Answer to Question #113009 in Operations Research for Suraj Singh

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), (\\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.


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