Question

Known that function f(x, y) is nonnegative and continuous on a closed rectangle area, and ∫∫Df(x, y) = 0.
Prove that f(x, y)=0.
What if ''nonnegative and intergrable function f(x, y)''?

f(x,y)do 0

Answer 1

"f(x, y) is continuous on R (closed rectangle are) . suppose f(x, y) notequalto 0 for some (x_0, y_0) elementof R than By continuity there exist an open ball B_r ((x_0, y_0)) of radius r round (x_0, y_0) such that f(x, y) notequalto 0 Forall (x, y) elementof B_r ((x_0, y_0)) Now as f greaterthanorequalto 0 Also consider B_r/2 ((x_0, y_0))^- Subset R then as B_r/2 ((x_0, y_0))^- closed and f continuous so, inf_(x, y) f(x, y) attain say m elementof B_r/2 ((x_0, y_0))^- Because f nonzero there = > m > 0 Now, integral integral_R f(x, y) d sigma = 0 = > integral integral f(x, y) /B_r/2 ((x_0, y_0)) d sigma + integral integral f(x, y) /B_r/2 ((x_0, y_0)) d sigma = 0 m

Contradiction because m and area of ball both positive so their product is also positive but the inequality shows that it's negative. Hence contradiction arise. So f must be identically zero.