The question arose what to do if the density of a random variable depends on $(\alpha-x)$ (maybe in some degree) on the interval from 0 to $\alpha$, for example. The likelihood function then turns out to depend on the product of $(\alpha-x)^n$. And when the MLE estimator is set to $\hat\alpha=max(X_i)$, it turns out that the likelihood function is reset to zero. What about the MLE estimator in such cases? For example, if r.v. with density $f(x)=\left ( \frac{2\left ( \alpha -x \right )}{\alpha^2}\right ) if x\in \left [ 0;\alpha \right ]$
or the second example in this message The Doctor (https://math.stackexchange.com/users/165662/the-doctor), Existence of Maximum Likelihood Estimator, URL (version: 2017-10-08): Existence of Maximum Likelihood Estimator
When substituting such an estimator $\hat{\mu}=X_{(1)}$, the likelihood function is equal to 0 too - one of the multipliers, where the minimum member of the sample comes across, resets it. What about this estimation then?
In your example, the lileihood is
$$L(X;a)=\frac{2^n}{a^{2n}} \prod_{i=1}^n (a-x_i) [0\le x_i \le a]$$
In terms of the variable $a$, this gives the domain $a \in [\max(x_i), +\infty)$. Inside this domain, $L(X;a)$ is differentiable. Hence, its global maximum is either a critical point inside the domain, or either lies in the extremes. But the latter is not possible, because $L(X; \max(x_i))=0$. Hence we can seek for critical points.
Using the log likelihood, deriving and equating to zero we get the following equation for the critical point(s):
$$ a =\frac{2n}{\sum \frac{1}{a-x_i}} =2\,{\rm Hm}(a-x_i) $$
where $Hm$ denotes the harmonic mean.
This should be solved numerically (inside the domain).