1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equality of functions

  1. Apr 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Let U [tex]\subset[/tex] [tex]\Re^{2}[/tex] be open, and f,g: U [tex]\rightarrow[/tex] [tex]\Re[/tex] are continuous, and [tex]\int^{b}_{a}[/tex] ( [tex]\int^{d}_{c}[/tex] f(x,y) dy ) dx = [tex]\int^{b}_{a}[/tex] ( [tex]\int^{d}_{c}[/tex] g(x,y) dy ) dx for every rectangle [a,b] x [c,d] in U. Show that f = g.


    2. Relevant equations



    3. The attempt at a solution
    Can someone tell me if I'm approaching this problem correctly? I know both integrals exist because f,g are assumed to be continuous. If I apply the fundamental theorem of calculus twice to each integral I get F(b,d) - F(a,d) - F(b,c) + F(a,c) = G(b,d) - G(a,d) - G(b,c) + G(a,c) for every (b,d),(b,c),(a,d),(a,c) in U (so does this imply that F(x,y) = G(x,y)?) where d/dx(d/dy F(x,y)) = f(x,y) and d/dx(d/dy G(x,y) = g(x,y). Because F(x,y) = G(x,y) I can conlude that f(x,y) = g(x,y). Is this correct or am I way off?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 2, 2009 #2
    anyone?
     
  4. Apr 3, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Suppose there exist [itex](x_0,y_0)[/itex] at which f and g are not equal:[itex]f(x_0,y_0)\ne g(x_0,y_0)[/itex] Take [itex]\epsilon= (1/2)|f(x_0,y_0)- g(x_0,y_0)|[/itex] and show that there exist some neighborhood of [itex](x_0,y_0)[/itex] in which |f(x,y)- g(x,y)|> [itex]\epsilon[/itex]. Integrate over a rectangle inside that neighborhood.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook