1

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.

Rodrigo de Azevedo
  • 1
sequence
  • 9,638
  • 1
    If you know that every supremum of convex functions is convex and that affine functions are convex, then you might want to use $$g=\sup{\alpha_{u,v}\mid (u,v)\in{-1,1}^2}$$ where, for every $(u,v)$, $$\alpha_{u,v}(x)=u\cdot(x-a)+v\cdot(x-b)$$ and if you do not, this is an occasion to learn these. As a corollary, for every integer $n$, every nonnegative $(\lambda_k)$ and every real numbers $(a_k)$,$$g(x)=\sum_{k=1}^n\lambda_k\cdot|x-a_k|$$ defines a convex function. – Did Sep 25 '16 at 22:17

1 Answers1

3

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.

Sangchul Lee
  • 167,468