Epigraphs and convexity I have a problem which is as follows:

Let the set $S$:
$$S = \left\{(x, y) \in \mathbb{R}^2 : |x|^{\frac 13} + |y|^{\frac 13} \leq 1\right\}$$
Find out if the set is convex or not, and sketch the set.

And then I found out that $S$ is not convex, and was able to sketch the set (by thinking that $S$ would have to be on or below the function).
But what I don't fully understand is the relation between the epigraph of the function and the set $S$. Is $S$ the epigraph of the function? If so, then $S$ is per definition the points that lie on or above the function but below $f(x,y)=1$? How is it then that $S$ is shaped like a four pointed star, if the epigraph is the points that lie above the function?
My second question is: if $S$ is the epigraph of the function, is it sufficient to show that since the function is not convex, that $S$ is not convex? Or can an epigraph be convex if its function is not convex?
I'm finding this somewhat confusing, so hope someone can clarify this for me.
 A: It is good to first get an empirical feeling of the problem. In Python, you'd do
x, y  = xy = randn(2, 100000)
data = xy[:, abs(x) ** (1./3) + abs(y) ** (1./3) <= 1]
subplots(figsize=(6, 6))
scatter(*data, s=1)
xlabel("$x$")
ylabel("$y$")
title("$|x|^{1/3} + |y|^{1/3} \\le 1$")
tight_layout()
grid("on")


Now it should be pretty clear why this function can't be convex. Indeed, the point $a = (0, 1)$ and $a = (1, 2)$ are both in $S$ but the coord $[a, b]$ is not a subset of $S$. For example $\frac{a+b}{2} \not \in S$ since $2(1/2)^{1/3} > 1$.
General case
In general $2(1/2)^p > 1$ for all $p \in (0, 1)$, from which you see that If $p \in (1, \infty]$,

then the $\ell_p$ unit ball is convex iff $p \ge 1$ (the if part is provided by Holder's inequality). 

A: A simple answer to your question is that:
$$ S = \{\textbf{x} \in \mathbb{R}^2: \lvert x_1 \rvert^{1/3} + \lvert x_2 \rvert^{1/3}  \leq 1\} = \{\textbf{x} \in \mathbb{R}^2: ( \lvert x_1 \rvert^{1/3} + \lvert x_2 \rvert^{1/3} )^3  \leq 1\} = \{\textbf{x} \in \mathbb{R}^2: \lVert \textbf{x}\rVert_{1/3} \leq 1\}.$$
This is basically the unit $q$-norm ball where $q=1/3$, so this is not a convex set.
[Note: A function is convex, if and only if, the epigraph of the function is convex.]
