I know $$f_\theta(\vec{x})=\frac{1}{\theta^n}\prod_{i=1}^n\mathbb{I}_{(\theta,2\theta)}(x_i)\hspace{0.5cm}\theta>0$$
and this function is equals to
$$f_{\theta}(\vec{x})=\frac{1}{\theta^n}\mathbb{I}_{\left(\frac{x_{(n)}}{2}, x_{(1)}\right)}(\theta)$$
Then the rate $$\frac{f_\theta(\vec{x})}{f_\theta(\vec{y})}=\frac{\frac{1}{\theta^n}\mathbb{I}_{\left(\frac{x_{(n)}}{2}, x_{(1)}\right)}(\theta)}{\frac{1}{\theta^n}\mathbb{I}_{\left(\frac{y_{(n)}}{2}, y_{(1)}\right)}(\theta)}$$ is independent of $\theta$ if and only if $x_{(1)}=y_{(1)}$ and $x_{(n)}=y_{(n)}$, so $$T(X)=(X_{(1)},X_{(n)})$$ is a minimal sufficient statistic for $\theta$
Now for the completness first I should finf $E_T[g(t)]$ and then if $E_T[g(t)]=0$ just when $g\equiv 0$ if means $T$ is complete statistic.
How ever for the case $X$ random sample from $U(\theta, 4\theta)$, my teacher first found $E[X_{(1)}]=\theta(n+4)(n+1)^{-1}$ and then $E[X_{n}]=\theta(4n+1)(n+1)^{-1}$, so for example if $$g(t)=(4n+1)x_{(1)}-(n+4)x_{(n)}$$ then $E[g(t)]=0$ but $g\not\equiv 0$ so $T$ is not complete for $\theta$
I don't know why the second method is equivalent to the first and for $U(\theta, 2\theta)$ I am almost sure that $T$ is not complete too. Should I do the same method of my teacher?
Thanks for your help.
First observe that $f(x|\theta)$ is the density of a "scale family" and the statistic
$$g(T)=\frac{X_{(n)}}{X_{(1)}}$$
is "scale invariant". Thus for a known theorem, $g(T)$ is ancillary for $\theta$ (its distributions does not depend on $\theta$) and thus, $\forall \theta$
$$\mathbb{E}\left[ \frac{X_{(n)}}{X_{(1)}} \right]=c$$
and therefore
$$\mathbb{E}\left[ \frac{X_{(n)}}{X_{(1)}} -c\right]=0$$
but clearly
$$\mathbb{P}\left[ \frac{X_{(n)}}{X_{(1)}} -c=0\right]\ne 1$$
This shows that $T$ is not complete