Real Analysis - Uniform Convergence

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SUMMARY

The discussion confirms that if sequences fn and gn converge uniformly to functions f and g on a set S, respectively, then their sum fn + gn converges uniformly to f + g on S. The proof utilizes the definition of uniform convergence, demonstrating that the triangle inequality allows for the combination of the two inequalities |fn(x) - f(x)| and |gn(x) - g(x)| to establish the desired result. This conclusion is established definitively through the application of the epsilon-delta criterion for uniform convergence.

PREREQUISITES
  • Understanding of uniform convergence in real analysis
  • Familiarity with the triangle inequality
  • Knowledge of epsilon-delta definitions in mathematical proofs
  • Basic concepts of sequences and functions
NEXT STEPS
  • Study the implications of uniform convergence on integration and differentiation
  • Explore examples of uniform convergence in real analysis
  • Learn about the Weierstrass M-test for uniform convergence of series
  • Investigate the relationship between pointwise and uniform convergence
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Students of real analysis, mathematicians focusing on convergence properties, and educators teaching advanced calculus concepts.

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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)| < \epsilon/2 for n > N_1.
gn -> g uniformly means that |gn(x) - g(x)| < \epsilon/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)| < \epsilon/2 + \epsilon/2 = \epsilon.

This implies |[fn(x) + gn(x)] - [f(x) + g(x)]| < \epsilon 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!
 
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