Bound a function with parameter involving logarithm

Question

In my master project I encounter the following function $$f_\varepsilon(x) = \ln\left(\frac{x^{1 + a} + \varepsilon^\beta}{\lambda(x)^2 + \varepsilon^2}\right)$$ for $a$ close to zero, $\beta \in (1, 2)$ and $$\lambda(x) = \frac{x}{|\ln(x)|^2}.$$ I aim to bound the following quantity $$g_\varepsilon(x) = \bigg|\frac{1}{f_{\varepsilon}(x)} - \frac{1}{f_{0}(x)}\bigg|$$ on a certain interval $[0, x_0]$, $x_0 \ll 1$, where $f_0$ is just $f_\varepsilon$ with the parameter $\epsilon = 0$. Clearly, for $x \to 0$, we get $$g_\varepsilon (0) = \frac{1}{(2 - \beta)|\ln \varepsilon|}$$ so I feel that we should get something like $$g_\varepsilon(x) \le \frac{C}{(2 - \beta)|\ln \varepsilon|}$$ for $C > 0$ independent of $\varepsilon$. I tried to bound $|\ln(\varepsilon)|g_\varepsilon(x)$ by such a constant but I couldn't prove rigourously that it didn't depend on $\varepsilon$. Any help would be greatly appreciated.

Answer 1

The existence of such a constant seems ok for me as both $1/f_0$ and $1/f_{\varepsilon}$ are continuous functions defined on a bounded interval $[0,x_0]$.

I think that you need to consider the case where there is $x_1$ satisfying both $f_0(x_1)=0$ and $0\lt x_1\lt x_0$.

If there is $x_1$ satisfying both $f_0(x_1)=0$ and $0\lt x_1\lt x_0$, then since $f_{\varepsilon}(x_1)$ is a non-zero constant, we get $\displaystyle\lim_{x\to x_1}g_{\varepsilon}(x)=\infty$.

---

Example :

If $a,x_1,x_0$ satisfy $$a=1+\frac{200}{41}\ln\bigg(\frac{41}{50}\bigg)\approx 0.032,\qquad x_1=e^{-41/50}\approx 0.440$$ and $x_1\lt x_0$, then we get $f_0(x_1)=0$. Since $f_{\varepsilon}(x_1)$ is a non-zero constant, we get $\displaystyle\lim_{x\to x_1}g_{\varepsilon}(x)=\infty$.

---

Added :

We have

$$g_\varepsilon(x) = \bigg|\frac{1}{f_{\varepsilon}(x)} - \frac{1}{f_{0}(x)}\bigg|\le \bigg|\frac{1}{f_{\varepsilon}(x)} \bigg|+\bigg|\frac{1}{f_{0}(x)}\bigg|\tag1$$

Since $\bigg(\dfrac{x^{1+a}}{\lambda^2}\bigg)'=\dfrac{4-(1-a)\ln x}{x^{2-a}}(\ln x)^3\lt 0$ with $\displaystyle\lim_{x\to 0^+}\dfrac{x^{1+a}}{\lambda^2}=\infty$, we see that $\dfrac{1}{f_0(x)}$ is positive and is increasing, so $$\bigg|\frac{1}{f_0(x)}\bigg|\le \frac{1}{f_0(x_0)}\tag2$$

Next, we may suppose that $\dfrac{x^{1+a}}{\lambda^2}\ge \varepsilon^{\beta-2}$, so $\dfrac{x^{1+a}+\varepsilon^{\beta}}{\lambda^2+\varepsilon^2}\ge \varepsilon^{\beta-2}$ implies $f_{\varepsilon}(x)\ge \ln(\varepsilon^{\beta-2})\gt 0$ (assuming that $0\lt \varepsilon\lt 1$), so $0\lt\dfrac{1}{f_{\varepsilon}(x)}\le\dfrac{1}{\ln(\varepsilon^{\beta-2})}$, therefore, we get $$\bigg|\frac{1}{f_{\varepsilon}(x)} \bigg|\le\frac{1}{\ln(\varepsilon^{\beta-2})}\tag3$$

It follows from $(1)(2)(3)$ that $$g_\varepsilon(x) \le \frac{1}{\ln(\varepsilon^{\beta-2})}+ \frac{1}{f_0(x_0)}$$