What is the mle of $\Theta$ given a sample $X_{1},...,X_{n}$ (iid) with pdf $f(x;\Theta)=e^{-(x-\Theta)},\Theta\leq x<\infty,-\infty<\Theta<\infty$? So this is what I have done:
$$L = \prod_{i=1}^{n} e^{-(x_{i}-\Theta)} H(x_{i}-\Theta)$$
$$L = e^{- \sum_{i=1}^{n} x_{i} + n\Theta} H(x_{(1)}-\Theta)$$
$$l = \ln(L) =  -\sum_{i=1}^{n} x_{i} + n\Theta +\ln( H(x_{(1)}-\Theta) )$$
$$\frac{dl}{d\Theta} =n + \frac{ d (\ln( H(x_{(1)}-\Theta) )) }{d\Theta}$$
Where $H(n)$ is the unit step function and and $x_{(1)}$ is the 1st order statistic or the minimum of the sample. Continuing:
$$\frac{ d (\ln( H(x_{(1)}-\Theta) )) }{d\Theta} = \frac{1}{H(x_{(1)}-\Theta)}  \delta (x_{(1)}-\Theta)(-1)$$
Where $\delta(n)$ is the delta function. Setting $\frac{dl}{d\Theta}=0$, we get:
$$n = \frac{1}{H(x_{(1)}-\Theta)}  \delta (x_{(1)}-\Theta)$$
Expanding the delta function and then the step function, we get:
$$n = \frac{1}{H(x_{(1)}-\Theta)}  \begin{pmatrix}
0 & \text{ if } x_{(1)}<\Theta \\ 
\infty & \text{ if } x_{(1)}=\Theta \\ 
0 & \text{ if } x_{(1)}>\Theta 
\end{pmatrix} = \begin{pmatrix}
\frac{0}{0} & \text{ if } x_{(1)}<\Theta \\ 
\infty & \text{ if } x_{(1)}=\Theta \\ 
0 & \text{ if } x_{(1)}>\Theta 
\end{pmatrix}$$
We know that $n\geq 1$. So the only way this can be true is if $x_{(1)}=\Theta$. Thus, the mle of $\Theta$ must be $x_{(1)}$. I believe this answer is correct, and it makes intuitive sense. However, I just want to make sure that my reasoning and the way I did it is valid.
 A: You can not take the log likelihood since the likelihood is null for $\Theta \geq x_{(1)}$. Simply observe that $L$ is increasing for $\Theta \in (- \infty,x_{(1)})$ and is null for $\Theta \geq x_{(1)}$. So the maximum is at $x_{(1)}$.
A: Your answer is correct, but several of your steps were technically not correct.
Firstly, when taking the logarithm $l = \log(L)$ you need to know that $L>0$. That is, you must assume $\Theta \leq x_{(1)}$. In order to make this step rigorous, you should remark that you are excluding the case $\Theta >x_{(1)}$ it is obvious that $L=0$ in this case, and so the maximum will not occur in this case.
From this point on, you may assume $H(x_{(1)}-\Theta)=1$.
As a side note, $H' = \delta$ is incorrect. This is a convenient abuse of notation to avoid introducing the (unnecessarily complex topic for this type of problem) of integration with respect to a discontinuous measure.  The derivative of $H$ does not exist at $0$, so setting the derivative to 0 and solving this way makes no sense since $H'$ makes no sense at 0.
However, as I remarked earlier, we have that $H=1$ in the nontrivial case.
Thus we find that $l=-\sum x_i + n\Theta$ and we are trying to maximize on the interval $(-\infty,x_{(1)}]$. And we proceed to find the maximum occurs when $\Theta = x_{(1)}$ using standard mathods.
