Let $X_1...X_n$ be from uniform ($\theta,\theta^2)$ where $\theta>1$ I seek the MLE. I've seen similar problems, but I think this one might be more interesting. Intuitively, we think $X_{(1)}$ should estimate theta. However I think since $\theta >1$ there is no MLE as this function blows up close to one. Is this correct?
Edit: bad picture explaining my reasoning:
No. Write likelihood function: $$ L(\theta, X_1,\ldots,X_n) = \frac{1}{(\theta^2-\theta)^n}\cdot \mathbb 1_{\{X_{(1)}\geq \theta, X_{(n)}\leq \theta^2\}} = \frac{1}{(\theta^2-\theta)^n}\cdot \mathbb 1_{\left\{\sqrt{X_{(n)}}\leq\theta\leq X_{(1)}\right\}} $$ So in your picture you should have a function that decreases on the interval $[\sqrt{X_{(n)}}, X_{(1)}]$ and is zero on the other parts of $[1,\infty)$. And the MLE is the left point : $\hat\theta=\sqrt{X_{(n)}}$ not the right since the function decreases inside interval.
Note also that $\sqrt{X_{(n)}}\leq X_{(1)}$ since $X_{(n)}\leq \theta^2$ and then $\sqrt{X_{(n)}}\leq \theta$, and $X_{(1)}\geq \theta$. So at your picture points on the axis should be swapped.