A multiparameter inequality

Question

Let $x\ge 2$ be a real number and $\nu=:\nu(x)$ an integer-valued function. Let $I$ be an interval of $\mathbb{R}$ such that $\lambda(I)<(\log\nu)^{-1/2}$ where $\lambda$ denotes the Lebesgue measure on $\mathbb{R}$. I am trying to find only one condition on $\nu,x$ and $\lambda(I)$ which guarantee that $$\frac{\log x}{x\lambda(I)^2}\le M\frac{\lambda(I)}{\nu \log\nu}$$ for some constant $M>0$. Can someone help with this concern? My guess: I think that the condition $\nu>C(\log x)^2$ (independent on $m$ $x$ and $\lambda(I)$) for some constant $C>0$ is enough to prove the above equality. But I failed to prove that.

Answer 1

First we must have $\nu>1$ due to $0\leq\lambda(I)<(\log\nu)^{-1/2}$. With this, rearrange the inequality to get $$(\nu\log\nu)\frac{\log x}x\leq M\lambda^3(I)<M(\log\nu)^{-3/2}$$ $$\Rightarrow\quad\nu(\log\nu)^{5/2}\leq\frac{Mx}{\log x}.$$ Using the log-bound $\nu(\log\nu)^{5/2}\leq4(\nu-1)^2$ for all $\nu>1$, it is sufficient to have $$1<\nu\leq1+\sqrt{\frac{Mx}{4\log x}}.$$ If we want to further simplify the bound, use $\log x\leq x^{1/2}$ to get $$1<\nu\leq 1+Cx^{1/4}$$ for some constant $C\leq\sqrt M/2$.

---

Proof of the log-bound:

Let $f(x)=4(x-1)^2-x(\log x)^{5/2}$, we show $f(x)\geq0$ for all $x\geq1$.

Since $f(1)=0$, we only need to show $f'(x)\geq0$ for $x\geq1$. To this end, find $$f'(x)=4(x-1)-(\log x)^{5/2}-(5/2)(\log x)^{3/2}.$$ We prove the following claim $$(\log x)^p\leq a(x-1)\quad\forall\,x\geq1,p\geq1,a>0\ \text{such that}\ \frac1{p-1}\left(\frac ap\right)^{\frac1{p-1}}\geq e^{-1}.$$ Let $g(x)=a(x-1)-(\log x)^p$, since $g(1)=0$ we need $g'(x)=a-p(\log x)^{p-1}/x\geq0$, or equivalently, $$\log x\leq(ax/p)^{1/(p-1)}. \tag{$\star$}$$ It is easy to verify $\alpha x^\beta-\log x$ attains minimum at $x=(\alpha\beta)^{-1/\beta}$ for any $\alpha,\beta>0$. Hence, we need $\alpha\beta\geq e^{-1}$ for $\alpha x^\beta\geq\log x$. Taking this into $(\star)$ with $\alpha=(a/p)^{1/(p-1)}$ and $\beta=1/(p-1)$, we get the claim.

Using the claim, we obtain the following bound $$(\log x)^{5/2}\leq2(x-1)\quad\text{and}\quad(\log x)^{3/2}\leq\frac45(x-1).$$ Taking these into $f'(x)$ and we obtain $f'(x)\geq0$ and thus the log-bound.