Upper bound $x\mapsto\frac{W\left(\frac{2\alpha}{x^2}\right)}{2\alpha}$ for $x\in[0,1]$ Lambert function

Question

Let $W$ be the Lambert function.

Let $\alpha>0$.

I am looking for an upper bound on the function $x\mapsto\frac{W\left(\frac{2\alpha}{x^2}\right)}{2\alpha}$ for $x\in[0,1]$.

Ideally, the bound should be of the form $\ln\left(\frac{1}{x}\right)$, but so far I have not be able to prove that $$\frac{W\left(\frac{2\alpha}{x^2}\right)}{2\alpha}\leq\beta\ln\left(\frac{1}{x}\right) $$ for all $x\in[0,1]$ and for some $\beta$.

Answer 1

Let $t=\frac{2 \alpha }{x^2}$. The simplest bound is, for $t >e$, $W(t) < \log(t)$. So, if $x <\sqrt{\frac{2 \alpha }{e}}$, $$\frac{W\left(\frac{2\alpha}{x^2}\right)}{2\alpha}< \frac {\log(2\alpha)}{2\alpha}-\frac {\log(x)}{\alpha}=\frac {\log(2\alpha)}{2\alpha}+\frac {1}{\alpha}\log \left(\frac{1}{x}\right)$$