Consider the function $\mathbb R^n \to \mathbb R, x \mapsto\|x\|_p ^\beta $, where $\|\cdot\|_p$ is the $p$-norm.
It can be easily verified that $\|\cdot\|_p^\beta$ is convex if $\beta = p \geqslant 1$ or $\beta = 1$. Moreover,
For every $\beta \geqslant 1, p \geqslant 1$:
$$ (\| \psi x + (1-\psi)y \|_p) ^\beta \leqslant ( \psi \|x\|_p + (1-\psi)\| y \|_p )^\beta \leqslant \psi\|x\|_p^\beta + (1-\psi)\|y\|_p^\beta.$$
Because $t \mapsto |t|^\beta$ is convex.
What about the other cases?
If $n=1$ then the convexity is iff $\beta\ge 1$. If $n\ge 2$ then the convexity is iff $p,\beta\ge 1$.
We assume that $p>0$, otherwise $\|\cdot \|_p$ is undefined for $n\ge 2$. Also we assume that $\beta>0$, otherwise $\|0\|_p^\beta$ is undefined.
Assume a function $f:\mathbb R^n \to \mathbb R$, $x \mapsto\|x\|_p ^\beta$ is convex. Let $e\in \mathbb R^n$ be any non-zero vector. Then
$$\|e\|_p^\beta=f(e)\le\frac 12\left(f(0)+f(2e)\right)= 2^{\beta-1}\|e\|_p^\beta,$$
so $\beta\ge 1$.
If $n=1$ then $f(x)=|x|^\beta$ is convex. Assume that $n\ge 2$. Let $e_1=(1,0,0,\dots,0)$ and $e_2=(0,1,0,\dots,0)$ be vectors of $\Bbb R^n$. Then
$$2^\frac{(1-p)\beta}{p}=f\left(\frac{e_1+e_2}2\right)\le \frac 12\left(f(e_1)+f(e_2)\right)=1=2^0,$$
so $(1-p)\beta\le 0$ that is $p\ge 1$.