= 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> are
$$
L(\mu,v)=(2\pi v)^{-n/2}\exp\left[-\frac{S+n(\mu-\bar x)^2}{2v}\right],\qquad
\ell(\mu,v)=-\frac n2\log(2\pi v)-\frac{S+n(\mu-\bar x)^2}{2v}.
$$
For 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 $S>0$,
$$
\boxed{\widehat\mu=\bar x,\qquad\widehat{\sigma^2}=S/n.}
$$
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 $\mu=\bar x$, $v\downarrow0$; there is no maximizer in the parameter space $v>0$. The freezing argument below uses the nondegenerate case $S>0$.
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