This is sort of related to (different hypothesis though): GLRT statistic for composite normal hypothesis, two unknowns GLRT statistic for composite normal hypothesis, two unknowns
Problem
I am struggling with defining a generalized likelihood ratio test for $X_1, \dots, X_n \sim N(\mu, \sigma^2)$ where $\mu$ is unknown and our hypothesis test is $H_0 : \sigma \leq \sigma_0$ vs $H_1 : \sigma > \sigma_0$.
Attempt
I know that I need to find a test statistic in this form:
$$ \lambda(x) = \dfrac{\sup_{H_0} L(\mu,\sigma | x)}{\sup_{H_0 \cup H_1} L(\mu,\sigma | x)} $$
I know that I can do this by way of MLEs. As shown in the link above, the 2 MLEs are $(\bar{X}, \frac{1}{n}\sum_{i=1}^n (X_i - \bar{X})^2)$. Specifically, for $\sigma$, the MLE would be $\hat{\sigma}=\sqrt{\frac{1}{n}\sum_{i=1}^n (X_i - \bar{X})^2)}$.
My thinking is that the MLE $\hat{\sigma}$ would be the denominator for $\lambda(x)$.
Issue
I am lost with what to do with the numerator. My suspicion is that it follows similarly to the link I posted where we have $\lambda(x) = 1$ when $\hat{\sigma} \leq \sigma_0$.
I am not sure what happens when $\hat{\sigma} > \sigma_0$.
My shot in the dark is that we have a test statistic defined as follows:
$$ \lambda(x) = \begin{cases} 1 & \hat{\sigma} \leq \sigma_0 \\ \dfrac{L(\mu, \sigma_0 | x)}{L(\mu, \hat{\sigma} | x)} & \hat{\sigma} > \sigma_0 \end{cases} $$
Or something like that.