Uniform bound of a function

Question

Let us consider the function $$\frac{\ln\left(1+ \frac{\varepsilon}{x^3}\right)}{\ln(x)^2}$$ for $x \in (0,T]$, $T \ll 1$ and $\varepsilon < T$. I would like to bound this function by a constant $C = C(\varepsilon)$ such that $C \to 0$ for $\varepsilon \to 0$. The graph of this function is given by

and for $\varepsilon$ getting smaller, the maximum of this function will become smaller as well. I tried to find an explicit form of this maximum in terms of $\varepsilon$ by taking the derivative and then see where it vanishes, but I haven't been able to solve the equation.

On the other hand, I tried to apply de l'Hospital rule, such that $$\frac{\ln ( 1 + \varepsilon/x^3 )}{\ln(x)^2}= \frac{f_\varepsilon(x)}{g(x)} \le \sup_{t \in (0,T)}\frac{f_\varepsilon'(t)}{g'(t)},$$ but $$\frac{f_\varepsilon'(t)}{g'(t)} = \frac{3\varepsilon}{\ln(x)(x^3 + \varepsilon)}$$ and I cannot bound this by something going to $0$ for $\varepsilon \to 0$.

Does one of you have an idea how to solve this ?

Answer 1

Define $f(x) = \frac{\log{\left(1+\frac{\epsilon}{x^3} \right)}}{\log{(x)}^2}$. Notice the following:

1. $\lim_{x\to 0^{+}} f(x) = 0$, and therefore $f$ is continuous in $[0,T]$ and a maximum $C(\epsilon)$ exists in $[0,T]$ ($T<1$).
2. After some computations, $f'(x) = -\frac{1}{x\log^2{(x)}}\left[\frac{3\epsilon}{x^3+\epsilon}+\frac{2}{\log{(x)}}\log{\left(1+\frac{\epsilon}{x^3} \right)}\right]$ for $x \in (0,T]$.
3. Since $\log{(1+\frac{1}{x})}-\log{(\frac{1}{x})} = \log{(1+x)} = x - \frac{x^2}{2}+O(x^3)$ for $x>0$, we obtain $$ \log{\left(1+\frac{1}{x}\right)} = \log{\left(\frac{1}{x}\right)}+x-\frac{x^2}{2}+O(x^3). $$
4. From 2 and 3, realise that $$f'(\epsilon^{\frac{2}{3}}) = -\frac{9}{4\epsilon^{\frac{2}{3}}\log^2{(\epsilon)}}\left[\frac{3}{1+\epsilon}+\frac{3}{\log{(\epsilon)}}\log{\left(1+\frac{1}{\epsilon} \right)}\right]\to 0$$ as $\epsilon \to 0^{+}$. Therefore $C(\epsilon) \approx f(\epsilon^{2/3}) = \frac{9}{4}\frac{\log{\left(1+\frac{1}{\epsilon} \right)}}{\log^2{(\epsilon)}}\approx -\frac{9}{4\log{\epsilon}} >0$ and $C\to 0$ as $\epsilon \to 0$.

Edit 1: Edited the computations, as function changed from $ \frac{\log{\left(1+\frac{\epsilon}{x^2} \right)}}{\log{(x)}^2}$ to $\frac{\log{\left(1+\frac{\epsilon}{x^3} \right)}}{\log{(x)}^2}$.

Edit 2: How did I know $f'(\epsilon^{2/3}) \to 0$ ? Plug in $\epsilon^{\alpha}$ for some $\alpha>1/3$. Then $$f'(\epsilon^{\alpha}) = -\frac{1}{2\alpha \epsilon^{\alpha}\log^2{(\epsilon)}}\left[\frac{3}{1+\epsilon^{3\alpha-1}}+\frac{2}{\alpha \log{(\epsilon)}}\log{\left(1+\frac{1}{\epsilon^{3\alpha-1}} \right)}\right].$$ Focusing on the numerator when $\epsilon$ is small, we see that it is approximately $3+\frac{2}{\alpha \log{(\epsilon)}}\log{\left(\frac{1}{\epsilon^{3\alpha-1}} \right)}$, which is $3+\frac{2}{\alpha}-6$. Thus $\alpha = \frac{2}{3}$ was the right choice. Now, if the numerator vanishes with an order greater than $\frac{2}{3}$, then the problem is solved.

Edit 3: From the expression of $f'(\epsilon^{2/3})$, we have \begin{align*} \lim_{\epsilon \to 0^{+}} f'(\epsilon^{2/3}) &= -\frac{27}{4}\lim_{\epsilon \to 0^{+}}\frac{1}{\epsilon^{\frac{2}{3}}\log^3{(\epsilon)}}\left[\log{(\epsilon)}+ (1+\epsilon)\log{\left(1+\frac{1}{\epsilon} \right)}\right]\\ &= -\frac{27}{4}\lim_{\epsilon \to 0^{+}}\frac{1}{\epsilon^{\frac{2}{3}}\log^3{(\epsilon)}}\left[\epsilon - \epsilon \log{(\epsilon)}+\frac{1}{2}\epsilon^2+O(\epsilon^3)\right] = 0 \end{align*}

Answer 2

Here is an attempt: let $$f(x) = \frac{\ln\left(1 + \frac{\epsilon}{x^3}\right)}{\ln^2x}.$$ Then $$f'(x) = \frac{\frac{-3\epsilon}{x^4 + \epsilon x}\ln x - \frac{2}{x}\ln\left(1 + \frac{\epsilon}{x^3}\right)}{\ln^3 x}.$$ Notice that $f'(x) = 0$ if and only if the numerator is zero, that is, $$0 = \frac{-3\epsilon}{x^4 + \epsilon x}\ln x - \frac{2}{x}\ln\left(1 + \frac{\epsilon}{x^3}\right).$$ We can then multiply through in the previous equality by $\frac{x}{\ln^2 x}$ to obtain $$0 = \frac{-3\epsilon}{x^4 + \epsilon x}\ln x \color{blue}{\left(\frac{x}{\ln^2 x}\right)} - \frac{2}{x}\ln\left(1 + \frac{\epsilon}{x^3}\right)\color{blue}{\left(\frac{x}{\ln^2 x}\right)} = \frac{-3 \epsilon}{x^3 + \epsilon}\frac{1}{\ln x} - 2 f(x).$$ This shows that at the point of maximum (namely $x_{\epsilon}$) we have that $$f(x_{\epsilon}) = -\frac{3}{2} \frac{\epsilon}{x_{\epsilon}^3 + \epsilon} \frac{1}{\ln x_{\epsilon}}.$$ Since $$f(x_{\epsilon}) \le - \frac{3}{2}\frac{1}{\ln x_{\epsilon}},$$ if $x_{\epsilon} \to 0$ as $\epsilon \to 0$ then we deduce that $f(x_{\epsilon}) \to 0$. If $x_{\epsilon}$ does not approach $0$ then there is a subsequence with the property that $x_{\epsilon} \to \bar{x} > 0$. But then for every such subsequence, we would have that $$\lim_{\epsilon \to 0}f(x_{\epsilon}) = \frac{\ln(1 + \frac{0}{\bar{x}})}{\ln^2 \bar{x}} = 0.$$

Answer 3

Some thoughts:

Notation: We replace $\varepsilon$ with $s$.

We deal with the case that $T < 1/2$.

Fact 1: For all $s, x\in (0, 1/2)$, $$\frac{\ln(1 + s/x^3)}{\ln^2 x} \le 2 \cdot \frac{\ln(1 + 1/s^3)}{\ln^2 s}. \tag{1}$$ (The proof is given at the end.)

Noting that $\lim_{s\to 0} 2 \cdot \frac{\ln(1 + 1/s^3)}{\ln^2 s} = 0$, by Fact 1, we are done.

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Proof of Fact 1:

It suffices to prove that $$f(x) := 2 \cdot \frac{\ln(1 + 1/s^3)}{\ln^2 s}\ln^2 x - \ln(1 + s/x^3) \ge 0.$$

We have $$f(1/2) = 2 \cdot \frac{\ln(1 + 1/s^3)}{\ln^2 s}\ln^2 2 - \ln(1 + 8s) \ge 0. \tag{2}$$ (The proof of (2) is omitted here.)

Also, we have \begin{align*} f'(x) &= 4 \cdot \frac{\ln(1 + 1/s^3)}{\ln^2 s}\frac{\ln x}{x} + \frac{3s}{x(x^3 + s)}\\[6pt] &= -\frac{s}{x} \left[4 \cdot \frac{\ln(1 + 1/s^3)}{s\ln^2 s}\ln \frac{1}{x} - \frac{3}{x^3 +s}\right]\\[6pt] &= -\frac{s}{x} \left[\frac43 \cdot \frac{\ln(1 + 1/s^3)}{s\ln^2 s}\ln\frac{1}{x^3} - \frac{3}{x^3 +s}\right]\\[6pt] &\le 0 \tag{3} \end{align*} where we use $$\frac43 \cdot \frac{\ln(1 + 1/s^3)}{s\ln^2 s}\ln\frac{1}{x^3} - \frac{3}{x^3 +s} \ge 0. \tag{4}$$ Using (2) and (3), we have $f(x) \ge 0$ on $(0, 1/2)$. The desired result follows. The proof of (4) is given below.

Letting $y = 1/x^3 \ge 8$, it suffices to prove that $$\frac43 \cdot \frac{\ln(1 + 1/s^3)}{s\ln^2 s}\ln y - \frac{3y}{1 +ys} \ge 0$$ or $$g(y) := \frac43 \cdot \frac{\ln(1 + 1/s^3)}{s\ln^2 s}(1 + ys)\ln y - 3y \ge 0.$$

We have $$g''(y) = \frac43\cdot \frac{\ln(1 + 1/s^3)}{\ln^2 s} \cdot \frac{y - 1/s}{y^2}.$$ Thus, $g(y)$ is concave on $(0, 1/s)$ and convex on $(1/s, \infty)$.

We split into two cases:

Case 1: $1/s \le 8$

We have $g''(y) \ge 0$ on $y \ge 8$. We have, for all $s\in (0, 1/2)$, $$g'(8) = \frac{(8s+1)\ln(1 + 1/s^3)}{6s\ln^2 s} + \frac{4\ln 2 \,\ln(1 + 1/s^3)}{\ln^2 s} - 3 > 0. \tag{5}$$ Thus, $g'(y) \ge 0$ on $y \ge 8$. Thus, we have $$g(y) \ge g(8) = \frac{4\ln 2\, (8s + 1)\ln(1 + 1/s^3)}{s\ln^2 s} - 24 > 0. \tag{6}$$ (The proof of (5) and (6) is omitted here.)

Case 2: $1/s > 8$

We have \begin{align*} g'(1/s) &= \frac{8\ln(1 + 1/s^3)}{3\ln^2 s} + \frac{4\ln(1 + 1/s^3)\ln(1/s)}{3\ln^2 s} - 3\\ &\ge \frac{4\ln(1 + 1/s^3)\ln(1/s)}{3\ln^2 s} - 3\\ &= \frac{4\ln(1 + 1/s^3)}{\ln(1/s^3)} - 3\\ &> 0. \end{align*} Thus, $g'(y) \ge 0$ on $y \ge 1/s$. Thus, for all $y \ge 1/s$, $$g(y) \ge g(1/s) = \frac{8\ln(1 + 1/s^3)\ln(1/s)}{3s\ln^2 s} - \frac{3}{s} > 0.$$

On the other hand, since $g(y)$ is concave on $(8, 1/s)$ and $g(1/s) \ge 0$ and $g(8) \ge 0$, we have $g(y) \ge 0$ on $(8, 1/s)$.

We are done.