Question #113009
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}
1
Expert's answer
2020-05-05T18:12:38-0400

II)(0,3),(3,0)S2(0,3), (\sqrt{3},0)\in S_2, but (32,32)∉S2\left(\frac{\sqrt{3}}{2},\frac{3}{2}\right)\not\in S_2. So S2S_2 is not a convex set.

I)Since (3x2)=2(3-x^2)''=-2, 3x23-x^2 concave downward. Then Jensen's inequality f(αx1+(1α)x2)αf(x1)+(1α)f(x2)f(\alpha x_1+(1-\alpha)x_2)\ge\alpha f(x_1)+(1-\alpha)f(x_2) is true, where f(x)=3x2f(x)=3-x^2, x1,x2[0,3]x_1,x_2\in[0,\sqrt{3}] and 0α10\le\alpha\le 1.

Let (x1,y1),(x2,y2)S1(x_1,y_1),(x_2,y_2)\in S_1 and prove that [(x1,y1),(x2,y2)]S1[(x_1,y_1),(x_2,y_2)]\subset S_1. Since (x1,y1),(x2,y2)S1(x_1,y_1),(x_2,y_2)\in S_1, we have 0y1f(x1)0\le y_1\le f(x_1) and 0y2f(x2)0\le y_2\le f(x_2).


Take arbitrary(x3,y3)[(x1,y1),(x2,y2)](x_3,y_3)\in[(x_1,y_1),(x_2,y_2)]. We have (x3,y3)=α(x1,y1)+(1α)(x2,y2)(x_3,y_3)=\alpha (x_1,y_1)+(1-\alpha)(x_2,y_2) for some α[0,1]\alpha\in[0,1].

Then by the Jensen's inequality we obtain f(x3)=f(αx1+(1α)x2)αf(x1)+(1α)f(x2)f(x_3)=f(\alpha x_1+(1-\alpha)x_2)\ge\alpha f(x_1)+(1-\alpha)f(x_2). Since (x1,y1),(x2,y2)S1(x_1,y_1),(x_2,y_2)\in S_1, we have 0y1f(x1)0\le y_1\le f(x_1) and 0y2f(x2)0\le y_2\le f(x_2), so f(x3)αf(x1)+(1α)f(x2)αy1+(1α)y2=y3f(x_3)\ge\alpha f(x_1)+(1-\alpha)f(x_2)\ge\alpha y_1+(1-\alpha)y_2=y_3 and y3=αy1+(1α)y2α0+(1α)0=0y_3=\alpha y_1+(1-\alpha)y_2\ge\alpha\cdot 0+(1-\alpha)\cdot 0=0.


We obtain 0y3f(x3)0\le y_3\le f(x_3), so (x3,y3)S1(x_3,y_3)\in S_1. Since we take arbitrary (x3,y3)[(x1,y1),(x2,y2)](x_3,y_3)\in[(x_1,y_1),(x_2,y_2)], we obtain [(x1,y1),(x2,y2)]S1[(x_1,y_1),(x_2,y_2)]\subset S_1. So S1S_1 is a convex set.

Answer: S1S_1 is a convex set, S2S_2 is not a convex set.


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