I wanted to prove that if $0<k<\theta_o$, we have $$\sup_{k<\theta\leq \theta_o}\left(1-\frac{k^{\theta+1}}{\theta^{\theta+1}}\right)=1-\frac{k^{\theta_o+1}}{\theta_o^{\theta_o+1}}$$
To prove this, it suffices to prove $f(\theta)=\frac{k^{\theta+1}}{\theta^{\theta+1}}$ is a decreasing function. Using that the logarithm is monotone, this is equivalent to proving $l(\theta)=\ln(f(\theta))=(\theta+1)\ln\left( k\right)-(\theta+1)\ln(\theta)$ is decreasing. Let us compute this derivative:
$$\frac{\partial}{\partial \theta}l=\ln(k)-\ln(\theta)-\frac{\theta+1}{\theta}$$
Let $0<k<\theta$. We need to prove that that $$\ln\left(\frac{k}{\theta}\right)-1-\frac{1}{\theta}\leq 0$$
Because of convexity, we neeed $\ln(x)\leq x-1$, which implies:
$$\ln\left(\frac{k}{\theta}\right)-1-\frac{1}{\theta}\leq \frac{k}{\theta}-2-\frac{1}{\theta}\leq0$$
Whenever $\theta\geq \frac{k-1}{2}$. When $\theta<\frac{k-1}{2}$ there are problems: $\ln(20/1)-1-1/1>0$, for instance. So this is really not the way to go. Any thoughts on another approach to compute the supremum?
You only need that your function is decreasing on $k<\theta\leq\theta_0$, since your supremum is over that interval.
However you will still have strict monotonicity if $k<e^2$. Because $\theta=1$ is a maximum of $\ln\left(\frac{k}{\theta}\right)-1-\frac{1}{\theta}$.