
#1
Apr2908, 07:47 PM

P: 159

1. The problem statement, all variables and given/known data
Prove that if f_{n} > f uniformly on a set S, and if g_{n} > g uniformly on S, then f_{n} + g_{n} > f + g uniformly on S. 2. Relevant equations 3. The attempt at a solution f_{n} > f uniformly means that f_{n}(x)  f(x) < [tex]\epsilon[/tex]/2 for n > N_1. g_{n} > g uniformly means that g_{n}(x)  g(x) < [tex]\epsilon[/tex]/2 for n > N_2. By the triangle inequality, we have f_{n}(x)  f(x) + g_{n}(x)  g(x) <= f_{n}(x)  f(x) + g_{n}(x)  g(x) < [tex]\epsilon[/tex]/2 + [tex]\epsilon[/tex]/2 = [tex]\epsilon[/tex]. This implies [f_{n}(x) + g_{n}(x)]  [f(x) + g(x)] < [tex]\epsilon[/tex] for n > N_1, N_2. Therefore f_{n} + g_{n} > f + g uniformly on S. Is this correct? I'm pretty confident it's right, but I just want to make sure. Thanks! 



#2
Apr3008, 12:13 AM

Sci Advisor
HW Helper
Thanks
P: 25,168

Of course, it's right. You knew that.



Register to reply 
Related Discussions  
Uniform Convergence  Calculus & Beyond Homework  7  
Real Analysis proof help, convergence  Calculus & Beyond Homework  11  
Real Analysis  Radius of Convergence  Calculus & Beyond Homework  9  
Real Analysis least upper bound and convergence  General Math  4  
Real analysis Convergence/l.u.b  Calculus & Beyond Homework  5 