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, but (23,23)∈S2. So S2 is not a convex set.
I)Since (3−x2)′′=−2, 3−x2 concave downward. Then Jensen's inequality f(αx1+(1−α)x2)≥αf(x1)+(1−α)f(x2) is true, where f(x)=3−x2, x1,x2∈[0,3] and 0≤α≤1.
Let (x1,y1),(x2,y2)∈S1 and prove that [(x1,y1),(x2,y2)]⊂S1. Since (x1,y1),(x2,y2)∈S1, we have 0≤y1≤f(x1) and 0≤y2≤f(x2).
Take arbitrary(x3,y3)∈[(x1,y1),(x2,y2)]. We have (x3,y3)=α(x1,y1)+(1−α)(x2,y2) for some α∈[0,1].
Then by the Jensen's inequality we obtain f(x3)=f(αx1+(1−α)x2)≥αf(x1)+(1−α)f(x2). Since (x1,y1),(x2,y2)∈S1, we have 0≤y1≤f(x1) and 0≤y2≤f(x2), so f(x3)≥αf(x1)+(1−α)f(x2)≥αy1+(1−α)y2=y3 and y3=αy1+(1−α)y2≥α⋅0+(1−α)⋅0=0.
We obtain 0≤y3≤f(x3), so (x3,y3)∈S1. Since we take arbitrary (x3,y3)∈[(x1,y1),(x2,y2)], we obtain [(x1,y1),(x2,y2)]⊂S1. So S1 is a convex set.
Answer: S1 is a convex set, S2 is not a convex set.
Comments