Uniform convergence of a series

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SUMMARY

The discussion centers on the uniform convergence of the series defined by fn(x) = f(x + 1/n), where f is uniformly continuous. It is established that as n approaches infinity, fn converges uniformly to f. The key argument involves demonstrating that for any ε > 0, there exists a δ > 0 such that |f(x + h) - f(x)| < ε whenever |h| < δ, leading to the conclusion that |fn(x) - f(x)| < ε for all n > δ-1.

PREREQUISITES
  • Understanding of uniform continuity
  • Familiarity with limits and convergence in analysis
  • Knowledge of epsilon-delta definitions in calculus
  • Basic concepts of sequences and series in mathematical analysis
NEXT STEPS
  • Study the properties of uniformly continuous functions
  • Learn about the implications of uniform convergence in functional analysis
  • Explore examples of uniform convergence using specific functions
  • Investigate the relationship between uniform convergence and pointwise convergence
USEFUL FOR

Mathematics students, particularly those studying real analysis, educators teaching convergence concepts, and researchers interested in functional analysis.

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Homework Statement



We know that f is uniformly continuous.

For each n in N, we define fn(x)=f(x+1/n) (for all x in R).

Show that fn converges uniformly to f.

Homework Equations



http://en.wikipedia.org/wiki/Uniform_convergence

The Attempt at a Solution



I know that as n approaches infinity, that fn(x)=f(x), which implies that fn converges to f.

I'm currently trying to apply the fact that if fn is uniformly convergent, then

limn->infinity Sup {fn(x): x in R}=0.

But I keep getting stuck on the fact that there's an function in the definition of fn i.e., fn(x)=f(x+1/n). Is there a way to work with it?
 
Last edited:
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It's actually pretty straightforward. Given [tex]\epsilon > 0[/tex], for each x there is a [tex]\delta>0[/tex] such that [tex]|f(x+h)-f(x)|<\epsilon[/tex] whenever [tex]|h|<\delta[/tex]. Hence, for all [tex]n > \delta ^{-1}[/tex] we have [tex]|f_n(x)-f(x)|<\epsilon[/tex].
 

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