 Solution

Put $v={\sigma }^2>0$, $\bar{x}=n^{-1}{\sum }_i{x}_i$ and $S={\sum }_i(x_i-\bar{x}{)}^2$. The likelihood function and log-likelihood areFor every fixed $v$, the unique maximum in $\mu$ is $\bar{x}$. At that value, $\partial \ell /\partial v=-n/(2v)+S/(2v^2)$ changes sign from positive to negative at $v=S/n$. Thus, for $S>0$,This is the global maximum-likelihood estimate, since the first maximization was global for each $v$ and the second has a unique peak. If $S=0$, the likelihood function is unbounded at $\mu =\bar{x}$, $v\downarrow 0$; there is no maximizer in the parameter space $v>0$. The freezing argument below uses the nondegenerate case $S>0$.