Let the $g: [-2, 2] \rightarrow \mathbb{R}$ be defined by $g(x) = \begin{cases} 2x^2 \quad \text{ if } -2 \leq x \leq 0 \\ x+2 \quad \text{if } 0 < x \leq 2 \end{cases}$
Prove by definition: "Epi $g$ is not convex"
I'm not sure how you prove the above. I tried by stating "Epi $g$ is not convex $\iff g$ is not convex''. I defined $g_1 \in g: [-2,0] \rightarrow \mathbb{R}$ by $g_1(x) = 2x^2$ and $g_2 \in g : (0,2] \rightarrow \mathbb{R}$ by $g_2(x) = x + 2$.
I know that $\lim_{x \uparrow 0} g_1(x) = 0$ and $\lim_{x \downarrow 0} g_2(x)= 2$. At point $x = 0$ we see that the $g(x)$ is not continuous. Applying the definition ''$g$ is convex $\implies g$ is continuous" yields that $g(x)$ is not convex, hence Epi $g$ is not convex.
What do you think? Is this an appropriate proof? I think it isn't. I think you have to prove it by contradiction. That you should prove the convexity of $g$ with the definition "for all $x, y \in \mathbb{R}$ and every $\lambda \in [0,1]$, $g(\lambda x + (1 - \lambda) y) \leq \lambda g(x) + (1 - \lambda) g(y)$" and that you will end with something like $2 \leq 0$ which isn't true and therefore not convex etc. But how do you construct such a proof with a splitted function like $g(x) = \begin{cases} 2x^2 \quad \text{ if } -2 \leq x \leq 0 \\ x+2 \quad \text{if } 0 < x \leq 2 \end{cases}$ ?
$$ {\rm epi}\ g=\{ (x,y)|g(x)\leq y \} $$
Note that $(0,0),\ (2,4)\in {\rm epi}\ g$ Hence if ${\rm epi}\ g$ is convex then $$ \frac{1}{2} \{ (0,0)+(2,4)\}=(1,2)\in {\rm epi}\ g$$
But $$ g(1)=3\Rightarrow (1,t)\in {\rm epi}\ g,\ t\geq 3$$
It is a contradiction.