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Uniform convergence of a series

  1. Jul 15, 2010 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations


    3. 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: Jul 15, 2010
  2. jcsd
  3. Jul 15, 2010 #2
    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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