Real Analysis - Uniform Convergence

  • #1
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Homework Statement


Prove that if fn -> f uniformly on a set S, and if gn -> g uniformly on S, then fn + gn -> f + g uniformly on S.

Homework Equations



The Attempt at a Solution


fn -> f uniformly means that |fn(x) - f(x)| < [tex]\epsilon[/tex]/2 for n > N_1.
gn -> g uniformly means that |gn(x) - g(x)| < [tex]\epsilon[/tex]/2 for n > N_2.

By the triangle inequality, we have |fn(x) - f(x) + gn(x) - g(x)| <= |fn(x) - f(x)| + |gn(x) - g(x)| < [tex]\epsilon[/tex]/2 + [tex]\epsilon[/tex]/2 = [tex]\epsilon[/tex].

This implies |[fn(x) + gn(x)] - [f(x) + g(x)]| < [tex]\epsilon[/tex] for n > N_1, N_2.

Therefore fn + gn -> f + g uniformly on S.

Is this correct? I'm pretty confident it's right, but I just want to make sure. Thanks!
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Of course, it's right. You knew that.
 

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