Show that the function $|t|^p$ is convex on $\Bbb{R}$

Question

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?

Hi Sam. Please look at how I edited your post, to see how to type math. It will be helpful if you need to make changes. — Chickenmancer, Jan 16 '19 at 20:14
I don't really understand how to actually compute this derivative, mainly due to the |t| part. But also what happens at t=0? since I am to show this property for all of t on R. — kam, Jan 16 '19 at 20:23
@Sam.S If $t > 0$ then $|t| = t$ and if $t < 0$ then $|t| = -t.$ At $t = 0$ there is no derivative. On a second thought, it may not be possible to follow your approach to prove it because $|t|$ has no derivative at zero. — William M., Jan 16 '19 at 20:28
@WillM. in that case, have you any idea how to show this result? — kam, Jan 16 '19 at 20:31
@Sam.S I've linked to the proof that the composition of an increasing convex function and a convex function is convex. More specifically it is this post https://math.stackexchange.com/a/473922/231327 I hope you might find it helpful. — BigbearZzz, Jan 16 '19 at 21:06

score 2 · Accepted Answer · edited Jun 12 '20 at 10:38

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.

Show that the function $|t|^p$ is convex on $\Bbb{R}$

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