MATH.20:5.3 - Enclose a set and improve the consequence

Let S be the disk {(x,y): x²+y²≤1}. The work needs the largest value of g(x,y)=x+y, but first seeks cheaper comparable sets.

The diamond L={(x,y): |x|+|y|≤1} lies inside S, because x²+y²≤(|x|+|y|)²≤1. The square U=[-1,1]×[-1,1] contains S, because each coordinate square is at most one. Thus L⊆S⊆U.

Over L, x+y≤|x|+|y|≤1, attained at (1,0). Over U, x+y≤2. Therefore the maximum over S lies between 1 and 2. Its existence follows, for example, from continuity of g on the compact disk.

The point (1,1) attaining the outer bound is outside S. To tighten that bound, retain the relation between the coordinates:

(x+y)²≤2(x²+y²)≤2,

where the first inequality follows from (x-y)²≥0. Hence x+y≤sqrt(2). The point (1/sqrt(2),1/sqrt(2)) belongs to S and attains this value, completing the comparison.

The inner set supplied feasible values; the outer set supplied a restriction on all feasible values. A better inequality and its equality case closed the gap. The same inclusion method can enclose a feasible region, a family of functions or another set, with the corresponding proof of membership and bound on the requested operation.