Assume $x\ge 0$ so that

$f(x; \theta) = \frac{1}{\theta}I(x \le \theta)$ and

$L(x; \theta) = \prod_{j=1}^J \theta^{-1}I(x_j\le \theta) = \theta^{-J}I(\max_j x_j \le \theta)$

Note that the LL is

1. Zero if $\theta$ is smaller than the largest observation. This is clearly not the maximum.
2. Decreasing in $\theta$.

So, the smallest allowed value for $\theta$ maximizes the likelihood and is given by: $\hat{\theta} = \max_j x_j$.

Assume $x\ge 0$ so that

$f(x; \theta) = \frac{1}{\theta}I(x \le \theta)$ and

$L(x; \theta) = \prod_{j=1}^J \theta^{-1}I(x_j\le \theta) = \theta^{-J}I(\max_j x_j \le \theta)$

Note that the LL is

1. Zero if $\theta$ is smaller than the largest observation. This is clearly not the maximum.
2. Decreasing in $\theta$.

So, the smallest allowed value for $\theta$ maximizes the likelihood and is given by: $\hat{\theta} = \max_j x_j$.

This makes sense: Given a uniform sample, it must be possible to generate the largest number and the most conservative estimate is that largest number. But, this underestimates the interval. Since $E[\hat{\theta}] = \frac{J}{\theta^J}\int_0^\theta y\cdot y^{J-1}\,dy=\theta\frac{J}{J+1}$ an unbiased estimate is $\hat{\theta}\frac{J+1}{J}$. This approaches the LL-estimate for large $J$.

Assume $x\ge 0$ so that

$f(x; \theta) = \frac{1}{\theta}I(x \le \theta)$ and

$L(x; \theta) = \prod_{j=1}^J \theta^{-1}I(x_j\le \theta) = \theta^{-J}I(\max_j x_j \le \theta)$

This function is clearly decreasing in $\theta$ so the smallest possible value for $\theta$ maximizes the likelihood. The largest observation is not allowed to exceed $\theta$ because then the LL is zero; this gives $\hat{\theta} = \max_j x_j$.

Assume $x\ge 0$ so that

$f(x; \theta) = \frac{1}{\theta}I(x \le \theta)$ and

$L(x; \theta) = \prod_{j=1}^J \theta^{-1}I(x_j\le \theta) = \theta^{-J}I(\max_j x_j \le \theta)$

Note that the LL is

1. Zero if $\theta$ is smaller than the largest observation. This is clearly not the maximum.
2. Decreasing in $\theta$.

So, the smallest allowed value for $\theta$ maximizes the likelihood and is given by: $\hat{\theta} = \max_j x_j$.