Answer

Solution

1. 1

   $f$ is convex (the maximum of two convex functions $|x|$ and $x^2$). Hence its epigraph $D$ is convex. So Option C is false.

2. 2

   $\mathbb R^2\setminus D=\{(x,y):y<f(x)\}$. This region is the hypograph of $f$ (open). For a convex $f$, the hypograph is generally not convex (and indeed here it isn't: take $(\pm 2,0)$; their midpoint is $(0,0)$ which has $y=0$ but $f(0)=0$, so $(0,0)\notin\mathbb R^2\setminus D$). So Option A is false.

3. 3

   Similarly Option B: restrict to $\mathbb R^2_+$. Take $(2,0)$ (in $\mathbb R^2_+$? if $\mathbb R^2_+=\{y\ge 0\}$, then $y=0$; $(2,0)$: $f(2)=4>0$, so $(2,0)\notin D$, i.e.\ $(2,0)\in\mathbb R^2_+\setminus D$). Similarly $(0.5,0)$: $f(0.5)=0.5>0$, so $(0.5,0)\in\mathbb R^2_+\setminus D$. Their midpoint $(1.25,0)$: $f(1.25)=1.5625>0$, so in $\mathbb R^2_+\setminus D$. But $(0,0)\in\mathbb R^2_+\setminus D$? $f(0)=0$, and ``$\setminus D$'' means $y<f(x)$, i.e.\ $0<0$ false; so $(0,0)\notin\mathbb R^2_+\setminus D$. Take midpoint of $(2,0)$ and $(-2,0)$: $(0,0)$, not in $\mathbb R^2_+\setminus D$. So Option B is false too.

4. 4

   Hence None of the above.

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