Bigg](1+\mu)^{1/2}K$

Question

Suppose $\mu\geq 0$ and define $$ F(\mu):=\Bigg[\frac{4}{\pi}+\frac{\sqrt{2}}{\pi}\bigg(8+\pi\log K+\frac{\pi}{2}\log(1+\mu)\bigg)^{1/2}\Bigg](1+\mu)^{1/2}K $$ where $K>1$ is some constant. Then there exists $c'>0$ and $c>0$ such that
$$ F(\mu)\leq cK^2\log K $$ for all $\mu\leq c'K^2\log(c'K)$.

The argument above is made without proof for estimating an integral in a paper I read. Could anyone elaborate how one should choose the constants $c$ and $c'$?

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I have tried to use Cauchy-Schwarz to isolate the factor $K$ and the other terms, but cannot match the exponents in $K^2\log K$.

Answer 1

WLOG we can assume that $\mu>1$ and $\log\mu>1$. Then \begin{align} F(\mu)=&c_1(1+\mu)^{1/2}K +\big[c_2+c_3\log K+c_4\log(1+\mu)\big]^{1/2}(1+\mu)^{1/2}K\\ \leq &c_1(2\mu)^{1/2}K +[c_2+c_3\log K]^{1/2}(2\mu)^{1/2}K+[c_4\log(2\mu)]^{1/2}(2\mu)^{1/2}K\\ \leq &c_5\mu^{1/2}K+c_6(\mu\log K)^{1/2}K+c_7(\mu\log\mu)^{1/2}K\\ \leq &c_8(\mu\log\mu)^{1/2}K+c_6(\mu\log K)^{1/2}K+c_7(\mu\log\mu)^{1/2}K \end{align} Suppose $\mu\geq cK^2\log(cK)$. All we need to do now is to estimate the following two terms

• $(\mu\log K)^{1/2}K$
• $(\mu\log\mu)^{1/2}K$

For the first one $$ (\mu\log K)^{1/2}K \leq (\log K)^{1/2}[cK^2\log(cK)]^{1/2}K\leq c_9K^2\log K. $$ For the second one, denote $G(\mu):=(\mu\log\mu)^{1/2}K$. Then $$ G(\mu)\leq \biggr[cK^2\log(cK)\cdot \log\big(cK^2\log(cK)\big)\biggr]^{1/2}K \leq \sqrt{c}K^2\biggr[\log(cK)\cdot \log\big(cK^2\log(cK)\big)\biggr]^{1/2}\tag{1} $$ Note that $$ \log\big(cK^2\log(cK)\big)=\log(cK^2)+\log\big(\log(cK)\big) \leq2\log(\sqrt{c}K)+\log(cK)\leq c_{10}\log(c_{10}K). $$ Thus $$ \log(cK)\cdot \log\big(cK^2\log(cK)\big)\leq c_{11}^2[\log(c_{11}K)]^2\tag{2} $$ Combining (1) and (2), we have $$ G(\mu)\leq c_{12}K^2\log(c_{12}K) =c_{12}K^2(\log c_{12}+\log K)\leq c_{13}K^2\log K. $$