Is the answer true? $\sup \frac{(\theta-\theta')^2}{\exp(n(\theta'-\theta))-1}=\frac{1}{n^2}$

Question

I am trying to calculate the supremum ($sup$) for some equation, Is the answer true?

$$\sup \frac{(\theta-\theta')^2}{\exp(n(\theta'-\theta))-1}=\frac{1}{n^2}$$

Answer 1

Using the change of variables $u=n(\theta'-\theta)$, this is $$ \sup_u\frac{u^2/n^2}{\exp(u)-1}=\frac1{n^2}\cdot\sup_u\frac{u^2}{\exp(u)-1}. $$ The last ratio is negative when $u\lt0$ hence one should study it on $(0,+\infty)$. Upon inspection, it is maximum around $u=1.5936$ where its value is about $0.6476$.

Thus, the answer to the question in the title is: No.

Answer 2

I assume your taking the supremum over $\theta, \theta^{\prime}$. I would use techniques from calculus to find a maximum value. One approach is to define sequence of multivariable functions $$F_{n}(\theta, \theta^{\prime})=\frac{(\theta-\theta^{\prime})^{2}}{e^{n(\theta^{\prime}-\theta)}-1}$$ Now for each $n$ calculate the gradient, find the critical points, text for local maximum. You should also check the behavior as the magnitude of $\vert(\theta, \theta^{\prime})\vert$ gets large. For this, I would switch to polar coordinates and take a limit as $r\to \infty$. If you get something unbounded, the supremum must be $\pm \infty$. If bounded, compare with the values of $F_{n}$ at critical points and see which is the largest.