I am trying to find the likelihood ratio test for $H_0:\alpha\leq1$ vs $H_1:\alpha>1$ from $X \sim Beta(\alpha,1)$.
As a property of the Beta function, I know that the PDF $f(x|\alpha,1)=\alpha x^{\alpha-1}, 0\leq x\leq 1,\alpha>0$.
So I want to find: \begin{align*} \lambda(x)&=\frac{\text{sup}_{\alpha_1}L(\alpha|X)}{\text{sup}_{\alpha_0}L(\alpha|X)}=\frac{\text{sup}_{\alpha_1}\prod_{i=1}^n{\alpha_1 x_i^{\alpha_1-1}}}{\text{sup}_{\alpha_0}\prod_{i=1}^n{\alpha_0 x_i^{\alpha_0-1}}} \end{align*}
For an unrestricted function of the given PDF, I can maximize the log likelihood as follows to derive an estimator for $\alpha$:
\begin{align*} \ln{(L({\alpha}|X))}&=\ln{\left(\prod_{i=1}^n\alpha x_i^{\alpha-1}\right)}\\ &=n\ln{(\alpha)}+\sum_{i=1}^n(\alpha-1)\ln{(x_i)}\\ \frac{d}{d\alpha}\ln{(L(\alpha|X))}&=\frac{n}{\alpha}+\sum_{i=1}^n\ln{(x_i)}\tag{set derivative to zero}\\ \therefore\hat\alpha&=-\frac{n}{\sum_{i=1}^n\ln{(x_i)}}\tag{verified that second derivative was negative}\\ &=-\frac{n}{\prod_{i=1}^n x_i} \end{align*}
This is where I am confused. I do not understand how to incorporate the hypotheses ($H_0:\alpha\leq1$ vs $H_1:\alpha>1$) with my result from the unrestricted MLE. Essentially, I can find $\text{sup}_\alpha L(\alpha|X)$, but not $\text{sup}_{\alpha_{1,2}} L(\alpha|X)$ in order to construct the LRT.
The likelihood ratio test statistic should be
$$\lambda(\boldsymbol x) = \frac{\sup_{\alpha \le 1} \mathcal L(\alpha \mid \boldsymbol x)}{\sup_{\alpha > 0} \mathcal L(\alpha \mid \boldsymbol x)}, \tag{1}$$ that is to say, the numerator is the maximum likelihood under the restricted parameter space of the null hypothesis, and the denominator is the unrestricted maximum likelihood. You already know that $$\mathcal L(\alpha \mid \boldsymbol x) = \prod_{i=1}^n \alpha x_i^{\alpha - 1} = \alpha^n P^{\alpha - 1},$$ where $P = \prod_{i=1}^n x_i$ is the sample product. Then the log-likelihood is $$\ell(\alpha \mid \boldsymbol x) = n \log \alpha + (\alpha - 1) \log P,$$ and its critical value occurs when $$0 = \frac{\partial \ell}{\partial \alpha} = \frac{n}{\alpha} + \log P, \tag{2}$$ or $$\hat \alpha = -\frac{n}{\log P}. \tag{3}$$ This is the denominator for $\lambda$. The numerator, however, requires you to find the maximum of $\ell$ subject to the constraint $\alpha \le 1$. In this case, how do we proceed? Well, if $\hat \alpha = -\frac{n}{\log P} > 1$, that is to say, if $$\prod_{i=1}^n x_i = P > e^{-n}, \tag{4}$$ then we cannot choose this value. Our intuition suggests that we should choose $\hat \alpha = 1$ whenever the sample product is too large, where by "too large" we mean Equation $(4)$. But under this condition, from Equation $(2)$, we have $$\frac{n}{\alpha} + \log P > \frac{n}{\alpha} + \log e^{-n} = \frac{n}{\alpha} - n = n \left(\frac{1}{\alpha} - 1\right) > 0$$ for all $\alpha \in (0,1]$, thus $\ell$ is strictly increasing; hence it is maximized when we can pick $\alpha$ as large as possible on this interval; i.e., $\hat \alpha = 1$.
All that is left is to evaluate $\mathcal L(-n / \log P \mid \boldsymbol x)$ for the denominator of $\lambda$, from which we then find the LRT statistic is $$\lambda(\boldsymbol x) = \begin{cases} 1, & P \le e^{-n}, \\ P \left(- \frac{e}{n} \log P \right)^n, & P > e^{-n}. \end{cases}$$
Now let's apply this test in practice. Suppose I have the data $$\boldsymbol x = (0.622053, 0.999547, 0.54362, 0.640624, 0.812553, 0.702147, 0.507879),$$ with $n = 7$. I first calculate $P = 0.0627435$. Then I calculate whether $P > e^{-n} = 0.000911882$, which is true. Then $$\lambda(\boldsymbol x) = P\left(-\frac{e}{n} \log P\right)^n = 0.104202.$$ The smaller this value, the more evidence the data is realized from a beta distribution with $\alpha > 1$. In fact, I generated this sample from a beta distribution with $\alpha = 3$.
Conversely, suppose we have this data: $$\begin{align} \boldsymbol x &= (0.680164, 0.272671, 0.0401028, 0.51946, 0.380103, \\ &\phantom{{} = (} 0.945771, 0.469441, 0.505287, 0.354717, 0.0788713).\end{align}$$ Then we calculate $P = 9.21698 \times 10^{-6}$ and this is smaller than $e^{-10} \approx 0.0000453999$, thus $\lambda(\boldsymbol x) = 1$. This data was generated from a beta distribution with $\alpha = 0.75$. However, what we still need to do is, for a given size of the test (i.e. Type I error control), we would need to compute a rejection criterion in the form of a critical value $c$, such that if $\lambda(\boldsymbol x) < c$, the null is rejected.