How does one show that $g(x):=\left|x-a\right|+\left|x-b\right|$ is convex, without employing the Dirac delta function? I've played with inequalities, but couldn't come up with something very meaningful.
It is also given that $a<b$, and $g(x)$ is real.
Here is a not-so-obvious shortcut: Consider $f_{\epsilon}(x) = \sqrt{x^2 + \epsilon}$. It is easy to check that $f''_{\epsilon}(x) > 0$ and hence $f_{\epsilon}$ is convex. Now
$$ g(x) = \lim_{\epsilon \downarrow 0} (f_{\epsilon}(x - a) + f_{\epsilon}(x - b)) $$
is the limit of convex functions. Since convexity is preserved under pointwise limit, $g$ is convex.
A more elementary way is to first show that $x \mapsto |x - x_0|$ is convex for any $x_0 \in \Bbb{R}$. This is easily done by considering a couple of cases. Then the convexity of $g$ is immediate.