TITLE: Show that the function $|t|^p$ is convex on $\Bbb{R}$
QUESTION [1 upvotes]: Let $f:\Bbb{R}\to\Bbb R$ be a function defined as $f(t) = |t|^p$ with $p\geq1$. I want to show that $f$ is convex by showing its second derivative is non-negative. I have tried calculating the second derivative but this led to confusion, is there a simpler way?

REPLY [2 votes]: The function $f:\Bbb R\to[0,\infty)$ defined by $$f(x)=|x|$$ is convex. Also, the function $g:[0,\infty)\to [0,\infty)$ defined by
$$
g(t)=t^p
$$ is convex and increasing for $p\ge 1$. The composition of two such functions is convex, hence
$$
|t|^p = g\circ f(t)
$$
is a convex function.
You can prove that $g$ is convex using derivative test as you want.

Showing that $f$ is convex: Take any $x,y\in\Bbb R$, we have
$$
|(1-\lambda)x + \lambda y| \le |1-\lambda||x| + |\lambda| |y| = (1-\lambda)|x| + \lambda |y|
$$
for any $\lambda\in[0,1]$, which show that $f(x)=|x|$ is convex.