Improve the bound of a Lipschitz function

Question

Let $\lambda$ be a function defined as $$\lambda(x) = \frac{x}{|\ln (x)|^2}$$ on an interval $[0,x_0]$ for some $x_0 \ll 1$ ($x_0$ can be chose arbitrarily small). Now let us consider $\varepsilon$ taken such that $\varepsilon \ll x_0/2$. I would like to know if there exists a constant $C>0$ (that might depend on $x_0$ but not $\varepsilon$) such that $$|\lambda(x + \varepsilon) - \lambda(x)| \le C \lambda(\varepsilon)$$ on the interval $[0,x_0/2]$. As our function $\lambda$ is continuously differentiable on $[0,x_0/2]$, it is in particular Lipschitz continuous so that we have $$|\lambda(x + \varepsilon) - \lambda(x)|\le C\varepsilon$$ for $C = \sup_{x \in [0,x_0/2]}\lambda'(x)$. Now this is not enough as I would like a bound like $\varepsilon/|\ln(\varepsilon)|^2$, and I miss this $1/|\ln(\varepsilon)|^2$. At $x = 0$ this bound is clearly satisfied, but I am not sure about the other value of $x \in (0,x_0/2]$. Any help would be greatly appreciated.

Answer 1

Some thoughts.

For convenience, replace $\varepsilon$ with $t$.

We have, for all $x, t> 0$ with $x + t < 1$, $$\frac{x + t}{\ln^2(x + t)} - \frac{x}{\ln^2 x} \ge \frac{\ln x - 2}{\ln^3 x} \cdot t. \tag{1}$$ (The proof of (1) is given at the end.)

Using (1), we have $$\frac{\lambda(x + t) - \lambda(x)}{\lambda (t)} \ge \frac{\ln x - 2}{\ln^3 x}\cdot \ln^2 t.$$

Proof of (1):

Let $$f(t) := \frac{x + t}{\ln^2(x + t)} - \frac{x}{\ln^2 x} - \frac{\ln x - 2}{\ln^3 x} \cdot t.$$

We have $$f'(t) = \frac{1}{\ln^2(x + t)} - \frac{2}{\ln^3(x + t)} - \frac{\ln x - 2}{\ln^3 x}$$ and $$f''(t) = \frac{6 - 2\ln(x + t)}{(x + t)\ln^4(x + t)} > 0.$$ Also, $f'(0) = 0$. Thus, $f'(t) \ge 0$. Also, $f(0) = 0$. Thus, $f(t) \ge 0$.

We are done.