I am looking to compute maximum likelihood estimators for $\mu$ and $\sigma^2$, given n i.i.d random variables drawn from a Gaussian distribution. I believe I know how to write the expressions for negative log likelihood (kindly see below), however before I take derivatives with respect to $\mu$ and $\sigma^2,$ I want to prove that the neg. log likelihood is a convex function in $\mu$ and $\sigma^2$. 

This is where I'm stuck - I'm unable to prove that the Hessian is Positive Semidefinite.

The negative log-likelihood function,
$$ l(\mu, \sigma^2) = \frac{n}{2}ln(2\pi) + \frac{n}{2}ln(\sigma^2) + \sum\_{i=1}^n \frac{(xi - \mu)^2}{2\sigma^2}$$
Let $\alpha = \frac{1}{\sigma^2}$ (The book Convex Optimization by Boyd & Vandenberghe notes in Section 7.1 that this transformation should make the neg. log-likelihood convex in $\alpha$). We now get,
$$ l(\mu, \alpha) = \frac{n}{2}ln(2\pi) - \frac{n}{2}ln(\alpha) + \sum\_{i=1}^n \frac{(x\_i - \mu)^2\alpha}{2}$$
$$ = \frac{n}{2}ln(2\pi) + \frac{1}{2}\sum\_{i=1}^n\left(-ln(\alpha) + \frac{(x\_i - \mu)^2\alpha}{2}\right)$$

Define,
$$g\_i(\mu, \alpha) = -ln(\alpha) + \frac{(x\_i - \mu)^2\alpha}{2} $$

Now my approach is to show that $g\_i(\mu, \alpha)$ is convex in $\mu$, $\alpha$ and use that to say that $l(\mu, \alpha)$ being a sum of convex $g\_i$'s is also convex in $\mu$, $\alpha$. The Hessian for $g\_i$ is:

$$ \nabla^2g\_i = 
\begin{pmatrix}
2\alpha & -2(x\_i - \mu)\\
-2(x\_i - \mu) & \frac{1}{\alpha^2} \\ 
\end{pmatrix}
$$

And the determinant of the Hessian is,
$$ \lvert \nabla^2g\_i \rvert = \frac{2}{\alpha} - 4(x\_i - \mu)^2$$
**This is where I'm stuck - I cannot show that this determinant is non-negative for all values of $\mu$ and $\alpha (>0)$. Kindly help figure out my conceptual or other errors.**

Kindly note I've consulted the following similar queries:
[How to prove the global maximum log likelihood function of a normal distribution is concave](https://math.stackexchange.com/questions/2656541/how-to-prove-the-global-maximum-log-likelihood-function-of-a-normal-distribution)

and [Proving MLE for normal distribution](https://math.stackexchange.com/questions/484806/proving-mle-for-normal-distribution)

**However both of them only show that the Hessian is non-negative at a point where $\mu$ and $\alpha$ equal their estimated values. The mistake I see is that the estimates were arrived in the first place by assuming the neg. log-likelihood is convex (i.e. by equating gradient to 0, which is the optimality criterion for a convex function).**

Thanks