Question

Check whether the following sets are convex or not: i) S1 = {( x, y)| y - 3 ≤ -(x^2), x ≥ 0, y ≥ 0}
ii) S2 = {(x, y)| y - 3 ≥ -(x^2), x ≥ 0, y ≥ 0}

Answer 1

II)$(0,3), (\sqrt{3},0)\in S_2$, but $\left(\frac{\sqrt{3}}{2},\frac{3}{2}\right)\not\in S_2$. So $S_2$ is not a convex set.

I)Since $(3-x^2)''=-2$, $3-x^2$ concave downward. Then Jensen's inequality $f(\alpha x_1+(1-\alpha)x_2)\ge\alpha f(x_1)+(1-\alpha)f(x_2)$ is true, where $f(x)=3-x^2$, $x_1,x_2\in[0,\sqrt{3}]$ and $0\le\alpha\le 1$.

Let $(x_1,y_1),(x_2,y_2)\in S_1$ and prove that $[(x_1,y_1),(x_2,y_2)]\subset S_1$. Since $(x_1,y_1),(x_2,y_2)\in S_1$, we have $0\le y_1\le f(x_1)$ and $0\le y_2\le f(x_2)$.

Take arbitrary$(x_3,y_3)\in[(x_1,y_1),(x_2,y_2)]$. We have $(x_3,y_3)=\alpha (x_1,y_1)+(1-\alpha)(x_2,y_2)$ for some $\alpha\in[0,1]$.

Then by the Jensen's inequality we obtain $f(x_3)=f(\alpha x_1+(1-\alpha)x_2)\ge\alpha f(x_1)+(1-\alpha)f(x_2)$. Since $(x_1,y_1),(x_2,y_2)\in S_1$, we have $0\le y_1\le f(x_1)$ and $0\le y_2\le f(x_2)$, so $f(x_3)\ge\alpha f(x_1)+(1-\alpha)f(x_2)\ge\alpha y_1+(1-\alpha)y_2=y_3$ and $y_3=\alpha y_1+(1-\alpha)y_2\ge\alpha\cdot 0+(1-\alpha)\cdot 0=0$.

We obtain $0\le y_3\le f(x_3)$, so $(x_3,y_3)\in S_1$. Since we take arbitrary $(x_3,y_3)\in[(x_1,y_1),(x_2,y_2)]$, we obtain $[(x_1,y_1),(x_2,y_2)]\subset S_1$. So $S_1$ is a convex set.

Answer: $S_1$ is a convex set, $S_2$ is not a convex set.