- #1
steelphantom
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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!